NCERT solution Class 10 pair of linear equations in two variables

NCERT Solutions

Class 10 Maths

Pair of Linear Equations in Two Variables

NCERT Questions

Ex. 3.1 Q.1
Ex.3.1 Q.2
Ex.3.1 Q.3
Ex.3.1 Q.4
Ex.3.1 Q.5
Ex.3.1 Q.6
Ex.3.1 Q.7
Ex.3.2 Q.1
Ex.3.2 Q.2
Ex.3.2 Q.3
Ex.3.3 Q.1
Ex.3.3 Q.2

Ex.3.2 Q.3

Form the pair of linear equations for the following problems and find their solution by substitution method.

(1) The difference between two numbers is 26 and one number are three times the other. Find them.

(2) The larger of two supplementary angles exceeds the smaller by 18 degrees. Find them.

(3) The coach of a cricket team buys 7 bats and 6 balls for Rs 3800.

Later, she buys 3 bats and 5 balls for Rs 1750. Find the cost of each bat and each ball.

(4) The taxi charges in a city consist of a fixed charge together with the charge for the distance covered.

For a distance of 10 km, the charge paid is Rs 105 and for a journey of 15 km, the charge paid is Rs 155.

What are the fixed charges and the charge per km? How much does a person have to pay for travelling a distance of 25 km?

(5) A fraction becomes , if 2 is added to both the numerator and the denominator.

If, 3 is added to both the numerator and the denominator it becomes . Find the fraction.

(6) Five years hence, the age of Jacob will be three times that of his son.

Five years ago, Jacob’s age was seven times that of his son. What are their present ages?

(1) Let the larger number be = x

Let the smaller number be = y

The difference between the two numbers is 26

x – y = 26

x = 26 + y

Given that one number is three times the other

So, x = 3y

Putting the value of x, we get

26 + y = 3y

–2y = – 26

y = 13

Putting value of y, we get

x = 3 × 13 = 39

Therefore, the numbers are 13 and 39.

(2) Let the first angle be= x

and the second angle be = y

As both angles are supplementary so that sum will 180

x + y = 180

x = 180 – y ... (1)

Given; difference is 18 degrees

Therefore, x – y = 18

Putting the value of x, we get

180 – y – y = 18

– 2y = – 162

y =

y = 81

Putting the value back in equation (1), we get

x = 180 – 81 = 99

Hence, the angles are 99° and 81°.

(3) Let the cost of each bat be = Rs x

Let the cost of each ball be = Rs y

Given: coach of a cricket team buys 7 bats and 6 balls for Rs 3800.

7x + 6y = 3800

6y = 3800 – 7x

Dividing by 6, we get

y = (3800 – 7x) ÷ 6 ………... (1)

Given that she buys 3 bats and 5 balls for Rs 1750 later.

3x + 5y = 1750

Putting the value of y

3x + 5 ((3800 – 7x) ÷ 6) = 1750

Multiplying by 6, we get

18x + 19000 – 35x = 10500

–17x =10500 – 19000

–17x = – 8500

x =

x = 500

Putting this value in equation (1) we get

y = (3800 – 7 × 500) ÷6

y = 300 ÷ 6

y = 50

Hence the cost of each bat = Rs 500 and the cost of each ball = Rs 50.

(4) Let the fixed charge for taxi = Rs x

and variable cost per km = Rs y

Total cost = fixed charge + variable charge

Given that for a distance of 10 km, the charge paid is Rs 105

x + 10y = 105 … (1)

x = 105 – 10y

Given that for a journey of 15 km, the charge paid is Rs 155

x + 15y = 155

Putting the value of x, we get

105 – 10y + 15y = 155

5y = 155 – 105

5y = 50

Dividing by 5, we get

y = 50 ÷ 5 = 10

Putting this value in equation (1) we get

x = 105 – 10 × 10

x = 5

Cost for traveling a distance of 25 km = x + 25y

= 5 + 25 × 10

= 5 + 250

=255

A person has to pay Rs 255 for 25 Km.

(5) Let the Numerator be = x

Let the Denominator be = y

Fraction =

A fraction becomes , if 2 is added to both the numerator and the denominator

(x + 2) ÷ y + 2 =

On Cross multiplying,

11x + 22 = 9y + 18

Subtracting 22 from both sides,

11x = 9y – 4

Dividing by 11, we get

x = (9y – 4) ÷ 11 … (1)

Given: if 3 is added to both the numerator and the denominator it becomes .

(x + 3) ÷ (y + 3) = … (2)

On Cross multiplying,

6x + 18 = 5y + 15

Subtracting the value of x, we get

6(9y – 4) ÷ 11 + 18 = 5y + 15

Subtracting 18 from both the sides

6(9y – 4) ÷ 11 = 5y – 3

54 – 24 = 55y – 33

–y = – 9

y = 9

Putting this value of y in equation (1), we get

x = (9y – 4) ÷ 11 … (1)

x = (81 – 4) ÷ 77

x = 77 ÷ 11

x = 7

Hence our fraction is .

(6) Let the present age of Jacob be = x year

and the present Age of his son be = y year

Five years from now,

Jacob’s age will be = x + 5 year

Age of his son will be = y + 5year

Given; the age of Jacob will be three times that of his son

x + 5 = 3(y + 5)

Adding 5 to both sides,

x = 3y + 15 – 5

x = 3y + 10 … (1)

Five years ago,

Jacob’s age = x – 5 years

His son’s age = y – 5 years

Jacob’s age was seven times that of his son

x – 5 = 7(y – 5)

Putting the value of x from equation (1) we get,

3y + 10 – 5 = 7y – 35

3y + 5 = 7y – 35

3y – 7y = – 35 – 5

–4y = – 40

y =

y = 10 year

Putting the value of y in equation we get,

x = 3 × 10 + 10

x = 40 years

Hence, Present age of Jacob = 40 years and present age of his son = 10 years.

Video solution:

Pair of Linear Equations in Two Variables Class 10 Maths NCERT Chapter 3 NCERT Solutions

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