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Students can Download Chapter 12 Introduction to Three Dimensional Geometry Questions and Answers, Plus One Maths Chapter Wise Questions and Answers helps you to revise the complete Kerala State Syllabus and score more marks in your examinations.

Kerala Plus One Maths Chapter Wise Questions and Answers Chapter 12 Introduction to Three Dimensional Geometry

Plus One Maths Introduction to Three Dimensional Geometry Three Mark Questions and Answers

Question 1.
Prove by using distance formula that the A(1, 2, 3), B(-1, -1, -1) and C(3, 5, 7) are collinear.
Answer:

Now BC = AB + AC
Thus A, B, C are collinear.

Question 2.
Verify the following: (3 score each)

1. (0, 7, -10), (1, 6, -6) and (4, 9, -6) are the vertices of an isosceles triangle.
2. (0, 7, 10), (-1, 6, 6) and (-4, 9, 6) are the vertices of a right angled triangle.
3. (-1, 2, 1), (1, -2, 5), (4, -7, 8) and (2, -3, 4) are the vertices of a parllelogram.

Answer:
1. Let A(0, 7, -10), B(1, 6, -6) and C(4, 9, -6) be the two points.

Now AB = BC, thus ABC is an isosceles triangle.

2. Let A(0, 7, 10), B(-1, 6, 6) and C(-4, 9, 6) be the two points.


Now AC² = AB² + BC², thus ABC is a right triangle.

3. Let A(-1, 2, 1), B(1, -2, 5), C(4, -7, 8) and D(2, -3, 4) be the two points.

Now AB = CD, BC = AD, AC ≠ BD, thus A, B, C, D are vertices of a parallelogram.

Question 3.
Find the equation of set points which are equidistant from the points (1, 2, 3) and (3, 2, -1).
Answer:
Let P(x,y,z) be any point which is equidistant from the point A(1, 2, 3) and B (3, 2, -1).
Given; PA = PB

(x – 1)² + (y – 2)² + (z – 3)²
= (x – 3)² + (y – 2)² + (z + 1)²
= x² – 2x + 1 + y² – 4y + 4 + z² – 6z + 9
= x² – 6x + 9 + y² – 4y + 4 + z² + 2z + 1 – 2x + 14 – 6z = -6x + 14 + 2z
⇒ 4x – 8z = 0
⇒ x – 2z = 0.

Question 4.
Find the coordinate of the point which divides the line segment joining the points (3, -2, 5) and (3, 4, 2) in the ratio 2:1 (3 score each)

1. 2:1 internally
2. 2:1 externally

Answer:
1. Let P(x, y, z) be any point which divides the line segment joining points A(3, -2, 5) and B (3, 4, 2) in the ratio 2:1 internally.

Therefore coordinates of P are (3, 2, 3).

2. Let P(x, y, z) be any point which divides the line segment joining points A(3, -2, 5) and B (3, 4, 2) in the ratio 2:1 externally.

Therefore coordinates of P are (3, 10, -1).

Question 5.
Find the ratio in which the line joining the points (1, 2, 3) and (-3, 4, -5) is divided by the xy-plane.
Answer:
Let the line joining the points A(1, 2, 3) and B(-3, 4, -5) is divided by the xy-plane in the ratio k:1.
Then the coordinate

Since the point lies on xy-axis, we have;

Thus the required ratio is \(\frac{3}{5}\); ie, 3:5.

Question 6.
Find the coordinates of the points which trisect the line segment joining the points P(4, 2, -6) and Q (10, -16, 6).
Answer:

Let R and S be two points which trisect the line join of PQ. Therefore PR = RS = SQ Then coordinate of R is

= (6, -4, -2)
Then coordinate of S is

= (8, -10, 2).

Plus One Maths Introduction to Three Dimensional Geometry Practice Problems Questions and Answers

Question 1.
Find the distance between the following pair of points: (1 score each)

1. (2, 3, 5) and (4, 3, 1)
2. (-3, 7, 2) and (2, 4, -1)
3. (-1, 3, -4) and (1, -3, 4)

Answer:
1. Let A(2, 3, 5) and B(4, 3, 1) be the two points.

2. Let A(-3, 7, 2) and B(2, 4, -1) be the two points.

3. Let A(-1, 3, -4) and B(1, -3, 4) be the two points.

Students can Download Chapter 5 Trial Balance and Rectification of Errors Questions and Answers, Plus One Accountancy Chapter Wise Questions and Answers helps you to revise the complete Kerala State Syllabus and score more marks in your examinations.

Kerala Plus One Accountancy Chapter Wise Questions and Answers Chapter 5 Trial Balance and Rectification of Errors

Plus One Accountancy Trial Balance and Rectification of Errors One Mark Questions and Answers

Question 1.
Which of the following statement is wrong regarding Trial Balance.
(a) Trial balance is a part of the double-entry system.
(b) It is not an account
(c) It is prepared on a specific date
(d) It is prepared to check the arithmetical accuracy of the books of accounts.
Answer:
(a) Trial balance is a part of double entry system.

Question 2.
Sale of machinery is credited to sales a/c is an error of
(a) Commission
(b) Omission
(c) Principle
(d) None of these
Answer:
(c) Principle

Question 3.
Debiting wages account with the amount of wages paid on erection of machinery is an errors of be debited with
(a) Rs. 400
(b) Rs. 800
(c) Rs. 200
(d) None of the above
Answer:
(b) Rs. 800

Question 4.
Which of the errors does not affect the trial balance?
(a) Wrong balancing
(b) Wrong totaling
(c) Writing an amount in the wrong account but in the correct side.
(d) None of the above.
Answer:
(c) Writing an amount in the wrong account but in the correct side.

Question 5.
Instead of debiting Sanoj’s A/c his account was credited by Rs. 400. To rectify this his account should
(a) Omission
(b) Principle
(c) Commission
Answer:
(b) Principles

Question 6.
Trial balance is
(a) An account
(b) A statement
(c) A subsidiary Book
(d) A Principal book
Answer:
(b) A Statement

Question 7.
Agreement of trial balance is affected by>
(a) One-sided errors only
(b) Two-sided errors only
(c) Both a and b
(d) None of these
Answer:
(c) Both a and b

Question 8.
A trial balance is prepared to check the …………… of ledger accounts.
Answer:
Arithmetical accuracy

Question 9.
If purchase of furniture is recorded in the purchase book, it is an error of …………..
Answer:
Principle

Question 10.
If purchase of goods on credit is not recorded in the books, it is an error of …………
Answer:
Omission

Question 11.
An account in which the difference of trial balance is temporarily put is ………… account.
Answer:
Suspense.

Question 12.
Error of …………. does not affect the trial balance.
Answer:
Principle.

Plus One Accountancy Trial Balance and Rectification of Errors Two Mark Questions and Answers

Question 1.
Match the following.

1. Installation charges debited to wages account.
2. Cash paid to ‘X’ is not posted to his account.
3. Cash paid to ‘A’ is posted, to B’s account.
4. Sales books and purchase books are undercast by same account.
5. The sales book is overcast by Rs. 100.

a. Error of commission
b. Compensating error
c. Error of principle
d. One side error
e. Error of omission

Answer:

• 1-c
• 2-e
• 3-a
• 4-b
• 5-d

Question 2.
Complete the following on the basis of hint given.

1. Errors of principle – Rule of accounting is violated
2. _____ – Wrong amount is written in subsidiary books.

Answer:
Error of commission.

Question 3.

1. One-sided error – Undercast in Sales returns book
2. ______ – Purchase of machinery entered in purchase book.

Answer:
Two-sided error.

Question 4.
What do you mean by Trial Balance?
Answer:
Trial balance is a statement that shows either the balance or total amounts of debit items and credit items of all accounts. It is prepared on a particular date to test the arithmetical accuracy of the books of accounts kept under the double-entry system.

Question 5.
Why a trial balance is prepared?
Answer:
A trial balance is prepared:

1. To check the arithmetical accuracy of the ledger accounts.
2. To help in locating errors
3. To provide a basis for preparing the financial statements.

Question 6.
Agreement of Tial balance is a proof of accuracy of books of accounts. Do you agree? Explain.
Answer:
No, the tallying of the trial balance does not mean that no errors have been committed in the accounting records. There can be errors which do not affect the equality of debits and credits and there can be errors which affect the equality of debits and credits.

Question 7.
Furniture purchased for Rs. 10,000 from Nirmal has been recorded as follows:

• Nirmal A/c Dr 10,000
• FurnitureA/c 10,000
• If it is incorrect, rectify it by passing rectification entry.

Answer:

• The journal entry is incorrect.
• Rectification entry is – FurnitureA/c Dr 20,000
• Nirmal A/c 20,000

Question 8.
On 31^st March 2005, when the trial balance of Abi stores was prepared it showed a difference. Inspite of his continuous efforts, the accountant could not locate errors. But the preparation of financial statements cannot be delayed. Is it possible to prepare financial statements with such a trial balance. Can you suggest a solution at the juncture?
Answer:
If the trial balance doesn’t agree even after the repeated efforts, it is better to place the difference under suspense A/c. Otherwise it will cause inordinate delay in tpe preparation of final accounts. However, the suspense a/c has to be later removed by closely scrutinising the books of accounts in due course.

Question 9.
‘Closing stock is normally given out the trial balance’. State the reason.
Answer:
Usually no separate ledger accounts are maintained for stock account. The valuation of stock will be done at the end of a particular period when financial statements are prepared. Therefore it appears out the trial balance.

Question 10.
What are the methods of preparing trial balance?
Answer:
There are three methods of preparing trial balance They are:

1. Totals method (Total of each side in the ledger)
2. Balance method (Showing the balances of all ledger accounts.
3. Totals – cum – balances method

Question 11.
Explain Error of principles with examples.
Answer:
If any accounting rules or principles are violated in recording a transaction, it is an error of principles. This error does not affect the agreement of trial balance.
Examples :-

1. Expenses paid for installation of machinery debited to expenses account.
2. The sale of building is credited to sales account.
3. Amount spent on repair of machinery debited to machinery account.
4. Purchase of furniture debited to purchase account.

Plus One Accountancy Trial Balance and Rectification of Errors Three Mark Questions and Answers

Question 1.
When do you open a suspense account? Explain its uses.
Answer:
When all attempts fails to locate errors and the preparation of the final accounts can not be further delayed, the difference in the trial balance is temporarily transferred to an account called “suspense account’’. Uses of suspense Account.

1. It facilitates the preparation of financial statements even when the trial balance has not tallied.
2. It helps in giving rectifying entries after the preparation of trial balance.

Question 2.
Mr. Murali’s trial balance showed a difference of Rs. 5,500 on the credit side. The following errors were revealed from the books of account.

• The purchase of books were overcast by Rs. 3000.
• The salary paid to Renjith Rs. 7000 has been posted twice.
• Received from Manu Rs. 5000 has been credited Rs. 500.
• Prepare a suspense account.

Answer:
Suspense A/c

Question 3.
Explain Error of omission with Examples.
Answer:
Error of omission:
When a transaction is not entered in the books of original entry or not posted from the books original entry to the ledger, an error of omission is caused. The omission may be complete or partial. If a transaction is not entered in the subsidiary books, it is a case of complete omission as the posting in the ledger accounts are also omitted.

In this both the debit and credit aspects go unrecorded, it does not affect the agreement of Trial Balance. If only one aspect of a transaction is recorded, it is a case of Partial omission. It happens while posting from day books to ledger accounts. This will affect the agreement of Trial balance.

Question 4.
Explain Error of Commission with example.
Answer:
Error of commission:
Errors committed when transactions are incorrectly recorded are called error of commission. These are the errors caused by wrong posting, wrong totaling, wrong balancing, wrong carryforwards, etc. For example- if Rs. 290 received from Ram is credited to his account as Rs. 209, it is an error of commission, Error in posting as to the side of account – Rs. 300 received from Kumar is posted his Debit side.

Question 5.
Explain Error of Principles Error with examples.
Answer:
Error of principles:
If any accounting rules or principles are violated in recording a transaction, it is an error of principles. This error does not affect the agreement of trial balance.
Examples:

1. An item of capital expenditure is wrongly debited to a revenue account.
2. Sale of building is credited to sales account.

Question 6.
Explain compensating error with examples.
Answer:
Compensating Error:
These errors arise when a mistake made in one direction is compensated by another mistake made in the opposite direction, to the extent of same amount. These errors do not affect the agreement of trial balance.

For example:
If Purchase of goods from ‘X’ for Rs. 800 is credited to his account as Rs.80 and a purchase from T for Rs. 80 is credited to y’s account as Rs. 800.

Question 7.
Name the errors which do not affect the Trial Balance.
Answer:

1. Error of complete omission
2. Error of principle,
3. Compensating Error
4. Error of recording in the book of original entry
5. Error of posting to wrong account on credit side with correct amount.

Question 8.
Name the errors which affect the Trial Balance.
Answer:

1. Errors due to partial omission
2. Error of casting
3. Posting an amount in the wrong side of an account.
4. Posting of a wrong amount
5. Wrong totaling or balancing of accounts

Question 9.
Rectify the following errors:
Cash sales Rs. 16,000
a) were not posted to sales a/c
b) were posted as Rs. 6000 in sales account
c) were posted to commission account
Answer:

Question 10.
Locate the types of errors involved in the following transactions.

1. Purchased goods for 10,000 debited to furniture a/c.
2. Carriage paid Rs. 1,000 debited to carriage a/c as Rs. 100.
3. Amount received from Rajesh Rs. 1,800 has not been entered in the books of accounts.

Answer:

1. Error of principle
2. Error of commission
3. Error of omission

Plus One Accountancy Trial Balance and Rectification of Errors Four Mark Questions and Answers

Question 1.
What are the objectives of preparing a trial balance?
Answer:
A Trial balance is prepared with the following objectives:
1. To ascertain the arithmetical accuracy of ledger accounts:
A Trial balance is prepared to check the arithmetical accuracy of ledger accounts. If the sum of the debit and credit columns of Trial balance is equal, it is assumed that the posting to the ledger accounts is accurate. This is because, for every debit we give an equal credit.

2. To help in ascertaining errors:
Some of the errors in the books of account can be detected by the trial balance. An untallied trial balance indicates that some errors have been committed.

3. To provide a basis for preparing final accounts:
The ultimate aim of maintaining books of accounts is to ascertain the financial result and position of the business. For this purpose profit and loss account and blance sheet is prepared on the basis of trial balance.

Question 2.
Check the arithmetical accuracy of the ledger account balances from the following.

Trial Balance as on …………..
Answer:

Question 3.
Pass journal entries to rectify the following transactions.
a) Purchase of machinery was debited to purchase a/c Rs. 8,000.
b) Goods sold on credit was recorded twice Rs. 1,500.
c) The repair of building was debited to building a/c Rs. 6,000.
d) Rs. 200 received from Jose posted to the debit of his account.
Answer:

Question 4.
Rectify the following errors:

1. Stationery purchased for ₹500 has been wrongly debited to drawings a/c.
2. A credit sale of ₹1,500 has been wrongly passed through the purchase book.
3. ₹4,000 received from Sidhu have been posted on the credit side of his account ₹4,400.
4. Salary ₹10,000 paid to Mr. Shaju was debited to his personal account.

Answer:
Journal

Plus One Accountancy Trial Balance and Rectification of Errors Five Mark Questions and Answers

Question 1.
Show how you would correct the following errors, write Journal entries.

1. Rs. 5,000 received on the sale of an asset has been credited to sales account.
2. A credit purchase of goods amounting to Rs. 500 from Raghavan had not been recorded in the books.
3. Sales book was undercast by Rs. 500.
4. Wages Rs. 750 paid for the installation of a new machine are debited to salaries and wages account.
5. An amount of Rs. 500 written off as depreciation has not been posted to depreciation account.

Answer:
Journal Proper

Question 2.
Identify the mistakes crept into the trial balance and redraft it in its proper form.
Trial Balance

Answer:

Question 3.
Rectify the following errors
(a) Furniture purchased for ₹10,000 wrongly debited to purchase account.
(b) Machinery purchased on credit from Raman for ₹20,000 was recorded through purchase book.
(c) Repairs on machinery ₹1.400 debited to machinery account.
(d) Repairs on overhauling of secondhand machinery purchased ₹2,000 was debited to repair the account.
(e) Sales of old machinery at book value of ₹3,000 was credited to sales account.
Rectification Entries in Journal
Answer:

Plus One Accountancy Trial Balance and Rectification of Errors Six Mark Questions and Answers

Question 1.
Rectify the following Mistakes.

1. The purchase day book is undercast (totaled less) by Rs. 180.
2. Sales day book is overcast (totaled more) by Rs. 100.
3. The purchase return book is totaled more by Rs. 80.
4. Sales return book is less by Rs. 110.
5. Rent paid Rs. 600 is omitted to post to rent account.
6. The commission received Rs. 40 is omitted and not posted to the commission account.
7. Salary paid Rs. 350 is posted twice to salary A/c.

Answer:
1. As the purchase account is less by Rs. 180, It should be debited to the purchase account. It should be shown in the debit side as : Undercast in purchase Day Book Rs. 180 Or Mistake in posting.

2. Sales Account is more by Rs. 100. To correct it, Rs. 100 be shown in the debit side of sales A/c as: Overcast in Day Book Rs. 100.

3. Purchase Returns Accounts is credited more by Rs. 80. Now it should be debited with the amount as Overcast in Day Book Rs. 100.

4. The effect is that the sales Return Account is less by Rs. 110. It should how be debited with the amount as Mistake in totaling Rs. 110.

5. Rent account should be debited with Rs. 600 as Omission in posting Rs. 600.

6. The commission account should be credited with Rs. 40 as Omission in posting Rs. 40.

7. The salary account is debited more by Rs. 350. It should now be credited with Rs. 350 as Mistake in polling Rs.350.

Question 2.
Veeran failed to balance his trial balance the credit side being more by Rs. 420. The difference is placed in a suspense a/c. Later on the following are discovered. Give rectifying entries and also prepare suspense A/c.

1. Sales Book was undercast by Rs. 100.
2. Goods for Rs. 300 purchased on credit from Raj was wrongly entered in the sales book. The account of Raj was correctly credited.
3. The sales return book was undercast by Rs. 30.
4. A credit item of Rs. 10 was wrongly debited to Renny’s account as Rs. 100.

Answer:
Journal

Suspense A/c

Question 3.
Rectify the following errors:-
a) Discount allowed to Ramesh Rs. 60 on receiving Rs.2000 from him was not recorded in the books.
b) Discount received from Ram Rs. 80 on paying 1900 to him was not posted at all.
c) Bill receivable from Narayan Rs. 1000 was dishonored and wrongly debited allowance account as Rs. 10,000.
d) Cash received from Mohan Rs. 3000 was posted to Naveen as Rs. 1000.
e) Cheques for Rs. 7800 received from Anu in full settlement of his account of Rs. 8000, was dishonored. No. entry was passed in the books on dishonor of the cheque.
f) Old machinery sold to Kannan at its book value of Rs.4000 was recorded through sales book.
g) Depreciation written off as the Machinery Rs.3000, was not posted at all.
Answer:

Plus One Accountancy Trial Balance and Rectification of Errors Eight Mark Questions and Answers

Question 1.
Rectify the following errors by giving correcting entries.

1. Credit purchase of goods for Rs. 850 from chand and sons has not been recorded in the daybook.
2. Rent paid to landlord is debited in landlord’s account Rs. 600.
3. Purchase of Machinery from Precision Machinery Ltd. for Rs. 28,000 is recorded in purchase daybook. j
4. Carriage paid on purchase of furniture Rs. 300 is debited in carriage account.
5. Private expenses Rs. 200 is debited in Trade Expenses Account.
6. Goods sold to Renjith for Rs. 500 has been wrongly recorded in purchase daybook.
7. Purchase of furniture for the personal use of the proprietor of Rs. 920 has been debited in furniture a/c.
8. Rs. 180 received from Salini has been credited in the account of Sajini.

Answer:

Students can Download Chapter 11 Conic Sections Questions and Answers, Plus One Maths Chapter Wise Questions and Answers helps you to revise the complete Kerala State Syllabus and score more marks in your examinations.

Kerala Plus One Maths Chapter Wise Questions and Answers Chapter 11 Conic Sections

Plus One Maths Conic Sections Three Mark Questions and Answers

Question 1.
Find the coordinate of the foci, the vertices, the length of major axis, the minor axis, the eccentricity and the length of the latus rectum of the ellipse. (3 score each)

1. \(\frac{x^{2}}{4}+\frac{y^{2}}{25}\) = 1
2. \(\frac{x^{2}}{16}+\frac{y^{2}}{9}\) = 1

Answer:
1. Since 25 > 4 the standard equation of the ellipse is \(\frac{x^{2}}{b^{2}}+\frac{y^{2}}{a^{2}}\) = 1 ⇒ a² = 25; b² = 4
c² = a² – b² = 25 – 4 = 21 ⇒ c = \(\sqrt{21}\)
Coordinate of foci are (0, ±\(\sqrt{21}\))
Coordinate of vertex are (0, ±5)
Length of major axis = 2a = 2 × 5 = 10
Length of minor axis = 2b = 2 × 2 = 4
Eccentricity = e = \(\frac{c}{a}=\frac{\sqrt{21}}{5}\)
Length of latus rectum = \(\frac{2 b^{2}}{a}=\frac{2 \times 4}{5}=\frac{8}{5}\).

2. Since 16 > 9 the standard equation of the ellipse is \(\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}\) = 1 ⇒ a² = 16; b² = 9
c² = a² – b² = 16 – 9 = 7 ⇒ c = \(\sqrt{7}\)
Coordinate of foci are (±\(\sqrt{7}\), 0)
Coordinate of vertex are (±4, 0)
Length of major axis = 2a = 2 × 4 = 8
Length of minor axis = 2b = 2 × 3 = 6
Eccentricity = e = \(\frac{c}{a}=\frac{\sqrt{7}}{4}\)
Length of latus rectum = \(\frac{2 b^{2}}{a}=\frac{2 \times 9}{4}=\frac{9}{2}\).

Question 2.
Find the ellipse satisfying the following conditions: (3 score each)

1. Vertex (±5, 0); foci (±4, 0)
2. Ends of the major axis (±3, 0), ends of minor axis (0, ±2)
3. Length of the major axis 26, foci (±5, 0)
4. b = 3, c = 4, centre at origin; foci on the x-axis

Answer:
1. Foci (±4, 0) lie on the x-axis. So the equation of the ellipse is of the form \(\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}\) = 1
Given; Vertex (±5, 0) ⇒ a = 5
Given; Foci(±4, 0) Foci ⇒ c = 4 = \(\sqrt{a^{2}-b^{2}}\)
⇒ 4 = \(\sqrt{25-b^{2}}\) ⇒ 16 = 25 – b² ⇒ b² = 9
Therefore the equation of the ellipse is
\(\frac{x^{2}}{25}+\frac{y^{2}}{9}=1\).

2. The ends of major axis lie on the x-axis. So the equation of the ellipse is of the form
\(\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}\) = 1
Given; Ends of the major
axis (±3, 0) ⇒ a = 3, ends of minor axis
(0, ±2) ⇒ b = 2
Therefore the equation of the ellipse is
\(\frac{x^{2}}{9}+\frac{y^{2}}{4}=1\).

3. Since foci (±5, 0) lie on x-axis, the standard form of ellipse is \(\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}\) = 1
Given; 2a = 26 ⇒ a = 13
Given; c = 5 = \(\sqrt{a^{2}-b^{2}}\)
⇒ 25 = 169 – b² ⇒ b² = 144
Therefore the equation of the ellipse is
\(\frac{x^{2}}{169}+\frac{y^{2}}{144}=1\).

4. The standard form of ellipse is
\(\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}\) = 1
Given; c = 4 = \(\sqrt{a^{2}-b^{2}}\)
⇒ 16 = a² – 9 ⇒ a² =25
Therefore the equation of the ellipse is
\(\frac{x^{2}}{25}+\frac{y^{2}}{9}=1\).

Question 3.
Find the coordinates of foci, the vertices, eccentricity and the length of latus rectum of the following hyperbolas. (3 score each)

1. \(\frac{y^{2}}{9}-\frac{x^{2}}{27}=1\)
2. 5y² – 9x² = 36

Answer:
1. The equation of the hyperbola is of the form
\(\frac{y^{2}}{a^{2}}-\frac{x^{2}}{b^{2}}=1\) ⇒ a² = 9; b² = 27
⇒ c² = a² + b² ⇒ c² = 9 + 27 = 36 ⇒ c = 6
Coordinate of foci are (0, ±6)
Coordinate of vertices are (0, ±a) ⇒ (0, ±3)
Eccentricity = \(\frac{c}{a}=\frac{6}{3}=2\)
Length of latus rectum = \(\frac{2 b^{2}}{a}=\frac{2 \times 27}{3}=18\).

2. Given; 5y² – 9x² = 36


The equation of the hyperbola is of the form

Coordinate of vertices are

Question 4.
Find the hyperbola satisfying the following conditions: (3 score each)

1. Vertices (+2, 0), foci (±3, 0)
2. Foci (±5, 0), the transverse axis is of length 8.
3. Foci(0, ±13), the conjugate axis is of length 24.
4. Foci (±3\(\sqrt{5}\), 0), the latus rectum is of length 8.
5. Vertices (±7, 0) , e = \(\frac{4}{3}\).

Answer:
1. Since Vertices are (±2, 0) the standard form of hyperbola is

⇒ a = 2
Given; foci (±3, 0) ⇒ c = 3
⇒ c² = a² + b² ⇒ 9 = 4 + b² ⇒ b² = 5
The equation of the hyperbola;

2. Since Foci are (±5,0) the standard form of hyperbola is

⇒ c = 5
Given; the transverse axis is of length 8.
⇒ 2a = 8 ⇒ a = 4
⇒ c² = a² + b² ⇒ 25 = 16 + b² ⇒ b² = 9
The equation of the hyperbola;

3. Since Foci are(0, ±13)the standard form of hyperbola is

⇒ c = 13
Given; the conjugate axis is of length 24.
⇒ 2b = 24 ⇒ b = 12
⇒ c² = a² + b² ⇒ 169 = a² + 144 ⇒ a² = 25.
The equation of the hyperbola;

4. Since Foci are(±3\(\sqrt{5}\), 0) the standard form of hyperbola is

⇒ c = 3\(\sqrt{5}\)
Given; the latus rectum is of length 8.
⇒ \(\frac{2 b^{2}}{a}\) = 8 ⇒ b² = 4a
⇒ c² = a² + b²
⇒ 45 = a² + 4a ⇒ a² + 4a – 45 = 0
⇒ (a + 9)(a – 5) = 0 ⇒ a = -9, 5
a = -9 is not possible
⇒ a = 5 ⇒ b² = 20
The equation of the hyperbola;

5. Since Vertices are (±7, 0) the standard form of hyperbola is

⇒ a = 7
Given;

⇒ c² = a² + b²

The equation of the hyperbola;

Question 5.
The line x – 1 = 0 is the directrix of a parabola, y² = kx then

1. Find the value of k. (1)
2. Find the vertex, focus, axis of parabola and length of latus rectum of the parabola. (3)

Answer:
1. Given directrix x – 1 = 0, so a = 1, then the equation of the parabola is
y² = -4ax ⇒ y² = -4x
Hence k = – 4

2. Vertex is (0, 0)
Focus of SS’ = (-1, 0)
Length of latus rectum = 4 × 1 = 4.

Question 6.
In the figure S and S’ are foci of the ellipse, \(\frac{x^{2}}{25}+\frac{y^{2}}{16}=1\) and P is a viable point on the ellipse.

1. Find the co-ordinate of foci. (2)
2. Find the distance between S and S1. (1)
3. What is the maximum area of the triangle PSS’. (1)

Answer:
1. Since 25 > 16 the standard equation of the ellipse is \(\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}\) = 1
⇒ a² = 25; b² = 16
c² = a² – b² = 25 – 16 = 9 ⇒ c = 3
Coordinate of foci are (±3, 0).

2. Distance between the two focus is 3 + 3 = 6.

3. Maximum area is attained when the point P reaches the point the ellipse meet the y
axis. Then area is = \(\frac{1}{2}\) × 6 × 4 = 12.

Plus One Maths Conic Sections Four Mark Questions and Answers

Question 1.
Find the equation of the circle passing through the points (4, 1) and (6, 5) and whose centre is on the line 4x + y = 16.
Answer:
Let the equation of the circle be
x² + y² + 2gx + 2fy + c = 0 ______(1)
Since (1) passes through (4, 1)
16 + 1 + 8g + 2f + c = 0
⇒ 8g + 2f + c = -17 ______(2)
Since (1) passes through (6, 5)
36 + 25 + 12g + 10f + c = 0
⇒ 12 g + 10f + c = -61 _______(3)
(1) – (2) ⇒ -4g – 8f = 16
⇒ -g – 2f = 4 ______(4)
Since centre is on the line 4x + y = 16, we have;
⇒ -4g – f = 16 ______(5)
Solving (4) and (5) We get; g = -3; f = -4
(2) ⇒ 8(-3) + 2(-4) + c = -17
⇒ -24 – 8 + c = -17 ⇒ c = 15
Then the equation of the circle is
x² + y² – 6x – 8y + 15 = 0.

Question 2.
Find the equation of the circle with radius 5 whose centre lies on x-axis and passes through the point (2, 3).
Answer:
Let the equation of the circle be
(x – h)² + (y – k)² = r²
centre lies on x axis so let the centre be (h, 0),
then (x – h)² + y² = 25
Since circle pass through (2, 3) we have;
(2 – h)² + 3² = 25
⇒ (2 – h)² = 16 ⇒ 2 – h = ±4
⇒ h = 6, -2
When h = 6 ; equation of circle is
(x – 6)² + (y – 0)² = 25
⇒ x² + 36 – 12x + y² = 25
⇒ x² + y² – 12x + 11 = 0
When h = -2; equation of circle is
(x + 2)² + (y – 0)² = 25
⇒ x² + 4 + 4x + y² = 25
⇒ x² + y² + 4x – 21 = 0.

Question 3.
Centre at (0, 0), major axis on the y-axis and passes through the points (3, 2) and (1, 6).
Answer:
Major axis lie on the y-axis so the standard equation of the ellipse is of the form

Since the ellipse passes through (3, 2)

Since the ellipse passes through (1, 6)

Solving (1) and (2), we have Since the ellipse passes through (3, 2)
a² = 40; b² = 10
Thus the equation of the ellipse is

Plus One Maths Conic Sections Six Mark Questions and Answers

Question 1.
Consider the point A (0, 0), B (4, 2) and C (8, 0)

1. Find the mid-point of AB. (1)
2. Find the equation of the perpendicular bisector of AB. (2)
3. Find the equation of the circum circle (Circle passing through the point A, B, and C) of triangle ABC. (3)

Answer:
1. Mid-point of AB is (2, 1).

2. Slope of line through AB
\(=\frac{2-0}{4-0}=\frac{1}{2}\)
Slope of perpendicular line is – 2
Equation of the perpendicular line to AB is
y – 1 = -2(x – 2) ⇒ 2x + y = 5.

3. The meeting point of perpendicular bisector of AB and AC will be the centre of the circum circle.
The line perpendicular to AC is x = 4
Solving and x = 4
We get y = 5 – 8 = -3 and x = 4
Hence centre is (4, -3) and radius is

Equation of the circle is
(x – 4)² + (y + 3)² = 5.

Plus One Maths Conic Sections Practice Problems Questions and Answers

Question 1.
Find the equation of the circle In each of the following cases. (2 score each)

1. Centre (0, 2) and radius 2.
2. Centre (-2, 3) and radius 4.
3. Center \(\left(\frac{1}{2}, \frac{1}{4}\right)\) and radius \(\frac{1}{12}\).

Answer:
1. The equation of the circle is
(x – 0)² + (y – 2)² = 2²
⇒ x² + y² – 4y + 4 = 4
⇒ x² + y² – 4y = 0.

2. The equation of the circle is
(x + 2)² + (y – 3)² = 4²
⇒ x² + 4x + 4 + y² – 6y + 9 = 16
⇒ x² + y² + 4x – 6y – 3 = 0.

3. The equation of the circle is

⇒ 144x² + 36 – 144x + 144y² + 9 – 72y = 1
⇒ 144x² + 144y² – 144x – 72y + 44 = 0
⇒ 36x² + 36y² – 36x – 18y + 11 = 0.

Question 2.
Find the centre and radius of the following circles. (2 score each)

1. x² + y² – 4x – 8y – 45 = 0
2. x² + y² – 8x – 10y -22 = 0
3. 2x² + 2y² – x = 0

Answer:
1. Comparing with the general equation we have
g = -2; f = -4; c = -45
Centre – (-g, -f) ⇒ (2, 4)
Radius – \(\sqrt{g^{2}+f^{2}-c}\)

2. Comparing with the general equation we have
g = -4; f = -5; c = -22
Centre – (-g, -f) ⇒ (4, 5)
Radius – \(\sqrt{g^{2}+f^{2}-c}\)

3. Convert the equation into standard form

Question 3.
Find the coordinate of the focus, axis of the parabola, the equation of the directrix and the length of the latus rectum. (2 score each)

1. y² = 20x
2. x² = 8
3. 3x² = -15

Answer:
1. Comparing the equation with the general form we get; 4a = 20 ⇒ a = 5
Coordinate of focus are (5, 0)
Axis of the parabola is y = 0
Equation of the directrix is x = -5
Length of latus rectum = 4 × 5 = 20.

2. Comparing the equation with the general form we get; 4a = 8 ⇒ a = 2
Coordinate of focus are (0, 2)
Axis of the parabola is x = 0
Equation of the directrix is y = – 2
Length of latus rectum = 4 × 2 = 8.

3. Convert the equation into general form, we get x² = -5y. Comparing the equation with the general form we get;
4a = 5 ⇒ a = \(\frac{5}{4}\)
Coordinate of focus are (0, \(-\frac{5}{4}\))
Axis of the parabola is x = 0
Equation of the directrix is y = \(\frac{5}{4}\)
Length of latus rectum = \(\frac{4 \times 5}{4}\) = 5.

Question 4.
Find the equation of the parabola satisfying the following conditions; (2 score each)

1. Focus(6, 0); directrix x = – 6
2. Vertex (0, 0); Focus (3, 0)
3. Vertex (0, 0) passing through (2, 3) and axis along x-axis

Answer:
1. Since the focus (6, 0) lies on the x-axis, therefore x-axis is the axis of parabola.
Also the directrix is x = – 6, ie; x = – a And focus (6, 0), ie; (a, 0)
Therefore the equation of the parabola is
y² = 4ax ⇒ y² = 24x.

2. The vertex of the parabola is at (0, 0) and focus is at (3, 0). Then axis of parabola is along x-axis. So the parabola is of the form y² = 4ax . The equation of the parabola is y² = 12x.

3. The vertex of the parabola is at (0, 0) and the axis is along x-axis. So the equation of parabola is of the torn y² = 4ax .
Since the parabola passes through point (2, 3)
Therefore, 3² = 4a × 2 ⇒ a = \(\frac{9}{8}\)
The required equation of the parabola is
y² = 4 × \(\frac{9}{8}\) x ⇒ y² = \(\frac{9}{2}\)x.

Question 5.
Find the Focus, vertex and latus rectum of the parabola y² = 8x.
Answer:
Given; y² = 8x, we have 4a = 8 ⇒ a = 2
Focus = (2, 0); Vertex = (0, 0)
Latus rectum = 4a = 8

Students can Download Chapter 10 Straight Lines Questions and Answers, Plus One Maths Chapter Wise Questions and Answers helps you to revise the complete Kerala State Syllabus and score more marks in your examinations.

Kerala Plus One Maths Chapter Wise Questions and Answers Chapter 10 Straight Lines

Plus One Maths Straight Lines Three Mark Questions and Answers

Question 1.
Find the equation of the line passing through the point (2, 2) and cutting off intercepts on the axis whose sum is 9.
Answer:
Let equation of the line is \(\frac{x}{a}+\frac{y}{b}\) = 1 ______(1)
Given a + b = 9
Since (1) passes through (2, 2) we have;
\(\frac{2}{b}+\frac{2}{b}\) = 1 ⇒ 2a + 2b – ab ⇒ 2 (a + b) = ab
⇒ 18 = ab. Then the numbers are 3 and 6.
Hence the equation of the line is
\(\frac{x}{3}+\frac{y}{6}\) = 1 ⇒ 6x + 3y = 18 ⇒ 2x + y = 6

OR

\(\frac{x}{6}+\frac{y}{3}\) = 1 ⇒ 3x + 6y = 18 ⇒ x + 2y = 6.

Question 2.
The perpendicular from the origin to a line meets it at the point (-2, 9), find the equation of the line.
Answer:
Slope of the line through the origin and (-2, 9)

Then the slope of the required line is \(\frac{2}{9}\).
Hence the equation is y – y₁ = m(x – x₁)
⇒ y – 9 = \(\frac{2}{9}\)(x – (-2))
⇒ 9y – 81 = 2x + 4 ⇒ 2x – 9x + 85 = 0.

Question 3.
Reduce the following into normal form. (3 score each)

1. \(\sqrt{3}\)x + y – 8 = 0
2. 3x + 3y – 1 = 0

Answer:
1. The given equation can be written in the form \(\sqrt{3}\)x + y = 8
Divide the above equation by

we get;

which is the normal form. On comparing with
x cos θ + y sin θ = p
Where; cosθ = \(\frac{\sqrt{3}}{2}\); sinθ = \(\frac{1}{2}\) and p = 4.

2. The given equation can be written in the form 3x + 3y = 1
Divide the above equation by

we get;

which is the normal form. On comparing with x cosθ + y sinθ = p
Where; cosθ = \(\frac{1}{\sqrt{2}}\); sinθ = \(\frac{1}{\sqrt{2}}\) and p = \(\frac{1}{3 \sqrt{2}}\).

Question 4.
Find the angle between the given lines. (3 score each)

1. y – \(\sqrt{3}\)x – 5 = 0 and \(\sqrt{3}\)y – x + 6 = 0
2. 3x – 2y + 9 = 0 and 2x + y – 9 = 0

Answer:
1. Given; y – \(\sqrt{3}\)x – 5 = 0 and \(\sqrt{3}\)y – x + 6 = 0

⇒ θ = 30°.

2. Given; 3x – 2y + 9 = 0 and 2x + y – 9 = 0
⇒ 2y = 3x + 9 and y = -2x + 9

Question 5.
Find the transformed equation of the straight line 2x – 3y + 5 = 0, when the origin is shifted to the point (3, -1) after translation of axes.
Answer:
Let coordinates of a point P changes from (x, y) to (X, Y) in new coordinate axes whose origin has the coordinates h = 3, k = -1. Therefore, we can write the transformation formulae as x = X + 3 and y = y – 1.

Substituting, these values in the given equation of the straight line, we get 2(X + 3) – 3 (Y – 1) + 5 = 0 or 2X – 3Y + 14 = 0. Therefore, the equation of the straight line in new system is 2x – 3y + 14 = 0.

Question 6.
Find what the following equations become when the origin is shifted to the point (1, 1) x² + xy – 3y² – y + 2 = 0.
Answer:
We can write the transformation formulae as x = X + 1 and y = Y+ 1.
The new equation is
(X + 1)² + (X + 1)(Y + 1) – 3(Y + 1)² – (Y + 1) + 2 = 0
X² + XY – 3Y² + 3X – 6Y + 1 = 0
Therefore, the equation of the straight line in new system is x² + xy – 3y² + 3x – 6y + 1 = 0.

Question 7.
1. Identify the figure in which the line has a positive slope. (1)
a.

b.

c.

d.

2. Find the x and y intercepts of the line 3x + 4y – 12 = 0 (2)
Answer:
1. Figure b

2. 3x + 4y = 12

x – intercept = 4; y – intercept = 3.

Plus One Maths Straight Lines Six Mark Questions and Answers

Question 1.
Consider the line 4x – 3y + 12 = 0

1. Find the equation of the line passing through the point (1, 2) and parallel to the given line. (2)
2. Find the distance between these two parallel lines. (1)
3. Which among the following lines is perpendicular to the line 4x – 3y + 12 = 0 (1)
   • 2x + 3y – 8 = 0
   • 4x – 3y + 5 = 0
   • x + y = 7
   • 3x + 4y + 9 = 0

Answer:
1. Equation of the parallel line 4x – 3y + k = 0
Passing through (1, 2)
4(1) – 3(2) + k = 0 ⇒ k = 2
⇒ 4x – 3y + 2 = 0.

2. Distance \(\left|\frac{12-2}{\sqrt{25}}\right|=\frac{10}{5}=2\).

3. 3x + 4y + 9 = 0.

Plus One Maths Straight Lines Practice Problems Questions and Answers

Question 1.
Find the slope of the lines passing through the points (1 score each)

1. (3, -2) and (-1, 4)
2. (4, -5) and (2, 1)
3. (0, -2) and (4, 3)

Answer:
1. Slope

2. Slope

3. Slope

Question 2.
Find a point on the x-axis, which is equidistant from the points (7, 6) and (3, 4).
Answer:
Let (x, 0) be the points on the x-axis. Then the distance will be same;
(x – 3)² + 16 = (x – 7)² + 36
⇒ x² – 6x + 9 + 16 = x² – 14x + 49 + 36
⇒ 14x – 6x = 49 + 36 – 9 – 16
⇒ 8x = 60 ⇒ x = \(\frac{15}{2}\)
Hence the point is (\(\frac{15}{2}\), 0).

Question 3.
Find the value of x for which the points (x, -1), (2, 1) and (4, 5) are collinear.
Answer:
The slope of the lines joining the points (x, -1) and (2, 1); (2, 1) and (4, 5) are same.

Question 4.
Line through the points (-2, 6) and (4, 8) is perpendicular to the line through the points (8, 12) and (x, 24). Find the value of x.
Answer:
The product of the slopes of the lines joining the points (-2, 6) and (4, 8); (8, 12) and (x, 24) will be equal to -1.

⇒ 12 = -3x + 24 ⇒ 3x = 12⇒ x = 4.

Question 5.
Find the equation the following lines satisfying the given conditions. (2 score each)

1. Passing through the point (-2, 3) with slope -4.
2. Passing through the point (-4, 3) with slope \(\frac{1}{2}\).
3. Line with y-intercept \(-\frac{3}{2}\) and slope \(\frac{1}{2}\)
4. Line with x-intercept – 3 and slope – 2.
5. Line which makes intercepts -3 and 2 on the x- and y-axis respectively.
6. Perpendicular distance from origin is 5 units and the angle the perpendicular makes with the positive direction of x-axis is 30°.
7. Passing through the point (-1, 1) and (2, -4).
8. Passing through the point (1, -1) and (3, 5).

Answer:
1. Equation of the line is y – y₁ = m(x – x₁)
⇒ y – 3 = -4(x – (-2))
⇒ y – 3 = -4x – 8 ⇒ 4x + y + 5 = 0.

2. Equation of the line is y – y₁ = m(x – x₁)
⇒ y – 3 = \(\frac{1}{2}\)(x – (-4))
⇒ 2y – 6 = x + 4 ⇒ x – 2y + 10 = 0.

3. Equation of the line is y = mx + c
⇒ y = \(\frac{1}{2}\) x – \(\frac{3}{2}\) ⇒ 2y = x – 3
⇒ x – 2y – 3 = 0.

4. Equation of the line is y = m(x – d)
⇒ y = -2(x – (-3)) ⇒ y = -2x + 6
⇒ 2x + y = 6.

5. Equation of the line is \(\frac{x}{a}+\frac{y}{b}\) = 1
⇒ \(\frac{x}{-3}+\frac{y}{2}\) = 1 ⇒ 2x – 3y = -6
⇒ 2x – 3y + 6 = 0.

6. Equation of the line is xcosθ + ysinθ = p
⇒ xcos30° + ysin30° = 5

7. Equation of the line is

8. Equation of the line is

Question 6.
Find the equation of the line passing through the point (-3, 5) and perpendicular to the line through the points (2, 5) and (-3, 6).
Answer:
Slope of the line through the points (2, 5) and (-3, 6)

Then the slope of the required line is 5. Hence the equation is y – y₁ = m(x – x₁)
⇒ y – 5 = 5(x – (-3)
⇒ y – 5 = 5x + 15 ⇒ 5x – y + 20 = 0.

Question 7.
Find the equation of the line that cut off equal intercepts on the coordinate axis and passes through the point (2, 3).
Answer:
Let the equal intercept is ‘a’, then equation of the required line is \(\frac{x}{a}+\frac{y}{a}=1\)
Since this line passes through the point (2, 3),
we have; \(\frac{2}{a}+\frac{3}{a}=1\) ⇒ \(\frac{5}{a}\) = 1 ⇒ a = 5
Therefore the equation of the line is \(\frac{x}{5}+\frac{y}{5}=1\)
⇒ x + y = 5.

Question 8.
P(a, b) is the mid-point of a line segment between axis. Show that equation of the line is \(\frac{x}{a}+\frac{y}{a}=2\).
Answer:
Since the P(a, b) is the mid-point of the line segment, then the x-intercept and the y-intercept will be 2a and 2b. Hence the equation of the line is

Question 9.
Find the slope, x-intercept and y-intercept of the following lines. (2 score each)

1. 3x – 4y + 10 = 0
2. 6x + 3y – 5 = 0
3. 4x – 3y = 6

Answer:
1. Given the equation of the line is 3x – 4y + 10 = 0 ⇒ 4y = 3x + 10

2. Given the equation of the line is 6x + 3y – 5 = 0 ⇒ 3y = -6x + 5

3. Given the equation of the line is
4x – 3y = 6 ⇒ 3y = 4x – 6

Question 10.
Find the distance between the parallel lines. (2 score each)

1. 3x – 4y + 7 = 0 and 3x – 4y + 5 = 0
2. 15x + 8y – 34 = 0 = 0 and 30x + 16y + 62 = 0

Answer:
1. The distance between the lines 3x – 4y + 7 = 0 and 3x – 4y + 5 = 0

2. The distance between the lines
15x + 8y – 34 = 0 = 0 and 15x + 8y + 31 = 0 is

Question 11.
Find the distance between the given point and the line. (2 score each)

1. Line 3x – 4y – 26 = 0 and point (3, -5)
2. Line 12(x + 6) = 5(y – 2)and point(-1, 1)

Answer:
1. The distance between the line and the

2. Express the line in standard form
12(x + 6) = 5(y – 2) ⇒ 12x + 72 = 5y – 10
⇒ 12x – 5y + 82 = 0
The distance between the line and the point is

Question 12.
Find the equation of the line parallel to the line 3x – 4y + 2 = Oand passing through the point (-2, 3).
Answer:
The equation of the line parallel to the line 3x – 4y + 2 = 0 is of the form 3x – 4y + k = 0.
Since it passes through (-2, 3), we have;
3(-2) – 4(3) + k = 0 ⇒ -6 – 12 + k = 0
⇒ -18 + k = 0 ⇒ k = 18
Hence the equation is 3x – 4y + 18 = 0.

Question 13.
Find the equation of the line x – 7y + 5 = 0 perpendicular to the line and having x-intercept 3.
Answer:
The equation of the perpendicular line will be 7x + y + k = 0.
Since x-intercept is 3, the line passes through the point (3, 0). So we have;
7(3) + 0 + k = 0 ⇒ 21 + 0 + k = 0 ⇒ k = -21
Therefore the equation is 7x + y – 21 = 0.

Question 14.
Find the new coordinates of point (3, -4) if the origin is shifted to (1, 2) by a translation.
Answer:
The coordinates of the new origin are h = 1, k = 2, and the original coordinates are given to be
x = 3, y = -4
X = x – h; Y = y – k
Substituting the values, we have
X = 3 – 1 = 2 and Y = -4 – 2 = -6
Hence, the coordinates of the point (3, -4) in the new system are (2, – 6).

Students can Download Chapter 1 The Discipline of Computing Questions and Answers, Plus One Computer Science Chapter Wise Questions and Answers helps you to revise the complete Kerala State Syllabus and score more marks in your examinations.

Kerala Plus One Computer Science Chapter Wise Questions and Answers Chapter 1 The Discipline of Computing

Plus One The Discipline of Computing One Mark Questions and Answers

Question 1.
Which is the base of Mayan’s Number System?
Answer:
Base 20

Question 2.
Greek Number System is known as _________.
Answer:
Ionian number system

Question 3.
Which was the first computer for basic arithmetic calculations?
Answer:
Abacus

Question 4.
Who invented logarithms?
Answer:
John Napier

Question 5.
What is the name of the machine developed by Blaise Pascal?
Answer:
Pascaline

Question 6.
Who was the first programmer in the world?
Answer:
Augusta Ada Lowelace

Question 7.
Computing machines recognizes and operates in ___________ language.
Answer:
Machine

Question 8.
What does EDVAC stand for?
Answer:
Electronic Discrete Variable Automatic Computer. It is designed by Von Neumann.

Question 9.
Give the name for a simple kind of theoretical computing machine.
Answer:
Turing Machine.

Question 10.
The Sumerian number system is also known as _________
Answer:
Sexagesimal

Question 11.
What are the features of Hindu Arabic Number system?
Answer:
The Hindu – Arabic number system had a symbol (0) for zero originated in India 1500 years ago. Its base is 1 0 and it is adopted by many countries.

Question 12.
How is the zero represented in the Babylonian Number System.
Answer:
Blank space

Question 13.
Write the number that is represented in the following Abacus.

Answer:

• 3 beads of the leftmost rod moved hence it represents 3.
• No beads (0) of the next rod moved hence it represents 0.
• 2 beads of the rightmost rod in the lower part is moved (2+) and one bead is moved in the upper part represents 5.

Hence 2 + 5 = 7
So the answer is 307

Question 14.
Which is the first automatic electromechanical computer?
(a) Pascaline
(b) Abacus
(c) Mark 1
(d) Analytical Engine
Answer:
(c) Mark 1

Question 15.
Find the correct match for each item in column A and B.

Answer:

Question 16.
Write the number represented in the abacus given below.

Answer:

• 2 Beads of the left most rod are moved hence it represents 2 – (a)
• 2 Beads of the next rod are moved (2) and one bead is moved in the upper part represents 5. Hence 2 + 5 = 7 – (b)
• 4 Beads of the next rod are moved (4) and one bead is moved in the upper part represents 5. Hence 4 + 5 = 9 – (c)

One bead of the right most rod in the lower part is moved (1) and one bead is moved in the upper part represents 5. Hence 1+5 = 6 – (d)
Join a, b, c and d hence the result is 2796

Question 17.
Who invented a machine to multiply any number by a number between 2 to 9? (1)
(a) Blaise Pascal
(b) John Napier
(c) G.W. Von Leibniz
(d) Joseph Marie Jacquard
Answer:
(c) G.W. Von Leibniz

Question 18.
Order the following technologies according to different generations of a computer (First to Fifth Generation) (1)

1. Transistor
2. Vacuum Tube
3. Artificial intelligence
4. Microprocessor

Answer:

1. Vacuum tube
2. Transistor
3. Microprocessor
4. Artificial intelligence

Question 19.
The number (158)10 can be represented in Hexadecimal number system as ______________
Answer:

Plus One The Discipline of Computing Two Mark Questions and Answers

Question 1.
Discuss the impact of the Hindu- Arabic numeral system in the world.
Answer:
It was originated in India around 1500 years ago and was a positional number system with symbol for zero. This greatest contribution is adopted by many of the countries.

Question 2.
Compare the Roman Number System and Mayan’s Number System.
Answer:
The Roman numerals consists of 7 letters such as I, V, X, L, C, D, M and the base is 7.
The Mayans used number system with base 20 because of the sum of the number of fingers and toes is 10 + 10 = 20. This was used for astronomical observations.

Question 3.
A full room sized computer in the first generation now becomes palm sized by the fourth generation. Explain the technological changes that made it possible.
Answer:
Vaccum tubes were used in first generation computers. Hence the size of computer was a size of full room. Instead of vaccum tubes transistors were used in the 2^nd generation hence size became smaller. In 3^rd generation, Integgted circuits (IC’s) were used. It reduced the size again in 4^th generation, microprocessors are used. It reduces the size again and again.

Plus One The Discipline of Computing Three Mark Questions and Answers

Question 1.
Discuss the developments of the number system from the Egyptian to the Chinese Era.
Answer:
Around 3000 BC Egyptians introduced a number system with base 10. They used unique symbols for 1 to 9, 10 to 90, 100 to 900 and 1000 to 9000. They write from right to left. The next era was of Sumerian/Babylonian number system its base was 60, the largest base also known as sexagesimal system.

They write from left to right. They use blank space for zero. Around 2500 BC Chinese introduced simplest and the most efficient number system. Its base was 10.

Question 2.
Discuss the features of Abacus.
Answer:
In 3000 BC Mesopotamians introduced this and it means calculating board or frame. It is considered as the first computer for basic arithmetical calculations and consists of beads on movable rods divided into two parts.

The Chinese improved the Abacus with seven beads on each wire. A horizontal divider separates the top two beads from the bottom five. The top two beads have a place value of 5 and the below five beads. The beads which are pushed against the horizontal bar represent the number.

Question 3.
Compare the Analytical Engine and Difference Engine of Charles Babbage.
Answer:
The intervention of human beings was eliminated by Charles Babbage in calculations by using Difference engine in 1822. It could perform arithmetic operations and print results automatically.

The Analytical Engine was a proposed mechanical general-purpose computer designed by English mathematician Charles Babbage. The Analytical Engine incorporated an arithmetic logic unit, control flow in the form of conditional branching and loops, and integrated memory.

Charles Babbage is considered as the “Father of computer” It is considered as the predecessor of today’s computer. This engine was controlled by programs stored in punched cards. These programs were written by Babbage’s assistant, Augusta Ada King, who was considered as the first programmer in the World.

Question 4.
Bring out the significance of Hollerith’s machine.
Answer:
In 1887, Herman Hollerith an American made first electromechanical punched cards with instructions for input and output. The card contained holes in a particular pattern with special meaning.

The Us Census Bureau had large amount of data to tabulate, that will take nearly 10 years. By this machine this work was completed in one year. In 1896, Hollerith started a company Tabulating Machine Corporation. Now it is called International Business Machines(IBM).

Question 5.
State the Moore’s Law and discuss its significance.
Answer:
The number of transistors on IC’s doubles approximately every two years. This law is called Moore’s Law, it is named after Gordon E Moore. It is an observation and not a physical or natural law. He predicted that the trend would continue for at least ten years. It is true and the trend continued for more than half a century.

Question 6.
Discuss the evolution of computer languages.
Answer:
The instructions to the computer are written in different languages. They are Low Level Language (Machine language), Assembly Language (Middle level language) and High Level Language (HLL). In Machine Language 0’s and 1’s are used to write , program. It is very difficult but this is the only language which is understood by the computer. In assembly language mnemohics (codes) are used to write programs.

Electronic Delay Storage Automatic Caleulator(EDSAC) built during 1949 was the first to use assembly language. In HLL English like statements are used to write programs. A – 0 programming language developed by Dr. Grace Hopper, in 1952, for UNIVAC-I is the first HLL.

A team lead by John Backus developed FORTRAN @IBM for IBM 704 computer and ‘Lisp’ developed by Tim Hart and Mike Levin at Massachusetts Institute of Technology. The other HLLs are C, C++, COBOL, PASCAL, VB, Java, etc. HLL is very easy and can be easily understood by the human being.

Usually programmers prefer HLL to write programs because of its simplicity. But computer understands only machine language. So there is a translation needed. The program which performs this job are language processors.

Question 7.
Discuss the working of Turing Machine.
Answer:
In 1936 Alan Turing introduced a machine, called Turing Machine. A Turing machine is a hypothetical device that manipulates symbols on a strip of tape according to a table of rules. This tape acts like the memory in a computer. The tape contains cells which starts with blank and may contain 0 or 1. So it is called a 3 Symbol Turing Machine.

The machine can read and write, one cell at a time, using a tape head and move the tape left or right by one cell so that the machine can read and edit the symbol in the neighbouring cells. The action of a Turing machine is determined by

1. the current state of the machine
2. the symbol in the cell currently being scanned by the head and
3. a table of transition rules , which acts as the program.

Question 8.
Explain Turing Test in detail?
Answer:
The Turing test is a test of a machine’s ability to exhibit intelligent behaviour equivalent to, or indistinguishable from, that of a human. The test involves a human judge engages in natural language conversations with a human and a machine designed to generate performance indistinguishable from that of a human being.

All participants are separated from one another. If the judge cannot reliably tell the machine from the human, the machine is said to have passed the test. The test does not check the ability to give the correct answer to questions; it checks how closely the answer resembles typical human answers. Turing predicted that by 2000 computer would pass the test.

Question 9.
Following are some facts related to evolution of computers :
(Usage of Transistors, Introduction of Very Large Scale Integrated Circuit, Construction of ENIAC) Categorize these under respective generations and briefly explain each. (3)
Answer:

• Transistor – Second generation
• VLSI – Fourth generation
• ENIAC – First generation

Question 10.
Why is it said “Turing machines are equivalent to modem electronic computers at a certain theoretical level”? (3)
Answer:
In 1887, Herman Hollerith an American made first electromechanical punched cards with instructions for input and output. The card contained holes in a particular pattern with special meaning. The Us Census Bureau had large amount of data to tabulate, that will take nearly 10 years.

By this machine this work was completed in one year. In 1896, Hollerith started a company Tabulating Machine Corporation. Now it is called International Business Machines(IBM).

Question 11.
Compare any three features of five generations of computers.
answer:

Plus One The Discipline of Computing Five Mark Questions and Answers

Question 1.
List out and explain the various generations of computers.
Answer:
There are five generations of computers from 16^th century to till date.

1. First generation computers (1940 – 1956):
Vacuum tubes were used in first generation computers. The input was based on punched cards and paper tapes and output was displayed on printouts. The Electronic Numerical Integrator and Calculator(ENIAC) belongs to first generation was the first general purpose programmable electronic computer built by J. Presper Eckert and John V. Mauchly.

It was 30 – 50 feet long, weight 30 tons, 18,000 vacuum tubes, 70,000 registers, 10,000 capacitors and required 1,50,000 watts of electricity. It requires Air Conditioner. They later developed the first commercially Successful computer, the Universal Automatic Computer(UNIVAC) in 1952.

The mathematician John Von Neumann designed a computer structure that structure is in use nowadays. Von Neumann structure consists of a central processing unit (CPU), Memory unit, Input and Output unit.

The CPU consists of arithmetic logical unit(ALU) and control unit(CU). The instructions are stored in the memory and follows the “Stored Program Concept”. Colossus is the secret code breaking computer developed by a British engineer Tommy Flowers in 1943 to decode German messages.

2. Second generation computers (1956 -1963):
Transistors, instead of Vacuum tubes, were used in 2nd generation computers hence size became smaller, less expensive, less electricity consumption and heat emission and more powerful and faster. A team contained John Bardeen, Walter Brattain and William Shockley developed this computer at Bell Laboratories.

In this generation onwards the concept of programming language was developed and used magnetic core (primary) memory and magnetic disk(secondary) memory.

These computers used high level languages(high level language means English like statements are used) like FORTRAN (Formula translation) and COBOL(Common Business Oriented Language). The popular computers were IBM 1401 and 1620.

3. Third generation computers (1964 – 1971):
Integrated Circuits(IC’s) were used. IC’s or silicon chips were developed by Jack Kilby, an engineer in Texas Instruments. It reduced the size again and increased the speed and efficiency. The high level language BASIC(Beginners All purpose Symbolic Instruction Code) was developed during this period The popular computers were IBM 360 and 370.

Due to its simplicity and cheapness more people were used . The number of transistors on IC’s doubles approximately every two years. This law is called Moore’s Law, it is named after Gordon E Moore. It is an observation and not a physical or natural law.

4. Fourth generation computers (1971 onwards):
Microprocessors are used hence computers are called microcomputers. Microprocessor is a single chip which contains Large Scale of IC’s(LSI) like transistors, capacitors, resistors etc due to this a CPU can place on a single chip. Later LSI were replaced by Very Large Scale Integrated Circuits(VLSI). The popular computers are IBM PC and Apple II.

5. Fifth generation computers (future):
Fifth generation computers are based on Artificial Intelligence(AI). AI is the ability to act as human intelligence like speech recognition, face recognition, robotic vision and movement etc. The most common Al programming language are LISP and Prolog.

Question 2.
Prepare a seminar report on evolution of positional number system.
Answer:
In positional number system, each and every number has a weight. Earlier sticks are used to count items such as animals or objects. Around 3000 BC the Egyptians use number systems with radix 10(base the number of symbols or digits used in the number system) and they write from right to left.

Later Sumerian/Babylonian use number system with largest base 60 and were written from left to right. They use space for zero instead of a symbol,0. In 2500 BC, the Chinese use simple and efficient number system with base 10 very close to number system used in nowadays.

In 500 BC, the Greek number system known as Ionian, it is a decimal number system and used no symbols for zero. The Roman numerals consists of 7 letters such as l, V, X, L, C, D, M.

The Mayans used number system with base 20 because of the sum of the number of fingers and toes is 10 + 10 = 20. It is called vigesimal positional number system. The numerals are made up of three symbols zero (shell shape, with the plastron uppermost), one (a dot) and five (a bar or a horizontal line).

To represent 1 they used one dot, two dots for 2, and so on. The Hindu – Arabic number system had a symbol (0) for zero originated in India 1500 years ago.

Question 3.
Discuss the various computing machines emerged till 1900’s.
Answer:
1. Abacus:
In 3000 BC Mesopotamians introduced this and it means calculating board or frame. It is considered as the first computer for basic arithmetical calculations and consists of beads on movable rods divided into two parts. The Chinese improved the Abacus with seven beads on each wire. Different Abacus are given below.

2. Napier’s bones:
A Mathematician John Napier introduced this in AP 1617.

3. Pascaline:
A French mathematician Blaise Pascal developed this machine that can perform arithmetical operations.

4. Leibniz’s calculator:
In 1673, a German mathematician and Philosopher Gottfried Wilhelm Von Leibniz introduced this calculating machine.

5. Jacquard’s loom:
In 1801, Joseph Marie Jacquard invented a mechanical loom that simplifies the process of manufacturing textiles with complex pattern. A stored program in punched cards was used to control the machine with the help of human labour. This punched card concept was adopted by Charles Babbage to control his Analytical engine and later by Hollerith.

6. Difference engine:
The intervention of human beings was eliminated by Charles Babbage in calculations by using Difference engine in 1822. It could perform arithmetic operations and print results automatically.

7. Analytical engine:
In 1833, Charles Babbage introduced this. Charles Babbage is considered as the “Father of computer” It is considered as .the predecessor of today’s computer. This engine was controlled by programs stored in punched cards. These programs were written by Babbage’s assistant, Augusta Ada King, who was considered as the first programmer in the World.

8. Hollerith’s machine:
In 1887, Herman Hollerith an American made first electromechanical punched cards with instructions for input and output. The card contained holes in a particular pattern with special meaning.

The Us Census Bureau had large amount of data to tabulate, that will take nearly 10 years. By this machine this work was completed in one year. In 1896, Hollerith started a company Tabulating Machine Corporation. Now it is called International Business Machines(IBM).

9. Mark -1:
In 1944 Howard Aiken manufactured automatic electromechanical computer in collaboration with engineers at IBM that handled 23 decimal place numbers and can perform addition, subtraction, multiplication and subtraction.

Students can Download Chapter 9 Sequences and Series Questions and Answers, Plus One Maths Chapter Wise Questions and Answers helps you to revise the complete Kerala State Syllabus and score more marks in your examinations.

Kerala Plus One Maths Chapter Wise Questions and Answers Chapter 9 Sequences and Series

Plus One Maths Sequences and Series Three Mark Questions and Answers

Question 1.
The 10^th term of an AP is 73 and the 20^th term is 43. Find the 44^th term.
Answer:
a + 9d = 73 ______(1)
a + 19d = 43 ______(2)
(1) – (2) ⇒ -10d = 30 ⇒ d = -3
(1) ⇒ a + 9(-3) = 73
⇒ a – 27 = 73 ⇒ a = 100
a₄₄ =100 + (44 – 1)(-3) = 100 + (43)(-3) = -29.

Question 2.
The 7^th term of an AP is 34 and the 15^th term is 74. Find the 40^th term.
Answer:
a + 6d = 34 ______(1)
a + 14d = 74 ______(2)
(1) – (2) ⇒ -8d = -40 ⇒ d = 5
(1) ⇒ a + 6(5) = 34 ⇒ a + 30 = 34 ⇒ a = 4
a₄₀ = 4 + (40 – 1)(5) = 4 + (39)(5) = 199.

Question 3.
Find the sum to 32 terms of an AP whose third term is 1 and the 6^th term is -11.
Answer:
a + 2d = 1 ______(1)
a + 5d = -11 ____(2)
(1) – (2) ⇒ -3d = 12 ⇒ d = -4
(1) ⇒ a + 2(-4) = 1 ⇒ a – 8 = 1 ⇒ a = 9
S₃₂ = \(\frac{32}{2}\)(2 × 9 + (32 – 1)(-4))
= 16(18 + (31)(-4)) = 16(18 – 124)
= 16(-106) = -1696.

Question 4.
The sum to n terms of a series is 7n² – 5n. Show that it is an AP and find the 15^th term.
Answer:
Given; S_n = 7n² – 5n
t_n = S_n – S_n-1
= 7n² – 5n – {7(n -1)² – 5(n – 1)}
= 7n² – 5n – {7n² – 14n + 7 – 5n + 5}
= 7n² – 5n – 7n² + 14n – 7 + 5n – 5}
= 14n – 12
Since t_n is linear the given series is an AP.
t₁₅ = 14(15) – 12 = 198.

Question 5.
Find three numbers in AP whose sum is -3 and whose product is 8.
Answer:
Let the three consecutive terms be a – d, a, a + d
a – d + a + a + d = -3 ⇒ 3a = -3 ⇒ a = -1
Given; (-1 – d)(-1)(-1 + d) = 8
⇒ (-1 – d)(-1)(-1 + d) = 8
⇒ (1 – d²)(-1) = 8 ⇒ 1 – d² = -8
⇒ d² – 9 ⇒ d = -3, 3
The AP is 2, -1, -4 or -4, -1, 2.

Question 6.
Find three numbers in AP whose sum is 21 and product is 231.
Answer:
Let the three consecutive terms be a – d, a, a + d
a – d + a + a + d = 21 ⇒ 3a = 21 ⇒ a = 7
Given; (7 – d)(7)(7 + d) = 231
⇒ (7 – d)(7 + d) = 33
⇒ 49 – d² = 33 ⇒ d² = 16 ⇒ d = -4, 4
The AP is 11, 7, 3 or 3, 7, 11.

Question 7.
Find the sum of all natural numbers between 100 and 1000 which are multiple of 5.
Answer:
105 will be the starting number and will end in 995.
⇒ 105 + (n – 1)5 = 995
⇒ (n – 1)5 = 890 ⇒ n – 1 = 178 ⇒ n = 179
S₁₇₉ = \(\frac{179}{2}\)(t₁ + t_n) = \(\frac{179}{2}\)( 105 + 995)
\(\frac{179}{2}\)(1100) = 98450.

Question 8.
If the AM and GM between two numbers are 34 and 16 respectively. Find the numbers.
Answer:
Let the numbers be a and b.

(2) ⇒ a(68 – a) = 256 ⇒ 68a – a² = 256
⇒ a² – 68a + 256 = 0
⇒ a² – 64a – 4a + 256 = 0
⇒ (a – 64)(a – 4) = 0 ⇒ a = 64, 4
When a = 64, b = 4 and when a = 4, b = 64.

Question 9.
If the p^th, q^th and r^th terms of a GP are a, b, c respectively, show that a^q-r b^r-p c^p-q = 1.
Answer:
t_p = AR^p-1 = a; t_q = AR^q-1 = b; t_r = AR^r-1 = c

Plus One Maths Sequences and Series Four Mark Questions and Answers

Question 1.
The sum of the first p,g,r terms of an AP are a,b,c respectively, prove that

Answer:
Given;

= 2A{q – r + r – p + p – q} + {(q – r)(p – 1) + (r – p)(q – 1) + (p – q)(r – 1)}D

= 2A{0} + {qp – rp – q + r + rq – pq – r + p + pr – qr – p – q}D
= 0 + {0}D = 0
⇒ \(\frac{a}{p}\)(q – r) + \(\frac{b}{q}\)(r – p) + \(\frac{c}{r}\)(p – q) = 0.

Question 2.
The ratio between the sums to n terms of two AP is 7n + 1: 4n + 27. Find the ratio of their 11^th terms.
Answer:

For 11^th term \(\frac{n-1}{2}\) = 10 ⇒ n = 21
Put n = 21 in (1), we have;

Question 3.
If the sum of p terms of an AP is the same as the sum of its q terms, show that the sum of its (p+q) terms is zero.
Answer:
Given; S_p = S_g
⇒ \(\frac{P}{2}\)[2a + (p – 1 )d] = \(\frac{q}{2}\)[2a + (q – 1 )d]
⇒ 2ap + (p – 1)pd = 2aq + (q – 1 )qd
⇒ 2a(p – q) + (p² – p – q² +q)d = 0
⇒ 2a(p – q) + (p² – q² – (p – q))d = 0
⇒ (p – q){2a + (p + q – 1)d} – 0
⇒ {2a + (p + q – 1 )d} = 0
⇒ \(\frac{p+q}{2}\){2a + (p + q – 1 )d} = 0
⇒ S_p+q = 0.

Question 4.
The sum of the first two terms of a GP is -4 and the fifth term is 4 times the third term.

1. Find the first term and the common ratio. (2)
2. Find the GP. (2)

Answer:
1. Given; S₂ = -4
a + ar = -4 ⇒ a(1 + r) = -4 _____(1)
Also given; t₅ = 4t₃ ⇒ ar⁴ = Aar²
⇒ r² = 4 ⇒ r = ±2
When r = 2(1) ⇒ a(1 + 2) = -4 ⇒ a = \(-\frac{4}{3}\)
When r = -2 (1) ⇒ a(1 – 2) = -4 ⇒ a = 4

2. When r= 2; GP is \(-\frac{4}{3},-\frac{8}{3},-\frac{16}{3}, \dots\)
When r = – 2; GP is 4, -8, 16,…

Question 5.
The sum of three numbers in GP is 38 and their product is 1728. Find the GP.
Answer:
Let the numbers be \(\frac{a}{r}\), a, ar, the given;

⇒ 6(1 + r + r²) = 19r ⇒ 6 + 6r + 6r² = 19r
⇒ 6r² – 13r + 6 = 0
⇒ 6r² – 9r – 4r + 6 = 0
⇒ 3r(2r – 3) – 2(2r – 3) = 0
⇒ (3r – 2)(2r – 3) = 0
⇒ r = \(\frac{2}{3}, \frac{3}{2}\)
Therefore GP is 8, 12, 18 or 18, 12, 8.

Question 6.
Find the three numbers in GP whose sum is 13 and the sum of whose squares is 91.
Answer:
Let the numbers be \(\frac{a}{r}\), a, ar, the given;

⇒ 3 + 3r + 3r² = 13r ⇒ 3r² – 10r + 3 = 0
⇒ 3r² – 9r – r + 3 = 0
⇒ 3r(r – 3) – (r – 3) = 0
⇒ (3 r – 1)(r – 3) = 0 ⇒ r = 3, \(\frac{1}{3}\)
Therefore the numbers are 1, 3, 9 and 9, 3, 1.

Question 7.
Find the sum to n terms of the following series. (4 score each)

1. 1 × 4 + 3 × 7 + 5 × 10 +……….
2. 1 × 2² + 2 × 3² + 3 × 4² +………..
3. \(\frac{1}{1 \times 2}+\frac{1}{2 \times 3}+\frac{1}{3 \times 4}+\ldots \ldots\)
4. 1² + 3² + 5²+……….

Answer:
1. The given series is the product of two AP
1 , 3, 5, with t_n = 1 + (n – 1)2 = 2n – 1
4, 7, 10, with t_n = 4 + (n – 1)3 = 3n + 1
Then the n^th term of the given series is
t_n = (2n – 1)(3n + 1) = 6n² – n – 1

2. The given series has two AP
1, 2, 3,……. with t_n = 1 + (n – 1)1 = n
2, 3, 4,…… with t_n = 2 + (n – 1)1 = n + 1
Then the n^th term of the given series is
t_n = n(n + 1)² = n(n² + 2n + 1) = n³ + 2n² + n

3.

4. The given series has an AP
1, 3, 5……… with t_n = 1 + (n – 1)2 = 2n – 1
Then the n^th term of the given series is
t_n = (2n – 1)² = 4n² – 4n + 1

Question 8.
(i) n^th term of some sequence are given below. Which term can be the n th term ofanAP? (1)
(a) a_n = n(n + 1)
(b) a_n = 2 + 5n
(c) a_n = 2ⁿ + 2
(d) a_n = n² + n + 1
(ii) If the sum of 12^th and 22^nd terms of an AP is 100. Find the sum of first 33 terms. (3)
Answer:
(i) (b) a_n = 2 + 5n.

(ii) t₁₂ + t₂₂ = 100 ⇒ a + 11d + a + 21d = 100
⇒ 2a + 32d = 100

Plus One Maths Sequences and Series Six Mark Questions and Answers

Question 1.
1. The product of first 3 terms of a GP is 1000. If 6 terms to the second term and 7 is added to the third term, the terms become an AP.
(a) Find the second term of GP. (1)
(b) Find the terms of the GP. (2)
2. Find the sum of n terms of the series 7 + 77 + 777 +…………… (3)
Answer:
1. (a) \(\frac{a}{r}\) × a × ar = 1000 ⇒ a = 10
(b) \(\frac{a}{r}\), a + 6, ar + 7 from an AP
(a + 6) – \(\frac{a}{r}\) = (ar + 7) – (a + 6)
16 – \(\frac{10}{r}\) = 10r + 7 – 16
⇒ 16r – 10 = 10r² – 9r
⇒ 10r² – 25r + 10 = 0 ⇒ r = 2; \(\frac{1}{2}\)
Hence, 20, 10, 5 and 5, 10, 20;

2. S_n = 7 + 77 + 777 +……………
= 7(1 + 11 + 111 +…………)
= \(\frac{7}{9}\)(9 + 99 + 999 +………..)
= \(\frac{7}{9}\)(10 – 1 + 100 – 1 + 1000 – 1 +………..)
= \(\frac{7}{9}\)(10 + 100 + 1000+……….-1 – 1 – 1-………)

Plus One Maths Sequences and Series Practice Problems Questions and Answers

Question 1.
Find the n^th term of the following sequence (2 score each)

1. 5, 2, -1, -4, -7,…
2. 12, 7, 2, -3, -8,…….

Answer:
1. a = 5; d = 2 – 5 = -3
t_n = a + (n – 1)d = 5 + (n – 1)(-3)
= 5 – 3n + 3 = -3n + 8

2. a = 12; d = 7 – 12 = -5
t_n = a + (n – 1)d = 12 + (n – 1)(-5)
= 12 – 5n + 5 = -5n + 17.

Question 2.
A sequence is given {a_n} by a_n = n² – 1, n ∈ N Show that it is not an AP.
Answer:
Common difference = a_n+1 – a_n
= (n + 1)² – 1 -(n² – 1)
= n² + 2n + 1 – 1 – n² + 1
= 2n + 1.
Common difference is not independent of n so not an AP.

Question 3.
Find the sum to

1. 15 terms of the AP 3, 7, 11,………
2. 20 terms of the AP 10, 7, 4,……..
3. 81 terms of the AP -1, \(\frac{1}{4}, \frac{3}{2}, \ldots \ldots .\) (2 score each)

Answer:
a = 3; d = 7 – 3 = 4

2. a = 10; d = 7 – 10 = -3

= 10(20 + 19(-3)) =10(-37) = -370

3. a = -1; d = \(\frac{1}{4}\) + 1 = \(\frac{5}{4}\)

Question 4.
Insert 6 arithmetic means between 3 and 24.
Answer:
Let t₁ = 3, t₂, t₃, t₄, t₅, t₆, t₇, t₈ = 24 be the sequence.
Then;
t₈ = 24 ⇒ a + 7d = 24 ⇒ 3 + 7d = 24
⇒ 7d = 21 ⇒ d = 3
Hence the arithmetic means between 3 and 24 are 6, 9, 12, 15, 18, 21.

Question 5.
If the n^th term of a GP -2, 4, -8, 16 is 1024. Find n.
Answer:
Given; r = \(\frac{4}{-2}\) = -2; t_n = 1024 ⇒ 1024 = ar^n-1
⇒ 1024 = -2(-2)^n-1 ⇒ 1024 = (-2)ⁿ
⇒ (-2)¹⁰ = (-2)ⁿ ⇒ n = 10.

Question 6.
If the n^th term of a GP 2, 2\(\sqrt{2}\), 4,………is 64. Find n.
Answer:
Given; r = \(\frac{2 \sqrt{2}}{2}=\sqrt{2}\);
t_n = 64
⇒ 64 = ar^n-1
⇒ 64 = 2(\(\sqrt{2}\))^n-1

⇒ n = 11.

Question 7.
Find the sum of first 20 terms of the GP \(\sqrt{3}, 2 \sqrt{3}, 4 \sqrt{3}, 8 \sqrt{3}, \ldots \ldots\).
Answer:
Given;

Question 8.
In a GP{a_n}, if a₁ = 3, a_n = 96 and S_n = 189. Find common ratio and n.
Answer:
Given; a₁ = 3, a_n = 96 ⇒ ar^n-1 = 96
⇒ 3 × r^n-1 = 96 ⇒ r^n-1 = 32 = 2⁵
⇒ r^n-1 = 2⁵
Then, r = 2; n = 6.

Question 9.
Find the sum to n terms of the following; (3 score each)

1. 9 + 99 + 999 +………..
2. 4 + 44 + 444 +………..

Answer:
S_n = 9 + 99 + 999 +………..
= 10 – 1 + 100 – 1 + 1000 – 1 +………..
= 10 + 100 + 1000 +…….-1-1-1-…….
= 10 + 10² + 10³ +………-n

2. S_n = 4 + 44 + 444 +………..
= 4(1 + 11 + 111 +……..)

Question 10.
The third term of a GP is 4. Find the product of the first five terms.
Answer:
Given; t₃ = 4 ⇒ ar² – 4
The product of the first five terms
= a × ar × ar² × ar³ × ar⁴
= a⁵r¹⁰ = (ar²)⁵ = 4⁵ = 1024.

Question 11.
Insert 4 geometric means between 4 and 972.
Answer:
Let t₁ = 4, t₂, t₃, t₄, t₅, t₆ = 972 be the sequence. Then;
t₆ = 972 ⇒ ar⁵ = 972 ⇒ 4r⁵ = 972
⇒ r⁵ = 243 = 3⁵ ⇒ r = 3
Hence the arithmetic means between 4 and 972 are 12, 36, 108, 324.

Question 12.
Find the sum to infinity in each of the following Geometric Progression. (2 score each)

Answer:

Students can Download Chapter 11 Chemical Coordination and Integration Questions and Answers, Plus One zoology Chapter Wise Questions and Answers helps you to revise the complete Kerala State Syllabus and score more marks in your examinations.

Kerala Plus One Zoology Chapter Wise Questions and Answers Chapter 11 Chemical Coordination and Integration

Plus One Chemical Coordination and Integration One Mark Questions and Answers

Question 1.
Listed below are the hormones of anterior pituitary origin. Tick the wrong entry.
(a) Growth hormone
(b) Follicle stimulating hormone
(c) Oxytocin
(d) Adrenocorticotrophic hormone
Answer:
(c) Oxytocin

Question 2.
Mary is about to face an interview. But during the first five minutes before the interview she experiences sweating, increased rate of heart beat, respiration etc. Which hormone is responsible for her restlessness?
(a) Estrogen and progesterone
(b) Oxytocin and vasopressin
(c) Adrenaline and noradrenaline
(d) Insulin and glucagon
Answer:
(c) Adrenaline and noradrenaline

Question 3.
The steroid responsible for balance of water and electrolytes in our body is
(a) Insulin
(b) Melatonin
(c) Testosterone
(d) Aldosterone
Answer:
(d) Aldosterone

Question 4.
Name the birth hormone.
Answer:
Oxytocin

Question 5.
A slow runner runs exceptionally fast when a dog chases him. Name the hormone involved in this situation.
Answer:
Adrenaline

Question 6.
Pars intermedia produce hormone. (GH, PRL.TSH, MSH)
Answer:
MSH

Question 7.
The excessive secretion of thyroxine hormone is followed by the enlargement of the thyroid glands. Name the disease occurs due to this,
Answer:
Exophthalmic goiter or Grave’s disease.

Question 8.
Thymosin is responsible for
(a) Raising the blood sugar level
(b) Raising the blood calcium level
(c) Increased production of T lymphocytes
(d) Decrease in blood RBC
Answer:
(c) Increased production of T lymphocytes

Question 9.
In the mechanism of action of a protein hormone, one of the second messengers is
(a) Cyclic AMP
(b) Insulin
(c) 73
(d) Gastrin
Answer:
(a) Cyclic AMP

Question 10.
Note the relationship between first two words and suggest a suitable word for fourth place.

1. Alpha cell : Glucagon – Beta cell : __________
2. Glucocorticoids : Cortisol – Mineralocorticoids: _________

Answer:

1. Insulin
2. Aldosterone

Question 11.
Expand the following.
ACTH
Answer:
ACTH – Adreno Cortico Trophic Hormone.

Question 12.
The excessive secretion of thyroxine hormone is followed by the enlargement of the thyroid glands. Name the disease occurs due to this.
Answer:
Exophthalmic goiter or Grave’s disease.

Plus One Chemical Coordination and Integration Two Mark Questions and Answers

Question 1.
Match the following.

Answer:

1. Hyperglycemic hormone – Glucagon
2. Pregnancy hormone – Progesterone
3. Inhibiting hormone – Somatostatin

Question 2.
Functions of certain hormones are given below. Identify the hormones.

1. Regulation of BMR
2. Differentiation of T cells
3. Stimulate gluconeogenesis, lipolysis and proteolysis.
4. Support pregnancy and act on mammary gland and stimulate milk secretion.

Answer:

1. Thyroxine
2. Thymosin
3. Glucocorticoids
4. Progesterone

Question 3.
Give example of

1. Hyper calcaemic hormone
2. Hyperglycemic hormone

Answer:

1. PTH (Parathyroid Hormone)
2. Glucagon

Question 4.
Name the hormone that regulate each of the following.

1. Storage of glucose as glycogen.
2. Sodium Potassium metabolism
3. Basal Metabolic Rate
4. Urinary elimination of water

Answer:

1. Insulin
2. Aldosterone
3. Thyroxine
4. ADH

Question 5.
The destruction of adrenal cortex leads to the low production of both glucocorticoids and mineralocorticoids.

1. Name the disease occurs due to this.
2. Give the symptoms of this disease.

Answer:

1. Addison’s disease.
2. loose weight, their blood glucose and sodium levels drop and potassium levels rise.

Question 6.
It is necessary to include iodised salt in our diet. Do you agree with this statement? Justify your answer.
Answer:
Yes. Deficiency of iodine in food causes simple goitre.

Question 7.
Diagrammatic representation of the mechanism of action of a steroid hormone is shown here.

1. Label A & B
2. Name any two protein hormones.

Answer:

1. Label A & B
   • A – MRNA
   • B – Proteins
2. FSH, LH

Question 8.
Match the following.

Answer:

Question 9.
Anitha saw a poisonous snake on her way to school. She was frightened and her heart rate and breathing rate increased.

1. Name the hormones which are dominant at that time in her blood.
2. Which endocrine gland produces the hormone?
3. Which is the name given to these hormones?
4. To which organ this endocrine gland is attached?

Answer:

1. Adrenaline and noradrenaline
2. Adrenal gland
3. Emergency hormones
4. Kidney

Question 10.
Read column A and using the terms given in bracket fill column B.
(Progesterone, Testosterone, Glucagon, Glucocorticoids, Prolactin, Vasopressin, Thymosin, Adrenaline, Follicle Stimulating hormone)

Answer:

Question 11.
Match the following.

Answer:

Question 12.
A boy witnessed an accident on his way to school. Following changes occurred in him. (Increased heart beat, profuse sweating, pupils dilated, shivering, pale face, increased respiration)

1. Name the hormone responsible for these changes?
2. Name the gland which secretes this hormone?

Answer:

1. Adrenaline
2. Adrenal gland

Question 13.
After delivery milk ejection starts in human females. It is due to the hormonal interaction. Name the hormones involved in this process.
Answer:
Estrogen, Progesterone, Prolactin and Oxytocin

Question 14.
Diagram shows the location of the various endocrine glands in human body. Identify the labelled parts.

Answer:

1. A – Hypothalamus
2. B – Thymus
3. C – Thyroid gland
4. D – Thymus
5. E – Pancreas
6. F – Adrenal gland
7. G – Ovary
8. H – Testis

Question 15.
Name a hormone that has the following action and give the source of the hormone.
Regulate blood sugar level – causes it to fall
Answer:
Insulin produced from Islets of Langerhans of pancreas.

Question 16.
Comment on the following flow chart.

Answer:
The figure explain the feed back mechanism between hypothalamus and pituitary by hypothalamo-pituitary axis. It also specifies the existence of feed back control mechanism between the hypothalamo-pituitary axis and other endocrine gland.

Question 17.
Note the relationship and suggest a suitable word in the gap.

1. Exocrine: ducted; Endocrine: ______
2. Adenohypophysis: Growth _______ hormone; Neurohypophysis: _________
3. Glucocorticoid: Cortisol; Mineralocorticoid: __________
4. Alpha cells: Glucagon; Beta cells: _________
5. Thymus: Infentile gland; Thymosin: __________
6. Diabetes mellitus: lnsulin; Diabetes insipidus: _________

Answer:

1. ductless gland
2. Vasopressin
3. Aldosterone
4. Insulin
5. Infentile hormone
6. Vasopressin

Question 18.
Complete the concept map.

Answer:

1. A – 80-120 mg/ml
2. B – Insulin
3. C – Glycogenesis
4. D – Glucagon

Question 19.
The male’s body is stronger and rigid than female’s body. Give reason.
Answer:
Androgens control the muscular development in males. This hormones are absent in females.

Question 20.
The sight and sound of a baby can induce secretion of a hormone in a nursing mother.

1. Name the hormone.
2. Give the functions of this hormone.

Answer:

1. Prolactin
2. Prolactin promote growth of mammary gland, secretion of milk and keeps corpus luteum functional.

Question 21.
Pancreas is found to be not functioning in a patient. How this condition will affect physiological functioning in him?
Answer:
Pancreas is a mixed gland, produce hormones and enzymes. Hormonal imbalance problem is due to insulin and glucagon and problems in digestion is due to absence of pancreatic juice.

Question 22.

The given bar diagram shows the relative amount of glucose and insulin in a normal man. Redraw the graph to show the conditions in a diabetic patient.
Answer:

Question 23.
Construct a table and arrange the following items in three columns with appropriate headings: Tetany, adenohypophysis, Insulin, Myxoedema, Parathyroid, Dwarfism, Growth hormone, Thyroid, Diabetes mellitus, Pancreas, Thyroxin, PTH.
Answer:

Question 24.
A person with a swelling in the thyroid gland consults a doctor.

1. What type of disease does this person have?
2. As a student of biology, what remedy would you suggest for these?

Answer:

1. Goitre
2. Iodine is essential for the functioning of the thyroid gland. It is normally obtained from drinking water and iodised table salt. Other sources are onion, seafood etc.

Deficiency of Iodine causes a disorder known as goitre. In this disorder thyroid glands swell up. So taking adequate amount of Iodised salt is a remedy for goitre.

Question 25.
The destruction of adrenal cortex leads to the low production of both glucocorticoids and mineralocorticoids.

1. Name the disease occurs due to this.
2. Give the symptoms of this disease.

Answer:

1. Addison’s disease.
2. Weight loss, blood glucose and sodium levels drop and potassium levels rise.

Question 26.
Match column B & C with column A.

Answer:

Question 27.

• Identify the gland in the picture.
• How many layers are present in the adrenal gland? Name them and write the hormones from these layers.

Answer:

1. Adrenal gland
2. Two layers are present in the adrenal gland. They are the outer cortex and inner medulla. Adrenal cortex secretes glucocorticoids, mineralocorticoids and sex corticoids. Adrenal medulla, secretes adrenalin and Noradrenalin.

Question 28.
1. A flowchart showing the process of production of testosterone, is given below.

2. Redraw the chart correctly if there are any mistakes.
Answer:

Question 29.
Find the odd one out in each group and give reason.

1. TSH, ACTH, FSH, LH, ADH
2. Cretinism, Myxoedema, Simple goitre, Gigantism
3. Thyrotropin releasing hormone, Corticotropin releasing hormone, Gonadotropin releasing hormone, Somatostatin.
4. Adenohypophysis, Neurohypophysis, Hypothalamus, pars intermedia.

Answer:

1. ADH, ADH is a hormone from neurohypophysis all others are from adenohypophysis.
2. Gigantism – Gigantism is due to the imbalance of growth hormone, all others are due to the imbalance of thyroid hormone.
3. Somatostatin – Somatostatin is an inhibitory hormone, all others are releasing hormone.
4. Hypothalamus – remaining three are different parts of pituitary gland.

Question 30.
Progesterone is called as pregnancy hormone. Do you agree? Give reason.
Answer:
Yes. Progesterone helps in

1. Maintenance of pregnancy
2. Implantation
3. Formation of Placenta
4. Growth of mammary glands

Question 31.
Fill in the blanks:
Hormones Target gland

1. Hypothalamic hormones _______
2. Thyrotrophin(TSH) ____
3. Corticotrophin (ACTH)
4. Gonadotrophins(LH, FSH)

Answer:

1. Hypothalamic hormones Pituitary
2. Thyrotroph in (TSH) Thyroid gland
3. Corticotrophin (ACTH) Adrenal gland
4. Gonadotrophins(LH, FSH) Gonads

Plus One Chemical Coordination and Integration Three Mark Questions and Answers

Question 1.
Can you identify us?
Clue – We are two diseases, one is related to sugar in urine. Our first name is same and second name is different. Distinguish us.
Answer:

• Diabetes Mellitus and Diabetes insipidus.
• Diabetes mellitus is due to Insulin deficiency and elimination of sugarthrough urine.
• Diabetes Insipidus is due to ADH deficiency and elimination of excess water through urine.

Question 2.
When checked the blood sugar of a person, it is found to be 220 mg/100ml of blood; Is this a disorder?

1. If this is a disorder, name the disorder.
2. Which hormonal imbalance is responsible for this condition?
3. Can you suggest any remedial measures to get rid of?

Answer:

1. Yes. Diabetes mellitus
2. Insulin
3. Diet control, Exercise etc.

Question 3.
During Sunbath the body colour is changed.

1. Why the body colour is changed?
2. Do you think any hormone is responsible for this? If so, name the hormone.
3. Explain the action of hormone.

Answer:

1. Melanocyte stimulating hormone produced during the sunbath is the reason for the change.
2. Melanocyte stimulating hormone from pituitary.
3. Melanocyte stimulating hormone (MSH) stimulate cutaneous pigmentation by dispersion of melanin pigments.

Question 4.
A pregnant lady is admitted in a hospital for delivery. But the delivery is delayed than the expected time. Doctor prescribed to take an injection.

1. Which hormone is injected to the lady?
2. Which glands in the body normally secretes this hormone?
3. How this hormone helps in this process?

Answer:

1. Oxytocin
2. Oxytocin is secreted from pituitary.
3. Oxytocin helps the contraction of uterine wall and thereby helps in parturition

Question 5.
Expand the following.

1. AMP
2. ANF
3. ICZN

Answer:

1. CAMP – Cyclic Adenosine Mono Phosphate
2. ANF – Atrial Natriuretic Factor
3. ICZN – International Code of Zoological Nomenclature

Question 6.
The diagram shows the interaction between the hypothalamus, pituitary and thyroid glands. Arrow indicates the probable pathway of direct influence. Answer the following questions.

1. Name the hormones A and B.
2. State the effect of hormone A on the thyroid.
3. State the effect of hormone B on the hypothalamus.
4. Describe the control on pituitary by the hypothalamus in this situation.

Answer:

1. A-Thyroid stimulating hormone B-Thyroxine
2. Thyroid stimulating hormone stimulates the thyroid follicle to synthesis two hormones, thyroxine and triiodothyronine.
3. The high level of thyroxin inhibits the production of thyrotropin releasing hormone.
4. Thyrotropin releasing hormone of the hypothalamus stimulates the pituitary to release thyrotropin stimulating hormone (TSH) which in turn stimulate thyroid gland to produce thyroxine

Question 7.
Copy the table and fill in the blanks using appropriate words.

Answer:

Plus One Chemical Coordination and Integration NCERT Questions and Answers

Question 1.
Give example of:

1. Hypercalcemic harmone and hypoglycemic hormone
2. Hypercalcemic hormone
3. Gonadotropic hormone
4. Progestational hormone
5. Blood Pressure lowering hormone
6. Androgens and estrogens

Answer:

1. Hyperglycemic hormone. Glucagon, Hypoglycemic Hormone. Insulin
2. Hype – calcemic hormone. Parathyroid Hormone (PTH)
3. Gonadotrophic hormones. Luteinizing Hormone and Follicle Stimulating Hormone
4. Progestrone and estrogen
5. Blood pressure lowering hormone. Atrial Natriuretic Factor.
6. Androgen, Testosterone Estrogens. Estrone, Estradiole, Estriol

Question 2.
Which hormonal deficiency is responsible for the following?

1. Diabetes mellitus
2. Goitre
3. Cretinism

Answer:

1. Insulin
2. Thyroid
3. Thyroid

Question 3.
Briefly mention the mechanism of action of FSH
Answer:
In males, FSH and androgens regulate spermatogenesis. FSH stimulates growth and development of the ovarian follicles in females.

Question 4.
Match the following.

Answer:
(a) – (ii)
(b) – (iv)
(c) – (i)
(d) – (iii)

Plus One Chemical Coordination and Integration Multiple Choice Questions and Answers

Question 1.
Endemic goitre is a state of
(a) increased thyroid function
(b) normal thyroid function
(c) decreased thyroid function
(d) moderate thyroid function
Answer:
(c) decreased thyroid function

Question 2.
Name the hormone that stimulates the secretion of gastric juice.
(a) Rennin
(b) enterokinase
(c) Enterogastrone
(d) gastrin
Answer:
(d) gastrin

Question 3.
Islets of Langerhans are found in
(a) anterior pituitary
(b) kidney cortex
(c) spleen
(d) endocrine pancreas
Answer:
(d) endocrine pancreas

Question 4.
The hormone responsible response flight, flight and fight response is
(a) adrenalin
(b) thyroxin
(c) ADH
(d) oxytocin
Answer:
(a) adrenalin

Question 5.
Foetal ejection reflex in human female is induced by
(a) pressure exerted by amniotic fluid
(b) release of oxytocin from pituitary
(c) fully developed foetus and placenta
(d) differentiation of mammary glands
Answer:
(b) release of oxytocin from pituitary

Question 6.
A health disorder that results from the deficiency of thyroxin in adults and characterized by
(i) a low metabolic rate
(ii) increase in body weight
(iii) tendency to retain water in tissues is
(a) hypothyroidism
(b) simple goitre
(c) myxoedema
(d) cretinism
Answer:
(c) myxoedema

Question 7.
Gland responsible for calcium metabolism is
(a) thymus
(b) thyroid
(c) parathyroid
(d) adrenal
Answer:
(c) parathyroid

Question 8.
Secretion is under control of neurosecretory nerve axons in
(a) pineal gland
(b) adrenal cortex
(c) anterior pituitary
(d) posterior pituitary
Answer:
(d) posterior pituitary

Question 9.
The smallest endocrine gland is
(a) thyroid
(b) parathyroid
(c) pituitary
(d) adrenal
Answer:
(c) pituitary

Question 10.
Gigantism and acromegaly are due to
(a) hypothyroidism
(b) hyperthyroidism
(c) hypopituitarism
(d) hyperpituitarism
Answer:
(d) hyperpituitarism

Question 11.
Spermatogenesis is under the regulatory influence of
(a) ADH
(b) FSH
(c) LH
(d) STH
Answer:
(b) FSH

Question 12.
Which hormone is secreted in a woman if pregnancy has occurred?
(a) Oestrogen
(c) Luteinizing hormone
(d) Chorionic gonadotrophin
Answer:
(d) Chorionic gonadotrophin

Question 13.
Insulin and glucagon are transported to target organ by
(a) lymph
(b) blood
(c) pancreatic duct
(d) cystic duct
Answer:
(d) cystic duct

Question 14.
Neurons of people suffering from diabetes insipidus do not secrete
(a) enzyme
(b) steroid
(c) fatty acid
(d) ADH
Answer:
(d) ADH

Question 15.
The hormone that increases the blood calcium level and decreases its excretion by kidney is
(a) parathormone
(b) calcitonin
(c) thyroxin
(d) insulin
Answer:
(a) parathormone

Question 16.
If the pituitary gland of an adult rat is surgically removed, which of the following endocrine glands will be less affected?
(a) Adrenal cortex
(b) Adrenal medulla
(c) Thyroid
(d) Gonads
Answer:
(b) Adrenal medulla

Question 17.
Oestrogen and testosterone are steroid hormones, and are most likely bind to
(a) membrane ions channels
(b) enzyme-linked membrane receptors
(c) G-protein linked membrane receptors
(d) cytoplasmic receptors
Answer:
(d) cytoplasmic receptors

Question 18.
Steroid hormones easily pass through the plasma membrane by simple diffusion because they
(a) are water soluble
(b) contain carbon and hydrogen
(c) enter through pores
(d) are lipid soluble
Answer:
(d) are lipid soluble

Question 19.
A hormone secreted by the endocrinal cells of duodenal mucosa which influences the release of pancreatic juice is
(a) relaxin
(b) cholecystokinin
(c) secretin
(d) progesterone
Answer:
(b) cholecystokinin

Question 20.
Which of the following hormones does not contain a polypeptide?
(a) Prostaglandin
(b) Oxytocin
(c) Insulin
(d) Antidiuretic hormone
Answer:
(b) Oxytocin

Students can Download Chapter 8 Binomial Theorem Questions and Answers, Plus One Maths Chapter Wise Questions and Answers helps you to revise the complete Kerala State Syllabus and score more marks in your examinations.

Kerala Plus One Maths Chapter Wise Questions and Answers Chapter 8 Binomial Theorem

Plus One Maths Binomial Theorem Three Mark Questions and Answers

Question 1.
Write the middle term in the expansion of the following; (3 score each)

Answer:
i) Here 7 is an odd number. Therefore there are two middle terms \(\left(\frac{7+1}{2}=4\right)^{t h}\) and \(\left(\frac{7+1}{2}+1=5\right)^{t h}\), ie; 4^th and 5^th terms in the above expansion.

ii) Here 10 is an even number. Therefore middle terms \(\left(\frac{10}{2}+1=6\right)^{t h}\) term in the above expansion.

iii) Here 17 is an odd number. Therefore there are two middle terms \(\left(\frac{17+1}{2}=9\right)^{t h}\), ie; 9^th and 10^th terms in the above expansion. \(\left(x+\frac{2}{\sqrt{x}}\right)^{17}\).

Question 2.
Find the term independent of x in the following expansion. (3 score each)

Answer:
i) General term = t_r+1 = (-1)^r12C_r(x)^12-r \(\left(\frac{1}{x}\right)^{r}\)
=(-1)^r12C_r(x)^12-r-r = (-1)^r12C_r(x)^12-2r
Term independent of x in the expansion will be the term in which the power of x is zero, ie; 12 – 2r = 0 ⇒ 12 = 2r
⇒ r = 6
t₇ = (-1)⁶12C⁶x^12-2(6)
= \(\frac{12 \times 11 \times 10 \times 9 \times 8 \times 7}{1 \times 2 \times 3 \times 4 \times 5 \times 6}\) = 924.

ii) General term = t_r+1 =(-1)^r9C_r(x²)^9-r\(\left(\frac{1}{x}\right)^{r}\)
= (-1)^r9C_r(x)^18-2r-r = (-1)^r9C_r(x)^18-3r
Term independent of x in the expansion will be the term in which the power of x is zero.
ie; 18 – 3r = 0 ⇒ 18 = 3r ⇒ r = 6
t₇ = (-1)⁶9C₆x^18-3(6)
= \(\frac{9 \times 8 \times 7}{1 \times 2 \times 3}\) = 84.

iii) General term = t_r+1


Term independent of x in the expansion will be the term in which the power of x is zero.

iv) General term = t_r+1


Term independent of x in the expansion will be the term in which the power of x is zero.

Question 3.
Find the coefficient of x¹⁰ in the expansion of \(\left(2 x^{2}-\frac{3}{x}\right)^{11}\).
Answer:
General term = t_r+1

Given; 22 – 3r = 10 ⇒ 12 = 3r ⇒ r = 4
t₅ = (-1)⁴11C₄2^11-4 x^22-3(4) 3⁴
= 11C₄2⁷3⁴x¹⁰
Therefore the coefficient of x¹⁰ is 11C₄2⁷3⁴.

Question 4.
Find the coefficient of a⁵b⁷ in the expansion of (a – 2b)¹².
Answer:
General term = t_r+1 = (-1)^r12C_r(a)^12-r(2b)^r
= (-1)^r12C_r(a)^12-r2^rb^r
The term containing a⁵b⁷ is obtained by putting r = 7
⇒ t₈ = (-1)⁷12C₇(a)^12-72⁷b⁷
Therefore the coefficient of a⁵b⁷ is
(-1)⁷12C₇2⁷ = -12C₇2⁷.

Question 5.
Find the coefficient of (3 score each)

1. x¹¹ in the expansion of \(\left(x-\frac{2}{x^{2}}\right)^{17}\)
2. x⁹ in the expansion of \(\left(3 x^{2}+\frac{5}{x^{3}}\right)^{12}\)
3. x²⁰ in the expansion of \(\left(3 x^{3}-\frac{2}{x^{2}}\right)^{40}\)

Answer:
1. General term

The term containing x¹¹ is obtained by
17 – 3r = 11 ⇒ 6 = 3r ⇒ r = 2
⇒ t₃ = (-1)²17C₂ (x)^17-3(2) 2² = 17C₂(x)¹¹ × 4
Therefore the coefficient of x¹¹ is 17C₂ × 4
= \(\frac{17 \times 16}{1 \times 2}\) × 4 = 544

2. General term

The term containing x⁹ is obtained by
24 – 5r = 9 ⇒ 15 = 5r ⇒ r = 3
⇒ t₄ = 12C₃(3)^12-3(x)^24-5(3)5³
= 12C₃(3)⁹(x)⁹5³
Therefore the coefficient of x⁹ is 12C₃(3)⁹5³.

3. General term = t_r+1
= (-1)^r40C_r(3x³)^40-r (\(\frac{2}{x^{2}}\))^r
= (-1)^r40C_r(3)^40-rx^120-3r (2)^r x^-2r
= (-1)^r40C_r(3)^40-rx^120-5r(2)^r
The term containing x²⁰ is obtained by
120 – 5r = 20 ⇒ 100 = 5r ⇒ r = 20
⇒ t₂₁ = (-1)²⁰40C₂₀(3)^40-20(x) ^120-5(20) 2²⁰
= 40C₂₀(3)²⁰(x)²⁰2²⁰
Therefore the coefficient of x²⁰ is 40C₂₀(3) ²⁰ 2²⁰.

Question 6.

1. Find the term independent of x in the expansion of \(\left(x^{2}+\frac{2}{x}\right)^{6}\) (3)
2. If the middle term in the expansion of \(\left(x^{m}+\frac{2}{x}\right)^{6}\) is independent of x, find the value of m.

Answer:
1. t_r+1 = ⁿC_ra^n-rb^r = ⁶C_r(x²)^6-r\(\left(\frac{2}{x}\right)^{r}\)
= ⁶C_rx^12-2rx^-r(2)^r = ⁶C_rx^12-3r(2)^r
For term independent of x;
12 – 3r = 0 r = 4
t₅ = ⁶C₄(2)⁴ = ⁶C₂ × 16 = \(\frac{6 \times 5}{1 \times 2}\) × 16 = 240

2. m = 1

Plus One Maths Binomial Theorem Four Mark Questions and Answers

Question 1.

1. Write the general term in the expansion \(\left(\frac{3 x^{2}}{2}-\frac{1}{3 x}\right)^{6}\) (2)
2. Find the term independent of x in the above expansion. (2)

Answer:
1. General term = t_r+1

2. Term independent of x in the expansion will be the term in which the power of x is zero.
ie; 12 – 3r = 0 ⇒ 12 = 3r ⇒ r = 4

Plus One Maths Binomial Theorem Practice Problems Questions and Answers

Question 1.
Expand the following. (2 score each)

1. (3a² – 2b)⁴
2. (3 – 4x²)⁵
3. \(\left(\frac{x}{2}-2 y\right)^{6}\)
4. \(\left(\frac{x}{2}-2 y\right)^{6}\)

Answer:
1.

2.

3.

4.

Question 2.
Write the general term in the expansion of the following; (2 score each)

Answer:
i) General term = t_r+1
= (-1)^r6C_r(x²)^6-r(y)^r
= (-1)^r6C_rx^12-ry^r.

ii) General term = t_r+1

iii) General term = t_r+1

iv) General term = t_r+1

Question 3.
If the coefficient of x² in the expansion of (1 + x)ⁿ is 6 then the positive value of n.
Answer:
t_r+1 = nC_rx^r, the term containing x² is obtained by putting r= 2.
nC₂ = 6 ⇒ \(\frac{n(n-1)}{2}\) = 6 ⇒ n(n -1) = 12
⇒ n(n -1) = 4 × 3 ⇒ n = 3.

Question 4.
Find the 13^th term in the expansion of \(\left(9 x-\frac{1}{3 \sqrt{x}}\right)^{18}\).
Answer:

= 18C₆(3)^12-12 = 18C₆ = 18564

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Kerala Plus One Accountancy Chapter Wise Questions and Answers Chapter 2 Theory Base of Accounting

Plus One Accountancy Theory Base of Accounting One Mark Questions and Answers

Question 1.
The rules and guidelines used in preparing accounting reports are called
(a) Accounting rules
(b) Basic rules
(c) Generally Accepted Accounting Principles
Answer:
(c) Generally Accepted Accounting Principles.

Question 2.
An accounting entity is an
(a) Accounting concept
(b) Accounting convention
(c) Modify Principle
Answer:
(a) Accounting concept

Question 3.
The Policy of ‘anticipate no profit and provide for all possible losses’ arises due to convention of
(a) Matching
(b) Conservatism
(c) Consistency
Answer:
(b) Conservatism.

Question 4.
A business unit is assumed to have an indefinite life comes under
(a) Going concern concept
(b) Business entity concept
(c) Money Measurment Concept
Answer:
(a) Going concern concept

Question 5.
Contingent liabilities are usually shown as a footnote in the balance sheet as per the following accounting principles.
(a) Consistency
(b) Disclosure
(c) Materiality
Answer:
(b) Disclosure

Question 6.
During the lifetime of an entity, accounting produce financial statements in accordance with which basic accounting concept.
(a) Conservation
(b) Matching
(c) Accounting period
(d) None of these
Answer:
(c) Accounting period.

Question 7.
Revenue is generally recognised at the point of sale denote the concept of ……….
(a) Consistency
(b) Objectivity
(c) Revenue Realisation
(d) None
Answer:
(c) Revenue Realisation

Question 8.
Revenue recognition is an /a
(a) Assumption
(b) Principle
(c) Accounting standard
Answer:
(b) Principle.

Question 9.
Accounting standard deals with depreciation accounting is
(a) As-5
(b) As-16
(c) As-6
(d) As-9
Answer:
(c) As-6

Question 10.
The ……….. Principle requires that the same accounting method should be used from one accounting period to the next.
Answer:
Consistency.

Question 11.
Companies must prepare financial statement at least yearly due to the ………….. assumption.
Answer:
Accounting Period.

Question 12.
Accounting standards are issued by ………… in India.
Answer:
Institute of Chartered Accounts of India.

Question 13.
Accounting standards in India are formulated and governed by …………. which was set up in ………
Answer:
Accounting Standard Board (ASB), 1977.

Question 14.
SEBI stands for
Answer:
Securities and Exchange Board of India.

Question 15.
ICAI stands for
Answer:
The Institute of Chartered Accountants of India.

Question 16.
………….. and ………….. generally referred to as the essence of financial accounting.
Answer:
The accounting concepts and accounting standards.

Question 17.
Find the odd one and give a reason,
(a) Dual aspect
(b) Historical cost
(c) Accounting period
(d) Verifiability and objectivity.
Answer:
(c) Accounting period, it is ah accounting assumption, But all others are accounting principles.

Question 18.
Revenue from sale of products is realized when
(a) the sale is made
(b) the cash is collected
(c) the production is completed
(d) the order placed to supply goods.
Answer:
(a) The sale is made.

Question 19.
Accounting principles are generally based on
(a) Practicability
(b) Subjectivity
(c) Convenience in recording
Answer:
(a) Practicability

Question 20.
Normally assets are recorded at cost price. This because
(a) Assets can be realized at the lime of winding up.
(b) Historical cost concept.
(c) Going concern concept
(d) All of these
Answer:
(d) All of these.

Plus One Accountancy Introduction to Accounting Two Mark Questions and Answers

Question 1.
Explain cash system of accounting and Mercantile system of accounting.
Answer:
Cash system or Receipt Basis:- Under this system, only actual cash receipts and payments are considered. Non-cash items such as outstandings, advances, and credit transactions are ignored. Mercantile system or Accrual Basis. Under this system, all items of income and expenditure, both cash items as well as non-cash items, such as outstanding and accrued incomes and expenses are taken into account.

Question 2.
1. Dual aspect concept – Two aspects of a transaction are recorded.
________________- Expected loss should be taken in to account.
2. Accounting Principles – Principles followed by accountants
___________________- Norms to be observed by the accountant
Answer:

1. Principle of prudence or conservatism.
2. Accounting Standards.

Question 3.
Explain IFRS.
Answer:
International Financial Reporting Standards (IFRS) are globally accepted accounting standards developed by the International Accounting Standard Board (IASB). IFRS is a set of accounting standards for reporting different types of business transactions and events in the financial statements. The objective is to facilitate international comparison for the true and fair valuation of a business enterprise.

Question 4.
Complete the following circle.

Answer:

Question 5.
What is GAAP?
Answer:
Generally Accepted Accounting Principles and Practices (GAAP), is a set of rules and practices that are followed while recording transactions and in preparing the financial statements.

The accounting assumptions, principles and modifying principles, as well as accounting standards, form the foundation upon which GAAP is built.

Question 6.
“Information delayed is information denied ”. State the principle applicable behind this statement.
Answer:
‘Timeliness principle’:
Timeliness implies that the financial statements are to be prepared and published in time. The relevance, dependability, and utility of the financial information depends on the timely publication of financial statements. The users of financial statements needs timely information.

Question 7.
Fixed Assets are depreciated over their useful life rather than over a shorter period. State the relevant accounting assumption. Explain.
Answer:
‘Going Concern Assumption’:
According to this concept, business will continue its operation long enough to allocate the cost of fixed assets over their useful lives against the income.

Question 8.
Financial Information should be neutral and free from bias. Comment on this statement with reference to the relevant accounting Principle.
Answer:
‘Verifiability and objectivity principle’:
This principle states that the accounting data provided in the books of accounts should be verifiable and dependable.

Question 9.
The sales achieved by a salesman and the commission payable to him is recorded in the books of accounts. But efficiency and intelligence of salesman is not recorded. Explain the reason with reference to the relevant accounting principle.
‘Money Measurement Concept’:
According to this concept, transactions that can be measured in terms of money only are recorded in the books of accounts. “The skill and intelligence of the salesman is not measurable in money terms.

Question 10.
Timely information, even if it is not cent percent reliable is better than reliable information which is late. Comment on this statement by quoting the accounting principle.
Answer:
‘Timeliness principle’:
Timeliness implies that the financial statements are to be prepared and published in time. The relevance, dependability, and utility of the financial information depends on the timely publication of financial statements. The users of financial statements need timely information.

Question 11.
‘Akash’ places an order on 1.1.2005 with Bino for the supply of machinery for Rs. 5,00,000/-. On receipt of the order, Bino purchases raw materials employs workers, produces machinery and delivers it to Akash on 1.2.2005. Akash makes the payment on 10.02.2005. On which date, the revenue is recognized? Substantiate your answer by quoting the relevant accounting concept. ‘Revenue Recognition Concept’.
Answer:
Revenue is recognized on 1.2.2005, i.e. when the title of goods passes from the seller to the buyer.

Question 12.
Star trading Co.Ltd. buys a piece of land for Rs. 50,00,000. After 2 years the value of land came to Rs. 70,00,000. But the accountant does not consider the increase in the value of Rs. 20,00,000 in the books of accounts and the land remains at Rs. 50,00,000 in the books. Do you agree with the accountant? If so, on what ground?.
Answer:
The accountant’s viewpoint is correct. This is based on the principle ‘Historical cost’. According to this principle, assets are to be recorded at their cost price and this cost is the basis for all subsequent accounting for those assets.

Question 13.
Mr. Muhammed, a sole trader, purchased a TV for Rs. 12,000/- for his domestic use and asks his accountant to record this as a business expense. But the accountant, argues that it is the violation of the accounting principle. Is he right? If so, prove your answer by highlighting the relevant accounting principle.
Answer:
The accountant’s viewpoint is correct because the business is separate from the businessmen as per the ‘‘Accounting Entity Concept”.

Question 14.
Provision for discount on debtors is accounted and provision for discount on creditors is not accounted. Why? State the relevant accounting principle.
Answer:
Accountant anticipates no profit but provide for all possible losses while recording business transaction. Conservatism principle or prudence.

Question 15
Name the systems of recording transactions in the book of accounts.
Answer:

1. Double Entry System.
2. Single Entry System.

Plus One Accountancy Introduction to Accounting Three Mark Questions and Answers

Question 1.
When a Proprietor purchased furniture at Rs. 10,000 for business purposes, he paid transportation and load¬ing charges of Rs. 1000 for bringing the furniture to the location of business premises. State whether it is possible to add the transportation and loading charges to the purchased price of furniture? What is the underlying principle behind it?
Answer:
1. Yes, it is possible to add transportation and loading charges to the purchased price of furniture.

2. “Historical cost principle” – This principle requires that all transactions should be recorded at their acquisition cost. The cost of acquisition refers to the cost of purchasing the asset and expenses incurred in bringing the assets to the intended condition and location of use.

Question 2.
Mr. Rajan Thomas invested Rs. 5,00,000 in his business. He is treated as a creditor to the extent of Rs. 5,00,000 by the business. Write the relevant accounting assumption and explain it.
Answer:
Accounting Entity Assumption:
This concept assumes that the entity of business is different from its owners. According to this concept, all the transactions of the business are recorded in the books of the business from the point of view of the business as an entity and even the owner is treated as a creditor to the extent of his capital.

Question 3.
An accountant followed a particular method of accounting in one year and in the next year he changes the method. Is it possible to get a better idea about the operation of the business?
Answer:
According to the Principle of “Consistency”, the frequent change in the accounting policies will adversely affect the reliability and comparability of financial information. The users of the financial statements assume that the business unit follows the same accounting principles and practices in preparing the financial statement.

If a change is adopted the business enterprises is required to record the fact as footnotes and to show the impact of such changes on financial affairs.

Question 4.
Classify the following into accounting assumptions, principles and modifying principles Accounting entity concept, dual concept, Money Measurement, Matching Principles, going concern concept, Accounting period concept, cost-benefit principle, consistency principle, Full disclosure principle, Timeliness, Historical cost principle.
Answer:

Question 5.
“Proprietor is treated as creditor to the extent of his capital.”

1. Write the relevent accounting assumption
2. Explain the concerned accounting assumption in relation to the statement given.

Answer:
Accounting Entity Assumption:
This concept assumes that the entity of business is different from its owners. According to this concept all the transactions of the business are recorded in the books of the business from the point of view of the business as an entity and even the owner is treated as a creditors to the extent of his capital.

Question 6.
“For every debit, there is an equal and corresponding Credit”.

1. Explain the statement by citing an example.
2. State the relevant accounting principle.

Answer:
1. “Every transaction has dual aspect i.e. debit and credit”.
For example Anish started business with Rs. 20,000. The effect of this transaction is that

• It increases cash (asset) Rs. 20,000.
• It increases capital Rs. 20,000.

The above transaction can be shown in the form of an accounting equation as follows Assets = Liabilities = Capital 20,000 = 0 + 20,000.

2. Duality Principle.

Question 7.
When should revenue be recognised? Are there exceptions to the general rule?
Answer:
Revenue is assumed to be realised when a legal right to receive it arises in the point of time when goods have been sold or services has been rendered. There are certain exceptions to the general rule of revenue realisation.

1. In the case of construction projects, revenue is realised before the contract is complete.
2. When the goods are sold on hire purchase, the amount collected in instalments is treated as realised.

Plus One Accountancy Introduction to Accounting Four Mark Questions and Answers

Question 1.
Mr. Damodar has confusion about the basic accounting assumptions. Can you help him to solve his confusion?

OR

Basic accounting assumption provides a foundation for the accounting process. Explain the Various accounting assumptions.
Answer:
Assumptions constitute the foundation of accounting. It lays down the general principles to be followed while preparing financial statements. There are four accounting assumptions. They are

1. Accounting Entity Assumption.
2. Money MeasurementAssumption
3. Going Concern Assumption
4. Accounting Period Assumption

1. Accounting Entity Assumption:
This concept assumes that the entity of business is different from its owners. The business is treated as a unit or entity separate from the person who control it. The proprietor is treated as a creditor to the extent of the amount invested by him on the assumption that he has given money and the business has received it.

2. Money MeasurementAssumption:
According to this concept, transaction that can be measured in terms of money only are recorded in the books of accounts. This helps to record different kinds of economic activities on a uniform basis. A business may have certain events that actually influence its working but is not capable of being expressed in monetary terms and hence, not record in the books of accounts. For eg: quality of products, sales policy, efficiency of M.D., etc.

3. Going Concern Assumption:
According to this concept, the business unit is assumed to have an indefinite life. There is no intention to wind up or end the business in the near future. Thus, considering the business as a perpetual one, its records are separately kept and maintained.

4. Accounting Period Assumption:
Under this concept, the accountings are done on a day-to-day basis are analysed for a particular period to find out the net results of the business as well as the financial position on a specific date.

Plus One Accountancy Introduction to Accounting Six Mark Questions and Answers

Question 1.
As an Accountant, explain the accounting principles you followed while preparing accounting records. Accounting principles are the general rules which govern the accounting techniques. The following are the major principles used in the accounting procedure.

1. Duality Principle:
According to this concept, each and every business transaction has two aspects- a giving aspect and a receiving aspect. The giving aspect of a transaction is called ‘credit’ and the receiving aspect of a transaction is called ‘debt’. Based on the duality principle, accounting equation is developed, ie; Asset = Liabilities + Capital.

2. Historical Cost Principle:
This principle requires that all transactions should be recorded at their acquisition cost. This principle is called historical, because the balance of assets and liabilities is carried forward from year to year to its acquisition cost, irrespective of increase or decrease in the market value of assets.

3. Matching Principle:
Under this concept, all the expenses, as well as the revenues of a particular period, should be accounted or otherwise it should be matched. In other words, it is the process of matching the revenue recognised during the period and the costs should be allocated to the period to obtain the revenue.

4. Full disclosure Principle:
This principle states that all information significant to the users of financial statements should be disclosed. It requires that all facts necessary to make financial statements not misleading must be disclosed.

5. Revenue Recognition Principle:
Revenue is the amount that a business earns through sale of goods or services. This principle helps in ascertaining the amount of revenue and the point of time at which it was realized. This principle is also called revenue realisation principle.

6. Verifiability and Objectivity Principle:
This principle states that the accounting data provided in the books of accounts should be verifiable and dependable. The figures exhibited in the financial statements should have supportive evidence such as bills, vouchers, etc. The evidence substantiating the business transaction should be objective, i.e., free from the bias of the accountants or others.

Question 2.
Explain modifying principles of accounting?
Answer:
There are certain general conventions or principles which supplement the basic principles for the preparation of accounting records and financial statements. They are called modifying conventions or principles. The important modifying principles are:

1. Cost-Benefit
2. Materiality
3. Consistency
4. Prudence or conservatism
5. Timeliness
6. Substance over legal form
7. Variation in accounting practices.

1. Cost-Benefit Principle:
This principle is a generally accepted norm that the cost of doing anything must not exceed the possible benefit that may be derived. This is applicable in the case of accounting also. Money spent for undertaking accounting work should definitely provide more benefit than the cost incurred.

2. Materiality Principle:
Materiality means relevance or importance or significance. As per this principle all material facts should be disclosed in the financial statements, but insignificant and immaterial facts need not be disclosed in details. For example, purchase of items like pen, pencil, scissors etc. are to be recorded as assets but practically these items are treated as expenses under the head stationery.

3. Consistency Principle:
Consistency means steadiness or unchanging nature. Accounting policies and practices adopted must be consistent for relatively reasonable period of time. The comparison of the financial statements of one year with that of another year will be effective and meaningful only if accounting practices and methods remain unchanged over year.

4. Conservatism or Prudence Principle This principle:
calls for losses while recording accounting information but at the same time does not permit anticipation of profits. This principle implies that while preparing financial statements all possible losses are to be provided for but incomes can be recognized only when there is certainty. It is base on the principle of prudence that stock is valued at market price or cost price whichever is les and provision is provided for doubtful debts.

5. Timeliness principle:
Timeliness implies that the financial statements are to be prepared and published in time. The relevance, dependability, and utility of the financial information depends on the timely publication of financial statements. The users of financial statements need timely information.

6. Principle of Substance over legal form:
This principle states that transactions and financial events are accounted for and presented in accordance with their substance and economic reality and not merely their legal form.

7. Variation in Accounting practices:
For the preparation of financial statements, the business enterprises are following certain specific guidelines and practices which are called generally accepted accounting principles and practices (GAAP). Certain industries may sometimes deviate from GAAP because of the peculiarity of its operation and practices.

Plus One Accountancy Introduction to Accounting Eight Mark Questions and Answers

Question 1.
What do you mean by Accounting standards? Name the accounting standards issued by ASB.
Answer:
An Accounting standard is a selected set of accounting policies or broad guidelines regarding the principles and methods to be chosen out of several alternatives. Standard conforms to applicable laws, customs, usage and business environment. In India, the Accounting standard Board (ASB) has the authority of issuing Accounting standards.

The sole objective of Accounting standards is to harmonise the diversified policies to make the system more useful and effective. They lay down the norms of accounting policies and practices byway of codes or guidelines to direct as to how the items appearing in the financial statements should be dealt with in the books of account and shown in the financial statements and annual reports.

The ASB has issued 29 accounting standards. They are as follows:

1. AS 1 – Accounting Standard 1 – Disclosure of Accounting Policies.
2. AS 2 – Accounting Standard 2 – Valuation of Inventories
3. AS 3 – Accounting Standard 3 – Cash Flow Statements
4. AS 4 – Accounting Standard 4 – Contingencies and Events occurring after the Balance sheet date.
5. AS 5 – Accounting Standard 5 – Net Profit or Loss for the period, prior period items and changes in accounting policies
6. AS 6 – Accounting Standard 6 – Depreciation Accounting.
7. AS 3 – Accounting Standard 7 – Accounting for Construction Contracts
8. AS 8 – Accounting Standard 9 – Accounting for Research and Development
9. AS 9 – Accounting Standard 9 – Revenue Recognition
10. AS 10-Accounting Standard 10-Accounting for Fixed Assets
11. AS 11 -Accounting Standard 11 -Accounting for the effects of changes in Foreign exchange rates
12. AS 12 – Accounting Standard 12 -Accounting for Government grants
13. AS 13-Accounting Standard 13-Accounting for Investments
14. AS 14 – Accounting Standard 14 – Accounting for Amalgamations
15. AS 15-Accounting Standard 15-Accounting for Retirement Benefit in the Financial statements of Employers
16. AS 16 – Accounting Standard 16 – Borrowing costs
17. AS 17 – Accounting Standard 17 – Segment Reporting
18. AS 18 – Accounting Standard 18 – Related party Disclosures
19. AS 19 – Accounting Standard 19 – Leases
20. AS 20 – Accounting Standard 20 – Earning per share
21. AS 21 – Accounting Standard 21 – Consolidated financial statements
22. AS 22 – Accounting Standard 22 – Accounting for taxes on income
23. AS 23 – Accounting Standard 23 – Accounting for investments in associates in consolidated financial statements
24. AS 24 – Accounting Standard 24 – Accounting for discontinued operations
25. AS 25 – Accounting Standard 25 – Interim Financial Reporting
26. AS 26 – Accounting Standard 26 – Intangible Assets
27. AS 27 – Accounting Standard 27 – Financial Reporting of interests in joint ventures.
28. AS 28 – Accounting Standard 28 – Impairment of assets
29. AS 29 – Accounting Standard 29 – Provisions, contingent liabilities, and contingent assets

Students can Download Chapter 10 Neural Control and Coordination Questions and Answers, Plus One zoology Chapter Wise Questions and Answers helps you to revise the complete Kerala State Syllabus and score more marks in your examinations.

Kerala Plus One Zoology Chapter Wise Questions and Answers Chapter 10 Neural Control and Coordination

Plus One Neural Control and Coordination One Mark Questions and Answers

Question 1.
Resting membrane potential is maintained by
(a) Hormones
(b) Neurotransmitters
(c) Ion pumps
(d) None of the above
Answer:
(c) Ion pumps

Question 2.
The function of our visceral organs is controlled by
(a) Sympathetic and somatic neural system
(b) Sympathetic and para sympathetic neural system
(c) Central and somatic nervous system
(d) None of the above
Answer:
(b) Sympathetic and para sympathetic neural system

Question 3.
Which of the following is not involved in Knee-jerk reflex?
(a) Muscle spindle
(b) Motor neuron
(c) Brain
(d) Inter neurons
Answer:
(c) Brain

Question 4.
Mark the vitamin present in Rhodopsin
(a) Vit A
(b) Vit B
(c) Vit C
(d) Vit D
Answer:
(a) Vit A

Question 5.
Human eyeball consists of three layers and it encloses
(a) Lens, iris, optic nerve
(b) Lens, aqueous humor and vitreous humor
(c) Cornea, lens, iris
(d) Cornea, lens, optic nerve
Answer:
(b) Lens, aqueous humor and vitreous humor

Question 6.
Name the structure which connects two cerebral hemisphere
Answer:
Corpus callosum

Question 7.
Name the region of keenest vision in our eye.
Answer:
Fovea or yellow spot

Question 8.
After playing on a giant wheel, we lost our balance. Why?
Answer:
Change in position of Otolith in ear gives mal information to the brain.

Question 9.
Arrange the ear ossicles in order from inner ear to the tympanum.
Incus Stapes Malleus
Answer:
Stapes → Incus → Malleus

Question 10.
Name the receptors respond to irritants such as ammonia, vinegar or hot chilly pepper.
Answer:
Pain receptors

Question 11.
The tissues of eye and ear contain photoreceptors and the auditory receptors. Likewise, some are found as film of liquid coating in the membranes of the receptor cells. Name it.
Answer:
Chemoreceptors

Plus One Neural Control and Coordination Two Mark Questions and Answers

Question 1.
Light ray → Lens → retina → pupil → cornea → brain → Vitreous chamber → Aqueous chamber → Optic nerve.
Correct the sequence.
Answer:
Light ray → Cornea → Aqueous chamber → Pupil → Lens → Vitreous chamber → Optic nerve → brain.

Question 2.
Classify the following into 3 groups and give appropriate headings.
Thalamus, Corporaquadrigerhina, Pons, hypothalamus, Cerebrum, Medulla Oblongata.
Answer:

Question 3.

1. Identify the above stages of nerve impulse conduction
2. Name the ions involved in this process

Answer:

1. Identification of above stages
   • I – Resting membrane potential
   • II – Action potential
2. Na⁺, K⁺

Question 4.
Differentiate Blindspot and Yellow spot.
Answer:
1. Blind spot:
The spot at the back of the eye where the optic nerve originates is known as blind spot. Rods and canes are absent here. So the image falling at this spot cannot be carried to the brain.

2. Yellow spot:
Lateral to the blind spot is a depressed area called yellow spot which contains only canes. It is the area of sharpest vision.

Question 5.
‘Sudden death may occur due to the damage of Medulla oblongata’. Why?
Answer:
The medulla oblongata contains several centres which regulate heartbeat, respiration, gastric, secretion, vomiting etc. It carries the nerve fibres which connect spinal cord and cerebrum.

Question 6.
Observe the figure given below.

1. Identify A & B
2. Write the main function of part A.

Answer:

1. Identification of A & B
   • A – Cochlea
   • B – Semicircular canals
2. Maintenance of balance of the body and posture.

Question 7.
Where do you find bipolar and multipolar neurons in our body?
Answer:

• Bipolar neuron – Retina of eye
• Multipolar neuron – Cerebral cortex

Question 8.
While playing cricket, the ball hit a boy’s head. He immediately vomited and felt difficulty to breath.

1. Identify the part of brain affected.
2. Give the function of the affected part.

Answer:

1. Medulla oblongata
2. The medulla oblongata contains several centres which regulate heartbeat, respiration salivation, vomiting, etc. It carries the nerve fibers which connect the spinal cord and cerebrum.

Question 9.
Nocturnal animals like bats and owls have vision during night. Give reason.
Answer:
Nocturnal animals like owls have only rods in the retina. The rods are sensitive to dim light and enable to see dim light and at night.

Question 10.
Observe the flow diagram of the pathway of a light ray entering the eye.
Lightray → lens retina →pupil → aqueous chamber → vitreous chamber → cornea → optic nerve → brain

1. Correct the sequence.
2. If the light ray falls on the blindspot, what will happen?

Answer:

1. Light ray → cornea → pupil → aqueous chamber → lens → vitreous chamber → retina → optic nerve → brain
2. Rods and cones’are absent in blind spot and so the image falling at this spot cannot be carried to the brain.

Question 11.
Arrange the following in the order of reception and transmission of the sound wave from the external auditory canal.
Answer:
Cochlear nerve, eardrum, stapes, incus, malleus, cochlea
Ear drum → malleus → incus → stapes → cochlea → cochlear nerve

Question 12.
Observe the given diagram.

1. Write any one difference between A and B.
2. Through which neurone the impulse conduction is faster. Justify.

Answer:

1. The difference between A and B
   • A – non-myelinated neuron
   • B – Myelinated neuron
2. Impulse conduction is faster through myelinated neuron.

Question 13.
The given diagram is a part of myelinated nerve fibre.

1. Identify the part where there is no myelin sheath.
2. Name the type of conduction going on in that type of nerve fibre.

Answer:

1. Nodes of Ranvier
2. Saltatory conduction

Question 14.
Observe the diagram.

Identify A, B, C and D.
Answer:

1. A – Nissl’s granule
2. B – Myelin sheath
3. C – NodeofRanvier
4. D – Synaptic knob

Question 15.

1. Identify the picture.
2. Write the peculiarity of the picture.

Answer:

1. Bipolar neuron
2. Bipolar neurons are the neurons with one axon and one dendrite

Question 16.
A patient approaches a doctor with a problem that he was not able to sleep for the last 6 months. The doctor said that it may be due to the defect in the relay centre of his brain and advised him to take sedative pills.

1. Which part of brain is described here as ‘relay centre’?
2. What are the actions of sedative pills in body?

Answer:

1. Thalamus
2. Sedative pills work in the Thalamus and prevent the transmission of impulses to cerebrum. Depress brain activity produce feelings of calmness, relaxation, drowsiness and deep sleep.

Question 17.
It is said that the number and pattern of convolutions are associated with the degree of intelligence.

1. Is it true?
2. If yes, give the scientific reason for it.

Answer:

1. True
2. More intelligent forms like mammals especially primates have more convolutions than lower forms.

Question 18.
Diagram below represents a neuron at resting membrane potential (RMP).

(a) How RMP is maintained?
(b) Draw the changes at the time of depolarization and explain how it happens.
Answer:
(a) RMP is maintained by
1. A resting membrane is poorly permeable to Na⁺ ions and Cl^– ions. But more permeable to K⁺ ions. The extra cellular fluid has a high concentration of Na⁺ ion and low concentration of K⁺ ions. But the intracellular medium has a reverse condition due to the permeability.

2. Sodium-Potassium pump maintain higher concentration of Na⁺ ions outside the membrane compared to the concentration of Na⁺ inside

(b)

When a neuron is stimulated, there is a momentary reversal of the resting potential. The reversal of polarity is known as depolarisation. When stimulated a resting membrane, sodium pump suddenly stops and sodium ions begin to enter the cells.

The presence of higher concentration of Na⁺ inside the cell cause the inside membrane +ve and outside become -ve. This condition is called depolarisation.

Question 19.
Copy the diagram and label A, B, C, D.

Answer:

1. A – Synaptic vesicles
2. B – Synaptic cleft
3. C – Presynaptic neuron
4. D – Post synaptic neuron

Question 20.
A diagram showing the chemical synaptic transmission is given below. Based on the diagram prepare a flow chart showing the process of synaptic transmission.

Answer:

Question 21.
While playing cricket the ball hit a boy’s head. He immediately vomited and felt difficulty to breath.

1. Identify the part of brain which may be affected by the incident.
2. Give functions of this particular part of brain.

Answer:

1. Medulla oblongata
2. Control heart beat, regulate respiration, control circulation, control digestion, control peristalsis etc.

Question 22.
Observe the schematic representation related to maintaining resting membrane potential of an axon.

1. What are the conditions exhibited in the figure for maintaining resting membrane potential?
2. What are the other conditions of resting membrane potential (not exhibited in the diagram)

Answer:

1. High permeability of axon wall for K⁺, Highly negative charge protein in axoplasm, Na⁺ – K⁺ pump.
2. High K⁺ concentration inside the axon, High Na⁺ concentration outside the axon, Low Na⁺ permeability the axon wall, 3Na⁺ ions for 2K⁺ ions.

Question 23.
Observe the portion of Brain and answer the following questions.

1. Name the covering of brain.
2. Identify A, B and C.

Answer:

1. Meninges
2. Identification of A, B, and C
   • A – Dura mater
   • B – Arachnoid mater
   • C – Pia mater

Question 24.
Identify the following figure and label the parts.

Answer:
Reflex arc

Question 25.
When we step on a thorn, we withdraw our legs suddenly, if we are taking this thorn out, we will not withdraw our legs. Comment on these two statements.
Answer:
1^st statement is a reflex action and 2^nd statement is a process controlled by Brain.

Question 26.
Observe the picture.

1. Identify the part labelled as (x).
2. In our retina there are more rod cells than cones. But our vision in darkness is poor. Give scientific explanation to this fact.

Answer:

1. Yellow spot or fovea
2. Image forms normally on yellow spot. In yellow spot rods are less and cones are more. So low dim light vision.

Question 27.
Find out the relationships and write the suitable word in the IVth place.

1. Cornea: Sclera; Yellow spot: _________
2. Incus: Middle ear; Cochlea: __________
3. Scala vestibuli Perilymph; Scala media: __________
4. Rods: Rhodopsin; Cones: ____________

Answer:

1. Retina
2. Inner ear
3. Endolymph
4. lodopsin

Question 28.
Arrange the organs according to the mechanism of hearing.
Oval window, Perilymph, Organ of Corti, Ear Ossicles, Pinna, Tectorial membrane, Endolymph, Auditory canal, auditory nerve, brain
Answer:
Pinna → Auditory canal → Ear ossicles → oval window → Perilymph → Endolymph → Organ of Corti → Auditory nerve → Brain

Question 29.
Copy the diagram and mark Tectorial membrane and sensory hair cell.

Answer:

Question 30.
Analyse the table and fill in the blanks given in the table with appropriate words.

Answer:

Question 31.
In animals like bats and owls photoreceptor cells of the retina have mainly rods.

1. What can you infer from this?
2. Write down the function of rods and cons.

Answer:

1. Rod cells bring about vision in night or dim light vision.
2. The function of rods and cons:
   • Cones are cone shaped sensory cells of retina that bring about vision in day light and also distinguish colours. There are three types of cone cells for sensing primary colours red, green and blue.
   • Rods are rod shaped sensory cells of the retina that bring about vision in night, but cannot distinguish colours.

Question 32.
Ear converts sound waves into neural impulses which are sensed and processed by the brain that enable to recognise sound. Construct a schematic diagram showing the mechanism of hearing the sound of a bell.
Answer:

Question 33.
Arrange the structures found in the retina from inside to outside.
Cone cells, optic nerve, Ganglion cells, Bipolar neuron.
Answer:
Optic nerve → ganglion cells → Bipolar neuron → Cone cells

Question 34.

1. Name the receptors of smell found as mucous coated thin, yellowish patch of modified pseudo stratified epithelium.
2. Where is these receptors located?

Answer:

1. Olfactory epithelium
2. It is located at the roof of the nasal cavity on either sides of the nasal septum.

Question 35.
The diagram shows a section through a part of the human ear.

1. Identify the parts labelled A, B and C.
2. Which parts are involved in the equilibrium of the body.

Answer:

1. Identification of A, B and C.
   • A – Semicircular canal
   • B – Vestibule
   • C – Cochlea
2. Semicircular canal and vestibule.

Question 36.
Why does red flower look black in dim light?
Answer:
Stimulation of cone cells require high intensity of light. In dim light the cone cells are not stimulated. That is why the red flower looks black in dim light.

Question 37.
The taste buds of Humans are located in pockets around the papillae on the surface and sides of the tongue, but some on the surface of the pharynx and the larynx.

1. What are the four basic taste senses?
2. Find out their location in tongue

Answer:

1. sweet, sour, salt, and bitter
2. sweet and salty on the front, bitter on the back, and sour on the sides.

Question 38.
The touch receptors are either free dendritic endings or encapsulated dendritic endings present in the skin.

1. What are the main functions of receptors of free dendritic endings?
2. Name the main receptors of encapsulated dendritic endings. Give its functions.

Answer:

1. They respond to pain and temperature
2. Meissner’s corpuscles, Pacinian corpuscles
   • Meissner’s corpuscles:
     These are found just beneath the skin epidermis in dermal papillae and abundant in fingertips and soles of the feet. These are light pressure receptors.
   • Pacinian corpuscles:
     These are scattered deep in the dermis and in the subcutaneous tissue of the skin .These are stimulated by deep pressure.

Plus One Neural Control and Coordination Three Mark Questions and Answers

Question 1.
The following steps are involved during synaptic transmission. Rearrange them in correct order.

1. Release of neurotransmitter at synaptic cleft.
2. Generation of a new potential at post synaptic neuron
3. Arrival of impulse at the axon terminal
4. Binding of neurotransmitter with specific receptor
5. Movement of synaptic vesicle towards the membrane.

Answer:

1. Arrival of impulse at the axon terminal
2. Movement of synaptic vesicle towards the membrane
3. Release of neurotransmitter at synaptic cleft.
4. Binding of neurotransmitter with specific receptor.
5. Generation of a new potential at post synaptic neuron.

Question 2.
A sharp tap is given at your knee cap with rubber hammer. You suddenly stretch your leg.

1. Give name of this response.
2. Which nerve centre is involved in this action?
3. Construct a flow chart of the pathway of impulses in this action.

Answer:

1. Reflex action
2. Spinal cord
3. Receptors in knee ® Sensory fibres ® Interneuron effector organs response → Motor fibres

Question 3.

1. Identify the organ and label A and B.
2. Gait of a drunkard is not normal. Why?
3. The death sentence given by the court is always by “hanging” in our country. Why this is preferred to other ways?

Answer:

1. Organ-Brain
   • A – Gray matter
   • B- White matter
2. Alcohol effects the cerebellum which control and co-ordinate voluntary muscular action.
3. Because death is less painful and fast as atlas pierces the medulla oblongata and it is smashed. Medullar oblongata is the control centre of respiratory and cardiac action. Stoppage of which cause death.

Question 4.
Make necessary correction in the flow chart given.

Answer:

Question 5.
Match the following.

Answer:

Question 6.
Suppose you dramatically escaped from a motor accident. Your heart beat and rate of respiration was increased at that moment.

1. Name the hormone involved in this change.
2. Which part of nervous system control these action?
3. Enlist other physiological changes that you may feel at that time.

Answer:

1. Adrenalin and Non-adrenaline.
2. Medulla oblongata.
3. High metabolic rate, High BP, High body temperature, High level of glucose in blood, Pupil diabetes.

Question 7.
Observe the figure below show two ion channels.

1. This pump does not work on part of the axon. Why?
2. Which are the ions participated in the process?
3. Suggest the use of this process in nerve impulse conduction?

Answer:

1. Myelin sheath: Myelin sheath act as insulator for axon and it increases the speed of nerve impulse conduction by saltatory conduction.
2. Na⁺ and K⁺
3. It creates action potential when stimulated by allowing the inward movement of Na⁺ ions through the Na channel and outward movement of K⁺ through K⁺ channel.

Question 8.
Neurons are the structural and functional unit of nervous system.

1. Based on the number of axon and dendrites, how the neurons are classified.
2. Give examples for each.
3. Which neuron receives signal from a sensory organ and transmit the impulse to CNS.

Answer:

1. Multipolar, bipolar and unipolar
2. Examples
   • Multipolar – found in cerebral cortex
   • bipolar – found in the retina of eye
   • unipolar-found in embryonic state
3. afferent neurons or sensory neuron.

Question 9.
Observe the figures a and b.

1. Mention the structural difference between the two neurons.
2. Name the type of impulse transmission through fig B.
3. Mention the functional difference between the two neutrons.

Answer:

1. Fig a is non-myelinated neuron and Fig B is myelinated neuron.
2. Saltatory conduction.
3. In myelinated impulse transmission is very fast.

Question 10.

1. Identify the diagrammatic representation
2. Name P & Q.
3. Mention the function of P & Q.

Answer:

1. Reflex action (knee jerk reflex)
2. Name P & Q
   • P – Afferent pathway
   • Q – Efferent pathway
3. The afferent pathway receives signal from a sensory organ and transmits the impulse into CNS. The efficient pathway carries signals from CNS to the effector.

Question 11.
A diagram of brain is given below.

1. Identify P Q R S
2. Which part is responsible forthinking, memory and reasoning
3. Name the nerve band which connects the two hemispheres of brain.

Answer:

1. Identification of P Q R S
   • P – Cerebrum
   • Q – Thalamus
   • R – Pons
   • S – Corpus callosum
2. Cerebrum
3. Corpus callosum

Plus One Neural Control and Coordination NCERT Questions and Answers

Question 1.
Compare the following:

1. Central neural system (CNS) and Peripheral neural system (PNS)
2. Resting potential and action potenial
3. Choroid and retina

Answer:
1. Central Neural System and Peripheral Neural System:
The CNS includes the brain and the spinal cord and is the site of information processing and control. The PNS comprises of all the nerves of the body associated with the CNS (brain and spinal cord).

2. Resting Potential and Action Potential:
The electrical difference across the resting plasma membrane is called as the resting potential. When a stimulus is applied at a site on the polarised membrane, the membrane at the site becomes freely permeable to Na⁺⁺.

The electrical potential difference across the plasma membrane at the site of stimulus is called the action potential, which is in fact termed as a nerve impulse.

3. Choroid and Retina.
The middle layer, choroid, contains many blood vessels and looks bluish in colour. The choroid layer is thin over the posterior two-third of the eye ball, but it becomes thick in the anterior part to form the ciliary body.

The inner layer is the retina and it contains three layers of cells – from inside to outside – ganglion cells, bipolar cells and photoreceptor cells.

Question 2.
Answer briefly:

1. How do you preceivethe colour of an object?
2. Which part of our body helps us in maintaining the body balance?
3. How does the eye regulate the amount of light that falls on the retina.

Answer:
1. Cones are responsible for color vision. They require brighter light to function than rods require. There are three types of cones, maximally sensitive to long wavelength, medium-wavelength, and short-wavelength light (often referred to as red, green, and blue, repectively, though the sensitivity peaks are not actually at these colors).

2. The inner ear has three semi-circular canals forming cochlea. Cochlea is responsible for maintaining the body balance.

3. The pupil in the eye functions like an aperture. This dilates in case of low light and constricts in case of intense light thereby regulating the amount of light falling on the retina.

Question 3.
The region of the vertebrate eye, where the nerve passes out of the retina is called the
(a) fovea
(b) iris
(c) blind spot
(d) optic chiasma
Answer:
(c) Blind spot

Plus One Neural Control and Coordination Multiple Choice Questions and Answers

Question 1.
Which of the following is not related to the autonomic nervous system?
(a) Peristalsis
(b) Digestion
(c) Excretion
(d) Memory and learning
Answer:
(d) Memory and learning

Question 2.
Comprehension of spoken and written words take place in the region of
(a) association area
(b) motor area
(c) Wernicke’s area
(d) Broca’s area
Answer:
(c) Wernicke’s area

Question 3.
How many laminae are present in the grey matter of spinal cord?
(a) Four
(b) Six
(c) Eight
(d) Ten
Answer:
(d) Ten

Question 4.
Animals possess nerve or nervous systems to respond to their environment. But the single celled Amoeba does not possesses any nerve cell. so, how it come to know whether a particle it encounters is a grain of sand and not its dinner by?
(a) thermotaxis
(b) skin
(c) hormones
(d) chemotaxis
Answer:
(d) chemotaxis

Question 5.
Thermoregulatory centre of human body is associate with
(a) cerebrum
(b) cerebellum
(c) hypothalamus
(d) medulla oblongata
Answer:
(c) hypothalamus

Question 6.
Sensation of stomach pain is due to
(a) interoceptors
(b) exteroceptors
(c) proprioceptors
(d) chemotactors
Answer:
(a) interoceptors

Question 7.
Bipolar neurons occur in
(a) vertebrate embryos
(b) retina of eye
(c) brain and spinal cord
(d) skeletal muscles
Answer:
(b) retina of eye

Question 8.
Which foramen is paired in mammalian brain?
(a) Foramen of Luschka
(b) Foramen of Magendie
(c) Foramen of Monro
(d) Inter-ventricular foramen
Answer:
(a) Foramen of Luschka

Question 9.
Which is thickened to form organ of Corti?
(a) Reissner’s membrane
(b) Basilar membrane
(c) Tectorial membrane
(d) All of these
Answer:
(b) Basilar membrane

Question 10.
Skeletal muscles are controlled by
(a) sympathetic nerves
(b) parasympathetic nerves
(c) somatic nerves
(d) autonomic nerves
Answer:
(c) somatic nerves

Question 11.
Which part of human brain is concerned with the regulation of body temperature?
(a) Medulla oblongata
(b) Cerebellum
(c) Cerebrum
(d) Hypothalamus
Answer:
(d) Hypothalamus

Question 12.
Alzheimer’s disease in humans is associated with the deficiency of
(a) dopamine
(b) glutamic acid
(c) acetyleholine
(d) Gamma Amino Butyric Acid (GABA)
Answer:
(c) acetyleholine

Question 13.
The posterior part of the retina, which is just opposite to the lens is
(a) cornea
(b) yellow spot
(c) fovea centralis
(d) Both (b) and (c)
Answer:
(b) yellow spot

Question 14.
In the central nervous system, myelinated fibres form the ______ while the non-myelinated fibre cells form the _________
(a) grey matter, white matter
(b) white matter, grey matter
(c) ependymal cells, neurosecretory cells
(d) neurosecretory cells, ependymal cells
Answer:
(b) white matter, grey matter

Question 15.
The potential difference across the membrane of nerve fibre when it does not shown any physiological activity is called resting potential. It is about
(a) -60mV
(b) -80mV
(c) +60mV
(d) +90mV
(e) -36mV
Answer:
(b) -80mV

Question 16.
Vomiting centre is located in the
(a) stomach and sometimes in duodenum
(b) gastro-intestinal tract
(c) hypothalamus
(d) medulla oblongata
Answer:
(d) medulla oblongata

Question 17.
The function of vagus nerve innervating the heart
(a) initiate the heart beat.
(b) reduce the heart beat
(c) accelerate the heart beat
(d) maintain constant heart heat
Answer:
(b) reduce the heart beat

Question 18.
The size of pupil is controlled by the
(a) ciliary muscles
(b) suspensory
(c) cornea
(d) iris muscles
Answer:
(d) iris muscles

Question 19.
An action potential in the nerve fibre is produced when positive and negative charges on the outside and the inside of the axon membrane are reversed because
(a) more potassium ions enter the axon as compared to sodium ions leaving it
(b) more sodium ions enter the axon as compared to potassium ions leaving it
(c) all potassium ions leave the axon
(d) all sodium ions enter the axon
Answer:
(b) more sodium ions enter the axon as compared to potassium ions leaving it

Question 20.
A 22 years student goes to his ophthalmologist. He has problem in reading books because he is not able to contract his
(a) suspensory ligament
(b) pupil
(c) iris
(d) ciliary muscles
Answer:
(d) ciliary muscles

Students can Download Chapter 7 Permutation and Combinations Questions and Answers, Plus One Maths Chapter Wise Questions and Answers helps you to revise the complete Kerala State Syllabus and score more marks in your examinations.

Kerala Plus One Maths Chapter Wise Questions and Answers Chapter 7 Permutation and Combinations

Plus One Maths Permutation and Combinations Three Mark Questions and Answers

Question 1.
Find the number of different signals that can be made by arranging at least three flags in order on a vertical pole, if 6 different coloured flags are available.
Answer:

Hence the number of different atleast 3 flag signals = 120 + 360 + 720 + 720 = 1920.

Question 2.
2. Find the value of n such that

1. nP₅ – 42 × nP₃, n>4
2. (n -1 )P₃: nP₄ = 1 : 9 (3 score each)

Answer:
1. Given; nP₅ = 42 × nP₃
⇒ n(n – 1 )(n – 2)(n – 3)(n – 4) = 42 × n(n – 1)(n – 2)
⇒ (n – 3)(n – 4) = 42
⇒ n₂ – 7n + 12 = 42
⇒ n₂ – 7n – 30 = 0
⇒ (n – 10)(n + 3) = 0
⇒ n = 10; n = -3
The acceptable value is n = 10.

2. Given; 9 × ^{(n – 1)}P₃ = ⁿP₄

Question 3.
Find the value of r if

1. 5 × 4P_r = 6 × 5P_{r – 1}
2. 5P_r = 6P_{r – 1} (3 score each)

Answer:
1. Given; 5 × 4P_r = 6 × 5P_{r – 1}

⇒ 30 – 11r + r² = 6
⇒ r² – 11r + 24 = 0
⇒ (r – 8)(r – 3) = 0
⇒ r = 8, 3
The acceptable answer is r = 3.

2. Given; 5P_r = 6P_{r – 1}

⇒ 42 – 13r + r² = 6
⇒ r² – 13r + 36 = 0
⇒ (r – 9)(r – 4) = 0
⇒ r = 9, 4
The acceptable answer is r = 4.

Question 4.
The letters of the word TUESDAY are arranged in a line, each arrangement ends in S.

1. How many different arrangements are possible? (2)
2. How many of them start with letter D? (1)

Answer:

1. In the word TUESDAY there are 7 letters. When word end in S, there are only 6 possible arrangements. This can be done in 6! = 720
2. The word start with D and end in S, this can be done in 5! = 120.

Question 5.
Consider the word ANNAMALAI

1. How many new words can be formed using the given words? (2)
2. Among the new words how many of them will begin with A and end with I. (1)

Answer:

1. In the word ANNAMALAI there are 9 letters, of which A appears 4 times, N appears 2 times and the rest ail are different. Therefore the total number of ways is \(\frac{9 !}{4! \times 2 !}\) = 7560.
2. The word start with A and end in I, this can be done in \(\frac{7 !}{3! \times 2 !}\) = 420.

Question 6.
Find the rank of the word NAAGI, if these words are written as in a dictionary.
Answer:
The order of the letters will be A, A, G, I, N

Therefore the position of ‘NAAGI’ is 24 + 12 + 12 + 1 = 49.

Question 7.
A committee of 3 persons is to be constituted from a group of 2 men and 3 women.

1. In how many ways can be done? (1)
2. How many of these committees would consist of 1 man and 2 women? (2)

Answer:
1. The required number of ways
= 5C₃ = 5C₂ = \(\frac{5 \times 4}{1 \times 2}\) = 10.

2. One man can be selected in 2C_1. 2 women can be selected in 3C₂
Therefore required number of ways
= 2C₁ × 3C₂ = 2 × 3 = 6.

Question 8.
It was found at a certain dinner meeting that after every member had shaken hand with every other members, 45 handshakes were interchanged. How many members were present at the metting?
Answer:
Let n be the member of person present in the meeting. The total number of handshakes is same as the number of ways of selecting 2 persons from among n persons.
The total number of handshakes = nC₁ = 45
⇒ \(\frac{n(n-1)}{1 \times 2}\) = 45 ⇒ n² – n – 90 = 0
⇒ (n – 10)(n + 9) = 0
⇒ n = 10, -9
n = 10 is acceptable.

Question 9.
In an exam, Arjun has to select 4 questions from each part. There are 6, 7 and 8 question in Part I, Part II and Part III, respectively. What is the number of possible combinations in which he can choose the question?
Answer:
Selecting 4 questions from Part I = 6C₄ = 6C₂
Selecting 4 questions from Part II = 7C₄ = 7 C₃
Selecting 4 questions from Part III = 8C₄
The required number of ways
= 6C₂ × 7C₃ × 8C₄ = 15 × 35 × 70 = 36750.

Plus One Maths Permutation and Combinations Four Mark Questions and Answers

Question 1.
Find the number of different 8-letter arrangements that can be made from the letters of the word DAUGHTER so that

1. all vowels occur together. (2)
2. all vowels do not occur together. (2)

Answer:
1. DAUGHTER this word has 8 different letters. A, U, E are the vowels. Treat these 3 as one unit, then there are 6 units and can be permuted in 6! ways. The above vowels can be permuted in 3! ways. Hence the total number of words is 6! × 3! = 4320.

2. Number of words all vowels do not occur together = Total number of different words – number of words in which vowels come together
= 8! – 6! × 3! = 6!(8 × 7 – 6) = 2 × 6!(28 – 3)
= 50 × 6! = 36000.

Question 2.
How many permutations are there of the 11 letters in MISSISSIPPI

1. taken all together? (2)
2. all the I’s not come together? (2)

Answer:
1. In the word MISSISSIPPI there are 11 letters, of which S appears 4 times, I appears 4 times, P appears 2 times and the rest all are different.
Therefore the total number of ways is \(\frac{11 !}{4 ! \times 4 ! \times 2 !}\) = 34650.

2. 4 I’s are kept together and should be counted as one unit, then there are 8 units. The number of ways is \(\frac{8 !}{4 \times 2 !}\) = 840. Therefore the I’s not come together = Total arrangements – 4 I’s together.
= 34650 – 840 = 33810.

Question 3.
Find the number of arrangements of 6 boys and 5 girls in a row so that

1. no two girls sit together. (2)
2. boys and girls occupy alternate positions. (2)

Answer:
1. Since no two girls sit together, we have first arrange the 6 boys among themselves. This can be done in 6! ways.

Now no two girls sit together if we place the girls in between boys. There are 7 places and it should be occupied by 5 girls, can be done in 7P₅ ways. Therefore the total number of ways is 6! × 7P₅ = 720 × 7 × 6 × 5 × 4 × 3 = 1814400.

2. Boys and girls occupy alternate position can be done as follows. First place the boys whose number is large.

Boys can be arranged in 6! ways. The place between boys can be filled by 5 girls, can be done in 5! ways. Therefore the total number of ways is 6! × 5! = 720 × 120 = 86400.

Question 4.
If the letters of the word DHRONA be permuted and arranged as in a dictionary, find the rank of the word.
Answer:
The order of the letters will be A, D, H, N, O, R


Therefore the position of ‘DHRONA’ is 120 + 24 + 6 + 6 + 6 + 2 + 2 + 1 + 1 = 168.

Question 5.
If the letters of the word MOTHER be permuted and arranged as in a dictionary, find the rank of the word.
Answer:
The order of the letters will be E, H, M, O, R, T

Therefore the position of ‘MOTHER’ is 120 + 120 + 24 + 24 + 6 + 6 + 6 + 2 + 1 = 309.

Question 6.
How many

1. Straight line (2)
2. Triangles can be formed by joining 12 points, 4 of which are collinear. (2)

Answer:
1. Number of straight lines that can be formed using 12 points = 12C₂ = \(\frac{12 \times 11}{1 \times 2}\) = 66. Number of straight lines that can be formed using 4 collinear points = 4C₂ = \(\frac{4 \times 3}{1 \times 2}\) = 6.  Since the 4 points are collinear, the required number of lines = 66 – 6 + 1 = 61.

2. Number of triangles that can be formed using 12 points = 12C₃ = \(\frac{12 \times 11 \times 10}{1 \times 2 \times 3}\) = 220.
Number of triangles that can be formed using 4 collinear points = 4C₃ = 4. Since the 4 points are collinear, the required number of triangles = 220 – 4 = 216.

Question 7.
From 7 men and 4 ladies a committee of 6 is to be formed. In how many ways can this be done when the committee contains

1. exactly two ladies. (2)
2. at least two ladies. (2)

Answer:
1. Exactly two ladies can be selected from 4 in 4C₂ = \(\frac{4 \times 3}{1 \times 2}\) = 6. The remaining 4 should be selected from 7 men in 7C₄ = \(\frac{7 \times 6 \times 5 \times 4}{1 \times 2 \times 3 \times 4}\) = 35. The required number of ways = 6 × 35 = 210.

2. Atleast 2 ladies can be selected as follows;

The required number of ways = 210 + 140 + 21 = 371.

Question 8.
A box contains 6 apples, 5 oranges, and 8 mangoes.

1. In how many ways a fruit is selected from the box. (1)
2. In how many different ways can an apple and an orange be selected. (1)
3. In how many different ways a person take one apple, one orange, and one mango. (2)

Answer:

1. The box contains 6 + 5 + 8 = 19 fruits, from this one fruit can be selected in 19C₁ = 19 ways.
2. An apple is to be selected from 6 apples and an orange be selected from 5 oranges. The required number of ways = 6C₁ × 5C₁ = 30.
3. An apple is to be selected from 6 apples, an orange is to be selected from 5 oranges and one mango is to be selected from 8 mangoes. The required number of ways = 6C₁ × 5C ₁ × 8C₁ = 240.

Question 9.
A student has to answer 6 out of 10 questions which are divided into two parts containing 5 questions each and he is permitted to attempt not more than 4 from any group. In how many ways can he make up his choice?
Answer:
The different possibilities are mentioned below;

The required number of ways
= 5C₁ × 5C₂ + 5C₃ × 5C₃ + 5C₂ × 5C₁
= 5 × 10 + 10 × 10 + 10 × 5 = 200.

Question 10.

1. How many chords can be drawn through 15 points on a circle? (2)
2. A bag contains 5 black and 6 red balls. Determine the number of ways in which 2 black and 3 red balls can be selected? (2)

Answer:

1. Number of chords = ¹⁵C₂ = \(\frac{15 \times 14}{1 \times 2}\) = 105
2. Selection of 2 black and 3 red balls = ⁵C₂ × ⁶C₃ = 10 × 20 = 200.

Question 11.

1. If ⁿC₂ = ⁿC₈ then find n
2. Find n if ⁿP₅ = 42 × ⁿP₃; n> 4

Answer:
1. n = 2 + 8 = 10.

2. DAUGHTER this word has 8 different letters. A, U, E are the vowels. Treat these 3 as one unit, then there are 6 units and can be permuted in 6! ways. The above vowels can be permuted in 3! ways. Hence the total number of words is 6! × 3! = 4320.

Plus One Maths Permutation and Combinations Six Mark Questions and Answers

Question 1.
Find the arrangements of letters of the word INDEPENDENCE. In how many of these arrangements,

1. do the words start with P. (2)
2. do all the vowels always occur together. (2)
3. do the vowels never occur together. (1)
4. do the words begin with I and end in P? (1)

Answer:
1. In the word INDEPENDENCE there are 12 letters, of which N appears 3 times, E appears 4 times, D appears 2 times and the rest all are different.

When the words start with P, then there are 11 letters to be filled in 11 spaces. Therefore the total number of ways is \(\frac{11 !}{3! \times 2 ! \times 4 !}\) = 138600.

2. The vowels EEEEI are to be kept together and should be treated as one unit. Then these vowels can be arranged in \(\frac{5 !}{4 !}\) ways. This single unit together with 7 letter will count to units, can be arranged in \(\frac{8 !}{3! \times 2 !}\). Therefore the total number of ways \(\frac{5 !}{4 !} \times \frac{8 !}{3 ! \times 2 !}\) = 16800.

3. Number of ways of arrangement with vowels do not come together = Total arrangement – vowels coming together.
= \(\frac{12 !}{3 ! \times 2 ! \times 4 !}\) – 16800 = 1663200 – 16800 = 1646400.

4. When the words start with I and ends with P, then there are 10 letters to be filled in 10 spaces. Therefore the total number of ways is \(\frac{10 !}{3 ! \times 2 ! \times 4 !}\) = 12600.

Question 2.
Consider the word ASSASSINATION.

1. How many permutations are there of the letters of the given word? (2)
2. How many different ways can be arranged so that the 4S’s come together? (2)
3. How many different ways can be arranged so that the 4S’s do not come together? (1)
4. How many begin with A? (1)

Answer:
1. In the word ASSASSINATION there are 13 letters, of which A appears 3 times, S appears 4 times, N appears 2 times, I appears 2 times and the rest all are different. Therefore the total number of ways is \(\frac{13 !}{3! \times 4 ! \times 2 ! \times 2 !}\) = 10810800.

2. 4 S’s are kept together and should be counted as one unit, then there are 10 units. The number of ways is \(\frac{10 !}{3! \times 2 ! \times 2 !}\) = 151200.

3. Number of words in which 4S’s do not come together = Total number of words – 4S’s together = 10810800 -151200 = 10659600.

4. The word will start with any one of the 4 A’s. Then total letter arrange will be 12. Number of words in which begin with A \(\frac{12 !}{2! \times 4 ! \times 2 ! \times 2 !}\) = 2494800.

Question 3.
A team of 11 cricket players is to be chosen from 15 players. In how many ways can this be done so as to:

1. Include a particular player A. (2)
2. Exclude a particular player B. (2)
3. Include A and exclude B. (2)

Answer:
1. A particular player A is to be included, then selection of 10 is to be made from 14 players.
The required number of ways 14C₁₀ = 14C₄ = \(\frac{14 \times 13 \times 12 \times 11}{1 \times 2 \times 3 \times 4}\) = 7 × 13 × 11 = 1001.

2. A particular player B is to be excluded, then selection of 11 is to be made from 14 players.
The required number of ways = 14C₁₁ = 14C₃ = \(\frac{14 \times 13 \times 12}{1 \times 2 \times 3}\)
= 14 × 13 × 2 = 364.

3. A particular player A is to be included and player B is to be excluded, then selection of 10 is to be made from 13 players. The required number of ways
= 13C₁₀ = 13C₃ = \(\frac{13 \times 12 \times 11}{1 \times 2 \times 3}\)
= 13 × 2 × 11 = 286.

Question 4.
What is the number of ways of choosing 4 cards from a pack of 52 playing cards? In how many of these

1. four cards are of the same suit, (2)
2. are face cards, (1)
3. two are red cards and two are black cards, (2)
4. cards are of the same colour? (1)

Answer:
Selection of 4 cards from 52 = 52C₄
1. There are 4 suits in a pack of 52 playing cards. They are Club, Spade, Diamond, and Heart. Selecting 4 from each can be done in,
= 13C₄ + 13C₄ + 13C₄ + 13C₄
= 4 × 13C₄ = 4 × \(\frac{13 \times 12 \times 11 \times 10}{1 \times 2 \times 3 \times 4}\) = 2860.

2. There are 12 face cards in a pack of 52 playing cards. Selection 4 cards can be done in 12C₄ = 495.

3. There are 26 red cards and 26 black in a pack of 52 playing cards. Selection of 2 cards should be done from each colour, this can be done in 26C₂ × 26C₂ = (325)² = 105625.

4. Selection of 4 cards from same colour = 26C₄ + 26C₄ = 29900

Plus One Maths Permutation and Combinations Practice Problems Questions and Answers

Question 1.
There are 3 routes from place A to place B and 2 routes from place B to place C. Find how many different routes are there from A to C.
Answer:
By fundamental principle of counting, there are 2 × 3 = 6 different ways.

Question 2.
How many 3 digit numbers can be formed from the digits 1, 2 and 3, assuming that the repetition of digits is not allowed.
Answer:

By fundamental principle of counting, there are 1 × 2 × 3 = 6 different 3 digit numbers.

Question 3.
How many two-digit even numbers with distinct digits can be formed from the digits 1, 2, 3, 4, 5.
Answer:

Hence by fundamental principle of counting, there are 4 × 2 = 8 different 2 digit even numbers.

Question 4.
Evaluate the following

1. \(\frac{7 !}{5 !}\)
2. \(\frac{12 !}{10 ! \times 2 !}\)
3. 6P₄
4. 9P₄
5. 10P₅

Answer:
1. \(\frac{7 !}{5 !}\) = \(\frac{7 \times 6 \times 5 !}{5 !}\) = 45

2.

3. 6P₄ = 6 × 5 × 4 × 3 = 360

4. 9P₄ = 9 × 8 × 7 × 6 = 3024

5. 10P₅ = 10 × 9 × 8 × 7 × 6 = 30240

Question 5.
Evaluate the following.

1. 10C₄
2. 21C₃
3. 19C₁₅
4. 31C₂₉ (1 score each)

Answer:

1. 10C₄ = \(\frac{10 \times 9 \times 8 \times 7}{1 \times 2 \times 3 \times 4}\) = 10 × 3 × 7 = 210
2. 21C₃ = \(\frac{21 \times 20 \times 19}{1 \times 2 \times 3}\) = 7 × 10 × 19 = 1330
3. 19C₁₅ = \(\frac{19 \times 18 \times 17 \times 16}{1 \times 2 \times 3 \times 4}\) = = 19 × 3 × 17 × 4 = 3876
4. 31C₂₉ = \(\frac{31 \times 30}{1 \times 2}\) = 465.

Question 6.
How many five digits telephone numbers can be constructed using the digits 0 to 9 if each number starts with 67 and no digit appears more than once?
Answer:
Digits 6, 7 already used, so only 8 digits available to fill remaining 3 places.

Hence by fundamental principle of counting there are 8 × 7 × 6 = 336 different 5 digit telephone numbers.

Question 7.
In how many ways can 5 persons sit in a car, two including the driver in the front seat and 3 in the back seat. If two particular person out of the 5 are to avoid the driver’s seat?
Answer:

Hence by fundamental principle of counting there are 3 × 4 × 3 × 2 × 1 = 72 different ways.

Question 8.
How many numbers can be formed from the digits 1, 2, 3 and 9 if repetition of digits is not allowed?
Answer:

Hence the number of total numbers = 4 + 12 + 24 + 24 = 64.

Question 9.
How many 3-digit even numbers can be formed from the digit 1, 2, 3, 4, 5, 6 if the digits can be repeated?
Answer:

Hence by fundamental principle of counting there are 6 × 6 × 3 = 108 different 3 digit even numbers.

Question 10.
How many 3-digit numbers can be formed from the digits 1, 2, 3, 4 and 5 assuming that

1. repetition of the digits is allowed? (1)
2. repetition of the digits is not allowed (1)

Answer:
1.

Hence by fundamental principle of counting there are 5 × 5 × 5 = 125 different 3 digit numbers.

2.

Hence by fundamental principle of counting there are 5 × 4 × 3 = 60 different 3 digit numbers.

Question 11.
How many 8 letter words, with or without meaning, can be formed using the word EQUATION, using each letter exactly once?
Answer:
EQUATION this word has 8 different letters. Different words that can be made from these letters is 81 = 1 × 2 × 3 × 4 × 5 × 6 × 7 × 8 =40320.

Question 12.
Find the number of ways in which the letters of the word ASSISTANT can be arranged among themselves.
Answer:
In the word ASSISTANT there are 9 letters, of which S appears 3 times, A appears 2 times, T appears 2 times and the rest all are different. Therefore the total number of ways is
\(\frac{9 !}{3! \times 2 ! \times 2 !}\) = 15120.

Students can Download Chapter 6 Linear Inequalities Questions and Answers, Plus One Maths Chapter Wise Questions and Answers helps you to revise the complete Kerala State Syllabus and score more marks in your examinations.

Kerala Plus One Maths Chapter Wise Questions and Answers Chapter 6 Linear Inequalities

Plus One Maths Linear Inequalities Three Mark Questions and Answers

Question 1.
Solve the following inequalities.

1. \(\frac{1}{2}\left(\frac{3 x}{5}+4\right) \geq \frac{1}{3}(x-6)\)
2. \(\left(\frac{2 x-1}{3}\right) \geq \frac{(3 x-2)}{4}-\frac{(2-x)}{5}\) (3 score each)

Answer:
1. Given;

⇒ 3(3x + 20) ≥ 10(x – 6)
⇒ 9x + 60 ≥ 10x – 60
⇒ 9x – 10x ≥ -60 – 60
⇒ -x ≥ -120 ⇒ x ≤ 120

2. Given;

⇒ 20(2x -1) ≥ 3[15x – 10 – 8 + 4x]
⇒ 40x – 20 ≥ 45x – 54 + 12x
⇒ 40x – 20 ≥ 57x – 54
⇒ 40x – 57x ≥ -54 + 20
⇒ -17x ≥ -34 ⇒ x ≤ 2

Question 2.
1. Which of the following sets of inequality represent the second quadrant? (1)
(a) x < 0, y < 0
(b) x > 0, y > 0
(c) x < 0, y > 0
(d) x > 0, y < 0
2. Write the system of inequalities that represents the shaded rectangle in the figure given below: (2)

Answer:
1. (a) x < 0, y < 0

2. The shaded figure is a rectangle. The side parallel to x axis are y = -1 and y = 1. The side perpendicular to x axis are x = 2 and x = -2. Hence the inequality that represent the shaded region are
-2 ≤ x ≤ 2; -1 ≤ y ≤ 1.

Question 3.
Find all pairs of consecutive even positive integers both of which are smaller than10 such that their sum is less than 23.
Answer:
Consecutive even positive integers be x and x + 2. Then, x + x + 2 < 23; x + 2 < 10
⇒ 2x < 23 – 2; x < 10 – 2
⇒ x < \(\frac{21}{2}\) = 10.5; x < 8
⇒ 8 < x ≤ 10 Therefore x can take values 9, 10. Hence the pairs are (9, 10), (10, 9).

Question 4.
The longest side of a triangle is 3 times the shortest side and the third side is 2 cm shorter than the longest side. If the perimeter of the triangle is at least 61 cm, find the minimum length of the shortest side.
Answer:
Let the shortest side is x, then;
Longest side = 3x;
Third side = Longest side – 2 = 3x – 2
Perimeter = 3x + 3x – 2 + x ≥ 61 ⇒ 7x – 2 ≥ 61
⇒ 7x ≥ 61 + 2 ⇒ x ≥ \(\frac{63}{7}\) = 9.

Plus One Maths Linear Inequalities Four Mark Questions and Answers

Question 1.
Solve the following system of inequalities graphically.

1. 2x – y > 1; x – 2y < -1
2. x + y ≤ 9; y > x; x ≥ 0
3. x – 2y ≤ 3; 3x + 4y ≥ 12; x ≥ 0, y ≥ 1
4. 2x + y – 3 ≥ 0; x – 2y + 1 ≥ 0; y ≤ 3 (4 score each)

Answer:
1. 2x – y > 1; x – 2y < -1

2. x + y ≤ 9; y > x ⇒ x – y < 0

3. x – 2y ≤ 3; 3x + 4y ≥ 12

4. 2x + y ≥ 3; x – 2y ≤ -1

Plus One Maths Linear Inequalities Practice Problems Questions and Answers

Question 1.
Solve the following inequalities.

1. 4x + 3 < 5x + 7
2. 3(x – 1) ≤ 2(x – 3) (1 score each)

Answer:
1. Given; 4x + 3 < 5x + 7
⇒ 4x – 5x < 7 – 3 ⇒ -x < 4 ⇒ x > -4.

2. Given; 3(x – 1) < 2(x – 3)
⇒ 3x – 3 ≤ 2x – 6 ⇒ 3x – 2x ≤ -6 + 3
⇒ x ≤ -3.

Question 2.
Solve the inequality \(\frac{3(x-2)}{5} \leq \frac{5(2-x)}{3}\).
Answer:
Given; \(\frac{3(x-2)}{5} \leq \frac{5(2-x)}{3}\)
⇒ 9(x – 2) ≤ 25(2 – x)
⇒ 9x – 18 ≤ 50 – 25x
⇒ 9x + 25x ≤ 50 + 18
⇒ 34x ≤ 68 ⇒ x ≤ 2.

Question 3.
Show the solution of each inequality on a number line.

1. 4x + 3 < 6x + 7
2. 5x – 3 ≥ 3x – 5
3. 3(1 – x) < 2(x + 4)
4. 2 – 3x < 2(x + 6)
5. -3 ≤ 3 – 2x < 6 (2 score each)

Answer:
1. Given; 4x + 3 < 6x + 7 ⇒ 4x – 6x < 7 – 3
⇒ -2x < 4 ⇒ x > -2

2. Given; 5x – 3 > 3x – 5 ⇒ 5x – 3x ≥ -5 + 3
⇒ 2x ≥ -2 ⇒ x ≥ 1.

3. Given; 3(1 – x) < 2(x + 4) ⇒ 3 – 3x < 2x + 8
⇒ -3x – 2x < 8 – 3 ⇒ -5x < 5 ⇒ x > -1.

4. Given; 2 – 3x < 2(x + 6) ⇒ 2 – 3x < 2x + 12
⇒ -3x – 2x < 12 – 2 ⇒ -5x < 10 ⇒ x > -2

5. Given; -3 ≤ 3 – 2x < 6
⇒ -3 ≤ 3 – 2x; 3 – 2x < 6
⇒ -3 – 3 ≤ -2x; -2x < 6 – 3
⇒ -6 ≤ -2x; -2x < 3 ⇒ 3 ≥ x; x > \(-\frac{3}{2}\) ⇒ \(-\frac{3}{2}\) < x ≤ 3

Question 4.
The marks obtained by a student of class XI in first and second terminal examination are 62 and 48, respectively. Find the minimum marks he should get in the annual examination to have an average of at least 60 marks.
Answer:
Let x denote the mark obtained by the student in Class XI examination, then;
\(\frac{62+48+x}{3}\) ≥ 60 ⇒ 110 + x ≥ 1800 ⇒ x ≥ 70.

Students can Download Chapter 4 Principle of Mathematical Induction Questions and Answers, Plus One Maths Chapter Wise Questions and Answers helps you to revise the complete Kerala State Syllabus and score more marks in your examinations.

Kerala Plus One Maths Chapter Wise Questions and Answers Chapter 4 Principle of Mathematical Induction

Plus One Maths Principle of Mathematical Induction Three Mark Questions and Answers

Question 1.
For all n ≥ 1 , prove that
1² + 2² + 3² +……….+ n² > \(\frac{n^{3}}{3}\)
Answer:
Let p(n): 1² + 2² + 3² + n²
Put n = 1 ⇒ p(1) = 1 > \(\frac{1}{3}\) which is true.
Assuming that true for p(k)
p(k): 1² + 2² + 3² +……….+ k² > \(\frac{k^{3}}{3}\)
Let p(k + 1): 1² + 2² + 3² +…….+ k² + (k + 1)² > \(\frac{k^{3}}{3}\) + (k + 1)²


Hence by using the principle of mathematical induction true for all n ∈ N.

Question 2.
For all n ≥ 1 , prove that 1 + 2 + 3 +…….+ n < \(\frac{1}{8}\)(2n + 1)²
Answer:
Let p(n): 1 + 2 + 3 +…….+ n ,
Put n = 1 ⇒ p(1) = 1 < \(\frac{9}{8}\) which is true.
Assuming that true for p(k)
p(k): 1 + 2 + 3 +…….+ k < \(\frac{1}{8}\)(2k + 1)²
Let p(k +1): 1 + 2 + 3 +……..+ k + (k +1) < \(\frac{1}{8}\) (2k + 1)² + (k + 1)

Hence by using the principle of mathematical induction true for all n ∈ N.

Question 3.
For all n ≥ 1, prove that p(n): 2³ⁿ – 1 is divisible by 7.
Answer:
p(1): 2³⁽¹⁾ – 1 = 8 – 1 = 7 divisible by 7, hence true. Assuming that for p(k)
p(k) : 2^3k – 1 is divisible by 7.
2^3k – 1 = 7M
P(k + 1): 2^{3(k + 1)} – 1 = 2^{3k + 3} – 1
= 2^3k2³ – 1 = 2^3k × 8 – 1
= 2^3k × 8 – 8 + 7 = 8(2^3k – 1) + 7
= 8(7M) + 7
Hence divisible by 7. Therefore by using the principle of mathematical induction true for all n ∈ N.

Question 4.
For all n ≥ 1, prove that p(n): n³ + (n + 1)³ + (n + 2)³ is divisible by 9.
Answer:
p(1): 1 + 2³ + 3³ = 1 + 8 + 27 = 36 divisible by 9, hence true. Assuming that true for p(k)
p(k): k³ + (k + 1)³ + (k + 2)³ is divisible by 9.
⇒ k³ + (k + 1)³ + (k + 2)³ = 9M
p(k +1 ): (k + 1)³ + (k + 2 )³ + (k + 3)³
= (k +1)³ + (k + 2)³ + k³ + 9k² + 27k + 27
= [(k + 1)³ + (k + 2)³ + k³] + 9[k² + 3k + 3]
= 9M + 9[k² + 3k + 3]
Hence p(k + 1)divisible by 9. Therefore by using the principle of mathematical induction true for all n ∈ N.

Plus One Maths Principle of Mathematical Induction Four Mark Questions and Answers

Question 1.
For all n ≥ 1, prove that

Answer:

Hence by using the principle of mathematical induction true for all n ∈ N.

Question 2.
For all n ≥ 1, prove that

Answer:
Let

Assuming that true for p(k)

Let p(k + 1):

Hence by using the principle of mathematical induction true for all n ∈ N.

Question 3.
For all n ≥ 1 , prove that 1.2.3 + 2.3.4 +………+ n(n + 1)(n + 2) = \(\frac{n(n+1)(n+2)(n+3)}{4}\).
Answer:
Let p(n): 1.2.3 + 2.3.4 +……..+ n(n + 1)(n + 2),
Put n = 1
p(1) = \(\frac{1(1+1)(1+2)(1+3)}{4}\) = 6 which is true.
Assuming that true for p(k)
p(k): 1.2.3 + 2.3.4 +……..+ k(k + 1)(k + 2) = \(\frac{k(k+1)(k+2)(k+3)}{4}\),
Let p(k + 1)


Hence by using the principle of mathematical induction true for all n ∈ N.

Question 4.
For all n ≥ 1, prove that

Answer:


Hence by using the principle of mathematical induction true for all n ∈ N.

Question 5.
For all n ≥ 1 , prove that p(n): n(n + 1 )(n + 5) is divisible by 3.
Answer:
p(1): 1(1 + 1)(1 + 5) = 12divisible by 3, hence true. Assuming that true for p(k)
p(k): k(k + 1)(k + 5) is divisible by 3.
k(k + 1)(k + 5) = 3M
p(k + 1): (k + 1)(k + 2)(k + 6)
= (k + 1)(k² + 8k + 12)
= (k + 1)(k² + 5k + 3k +12)
= (k + 1)[k(k + 5) + 3(k + 6)]
= [k(k + 1)(k + 5) + 3(k + 1)(k + 6)]
= [3M + 3(k + 1)(k + 6)]
= 3[M + (k + 1)(k + 6)]
Hence divisible by 3. Therefore by using the principle of mathematical induction true for all n ∈ N.

Question 6.
For all n ≥ 1 , prove that p(n): 2.7ⁿ + 3.5ⁿ – 5 is divisible by 24.
Answer:
p(1): 2.7¹ + 3.5¹ – 5 = 14 + 15 – 5 = 24 divisible by 24, hence true. Assuming that true for p(k)
p(k): 2.7^k + 3.5^k – 5 is divisible by 24.
⇒ 2.7^k + 3.5^k – 5 = 24M
p(k + 1): 2.7^{k + 1} + 3.5^{k + 1} – 5
= 2.7^k.7 + 3.5^k.5 – 5
= 2.7^k.(6 + 1) + 3.5^k.(4 + 1) – 5
= 12.7^k + 2.7^k + 12.5^k + 3.5^k – 5
= 12(7^k + 5^k) + (2.7^k + 3.5^k) – 5
= 12(7^k + 5^k) + 24M
7^k And 5^k are odd numbers, therefore (7^k + 5^k) will be an even and will be divisible by 24, Hence p(k + 1)divisible by 24. Therefore by using the principle of mathematical induction true for all n ∈ N.

Students can Download Chapter 9 Locomotion and Movement Questions and Answers, Plus One zoology Chapter Wise Questions and Answers helps you to revise the complete Kerala State Syllabus and score more marks in your examinations.

Kerala Plus One Zoology Chapter Wise Questions and Answers Chapter 9 Locomotion and Movement

Plus One Locomotion and Movement One Mark Questions and Answers

Question 1.
Ribs are attached to
(a) Scapula
(b) Sternum
(c) Clavicle
(d) llium
Answer:
(d) llium

Question 2.
What is the type of movable joint present between the atlas and axis?
(a) Pivot
(b) Saddle
(c) Hinge
(d) Gliding
Answer:
(a) Pivot

Question 3.
ATPase of the muscle is located in
(a) Actinin
(b) Troponin
(c) Myosin
(d) Actin
Answer:
(c) Myosin

Question 4.
Macrophages and leucocytes exhibit
(a) Ciliary movement
(b) Flagellar movement
(c) Amoeboid movement
(d) Gliding movement
Answer:
(c) Amoeboid movement

Question 5.
Which one of the following is not a disorder of bone?
(a) Arthritis
(b) Osteoporosis
(c) Rickets
(d) Atherosclerosis
Answer:
(d) Atherosclerosis

Question 6.
Suggest a suitable word for the fourth place

1. thin filament : Actin :: Thick filament : ___________
2. Pelvic girdle : humerus :: Pelvic girdle : ___________

Answer:

1. myosin
2. Femur

Question 7.
Copy the paragraph below about the structure of a striated muscle. Choose the words from the following list & fill up the gaps.

Under lightmicroscope, the striated muscle shows ________ (a) band & _______ (b) band. The distance between 2 adjacent Z lines is known as __________ (c).
(sarcomere, lightband, A band, H-zone, M line, darkband)
Answer:
(a) Ligthband
(b) Darkband
(c) Sarcomere

Question 8.
A person is suffering from joint pain. His blood test shown increased amount of Uric acid. What will be the diagnosis?
Answer:
Gout

Question 9.
The infants have 33 vertebrae in the vertebral column. But an adult has only 26 vertebrae. What will happen to the remaining vertebrae.
Answer:
Ulna

Plus One Locomotion and Movement Two Mark Questions and Answers

Question 1.
The axial skeleton contains 80 bones. Make a table according to where it is seen and number of bones present in each section.
Answer:

Question 2.
One of the following statements is incorrect. Find and correct it.

1. The number of cervical vertebrae is seven in all mammals except human beings.
2. Thoracic vertebrae, ribs and sternum together make ribcage.
3. Accumulation of uric acid in joints leads to gout.

Answer:
Statement 1 is incorrect. The number of cervical vertebrae is seven in all mammals including human beings.

Question 3.

1. Name the process shown here, which theory explain this process ?
2. Draw and complete the process.

Answer:
1. Muscle contraction, Sliding – filament theory

2.

Question 4.
Red muscle fibers have greater capacity to do work for a prolonged period where as white muscles suffer from fatigue after a short time. Give reason.
Answer:
Red muscle fibers are red in colour due to the presence of large amount of myoglobin. Myoglobin is an oxygen storing pigment. These muscle also contain plenty of mitochondria which can utilise the large amount of oxygen stored in them for ATP production.

So red muscle fibers have greater capacity to do work for a prolonged period. White muscle fibers possess very less quantity of myoglobin and mitochondria.

Question 5.
Shoulder joints are not very stable. But they are freely movable. Give reason?
Answer:
Shoulder joints are freely movable joint or synovial joint. In this type of joint there is presence of a fluid filled synovial cavity between the articulating surfaces of the two bones. Such an arrangement allows considerable movement.

Question 6.
Name the types of joint between the following.

1. Atlas/axis
2. Between Cranial bones
3. Carpal/Meta carpel of thumb
4. Between Humerus and pectoral girdle

Answer:

1. Pivot joint
2. Fibrous joint
3. Saddle joint
4. Ball and socket joint

Question 7.
Analyse the table & fill in the blanks with appropriate words.

Answer:

Question 8.
The last two pairs of ribs are called floating ribs.

1. What do you meant by the term floating ribs?
2. What are true ribs?

Answer:

1. Last 2 pairs of ribs are not connected ventrally to the sternum. There ribs are called floating ribs.
2. First seven pairs of ribs are called true ribs. Dorsally, they pre attached to the thoracic vertebrae and ventrally connected to the sternum with the help of hyaline cartilage.

Question 9.
Some joints in the body are characterized by pressure of a fluid filled cavity between articulating surfaces of two bones. Mention the peculiarity of these types of joint and give examples?
Answer:
This type of joint in called synovial joint or freely movable joint and allows considerable movement.
eg:

1. Ball and socket joint (between humerus and pectoral girdle)
2. Hinge joint (knee joint)

Question 10.
A muscle cell viewed under a microscope shown the following characters.

• Sarcolemma – Present
• Shape – Cylindrical
• Nucleus – Multinuclear
• Striation – Present

1. Identify the muscle cell.
2. Calcium ions are necessary for muscle contraction.

Answer:

If a muscle is placed in a solution containing calcium ions. Does the muscle contracts? Substantiate.
Answer:

1. Skeletal muscle
2. No. For muscle contraction to occur, presence of Ca⁺⁺ ions intracellularly is needed.

Question 11.
Pictorial representation of tissue is given in a lab diary with only one labelling – intercalated discusing your knowledge about tissue.

1. Identify the tissue.
2. Draw the structure of tissue and label the parts.

Answer:
1. Cardiac muscle

2.

Question 12.
You can move your hands at your will. But you can’t move your heart at your will. Comment.
Answer:
Muscles of the hand are skeletal muscles and they are voluntary muscle. Muscles of the heart are cardiac muscles and they are involuntary muscles.

Question 13.

1. Identify the muscle cell
2. Mention its function.

Answer:

1. Non-striated muscle (smooth muscle)
2. Smooth muscles are involuntary muscles located in the inner visceral organs like alimentary canal, reproductive tract etc. So they help the transportation of food through digestive tract and gamete through the genital tract.

Question 14.
An investigation was arrived out to find the effect of the temperature on muscle contraction. The results are shown in the graph.

1. Describe and suggest an explanation for one effect of temperature on muscle contraction.
2. Name the filaments involved in muscle contraction.

Answer:

1. When temperature increases muscle contraction increases, but when temperature decreases the extend of muscle contraction decreases, but the time remain in the contracted state increases,
2. Actin and myosin

Question 15.
The red colour of the blood is due to the presence of hemoglobin in it. Certain skeletal muscles are red in colour even though they lack hemoglobin. Now explain how the red muscles have that particular colour.
Answer:
Muscle contains a red coloured oxygen storing pigment called myoglobin. Myoglobin content is high in some skeletal muscles which gives a reddish appearance.

Question 16.
Actin, Troponin, Biceps, Muscles of blood vessels, Muscles of heart, Myosin, Muscles of reproductive tract, Tropomyosin, Muscles of Alimentary canal, Triceps.
Rearrange the terms in four columns on the basis of their similarity and give appropriate headings for each columns.
Answer:

Question 17.
Identify the myofibril and label the parts given below.

Answer:

Question 18.
Length of A band remains unchanged during muscular contraction. Is this statement true or false? Justify.
Answer:
True. A band or anisotropic band is the region where both thick filaments and thin filaments are present. During muscular contraction, the length of thick filament or thin filament does not change.

Question 19.
Match the column I with column II.

Answer:

Question 20.
Complete the division of human skeletal system by filling the blanks.

Answer:
(a) Axial
(b) Vertebral column
(c) 12
(d) One

Question 21.
Arrange the following bones into two columns and give proper heading to each columns.
Scapula, Carpals, Femur, Fibula, Tibia, Tarsals, Acetabulum, Metatarsals, Radius, Humerus, Glenoid cavity, Clavicle, Metal Carpals, Phalanges, Patella, Ilium, Ischium, Pubis.
Answer:

Question 22.
Your lungs and Heart are well protected in a cage. Which bone contributes to it?
Answer:
In an infant, there are 33 vertebrae in the vertebral column. Five of these bones fuse to form the sacrum and four of other join to become the coccyx. As a result, an adult vertebral column has 26 vertebrae.

Question 23.
There are joints at your shoulders and elbows. But their movement is different. Give reason.
Answer:

• Shoulder joint – Ball and socket joint
• Elbow joint – Hinge joint

Question 24.
What are the different type of movement shown by human cells?
Answer:
Amoeboid, ciliary and muscular and flagellar movement.

Question 25.
Fill the gaps.

Answer:

Question 26.
Draw a flow chart showing the flow of stimulus that results in muscle contraction. (Start from Neuromuscular junction)
Answer:

Question 27.
Match the following.

Answer:

Question 28.
Write the odd one out and give the reason for your answer.

1. Fibrous joints, Ball and socket joints, hinge joints, pivot joints
2. Humerus, Femur, Radius, Ulna.

Answer:

1. Fibrous joints: They are immovable joints of the bones of skull whereas all others are freely movable joints.
2. Femur: It is the thigh bone whereas all other are the bones of the forelimb

Question 29.
Bones of the older people turn brittle and break quickly.

1. Name the bone disorder stated above.
2. List any two reasons for the above disorder.

Answer:

1. Osteoporosis
2. reasons for the above disorder
   • Imbalance of hormones like thyrocalcitonin, parathyroid and sex hormones.
   • Deficiency of calcium and Vitamin D.

Question 30.
When you ride a bicycle two majorjointsofyourleg should perform properly.

1. Name the joints.
2. To which type of synovial joints these belong?

Answer:

1. Knee joint, Hip Joint
2. Hinge joint, Ball and Socket joint

Question 31.
In human body different type of movement shown by some cells.

1. Name the type of movement shown by human sperm.
2. Which is the part of sperm help this movement?

Answer:

1. Flagellar movement
2. Whip like movement of the tail and the middle piece of the sperm

Plus One Locomotion and Movement Three Mark Questions and Answers

Question 1.
State one difference in each of the following pairs on the basis of what is indicated in brackets.

1. Glenoid cavity and acetabulum (location and function)
2. Osteoarthritis and Gout (cause)

Answer:
1. Glenoid cavity:
It is a cap like depression in the pectoral girdle to which the head of the humerus fits in.

Acetabulum:
It is a cap like depression present in the pelvic girdle to which head of the femur bone fits in.

2. Osteoarthritis:
It is caused by the degeneration of the articular cartilage.

Gout:
It is caused by accumulation of uric acid crystals.

Question 2.

1. Observe the diagram and identify the type of joint.
2. Where can you find this type of joints in your body?
3. Comment on its ability to move.

Answer:

1. Ball and socket joint
2. Shoulder joints and Hip joints
3. In ball and socket joint, one of the bones forms a ball like head while the other forms a cup like socket into which head fits in. It allows the free movement of the bone with the ball like head.

Question 3.
X-ray photography of forelimb of a person is given here. One of the bones found to be fractured. Name the bone with fracture.

Answer:
a. (i) Across:
1. Hyoid
3. Cranium
4. Pelvic girdle

(ii) Down:
2. Coccyx
5. Ulna

b. Hyoid and cranium

Question 4.
According to colour, muscles are of two types.

1. Name the two muscles.
2. Why are they called so?
3. Give examples.

Answer:

1. Red muscle and white muscle.
2. Red muscles contain large amount of mitochondria and myoglobin whereas the white muscles contain only less amount of mitochondria and myoglobin.
3. Examples
   • Red muscles – Extensor muscles on the back.
   • White muscles – muscles of the eye ball.

Question 5.

1. Observe the above graph and find out the region of graph where muscle show maximum contraction. Justify your answer.
2. Name the different types of muscle proteins present in thick and thin filaments.
3. Ca²⁺ is necessary for muscle contraction. Why?

Answer:
1. Region 2 During Muscular contraction amount of Ca²⁺ released from ER is very high.

2. Thick filament – Muscle protein – myosin
Thin filament – Muscle protein – actin, troponin, tropomyosin.

3. Muscle contraction is regulated by a regulator protein troponin. Troponin mask the active site of actin molecule from myosin head, Action potential reaching a muscle cause the ER to release Ca²⁺.

This Ca²⁺ bind with the specific site of troponin. It causes a conformational change by which the active site of actin molecule is exposed and it results in cross bridge activity and muscle contraction.

Question 6.
Will muscle contraction occur in the following conditions. Justify your answer.

1. Troponin is removed from the thin filament.
2. No ATP is supplied to the muscle.
3. All endoplasmic reticulum are removed from muscle cells.

Answer:

1. Muscle contraction occur.
2. Muscle contraction does not occur. Attachment and detachment of myosin head to actin molecule requires ATP.
3. No muscle contraction. Ca²⁺ necessary for muscular contraction is released from endoplasmic reticulum.

Question 7.
The given statements are about the structure of muscles. Categorise them into a table giving appropriate headings.

1. These muscles have striation.
2. These are found inside the wall of hollow internal organs.
3. These muscles have no striations.
4. They are voluntary muscles.
5. They are muscles of the arms and legs.
6. They are involuntary muscles

Answer:

Question 8.
Observe the following diagram.

1. Identify A, B, C, D.
2. What is Sarcomere?
3. List out the changes takes place during muscle contraction to ABC and D.

Answer:

1. Identification
   • A – Sarcomere
   • B – Hzone
   • C – Aband
   • D – Iband
2. A sarcomere is the region between two adjacent Z line. It is formed of a complete A band in the middle and halves of two I bands present on either side.
3. the changes takes place during muscle contraction to ABC and D:
   • A – Two Z lines of the sarcomere come close together resulting in the shortening of the sarcomere.
   • B – H – Zone completely disappears.
   • C – No change in the length of A band
   • D – I – band get reduced

Question 9.
Give any example of synovial joint.
Answer:

1. Ball and socket joint (between humerus and pectoral girdle)
2. Hinge joint – knee joint

Question 10.
Suppose a person is suffering from calcium deficiency for a prolonged time. How does it affect muscular contraction? (Note: Description necessary)
Answer:
Muscle contraction begins when a nerve impulse reaches at the neuromuscular junction. During this time the sarcoplasmic reticulum releases Ca²⁺. The Ca²⁺ binds with the specific site of the troponin of the actin filament.

This causes a conformational change in the troponin molecule and expose the active site on the actin molecule. Myosin filament binds with the active site of actin and there is formation a cross bridge. By the movement of cross bridge contraction occurs.

The contraction is followed by relaxation and it occurs when Ca²⁺ in pumped back into the sarcoplasmic reticulum. As a result, the troponin molecules became free to mask the active site of the myosin head.

Question 11.
Identify the following diseases noting the symptoms given below

1. The liver is affected, skin and eyes turn yellow due to the deposit of bile pigments.
2. Difficulty in breathing causing wheezing due to the inflammation of bronchi and bronchioles.
3. Chronic disorder in which alveolar walls are damaged mainly due to cigarette smoking.
4. Auto immune disorder affecting neuro muscular junction leading to fatigue, weakening and paralysis of skeletal muscles.
5. Inflammation of joints due to deposition of uric and crystals.
6. Age related disorder due to decreased level of estrogen, characterised by decreased bone mass and increased chances of fractures.

Answer:

1. Jaundice
2. Asthma
3. Emphysema
4. Myasthenia gravis
5. Gout
6. Osteoporosis

Question 12.
Identify the following tissues and write down their location.

1. Loose connective tissue which stores fat.
2. Dense connective tissue which connect bone to bone.
3. A contractile tissue which possess intercalatory disc.

Answer:

1. Adipose tissue – beneath the skin
2. Ligament – attach one bone to another
3. Cardiac muscle – heart

Plus One Locomotion and Movement NCERT Questions and Answers

Question 1.
Define sliding filament theory of muscle contraction.
Answer:
Sliding Filament Theory of Muscle Contraction Mechanism of muscle contraction is best explained by the sliding filament theory which states that contraction of a muscle fibre takes place by the sliding of the thin filaments over the thick filaments.

Question 2.
Describe the important steps in muscle contraction.
Answer:
Muscle contraction is initiated by a signal sent by the central nervous system (CNS) via a motor neuron. A motor neuron alongwith the muscle fibres connected to it constitute a motor unit. The junction between a motor neuron and the sarcolemma of the muscle fibre is called the neuromuscular junction or motor end plate.

1. A neural signal reaching this junction releases a neurotransmitter (Acetylcholine) which generates an action potential in the sarcolemma. This spreads through the muscle fibre and causes the release of calcium ions into the sarcoplasm.

2. Increase in Ca²⁺ level leads to the binding of calcium with a subunit of troponin on actin filaments and thereby remove the masking of active sites for myosin.

3. Utilising the energy from ATP hydrolysis, the myosin head now binds to the exposed active sites on actin to form a cross bridge. This pulls the attached actin filaments towards the centre of ‘A’ band.

4. The 7’ line attached to these actions are also pulled inwards thereby causing a shortening of the sarcomere, i.e., contraction.

5. It is clear from the above steps, that during shortening of the muscle, i.e., contraction, the T bands get reduced, whereas the A’ bands retain the length.

6. The myosin, releasing the ADP and PI goes back to its relaxed state. A new ATP binds and the cross-bridge is broken. The ATP is again hydroysed by the myosin head and the cycle of cross bridge formation and breakage is repeated causing further sliding.

7. The process continues till the Ca²⁺ ions are pumped back to the sarcoplasmic cisternae resulting in the masking of actin filaments. This causes the return of 7 lines back to their original position, i.e., relaxation.

Question 3.
Write true or false. If false change the statement so that it is true.

1. Actin present in thin filament
2. H-zone of striated muscle fibre represents both thick and thin filaments
3. The human skeleton has 206 bones
4. There are 11 pairs of ribs in man.
5. Stenum is present on the ventral side of the body.

Answer:

1. True
2. False, H-zone represents thick filaments
3. True
4. False, There are 12 pairs of ribs in man
5. True

Question 4.
Match column I with column II

Answer:
(a) – (iv)
(b) – (ii)
(c) – (i)
(d) – (iii)

Question 5.
Name the type of joint between the following

1. Atlas/axis
2. Carpal/metacarpal of thumb
3. Between phalanges
4. Femur/acetabulum
5. Between carnival bones
6. Between public bones in the pelvic girdle

Answer:

1. Pivot joint
2. Saddle joint
3. Gliding joint
4. Ball and socket joint
5. Fibrous joint
6. Cartilaginous

Question 6.
Fill in the blank spaces:

1. All mamnals (expect a few) have __________ cervical vertebra.
2. The number of phalanges in each limb of human is __________
3. Thin filament of myofibril contains 2 ‘F’ actins and two other proteins namely __________ and _________
4. In a muscle fibre Ca²⁺ is stored in __________
5. _______and ________ pairs of ribs are called floating ribs.
6. The human cranium is made of _________ bones.

Answer:

1. 7
2. 14
3. tropomyosin, troponin
4. sarcoplasmic reticulum
5. 11^th, 12^th
6. 8

Plus One Locomotion and Movement Multiple Choice Questions and Answers

Question 1.
The muscle band that remains unchanged during contraction and relaxation of the skeletal muscle is
(a) I
(b) H
(c) A
(d) A-line
Answer:
(d) A-line

Question 2.
Intervertebral disc consists of a shock absorber connective tissue known as
(a) hyaline cartilage
(b) elastic cartilage
(c) fibrocartilage
(d) reticulo cartilage
Answer:
(c) fibrocartilage

Question 3.
Common among all mammals is
(a) ventral nerve cord
(b) seven cervical vertebrae
(c) all are carnivores
(d) all are producers
Answer:
(b) seven cervical vertebrae

Question 4.
The lactic acid generated during muscle contraction is converted to glycogen in
(a) muscles
(b) kidney
(c) pancreas
(d) liver
Answer:
(d) liver

Question 5.
Folding and unfolding of actin and myosin leads to amoeboid movement. This is hypothesised by
(a) Allen
(b) Goldacre and Lasch
(c) Berthold
(d) Jennigs
Answer:
(b) Goldacre and Lasch

Question 6.
Muscle fatigue is due to
(a) lactic acid
(b) citric acid
(c) Na
(d) K
Answer:
(a) lactic acid

Question 7.
Which of the following is not syncytial?
(a) Cardiac muscle
(b) Skeletal muscle
(c) Smooth muscle
(d) Interstitial muscle
Answer:
(c) Smooth muscle

Question 8.
Humerus fits into glenoid cavity is example of
(a) ball and socket joint
(b) pivot joint
(c) peg and socket joint
(d) condyloid join
Answer:
(a) ball and socket joint

Question 9.
Human vertebral column is formed by
(a) 21 vertebrae
(b) 30 vertebrae
(c) 26 vertebrae
(d) 33 vertebrae
Answer:
(d) 33 vertebrae

Question 10.
When body part moves towards the median axis is called
(a) abductor
(b) adductor
(c) supinator
(d) pronator
Answer:
(b) adductor

Question 11.
The thin filaments of a muscle fibre are made up of
(a) actin, troponin, tropomyosin
(b) actin, troponin
(c) niyosin, troponin
(d) actin, tropomyosin
Answer:
(a) actin, troponin, tropomyosin

Question 12.
Which of the following pairs is correctly matched?
(a) Cartilaginous joint Skull bones
(b) Hinge joint – Between vertebrae
(c) Fibrous joint – Between phalanges
(d) Gliding joint – Between zygapophyses of the successive vertebrae
Answer:
(d) Gliding joint – Between zygapophyses of the successive vertebrae

Question 13.
Nucleus purposes is found in
(a) brain
(b) nucleus
(c) intervertebral disc
(d) liver
Answer:
(c) intervertebral disc

Question 14.
Total number of bones found in right upper limb is
(a) 25
(b) 26
(c) 30
(d) 60
Answer:
(c) 30

Question 15.
In a vertebrate, which germ layer forms the skeletor muscles?
(a) Ectoderm
(b)Endoderm
(c) Mesoderm
(d) Both (a) and (c)
Answer:
(c) Mesoderm

Question 16.
Where did an epidemic bone softening disease itai- itai occurred first in?
(a) South Korea
(b) Japan
(c) China
(d) Burma
Answer:
(b) Japan

Question 17.
Muscle pump is
(a) beating of heart
(b) squeezing effect of muscles upon veins running through them
(c) peristaltic wave that travel along the alimentar canal
(d) None of the above
Answer:
(b) squeezing effect of muscles upon veins running through them

Question 18.
Slow muscle fibres are found in
(a) eye
(b) leg
(c) stomach
(d) heart
Answer:
(b) leg

Question 19.
The gliding joints are important for gliding movements. One example of such a joint is between the
(a) zygapophyses of adjacent vertebrae
(b) humerous and glenoid cavity
(c) occipital condyle and odontoid process
(d) femur and tibio fibula
Answer:
(a) zygapophyses of adjacent vertebrae

Question 20.
Aqueduct of Sylvius (iter) connects
(a) 1^st and 2^nd ventricles
(b) 3^rd and 4^th ventricles
(c) 2^nd and 3^rd ventricles
(d) 4^th and 1^st ventricles
Answer:
(b) 3^rd and 4^th ventricles

Question 21.
Volkmann’s canals occur in
(a) internal ear
(b) liver
(c) cartilage
(d) bone
Answer:
(d) bone

Question 22.
For muscle contraction, in myofibrils the formation, of a protein is essential, such protein discovered by
(a) Jean Hanson
(b) Con and Con
(c) lbert Szent Gyorgyi
(d) Hugh Huxley
Answer:
(c) lbert Szent Gyorgyi

Students can Download Chapter 11 Plant Growth and Development Questions and Answers, Plus One Botany Chapter Wise Questions and Answers helps you to revise the complete Kerala State Syllabus and score more marks in your examinations

Kerala Plus One Botany Chapter Wise Questions and Answers Chapter 11 Plant Growth and Development

Plus One Botany Plant Growth and Development One Mark Questions and Answers

Question 1.
Ethylene is used for
(a) Retarding ripening of tomatoes
(b) Hastening of ripening of fruits
(c) Slowing down ripening of apples
(d) Both b and c
Answer:
(b) Hastening of ripening of fruits.

Question 2.
Coconut milk contains
(a) ABA
(b) Auxin
(c) Cytokinin
(d) Gibberellin
Answer:
(c) Cytokinin

Question 3.
The affect of apical dominance can be overcome by which of the following hormone:
(a) IAA
(b) Ethylene
(c) Gibberellin
(d) Cytokinin
Answer:
(d) Cytokinin.

Question 4.
Growth can be measured in various ways. Which of these can be used as parameters to measure growth
(a) Increase in cell number
(b) Increase in cell size
(c) Increase in length and weight
(d) All the above
Answer:
(d) All the above

Question 5.
ABA acts antagonistic to
(a) Ethylene
(b) Cytokinin
(c) Gibberlic acid
(d) IAA
Answer:
(c) Gibberlic acid.

Question 6.
Name the growth regulator which was isolated from the endosperm of maize.
Answer:
Cytokinin(Zeatin).

Question 7.
What can induce bolting cabbage plant?
Answer:
Gibberellins

Question 8.
Certain plants flower only when they are exposed to low temperature for a few weeks . What do you call for this requirement?
Answer:
Vernalisation.

Question 9.
Some volatile substance from the ripened oranges that hastens the ripening of stored unripened bananas. Name the hormone which induces ripening.
Answer:
Ethylene

Question 10.
In a wheat field, some broad-leaved weeds were found by a farmer. Which phytohormone can be used to eradicate them.
Answer:
2, 4 – D (Dichlorophenoxy acetic acid).

Question 11.
Name the phytohormone that can cause the development of seedless fruits.
Answer:
Gibberellic acid.

Plus One Botany Plant Growth and Development Two Mark Questions and Answers

Question 1.
Match the following. (Column I with Column II)

Answer:

Question 2.
Match the following.

Answer:

Question 3.
In plants, the adverse environmental conditions such as severe drought can be overcome by the production of a hormone.

1. Name the hormone.
2. Write the role of the hormone.

Answer:

1. ABA
2. Stress hormone – Closure of stomata.

Question 4.
Auxin is a growth-promoting phytohormone. Write any two functions of auxin,
Answer:

1. Production of Seedless fruit
2. Used as Weedicide

Question 5.
Growth pattern of a plant is displayed in graph. Observe the figure.

1. What kind of growth form is this?
2. Why does the graph show a decline to a near-constant level?

Answer:

1. Sigmoid curve
2. Stationary phase

Question 6.
Some important functions of a Phytohormone are given below:
Production of Parthenocarpic Fruits Eradication of weeds

1. Identify the hormone.
2. Mention any other two important functions of this hormone.

Answer:

1. Auxin
2. Root initiation, Apical dominance

Question 7.
In Plants, adverse environmental conditions such as severe drought can be overcome by the use of a hormone.

1. Name the hormone
2. Write the role of this hormone.

Answer:

1. ABA (Abscisic Acid)
2. It causes the Closure of stomata. This reduces the rate of transpiration.

Question 8.

1. A farmer is going to plant Tapioca cuttings. Which hormone you will suggest for early rooting of tapioca cuttings?
2. A defoliated short day plant is kept under short-day conditions for 15 days and another short-day plant with single leaf is kept under short-day conditions for 1 day. Which plant will possibly flower first and give reasons?

Answer:

1. Auxin
2. A short day plant with single leaf will flower first Leaf recives the light stimulus and produce Florigen, It is then transported to the vegetative apex. As a result vegetative apex is converted into reproductive apex and bears floral leaves.

Question 9.
Synthetic auxins are of much importance in agriculture. Name any two synthetic auxins.
Answer:

• 2, 4 – D (2,4 – Dichlorophenoxy acetic acid)
• 2, 4, 5 – T. ( 2,4,5 – Trichlorophenoxy acetic acid).

Question 10.
Which hormone will add, if you are asked to help a farmer to

1. Quickly ripens fruit
2. Bolt a resette plant
3. Initiate rooting in a twig
4. Induce stomatal closure in leaves.

Answer:

1. a) Ethylene
2. Gibberelline
3. Auxin
4. Abscisic acid (ABA)

Question 11.
A gardener finds some broad-leaved dicot weeds growing in his lawns. What can be done to get rid of the weeds efficiently?
Answer:
The dicotyledonous plant grow by their apical shoot meristems while grasses (which make lawns) possess intercalary meristem. Certain auxins, such as synthetic 2, 4-Dichlorophenoxyacetic acid (2,4- D) when applied in excess can damage the shoot apical meristems but they do not cause any damage to the intercalary meristems. Thus, when 2, 4-D is sprayed on lawns, only the dicots get killed and the lawns become free of weeds.

Question 12.
What is the mechanism underlying the phenomenon by which the terminal/apical bud suppresses the growth of lateral buds? Suggest measures to overcome this phenomenon.
Answer:
The phenomenon by which the terminal apical bud suppresses the growth of lateral buds is referred to as apical dominance. This js because of the hormone auxin synthesised in the apical bud that inhibits lateral bud development.

This can be overcome by removing the apical bud (decapitating) and young leaves which will increase branching. It may also be possible to overcome this phenomenon by application of cytokinin and antiauxins like ethylene chlorohydrin, DCA (trichloroanisole), etc.

Question 13.
Ethylene is otherwise known as a fruiting hormone. So this is widely used in agricultural fields. Which compound is used as a source of ethylene and what are its merits?
Answer:
Ehephone-Merits:

1. It promotes female flowers in cucumbers.
2. It hastens fruit ripening in tomatoes and apples.
3. It promotes abscission of flowers and fruits.

Question 14.
Plant hormones are organic compounds influencing growth and development. One of the hormones was first isolated from human urine and other is a gaseous hormone.
a) Name the two hormones.
b) Write any two functions of these hormones.
Answer:

Question 15.
How will you induce lateral branching in a plant which normally does not produce them ? Give reason in support of your answer.
Answer:
When apical bud is removed, it promotes lateral branching. It is due to the removal the auxin from the tip and growth of apical bud is inhibited.

Question 16.
Describe how auxins are related with the bending of shoot towards the source of light.
Answer:
When unilateral light is given, auxin from the illuminated side shifted towards the shaded side and more growth occurs on that side. This causes the bending of the shoot.

Question 17.
How does abscisic acid act antagonistically to auxins and gibberellins?
Answer:
Gibberellin causes the breaking of bud dormancy but auxin check the fruit fall and leaf fall. These two physiological functions are inhibited by ABA.

Question 18.
What is meant by abscission? Name the phytohormone involved in it.
Answer:
It is the shedding of leaves, flowers, fruits, and bark. Abscisic acid(ABA)

Question 19.
What does a stationary phase of sigmoid growth curve indicate?
Answer:
It is otherwise called a declining phase because growth slows down.

Question 20.
What is the full form of I BA? Also, mention its one use in agriculture.
Answer:
Indole – 3 -butyric acid
It initiates root formation on stem cuttings by activating root primordia

Question 21
Fill in the places with appropriate word/words.

1. A phase of growth which is the maximum and fastest is _____.
2. Apical dominance as expressed in dicotyledonous plants is due to the presence of more _____ in the apical bud than in the lateral ones.
3. In addition to auxin, _____ must be supplied to culture medium to obtain a good callus in plant tissue culture.
4. ______ of vegetative plants are the sites of photoperiodic perception.

Answer:

1. exponential/log phase of an S – curve.
2. auxin/IAA
3. cytokinin/Kinetin/6 BAP/Zeatin/etc.
4. leaves.

Question 22.
Why is abscisic acid called stress hormone?
Answer:
Abscisic acid is also called stress hormone because the synthesis of abscisic acid is stimulated by drought, waterlogging and other adverse environmental conditions. It causes the closure of stomata.

Question 23.
Root and shoot elongation takes place at constant rate in one type of growth but in others zygote develops into embryo. What are the types of growth in both?
Answer:
In the former growth is arithematic type while in the latter initial geometric then arithematic type.

Question 24.
Give the term for the process given below.

1. formation of vascular cambium and cork cambium
2. Formation of secondary tissues in dicot stem

Answer:

1. Dedifferentiation
2. Redifferentiation

Question 25.

1. Name the phenomenon of the influence of day length on the flowering of long day and short-day plants
2. Which is the part of plant shows response to such phenomenon

Answer:

1. Photoperiodism
2. Leaf.

Question 26.
Identify the true or false statements from the following.

1. ABA is known as Anti- gibberellin
2. Auxin promotes flowering in pineapple
3. Low-temperature treatment not promote flowering in varieties of wheat
4. Bakane disease associated with ethylene

Answer:

1. True
2. True
3. False
4. False

Plus One Botany Plant Growth and Development Three Mark Questions and Answers

Question 1.
Identify the hormone have important role from the statement given below

1. In Tea plantation and hedge making
2. Increase stem length of sugar cane
3. ripening of fruits

Answer:

1. Auxin
2. Gibberellin
3. Ethylene

Question 2.
Which one of the plant growth regulators would you use if you are asked to:

1. Induce rooting in a twig
2. Quickly ripen a fruit
3. Delay leaf senescence
4. ‘bolt’ a rosette plant
5. Induce immediate stomatal closure in leaves.

Answer:

1. auxin
2. ethylene
3. cytokinin
4. gibberellins
5. ABA

Question 3.
The S-shaped growth curve is shown in the diagram. Label ‘a’ to ‘c’ and also write short notes on ‘a’ and ‘b’.

Answer:

• a – lag phase (slow phase)
• b – exponential phase (rapid phase)
• c – stationary phase (stagnant phase)

Question 4.
Plant growth substances (PGS) have innumerable practical Applications. Name the PGS you should use to

1. Increase yield of sugar cane
2. Promote lateral shoot growth
3. Inhibit seed germination

Answer:

1. GA3/gibberellinfgibberellic acid
2. Cytokinin zeatin/kinetin
3. ABA/Abscissic acid.

Plus One Botany Plant Growth and Development NCERT Mark Questions and Answers

Question 1.
Why is abscisic acid also known as stress hormone?
Answer:
Abscisic acid increases the tolerance of plants to various kinds of stresses. Therefore, it is also called the stress hormone.

Question 2.
‘Both growth and differentiation in higher plants are open’. Comment.
Answer:
Theoretically, growth and differentiation in higher plants are open. This means that there is no limit to the ‘extent a plant part can grow. But it is more correct to say that development and differentiation is open in higher plants. Once a cell loses its capacity to divide then it differentiates Differentiation is the process by which a particular plant starts doing the job it is meant to do.

For example, the job of a leaf is to make food for plant. Sometimes environment or a particular phase of growth can dictate a particular part to behave differently. This is the phase when redifferentiation occurs and the plant part takes on a new role. Thus, it can be said that development and differentiation are open to change under the given environmental conditions and demands of those conditions.

Question 3.
Which one of the plant growth regulators would you use if you are asked to:

1. induce rooting in a twig
2. quickly ripen a fruit
3. delay leaf senescence
4. induce growth in axillary buds
5. ‘bolt’ a rosette plant
6. induce immediate stomatal closure in leaves

Answer:

1. Auxins
2. Ethylene
3. Cytokinins
4. Auxins
5. Gibberellins
6. Abscisic Acid

Question 4.
Would a defoliated plant respond to the photoperiodic cycle? Why?
Answer:
The hormone responsible for photoperiodism is found in leaves. So a defoliated plant will be devoid of such hormones and will not respond to the photoperiodic cycle.

Question 5.
What would be expected to happen if:

1. GA is applied to rice seedlings
2. dividing cells stop differentiating
3. a rotten fruit gets mixed with unripe fruits
4. you forget to add cytokinin to the culture medium

Answer:
1. GAj increases the length of axis. This property will help increase the length of axis so that yield of rice can be increased.

2. When dividing cells stop differentiating then it is the maturity stage of that part of the plant. Further growth of that particular region will be stopped.

3. The ethylene present in rotten fruit will hasten the ripening process of unripe fruit and may lead to premature ripening.

4. There will be lesser cell division and culture will not grow as per the target.

Question 6.
Why is not anyone parameter good enough to demonstrate growth throughout the life of a flowering plant?
Answer:
Like most of the organisms plants also go through various phases of growth. When a seed is germinating then the parameters of growth will be different compared to growth of a mature plant. Moreover, plants can be of huge variety, right from perennial plants to plants living hundreds of years. Because of sheer diversity in size and life span any single parameter to define and measure the growth of a plant can never be sufficient.

Plus One Botany Plant Growth and Development Multiple Choice Questions and Answers

Question 1.
The ripening of fruits can be fastened by treatment with
(a) gibberellins
(b) cytokinins
(c) ethylene gas
(d) auxin
Answer:
(c) ethylene gas

Question 2.
Which of the following is the effect of a plant hormone, which is synthesized more in the absence of light?
(a) Inhibits the development of seedless fruits
(b) Responsible for closing of stomata
(c) Induces the dormancy of seeds
(d) Length of intemodes increases
Answer:
(d) Length of intemodes increases

Question 3.
Leaf abscission is caused by
(a) ABA
(b) cytokinin
(c) auxin
(d) gibberellin
Answer:
(a) ABA

Question 4.
A hormone delaying senescence is
(a) auxin
(b) cytokinins
(c) ethylene
(d) gibberellin
Answer:
(b) cytokinins

Question 5.
Which of the following induces bolting in rosette plants?
(a) Gibberellins
(b) Cytokinin
(c) Auxins
(d) Ethylene
Answer:
(a) Gibberellins

Question 6.
Sprouting of potato under storage condition can be prevented by
(a) auxin
(b) gibberellin
(c) ethylene
(d) cytokinin
Answer:
(a) auxin

Question 7.
The plant hormone produced by Rhizobium for nodulation is
(a) IBA
(b) NAA
(c) 2, 4-D
(d) IAA
Answer:
(d) IAA

Question 8.
Cell elongation in internodal regions of the green plants takes place due to
(a) indole acetic acid
(b) cytokinins
(c) gibberellins
(d) ethylene
Answer:
(c) gibberellins

Question 9.
How does pruning help in making the hedge dense?
(a) It induces the differentiation of new shoots from the rootstock
(b) It frees axillary buds from apical dominance
(c) it promotes adventitious root growth
(d) it promotes the growth of apical buds
Answer:
(b) It frees axillary buds from apical dominance

Question 10.
Name of a gaseous plant hormone is
(a) IAA
(b) gibberellin
(c) ethylene
(d) abscisic acid
Answer:
(c) ethylene

Question 11.
The maximum growth rate occurs in
(a) stationary phase
(b) senescent phase
(c) lag phase
(d) exponential phase
Answer:
(d) exponential phase

Question 12.
Growth promoting hormone is
(a) gibberellins
(b) ABA
(c) auxins
(d) both a and c
Answer:
(d) both a and c

Question 13.
The discovery of gibberellins is related with one of the following
(a) blast disease of rice
(b) foolish seedling disese
(c) bakane disease of rice
(d) early blight disease of potato
Answer:
(c) bakane disease of rice

Question 14.
Which of the following movements in plants is due the increased concentration of auxin?
(a) Movement of shoot towards the source of light
(b) Nyctinasty
(c) Movement of sunflower towards sun
(d) root differentiation
Answer:
(a) Movement of shoot towards the source of light

Question 15.
The problem of necrosis and gradual senescence, while performing tissue culture can be overcome by
(a) spraying auxins
(b) spraying cytokinins
(c) suspension culture
(d) subculture
Answer:
(b) spraying cytokinins

Question 16.
The following statements are given about plant growth hormones
I. Cytokinins especially help in delaying senescence
II. Auxins are involved in regulating apical dominance
III. Ethylene is especially useful in enhancing seed germination
IV. Gibberellins are responsible for immature falling of leaves
Which of the above statements are correct
(a) I and II only
(b) I and III only
(c) II and III only
(d) II, III, and IV only
Answer:
(a) I and II only

Question 17.
Which plant hormone promotes seed dormancy, bud dormancy and causes stomatal closure?
(a) auxin
(b) Abscisic acid
(c) GA
(d) CH₂=CH₂
Answer:
(b) Abscisic acid

Question 18.
Abscisic acid is primarily synthesized in
(a) lysosomes
(b) Golgi complex
(c) chloroplast
(d) ribosomes
Answer:
(c) chloroplast

Question 19.
Auxin in plant means for
(a) cell elongation
(b) fruit ripening
(c) cell division
(d) cell differentiation
Answer:
(c) cell division

Question 20.
The hormone present in the liquid endosperm of coconut is
(a) cytokinin
(b) gibberellin
(c) ethylene
(d) auxin
Answer:
(a) cytokinin