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## Kerala State Syllabus 9th Standard Maths Solutions Chapter 3 Pairs of Equations

### Kerala Syllabus 9th Standard Maths Pairs of Equations Text Book Questions and Answers

Textbook Page No. 36

Do each problem below either in your head, or using an equation with one letter, or two equations with two letters:

**Pairs Of Equations Class 9 Questions And Answers Kerala Syllabus Question 1.**

In a rectangle of perimeter one metre, one side is five centimetres longer than the other. What are the lengths of the sides?

Answer:

Shortest side = x

Longest side = x + 5

Perimeter = 1 m = 100 cm

2(x + x + 5) = 100

2x + 5 = 50; 2x = 45;

x = 22.5

∴ Shortest side = 22.5

Longest side = 22.5 + 5 = 27.5

**Pairs Of Equations Questions And Answers Kerala Syllabus Question 2.**

A class has 4 more girls than boys. On a day when only 8 boys were absent, the number of girls was twice that of the boys. How many girls and boys are there in the class?

Answer:

Number of boys = x

Number of girls = x + 4

2(x – 8) = x + 4

2x – 16 = x + 4

2x – x = 4 +16; x = 20

∴ Number of boys = 20

Number of girls = 24

**Pairs Of Equations Problems Kerala Syllabus Question 3.**

A man invested 10000 rupees, split into two schemes, at annual rates of interest 8% and 9%. After one year he got 875 rupees as interest from both. How much did he invest in each?

Answer:

If one part is x then

the remaining part is 10000 – x

\(x\times \frac { 8 }{ 100 } +\left( 10000-x \right) \times \frac { 9 }{ 100 } =100\)

8x + 90000 – 9x = 87500

90000 – 87500 = x

2500 = x

one part = 2500 and

remaining part = 7500

**Kerala Syllabus 9th Standard Maths Chapter 3 Question 4.**

A three and a half metre long rod is to be cut into two pieces, one piece is to be bent into a square and the other into an equilateral triangle. The length of their sides must be the same. How should it be cut?

Answer:

Total length = 3½ m

Since the sides of a square and equilateral triangle are equal, all these 7 sides are equal.

∴ Length of one side

\( 3\frac { 1 }{ 2 } \div 7=\frac { 7 }{ 2 } \div 7=\frac { 1 }{ 2 } \)m

Length of the rod for the square

\(= 4\times \frac { 1 }{ 2 } \) = 2m

Length of the rod for the equilateral triangle = \(3\times \frac { 1 }{ 2 } \) = \(1\frac { 1 }{ 2 } \)m

**Class 9 Maths Chapter 3 Kerala Syllabus Question 5.**

The distance travelled in t seconds by an object starting with a speed of u metres/second and moving along a straight line with speed increasing at the rate of a metres/second every second is given by ut + \(\frac { 1 }{ 2 } \) at² metres. An object moving in this manner travels 10 metres in 2 seconds and 28 metres in 4 seconds. With what speed did it start? At what rate does its speed change?

Answer:

If t = 2

ut + \(\frac { 1 }{ 2 } \)at²= 10

2u + 2a= 10

u + a = 5 — (1)

If t = 4

4u + 8a = 28

u + 2a = 7 — (2)

from (1) and (2)

a = 2

∴ u = 3

Textbook Page No. 40

**Pairs Of Equations Class 9 Questions And Answers Pdf Question 1.**

Raju bought seven notebooks of two hundred pages and five of hundred pages, for 107 rupees. Joseph bought five notebooks of two hundred pages and seven of hundred pages, for 97 rupees. What is the price of each kind of notebook?

Answer:

Cost of 200 page note book = x

Cost of 100 page note book = y

7x + 5y= 107 …………(1)

5x + 7y = 97 …………(2)

(1) × 5 \(\Rightarrow \) 35x + 25y = 535 …………(3)

(2) × 7 \(\Rightarrow \) 35x + 49y = 679 …………(4)

(4) – (3) \(\Rightarrow \) 24y = 144

y = \(\frac{144}{24} \) = 6

Substitute y = 6 in equation (1)

7x + 30 = 107; 7x = 77

x = \(\frac{77}{7} \) = 11

Price of the 200 pages notebook = Rs. 11

Price of the 100 pages notebook = Rs. 6

**Pairs Of Equations Class 9 Extra Questions And Answers Question 2.**

Four times a number and three times number added together make 43. Two times the second number, subtracted from three times the first give 11. What are the numbers?

Answer:

Let the first number = x and

the second number = y

4x + 3y = 43 …………(1)

3x – 2y = 11 …………(2)

(1) × 3 \(\Rightarrow \) 12x + 9y= 129 …………(3)

(2) × 4 \(\Rightarrow \) 12x – 8y = 44 …………(4)

(3) -(4) \(\Rightarrow \) 17y = 85; y = \(\frac{85}{17} \) = 5

Substitute y = 5 in equation (1)

4x + 3y = 43

4x + 15 = 43

4x = 43 – 15 = 28

∴ x = \(\frac{28}{4} \) = 7, y = 5

First number = 7

Second number = 5

**9th Standard Maths Chapter 3 Kerala Syllabus Question 3.**

The sum of the digits of two – digit number is 11. The number got by interchanging the digits is 27 more than the original number. What is the number?

Answer:

If the numbers are x and y

x + y = 11 …………(1)

10x + y + 27 = 10y + x

9x – 9y = -27

X – y = -3 …………(2)

(1) + (2) 2x = 8; x = 4

x + y = 11

4 + y = 11

y = 7

∴ Required number is 47.

**Kerala Syllabus 9th Standard Maths Notes Question 4.**

Four years ago, Rahim’s age was three times Ramu’s age. After two years, it would just be double. What are their ages now?

Answer:

Ramu’s present age = x

Rahim’s present age = y

4 years back,

Ramu’s age = x – 4

Rahim’s age = y – 4

3(x – 4) = y – 4

3x – 12 = y – 4

3x – y = 8 ……….(1)

After 2 years,

Ramu’s age = x + 2

Rahim’s age = y + 2

2(x + 2) = y + 2

2x + 4 = y + 2

2x – y = -2 ……….(2)

(1) – (2) \(\Rightarrow \) x = 10

3x – y = 8; 30 – y = 8; y = 22

x = 10, y = 22

Ramu’s present age = 10

Rahim’s present age = 22

**Pair Of Equations Class 9 Kerala Syllabus Question 5.**

If the length of a rectangle is in-creased by 5 metres and breadth decreased by 3 metres, the area would decrease by 5 square metres. If the length is increased by 3 metres and breadth increased by 2 metres, the area would increase by 50 square metres. What are the length and breadth?

Answer:

length = x; breadth = y

(x + 5)(y – 3) = xy – 5

xy – 3x + 5y – 15 = xy – 5

– 3x + 5y = + 10

3x – 5y = -10 ………..(1)

(x + 3)(y + 2) = xy + 50

xy + 2x + 3y + 6 = xy + 50

2x + 3y = 44 ………..(2)

(2) × 1 \(\Rightarrow \) 6x-10y = -20 ……….(3)

(3) × 2 \(\Rightarrow \) 6x + 9y = 132 …………(4)

(3)- (4) \(\Rightarrow \) -19y = -152

y = \(\frac{-152}{-19} \) = 8, 2x + 3y = 44

2x + 24 = 44; 2x = 20; x = 10

∴ x = 10, y = 8

Length of the rectangle = 10 m

Breadth of the rectangle = 8m

Textbook Page No. 42

**Hsslive Guru Maths 9th Kerala Syllabus Question 1.**

A 10 metre long rope is to be cut into two pieces and a square is to be made using each. The difference in the areas enclosed must be 1\(\frac{1}{4} \) square metres. How should it be cut?

Answer:

Length of one piece = x m

Length of other piece = (10 – x) m

∴ Rope is divided into 6 m and 4 m.

**9th Maths Notes Kerala Syllabus Question 2.**

The length of a rectangle is 1 metre more than its breadth. Its area is 3\(\frac{3}{4} \) square metres. What are its length and breadth?

Answer:

Length = x

Breadth = y

x = y + 1; x – y = 1

\(x y = 3 \frac{3}{4} =\frac{15}{4}\)

(x + y)² = (x – y)² + 4xy

1² + 4 × \(\frac{15}{4}\) = 1 + 15 = 16

x – y = 1; x + y = 4

2x = 5; x = 5/2 = 2.5

y=1.5

∴ Length = 2.5 m

Breadth = 1.5 m

**Hsslive Maths Class 9 Kerala Syllabus Question 3.**

The hypotenuse of a right triangle is 6\(\frac{1}{2} \) centimetres and its area is 7\(\frac{1}{2} \) square centimetres. Calculate the lengths of its perpendicular sides.

Answer:

The perpendicular sides are x and y. Given that,

From (3) and (4)

2x = 24/2 = 12

∴ x = 6

6 – y = 7/2

∴ y = 5/2 = 2.5

∴ Perpendicular sides = 6 and 2.5

### Kerala Syllabus 9th Standard Maths Pairs of Equations Exam Oriented Text Book Questions and Answers

**Kerala Syllabus 9th Standard Notes Maths Question 1.**

There are some oranges in a bag. When 10 oranges more added in the bag; the numbers become 3 times of the oranges initially had. Then how many oranges were there in the bag initially.

Let the number of oranges initially taken = x

X + …….. = 3X

3X – X = ……..

2X = ……..

X = ……… /2 = ……..

Answer:

x + 10 = 3x; 3x – x = 10 ; 2x = 10;

x = \(\frac{10}{2} \) = 5

**Kerala Syllabus 9th Standard Maths Notes Pdf Question 2.**

A box contains some white balls and some black balls. The number of black balls is 8 more than the number of white balls. The total number of balls is 4 times the number of white balls. Find out the number of white balls and the number of black balls.

Number of white balls = x

Number of black balls = ……… + 8

Total number of balls = ……. ‘ x;

i. e. (x) + (x + 8) = ………. x;

2x + 8 = ……. x;

8 = …… x – 2x

= …… x; x = 8/ …….

white balls = ……….

black balls = ……… + 8 = ……..

Answer:

Number of white balls = x

Number of black balls = x + 8

Total number of balls = 4 ‘ x;

(x) + (x + 8) = 4x ; 2x + 8 = 4x;

8 = 4x – 2x = 2x ; x = \(\frac{8}{2} \) = 4

white balls = 4

black balls = 4 + 8 = 12

**Pairs Of Equations Questions Kerala Syllabus Question 3.**

The sum of two numbers is 36 and the difference is 8. Find the numbers.

Let x, y be the numbers

x + y = 36

x – y = 8

(x + y) + (x – y) = ……. + ……

2x = …..

x = …… /2 = ……..

x – y = 8

……. – y = 8

…….. – 8 = y

Answer:

x + y = 36; x – y = 8

(x + y) + (x – y) = 36 + 8

2x = 44; x = \(\frac{44}{2} \) = 22

x – y = 8; 22 – y = 8;

22 – 8 = y; y = 14

numbers 22, 14

**Kerala Syllabus 9th Std Maths Notes Question 4.**

The cost of 2 pencils and 5 pens is Rs 17, two pencils and 3 pens of the same rate is Rs 11. Find out the prices of a pencil and a pen.

Let the price of pencil = x;

Price of pen = y;

∴ 2x + 5y = …….. 2x +….. y = …..

2x + …… y = 11

(2x + 5y) (-2x + ….. y) = -11

…… y = …….

y = \(\frac{…..}{……} \)

2x + 5x ….. = 17;

2x = 17 – ……..; x = ……. /2; = ……..

Answer:

2x + 5y = 17; 2x + 3y = 11;

(2x + 5y) – (2x + 3y) = 17 – 11; 2y = 6

y = \(\frac{6}{2} \) = 3; 2x + 5 × 3 = 17;

2x = 17 – 15;

x = \(\frac{2}{2} \) = 1

Price of pencil = Rs 1

Price of pen = Rs 3

Question 5.

Twice of a number added with thrice of another number gives 23. Four times the first number and 5 times the second number when added gives 41. Find out the numbers.

First number = x

Second number = y

∴ 2x + 3y = ……

4x + 5y = ……

2(2x + 3y) = 2 …….

4x + 6y = …….

(4x + 6y) – (4x + 5y) = ( …… ) – ( ….. )

y = …….; 2x + 3x ……. = …….

2x = ( …… ) – ( ……. ); x = ………./2

Answer:

2x + 3 y = 23

4x + 5y = 41

2(2x + 3y) = 2 × 23; 4x + 6y = 46

(4x + 6y) – (4x + 5y) = 46 – 41;

y = 5

2x + 3 × 5 = 23

2x = 23 – 15;

x = \(\frac{8}{2} \) = 4

Price of pencil = Rs 4

Price of pen = Rs 5

Question 6.

Rama spends Rs 97 to buy 4 two hundred page note books and 5 hundred page note books. Geetha spends Rs 101 to buy 5 two hundred page note book and 4 one hundred page note books. What is the prices of two types of note books?

Let the cost of two hundred page note books = x

The cost of one hundred page note books = y

(1) 4x + 5y = 97

(1) × 5 \(\Rightarrow \) 20x + ……..y = ……..

(2) 5x + 4y = 101

(2) × 4 \(\Rightarrow \) 20x + …….y = …….

……y = ( ….. ) – ( ……. );

y = …….. /……..

4x + 5x( ……. ) = 97

4x = 97 – ( ……. )

x = ……../4

Answer:

(1) 4x + 5y = 97

(2) 5x + 4y = 101

(1) × 5 → 20x + 25y = 485

(2) × 4 → 20x + 16y = 404

9y = 485 – 404

y = \(\frac{81}{9} \) = 9

4x + 5 × 9 = 97

4x = 97 – 45 = 52

Cost of two hundred page note book = Rs 13

Cost of one hundred page note book = Rs 9

Question 7.

6 years back the age of Muneer was 3 times the age of Mujeeb. After 4 years the age of Muneer becomes twice the age of Mujeeb. Find the age of two of them now.

Answer:

Age of Mujeeb 6 years back = x

Age of Muneer 6 years back = 3x

After 4 years

3x + 4 = 2(x + 4)

3x + 4 = 2x + 8

3x – 2x = 8 – 4;

x = 4

Age of Mujeeb 6 years back = 4 + 6 = 10

Age of Muneer 6 years back = 3(4 + 6) = 18 years.

Question 8.

The cost of 4 chairs and 5 tables is Rs 6600 and the cost of 5 chairs and 3 tables is Rs 5000 at the same prices. What are the prices of a table and a chair?

Answer:

Cost of a chair = Rs a

Cost of a table = Rs b

4a + 5b = 6600 ¾ …………(1)

5a + 3b = 5000 ¾ …………(2)

(1) × 5 → 20a + 25b = 33000

(2) × 4 → 20a + 12b = 20000

(20a + 25b) – (20a + 12b) = 33000 – 20000

(1) \(\Rightarrow \) 13b = 13000

b = \(\frac{13000}{13} \) = Rs 1000

4a + 5b = 6600;

4a + 5 × 1000 = 6600

4a = 6600 – 5000 = 1600

a = \(\frac{1600}{4} \) = Rs 400

Cost of a table = Rs 1000

Cost of a chair = Rs 400