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

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

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

(iii) The coach of a cricket team buys 7 bats and 6 balls for ₹ 3800. Later, she buys 3 bats and 5 balls for ₹ 1750. Find the cost of each bat and each ball.

(iv) 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 ₹ 105 and for a journey of 15 km, the charge paid is ₹ 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?

(v) A fraction becomes 9/11, if 2 is added to both the numerator and the denominator. If, 3 is added to both the numerator and the denominator it becomes 5/6. Find the fraction.

(vi) 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?

**Solution:**

(i)

Let the first (larger) number = x

And the second number = y

The difference between two numbers is 26.

x - y = 26 ....(1)

One number is three times the other

x = 3y ....(2)

Substituting x = 3y in equation (1), we get

3y - y = 26

2y = 26

y = 13

Substituting y = 13 in equation (2)

x = 3 × 13

x = 39

The two numbers are 39 and 13.

(ii)

Let the larger angle = x

and smaller angle = y

Since the angles are supplementary

x + y = 180 ....(1)

Larger angle exceeds the smaller by 18

x = y + 18 ....(2)

Substituting x = y + 18 in equation (1), we get y +18 + y = 180

2y = 180 - 18

y = 162/2

y = 81

Substituting y = 81 in equation (2), we get

x = 81 + 18

x = 99

The angles are 99 and 81.

(iii)

Let the cost of 1 bat = ₹ x And the cost of 1 ball = ₹ y Then,

7x + 6 y = 3800 ....(1)

3x + 5y = 1750 ....(2)

By solving the equation (1)

7x + 6 y = 3800

6 y = 3800 - 7x

y = (3800 - 7x)/6 ....(3)

Substituting y = (3800 - 7x)/6 in equation (2), we get

3x + 5 [(3800 - 7x)/6] = 1750

(18x + 19000 - 35x)/6 = 1750

-17x +19000 = 1750 × 6

17x = 19000 - 10500

x = 8500/17

x = 500

Substituting x = 500 in equation (3), we get

y = (3800 - 7 × 500)/6

y = 300/6

y = 50

Cost of 1 bat is ₹ 500

Cost of 1 ball is ₹ 50

(iv)

Let the fixed charge = ₹ x

And charge per km = ₹ y

Charge for a distance of 10 km

x + 10 y = 105 ....(1)

Charge for a distance of 15 km

x + 15y = 155 ....(2)

By solving the equation (1)

x +10 y = 105

x = 105 - 10 y ....(3)

Substituting x = 105 - 10y in equation (2), we get

105 - 10 y + 15 y = 155

5 y = 155 - 105

y = 50/5

y = 10

Substituting x = 5 in equation (3)

x = 105 -10 × 10

x = 105 - 100

x = 5

Now, charge for a distance of 25 km = x + 25y

= 5 + 25 × 10

= 5 + 250

= 255

Fixed charge = ₹ 5

Charge per km = ₹ 10

Charge for 25 km = ₹ 255

(v)

Let the numerator = x

And denominator = y

Then fraction = x/y

When 2 is added to both numerator and denominator

(x + 2)/(y + 2) = 9/11

11(x + 2) = 9( y + 2)

11x + 22 = 9y + 18

11x - 9y + 22 - 18 = 0

11x - 9y + 4 = 0 ...(1)

When 3 is added to both numerator and denominator

(x + 3)/(y + 3) = 5/6

6(x + 3) = 5( y + 3)

6x + 18 = 5y + 15

6x - 5y + 18 - 15 = 0

6x - 5y + 3 = 0 ....(2)

5 y = 6x + 3

y = (6x + 3)/5 ....(3)

Substituting y = (6x + 3)/5 in equation (1)

11x - 9 [(6x + 3)/5] + 4 = 0

[55x - 9(6x + 3) + 20]/5 = 0

55x - 54x - 27 + 20 = 0

x - 7 = 0

x = 7

Substituting x = 7 in equation (1) we get,

y = (6 × 7 + 3)/5

y = (42 + 3)/5

y = 45/5

y = 9

The fraction is 7/9

(vi)

Let the present age of Jacob = x years and his son = y years

5 years from now,

Jacob’s age = ( x + 5) years

Son’s age = ( y + 5) years

( x + 5) = 3( y + 5)

x + 5 = 3y + 15

x - 3y + 5 - 15 = 0

x - 3y - 10 = 0 ...(1)

5 years ago, Jacob’s age = ( x - 5) years

Son’s age = ( y - 5) years

( x - 5) = 7( y - 5)

x - 5 = 7 y - 35

x - 7 y - 5 + 35 = 0

x - 7 y + 30 = 0 ...(2)

7 y = x + 30

y = (x + 30)/7 ....(3)

Substituting y = (x + 30)/7 in equation (1)

x - 3[(x + 30)/7] - 10 = 0

[7x - 3( x + 30) - 70]/7 = 0

7x - 3x - 90 - 70 = 0

4x - 160 = 0

x = 160/4

x = 40

Substituting x = 40 in equation (3)

y = (40 + 30)/7

y = 70/7

y = 10

Present age of Jacob is 40 years and his son is 10 years.

**Video Solution:**

### NCERT Solutions for Class 10 Maths - Chapter 3 Exercise 3.3 Question 3:

**Summary:**

1) The two numbers are 39 and 13, 2) The angles are 99 and 81, 3)Cost of 1 bat is ₹ 500, Cost of 1 ball is ₹ 50, 4) Fixed charge = ₹ 5, Charge per km = ₹ 10, Charge for 25 km = ₹ 255, 5) The fraction is 7/9, 6) Present age of Jacob is 40 years and his son is 10 years.