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## Kerala State Syllabus 10th Standard Maths Solutions Chapter 4 Second Degree Equations

### Second Degree Equations Text Book Questions and Answers

Textbook Page No. 81

**Second Degree Equation Class 10 Kerala Syllabus Question 1.**

When each side of a square was reduced by 2 metres, the area became 49 square metres. What was the length of a side of the original square?

Answer:

Let the length of each side of the original square be x, then

The length of each side of the original square = 9 cm

**Second Degree Equation Class 10 Questions And Answers Kerala Syllabus Question 2.**

A square ground has a 2 metre wide path all around it. The total area of the ground and the path is 1225 square metres. What is the area of the ground alone?

Answer:

’Each side of the ground = x

Each side of ground+path = x + 4

(x + 4)^{2} = 1225 = 35^{2};

x + 4 = 35

x= 35 – 4 = 31

Area of the ground alone = 31^{2} = 961 m^{2}

**Second Degree Equation Class 10 Extra Questions Kerala Syllabus Question 3.**

The square of a term in the arithmetic sequence 2, 5, 8, ……., is 2500. What is its position?

Answer:

Let 2500 is the square of the n^{th} term of the arithmetic sequence 2, 5, 8, …………

It is the 17^{th} term.

To solve a system of linear equations with steps, use the system of linear equations calculator.

**Sslc Maths Second Degree Equations Kerala Syllabus Question 4.**

2000 rupees was deposited in a scheme in which interest is compounded annually. After two years the amount in the account was 2205 rupees. What is the rate of interest?

Answer:

Amount (P) = 2000

Compound interest (r %)

Year (n)

Textbook Page No. 86

**Sslc Second Degree Equation Questions Kerala Syllabus Question 1.**

1 added to the product of two consecutive even numbers gives 289. What are the numbers?

Answer:

Let the two consecutive even numbers be x, x+2

x, x+2

x(x+2) + 1 = 289

x^{2} + 2x = 288

x^{2}+ 2x – 288 = 0

x^{2}+ 2x = 288

(x+1)^{2} = 288 + 1

(x+1)^{2} = 289

x+1 = ± 17

x = 16, – 18

The numbers are 16, 18

**Second Degree Equation Questions And Answers Kerala Syllabus Question 2.**

9 added to the product of two consecutive multiples of 6 gives 729. What are the numbers?

Answer:

Let the two consecutive multiples of 6 be x, x+6

x, x+6

x(x + 6)+ 9 = 729

x^{2} + 6 x + 9 = 729

x^{2} + 6x = 720

x^{2}+ 6x – 720 = O

x^{2} +6x = 720

(x+3)^{2}= 720+9

(x+3)^{2} = 729

x+3 = ± 27

x = 24, – 30

The numbers are 24, – 30

**10th Maths Second Degree Equation Kerala Syllabus Question 3.**

How many terms of the arithmetic sequence 5, 7, 9, …, must be added to get 140?

Answer:

First term f= 5,

Common difference = 2

10 terms should be added to get 140

**Second Degree Equation Class 10 Model Question Paper Kerala Syllabus Question 4.**

16 added to the sum of the first few terms of the arithmetic sequence 9, 11, 13, gave 256. How many terms were added?

Answer:

9, 11, 13,…………

First term f= 9,

Common difference d = 2

Sum of first n terms

**Second Degree Equation Class 10 Notes Kerala Syllabus Question 5.**

An isosceles triangle has to be made like this

The height should be 2 meters less than the base. The area of the triangle should be 12 square meters. What should be the length of its sides?

Answer:

Base = AB = x

Height= CD = x – 2

**Sslc Maths Chapter 4 Questions And Answers Kerala Syllabus Question 6.**

A 2.6-meter long rod leans against a wall, its foot 1 meter from the wall. When the foot is moved a little away from the wall, its upper end slides the same length down. How much farther is the foot moved?

Answer:

Textbook Page No. 91

**Sslc Maths Chapter 4 Solutions Kerala Syllabus Question 1.**

The product of a number and 2 more than that is 168, what are the numbers?

Answer:

Let the number be x

x(x+2) = 168

x^{2} + 2x = I68

x^{2} + 2x + 1 = I68 + 1

(x+1)^{2} = 169

x + 1 = ± 13

x + 1 = 13

x + 1= – 13

x = 13 – 1 = 12

x = -13 – 1 = – 14

The number is 12 and 14 or – 12 and – 14

**Maths Chapter 4 Class 10 Kerala Syllabus Question 2.**

Find two numbers with sum 4 and product 2.

Answer:

Sum of the numbers = 4

If, First number = x

Then second number = 4 – x

**Class 10 Maths Second Degree Equation Kerala Syllabus Question 3.**

How many terms of the arithmetic sequence 99, 97, 95, … must be added to get 900?

Answer:

The spectrally extended signal produced by nonlinear function calculator 520 is likely to have a pronounced dropoff in amplitude as frequency increases.

**Class 10 Maths Chapter 4 Kerala Syllabus Kerala Syllabus Question 4.**

A rod 28 centimeters long is to be bent to make a rectangle.

i. Can a rectangle of diagonal 8 centimeters be made?

ii. Can a rectangle of diagonal 10 centimeters be made?

iii. How about a rectangle of diagonal 14 centimeters?

Calculate the lengths of the sides of the rectangles that can be made.

Answer:

∴ Cannot make a rectangle of diagonal 14 centimeters.

Textbook Page No. 97

**Maths Second Degree Equation Kerala Syllabus Question 1.**

The perimeter of a rectangle is 42 meters and its diagonal is 15 meters. What are the lengths of its sides?

Answer:

**Sslc Second Degree Equation Questions And Answers Kerala Syllabus Question 2.**

How many consecutive natural numbers starting from 1 should be added to get 300?

Answer:

Sum of natural numbers starting from 1:

**Second Degree Equation Class 10 Kerala Syllabus Kerala Syllabus Question 3.**

What number added to 1 gives its own square?

Answer:

Let the number be x

**Second Degree Equation Kerala Syllabus Question 4.**

In writing the equation to construct a rectangle of specified perimeter and area, the perimeter was wrongly written as 24 instead of 42. The length of a side was then computed as 10 meters. What is the area in the problem? What are the lengths of the sides of the rectangle in the correct problem?

Answer:

If length is 10 and perimter is 24

2 (l + b) = 24

2(10 + b) = 24

10 + 5 = 12

b = 2

Width= 2

Area= lb =10 x 2 = 20 sq.cm

Area in the correct problem =20

lb = 20

2(l + b) = 42

l + b = 21

l + \(\frac { 20 }{ l }\) = 21

**Kerala Syllabus 10th Standard Maths Chapter 4 Kerala Syllabus Question 5.**

In copying a second-degree equation to solve it, the term without x was written as 24 instead of – 24. The answers found were 4 and 6. What are the answers of the correct problem?

Answer:

### Second Degree Equations Orukkam Questions and Answers

Worksheet 1

**Second Degree Equations Exercises Kerala Syllabus Question 1.**

Write two numbers whose square is 25?

Answer:

Let x be the number

x^{2} = 25

x= √25 = ±5

numbers are –5, +5

**Second Degree Equation Class 10 Formulas Kerala Syllabus Question 2.**

When the square of a number is added to the number we get 30. What are the numbers?

Answer:

Let x be the number ,

x^{2} + x = 30

x^{2} + x – 30 = 0, (x + 6) (x – 5) = 0

x = – 6, 5

numbers are – 6, +5

**Second Degree Equations Class 10 Kerala Syllabus Question 3.**

Find the side of the square whose area and perimeter are numerically equal.

Answer:

Let x be the length of side.

Perimeter = Area

x^{2} = 4x

x = 4

Length of side = 4

Question 4.

How many odd numbers from 1 makes the sum 961?

Answer:

Sum of continuous n odd numbers = n^{2}

n^{2 }= 961

n = √961 = ± 31

Sum of continuous 31 odd numbers

Question 5.

A man’s age after 15 years will be the square of his age 15 years ago. What is his present age?

Answer:

Worksheet 2

Form the equation

Question 6.

The sum of a number and its square is ten times that number

Answer:

Let x be the number

x^{2} + x = 10 x

x^{2} – 9x = 0

Question 7.

The sum of a number and its square root is 6.

Answer:

Let x be the number

x + √x = 6

x – 6 = √x

(x – 6)^{2} =(–√x)2

x^{2} – 12x + 36 = x

x^{2} – 13 x + 36 = 0

Question 8.

The sum of first n natural numbers is 210.

Answer:

\(\frac { n(n+1) }{ 2 }\) = 210

n^{2} + n = 420

n^{2} + n – 420 = 0

Question 9.

The area of a rectangle whose length is 5 more than its width

Answer:

Let x be the length of side.

other side = x + 5

x (x + 5) = 150

x^{2} + 5x – 150 = 0

Question 10.

The sum of a number and its reciprocal is \(\frac { 5 }{ 2 }\)

Answer:

Let x be the number

Question 11.

The sum of even numbers from 2 in an order is 240

Answer:

2 + 4 + 6 + + 2n = 420

2(1 + 2 + 3 + ….+ n) = 420

Question 12.

A man’s age after 15 years will be the square of his age 15 years ago.

Answer:

Let x be the present age

(x – 15)^{2} = x + 15

x^{2} – 30 x + 225 = x +15

x^{2} – 31x + 210 = 0

Worksheet 3

Question 13.

When 8 times a number is added to its square we get 8. Find the number by making the equation properly.

Answer:

x^{2} + 2 x = 8

x^{2} + 2x + 1 = 8 + 1

(x + 1)^{2} =9

x + 1 =3

x = 2

Number = 2

Question 14.

Which term in the sequence 2, 5, 8 …….. gives its square 2500?

Answer:

nth term = 3n + (2 – 3) = 3n – 1

(3n – 1)^{2} = 2500,

3n – 1 = 50,

n = 17

square of 17^{th} term is 2500

Question 15..

A man’s age after 15 years will be the square of his age 15 years ago. Find the age

Answer:

Let x be the age

x + 15 = (x – 15)^{2}

x + 15 = x^{2} – 30 x + 225

x^{2} – 31 x = – 210

Question 16.

The length of a rectangle is 2 more than its width. The area of the rectangle is 80. Find length and breadth.

Answer:

Let x be the width, then length = x + 2

x (x + 2) = 80

x^{2} + 2x = 80

x^{2} + 2x + 1 = 80 + 1

(x + 1)^{2} = 81

x + 1 =9

x = 8

length = 10

width = 8

Question 17.

The sum of a number and its reciprocal is \(\frac { 5 }{ 2 }\) Find the number.

Answer:

Let x be the number

Question 18.

The sum of some even numbers starting from 2 is 420. Find the number of even numbers added.

Answer:

Worksheet 4

Question 19.

Sum of the squares of three consecutive natural numbers is 110.

Answer:

If the numbers are x, x + 1, x + 2, then

x^{2} + ( x + 1)^{2} + (x + 2)^{2} = 110

x^{2} + x^{2} + 2x + 1 + x^{2} + 4x + 4 = 110

3x^{2} + 6x – 105 = 0

Question 20.

The product of the digits of a two-digit number is 12.When 36 is added to the number we get a two-digit number in which the digits are reversed. Find the two-digit number .

Answer:

Let the two digit number be 10 x + y, then

x y = 12

y = \(\frac { 12 }{ x }\)

If we change the positions of digits 10 y + x

10 y + x = 36+ 10 x + y

Let’s take x be the digits in the position of 10.

Let’s take \(\frac { 12 }{ x }\) as the digits in the position of unit.

Number which digits are reversed \(\frac { 120 }{ x }\) +x

Question 21.

Serena and Johan had 45 diamond stones. They sold 5 stones. The product of the remaining stones is 124. Find the number of stones each had

Answer:

Let x be the stones Serena has and 45 – x be the stone Johan has. After selling 5

(x – 5)(45 – x – 5) = 124

(x – 5)(40 – x) = 124

x^{2} – 45x + 200 = – 124

x^{2} – 45x + 324 = 0

Number of diamond stones Serena has = 36

Number of diamond stones Johan has = 9

Question 22.

The sum of a number and its reciprocal is \(1 \frac{1}{2}\) . Find the number.

Answer:

Let x be the number

Question 23.

The sum of two numbers is 15. Sum of its reciprocals is \(\frac { 3 }{ 10 }\). Find the numbers.

Answer:

Let x, 15 – x be the numbers

Question 24.

A two-digit number is four times sum of its digits. The number is three times product of the digits. Find the number

Answer:

Let x, y be the numbers

10 x + y = 4 (x + y)

6 x = 3 y

Worksheet 5

Question 25.

A train travels a distance of 300km constant speed. If the speed of the train is increased by 5 km, the journey would have taken 2 hours less. Find the original speed of the train

Answer:

Let x km/hr be the speed.

Let the time required to travel in the same speed be \(\frac { 300 }{ x }\) hour. If the speed is increased by 5 km/ hr the time will decreased by 2 hours.

Question 26.

An express train takes 3 hours less than a passenger train for a journey of 600 km. If the speed of the passenger train is 10 less than the speed of the express train find the speeds of both trains (Use Pythagoras theorem in distance, not in speeds)

Answer:

Let x km/hr be the speed of the express.

Speed of the express 50 km/hr

Speed of the passenger 50 km/hr

Worksheet 6

Question 27.

One year ago a man’s age is eight times the age of his son. At present man’s age is the square of son’s age. Find the present age.

Answer:

Let x be the present age of son

Question 28.

A man’s age after 15 years will be the square of his age 15 years ago. Find the present age by forming a second degree equation

Answer:

Let x be the present age .

x + 15 = (x – 15)2

x + 15 = x^{2} – 30x + 225

x^{2} – 31x + 210 = 0

Present age = 21

Question 29.

The product of Layas’s age before 5 years and after 8 years is 30. Find the present age.

Answer:

Let x be the present age of Laya

(x – 5)(x + 8) = 30

x^{2} – 3x – 40 = 30,

x^{2} – 3x – 70 = 0

Present age of Laya = 7

Worksheet 7

Question 30.

Sravani teacher asked the students to construct a rectangle having area 5 square unit and perimeter 8. Jeevan, a wise student of the class, after making some calculations told that it is not possible to construct such a rectangle. Can you agree with him. Justify reasonably

Answer:

Let x be the length, then the area will be 5 cm^{2}, So the width will be \(\frac { 5 }{ x }\) cm. When the perimeter is 8 cm, 4 – x will be the width. The width which was obtained first will be \(\frac { 5 }{ x }\) and the width obtained now i.e., 4 – x will be equal.

– 4 < 0, x will not be obtained. Hence it is not possible to draw a rectangle with the given values.

Question 31.

The perimeter of a rectangle is 34 cm, area 60 square centimeter. Find the sides

Answer:

Let x be the length

Perimeter 34 cm, therefore width will be 17-x cm

Perimeter = 60 cm, therefore width be \(\frac { 60 }{ x }\) cm

x = 5 cm

length = 12 cm therefore width = 17 – 12 = 5 cm

length = 12 cm ,

width = 5 cm

Question 32.

The length of the rectangle is 4 more than its breadth. Area of the rectangle is 140 square centimeter. Calculate length and breadth

Answer:

Let x be the width of rectangle

length = x + 4 cm

x(x + 4) = 140

x^{2} – 4 x – 140 = 0

width =10 cm

length = 14 cm

Question 33.

When the sides of a square are increased by 4, area become 256. Find the length of the first square.

Answer:

Let x be the side of square

x will not -20

The side of the first square = 12 cm

Question 34.

The area of a right-angled triangle is 60 square unit. The one of the perpendicular sides is 10 more than other. Find the sides of the triangle.

Answer:

Let x cm, x + 10 cm be the perpendicular sides

Question 35.

The area of an isosceles triangle is 60 square meters. One of the equal sides is is 13 cm. Find the third side. Take base x then h = \(\sqrt{13^{2}-x^{2}}\)

Answer:

### Second Degree Equations SCERT Question Pool Questions and Answers

Question 36.

When the sides of a square are increased by 8 cm each, its area becomes 1225 sq. cm Frame an equation using the above data by taking the side of the smaller square as x cm. Find the sides of both the squares. [Score: 3, Time: 5 Minutes]

Answer:

One side of smaller square = x

One side of bigger square = x + 8

Area = (x + 8)^{2} = 1225 (1)

x + 8 = 35

x = 35 – 8 = 27 (1)

Side of smaller square = 27 cm

Side of bigger square = 35 cm (1)

Question 37.

The difference of two positive numbers is 6. Their product is 216. Find the numbers. [Score: 3, Time: 4 Minutes]

Answer:

Let the numbers be x, x + 6

Products : x (x+6) = 216 (1)

x^{2} + 6x = 216

x^{2} + 6x + 9 = 216 + 9 = 225

(x + 3)^{2} = 225 (1)

x + 3 = 15

x = 15 – 3 = 12

Numbers : 12, 18 (1)

Question 38.

In a right triangle one of the perpendicular sides is one less than 2 times the smaller side. Hypotenuse is one more than 2 times the same smaller side. By taking the smaller side as x cm, write the algebraic expression for the other two sides. Compute all the 3 sides of the right triangle. [Score: 4, Time: 5 Minutes]

Answer:

Smaller side = x

Perpendicular side = 2x – 1

Hypotenus = 2x + 1 (1)

x^{2} + (2x – 1)^{2} = (2x + 1)^{2} (1)

x^{2} + 4x^{2} – 4x + 1 = 4x^{2} + 4x + 1

x^{2} – 8x = 0 (1)

x (x – 8) = 0

x = 0 or x = 8

Side=8 cm, 15 cm, 17 cm (1)

Question 39.

Find the sides of a rectangle whose perimeter is 100 metres and area 600 sq. metres. [Score : 4, Time : 5 Minutes]

Answer:

Perimeter = 100 m .

Length+Breadth = 50 m.

Length = 25 + x.

Breadth = 25 – x

Area = (25 + x)(25 – x) = 600 (1)

25^{2} – x^{2} = 600 (1)

x^{2} = 625 – 600 = 25, x = 5 (1)

Sides, 25 + 5 = 30 m

25 – 5 = 20 m (1)

Question 40.

The one’s place of a two-digit number is 4. The product of the number and digit sum is 238.

a. If ten’s place digit is taken as x, Write the number.

b. Frame a second-degree equation and find the number. [Score : 4, Time : 6 Minutes]

Answer:

Digit in the ten’s place = x

Number = 10 x + 4 (1)

The product of the number and digit sum

= (x + 4) (10 x + 4) = 238 (1)

Question 41.

How many consecutive natural numbers from 1 should be added to get 465? [Score : 4, Time : 5 Minutes]

Answer:

Sum of first n natural numbers

Question 42.

The product of the digits of a two-digit number is 12. When 36 is added to this number, got a new number with digits reversed. Find the number. [Score : 5, Time : 7 Minutes]

Answer:

Question 43.

For a two-digit number, one’s place is 3 more than its ten’s place. The product of this number and its digit sum is square of double the digit sum. What is the number? [Score : 4, Time : 6 Minutes]

Answer:

Question 44.

A pavement of Width 2 metres is built around a square shaped garden. The area of the pavement alone is 116 square metres. Find one side of the garden.

[Score : 3, Time: 4 Minutes]

Answer:

Side of garden = x

Area of the pavement Area of the pavement Area of the pavement Side of garden including pavement = x + 4 (1)

Area of the pavement = (x + 4)^{2} – x^{2} = 116

x^{2} + 8x + 16 – x^{2} = 116 (1)

8x + 16 = 116,

\(x=\frac{116-16}{8}=12.5 \mathrm{m}\)

Side of garden = 12.5m (1)

Question 45.

A rectangle of Width 8 centimetres is cut off from a square sheet along its side. The remaining rectangular portion has an area 84 sq. metres. Calculate the side of the square. [Score : 4, Time : 6 Minutes]

Answer:

Let the side of square = x

Sides of rectangle = x, x – 8 (1)

Area = x(x – 8) = 84

x^{2} – 8x = 84 (1)

x^{2} – 8x + 16 = 84 + 16

(x – 4)^{2}= 100, (1)

x – 4= 10, x= 14

Side of square = 14 centimetre

Question 46.

The lengh of a rectangle is 3 metre more than 3 times its breadth. Its diagonal is 1 metre more than the length. Find the lengh and breadth of the rectangle. [Score: 4, Time: 7 Minutes]

Answer:

Sidc of rectangle = x

Length =3x +3, Diagonal = 3x + 4 (1)

(3x + 4)^{2} = x^{2} + (3x + 3)^{2} (1)

9x^{2} + 24 x + 16 = x^{2} + 9x^{2} + 18x + 9

x^{2} – 6x = 7, x^{2} – 6x = 7 (1)

x^{2} – 6x + 9 = 16, (x – 3)^{2} = 16

x – 3 = 4, x = 4 + 3 = 7 (1)

length of rectangle = 3 × 7 + 3 = 24 metre breadth of rectangle = 7 metre

Question 47.

In the figure AB is the diameter of the circle. CD = 10 cm. BC is 15 cm less than Ac. Find AB?

Answer:

Question 48.

The chords AB and CD of a circle intersect at. If MA = 6 cm, MB = 8 cm and CD = 16 cm. Find MC and MD.

Answer:

CD = 16 centimetre MC = 8 – x, MD = 8 + x

then, MA x MB = MC x MD (1)

6 x 8 = (8 – x)(8 + x)

48 = 64 – x^{2}, x^{2} = 16, x = 4 (1)

MC = 8 – 4 = 4 centimetre (1)

MD = 8 + 4 = 12 centimetre (1)

Question 49.

In the figure, AD is drawn perpendicular to the side opposite to the right angled vertex A. BC-13 cm and AD=6 cm.

a. Take BD = x and express DC in terms of x.

b. Frame a second-degree equation and find the lengths BD and DC. [Score : 4, Time: 5 Minutes]

Answer:

Question 50.

Can the sum of a number and it’s reciprocal be 2/3? [Score: 4, Time: 5 Minutes]

Answer:

Number = x , then its reciprocal = 1/x

Since the discriminant is negative. We do not get a solution.

∴ The sum of a numbrer and it’s reciprocal never gives 2/3. (1)

Question 51.

The sum of a number and its reciprocal is \(\frac { 13 }{ 6 }\) What is the number ? [Score: 4, Time: 5 Minutes]

Answer:

Let number = x, then its reciprocal = 1/x

Question 52.

The sum of a number is 12 and sum of its reciprocal is \(\frac { 3 }{ 8 }\) Find the numbers. [Score: 4, Time: 5 Minutes]

Answer:

If we take numbers as 6 + x, 6 – x (1)

Question 53.

Consider an arithmetic sequence with common difference 20. If the sum of reciprocals of two consecutive terms of this sequence is 1/24, find the first term of the arithmetic sequence. [Score: 4,Time: 6 Minutes]

Answer:

’Terms x – 10, x + 10 (1)

Question 54.

Prove that the difference of a number and its reciprocal will be always positive. [Score: 3, Time : 5 Minutes]

Answer:

Number = x, its reciprocal = 1/x (1)

then, \(x-\frac{1}{x}=k\) (1)

Since k being a positive number, k^{2} + 4 also negative. (1)

Question 55.

In copying a second-degree equation, the number without x was written as -30 in¬stead of 30. The answers found were 15 and -2. What are the answers of the correct problem? [Score: 5, Time: 8 Minutes]

Answer:

### Second Degree Equations Exam oriented Questions and Answers

Short Answer Type Questions (Score 2)

Question 56.

Find two numbers whose sum is 6 and product is 9.

Answer:

’First number = x

Second number = 6 – x

x ( 6 – x ) = 9

6x – x^{2} = 9

0 = 9 – 6x + x^{2}

x^{2} – 6x + 9 = 0

(x – 3)^{2} = 0

(x – 3 ) (x – 3) = 0

x = 3

Numbers are 3, 3.

Question 57.

Ammu is 7 years younger than Divya. If 2 is added to the product of their ages, we get 200. Find their ages.

Answer:

Age of Ammu = x

Age of Divya = x – 7

x(x – 7)+2 = 200

x^{2} – 7x + 2 = 200

x^{2} – 7x – 198 = 0 (x – 18) (x + 11) = 0

x= 18, x = – 11

(age will not be a negative number)

Age of Ammu = 18,

Age of Divya =18 – 7 = 11

Question 58.

In order to solve a quadratic equation, Arun did the following steps. Find the values of x by completing each step.

Answer:

Question 59.

The perimeter of a rectangle is 42 cm and its diagonal is 15 cm. Find the dimensions of the rectangle.

Answer:

Let breadth = x; perimeter = 2 (length + breadth) length + breadth = 42/2 = 21

∴ length = 21 – x

x^{2} + (21 – x) = 152

x^{2} + 441 – 42x + x^{2} = 225

2x^{2} + 216 – 42x =0

x^{2} – 21x + 108 = 0

(x – 12) (x – 9) = 0

x = 12 or x = 9

breadth = 9 unit

length = 12 unit

Question 60.

If the equation 4x^{2} – 5 x + k = 0 has two equal solutions. Find the value of k.

Answer:

If there is one solution, the discriminant = 0

Short Answer Type Questions (Score 3)

Question 61.

The product of a number and the number 8 more than it is 105.

a. What is the least number to be added to make the product a perfect square?

b. What are the numbers in the problem?

Answer:

a. Let the number be x, then second number is x+ 8

x(x + 8) = 105

x^{2} + 8x = 105

The number to be added to get a perfect square is 16

Question 62.

On Arts day, 180 sweets were distributed equally among the students. Tasting one sweet Deepa said, “It is a pity that 9 of our friends are absent today.”

Hearing this Deepu said, “Because of that we got one sweet more.”

a. Find out the total number of students,

b. How many students were present on that day?

Answer:

a. Let the number of students = n

Number of sweets given to one student = \(\frac { 180 }{ n }\)

Since 9 students were absent = \(\frac { 180 }{ n-9 }\)

b. Total present = 45 – 9 = 36

Question 63.

One of the perpendicular side of a right angled triangle has length 4 cm more than twice of the other side. It has surface area 80 cm^{2}. Find out the lengths of the perpendicular sides.

Answer:

Let one of the perpendicular side = x

Second side = 2x + 4, Area = 80

1/2 × (2x + 4) = 80

2x^{2} + 4x = 160

x^{2} + 2x = 80

x^{2 }+ 2x + 12 = 80 +12

(x + 1)^{2} = 81

x +1 = ± 9

x + 1 = 9

x = 8

x = – 10 cannot be accepted as – ve sign cannot exist for the length of the side

∴ One side = 8cm

Second side = 2 × 8 + 4 = 20 cm

Long Answer Type Questions (Score 4)

Question 64.

Reji’s father bought several note books of the same price. Total price Rs. 360. If the price of each book were less by 2 rupees, he would get 2 books more.

a. Find the number of books if the price of one book is x.

b. Find the number of books if the price of one books is less by 2 rupees.

c. Form a quadratic equation.

d. Find the number of books bought.

Answer:

(a) Price of each book be x rupees,

No. of Notebooks = \(\frac { 360 }{ x }\)

(b) If the cost of each book is Rs (x – 2)

No. of books brought for Rs. 360 = \(\frac { 360 }{ x – 2 }\)

Question 65.

1235 Sacks of rice is to be brought from Trivandrum to Kottayam. For this a minitruck takes 6 trips more than that of an ordinary truck. An ordinary truck can carry 30 sacks more than that of a mini truck. Find the capacity of each truck.

Answer:

Let the capacity of mini truck be x sacks and ordinary truck be x + 30 sacks

Long Answer Type Questions (Score 5)

Question 66.

a. The sum of a number and it’s reciprocal is \(\frac { 25 }{ 12 }\) What is the number?

b. Prove that the sum of a positive number and it’s reciprocal is always greater than or equal to 2.

Answer:

Free Limit using Substitution Calculator – Find limits using the substitution method step-by-step.

Question 67.

Sum of the area of two squares is 500 m^{2}. If the difference of their perimeters is 40m, find the sides of the two squares.

Answer:

Let the side of the squares be x and y meters.

According to the condition,

x^{2} + y^{2} = 500 (1)

4x – 4y = 40

(x – y) = 10

y = x – 10

Substituting the value of y in (1), we get

x^{2} + (x – 10)^{2} – 500

2x^{2} – 20x – 400 = 0

x^{2} – 10x – 200 = 0

x = 20 or x = – 10

As the side cannot be negative, x = 20

Hence, side of the first square, x = 20 m

Side of the second square, y = 20 – 10 = 10 m

Question 68.

a By increasing the speed of a bus by 10 km/hr, it takes one and half hours less to cover a journey of 450km. Find the original speed of the bus.

b. 250 Rupees is divided equally among a certain number of children. If there were 25 children more, each would have received 50 paise less. Find the number of children.

Answer:

Let speed of the bus be x km/hr

### Second Degree Equations Memory Map

The value of x in the second-degree equations can be found out by mainly 3 ways.

1. Factorization

2. Completing the square

3. By using equation