# Form the pair of linear equations in the following problems, and find their solutions (if they exist) by the elimination method:

(i) If we add 1 to the numerator and subtract 1 from the denominator, a fraction reduces to 1. It becomes 1/2 if we only add 1 to the denominator. What is the fraction?

(ii) Five years ago, Nuri was thrice as old as Ten years later, Nuri will be twice as old as Sonu. How old are Nuri and Sonu?

(iii) The sum of the digits of a two-digit number is 9. Also, nine times this number is twice the number obtained by reversing the order of the digits. Find the number.

(iv) Meena went to a bank to withdraw ₹ 2000. She asked the cashier to give her ₹ 50 and ₹ 100 notes Meena got ₹ 25 notes in all. Find how many notes of ₹ 50 and ₹ 100 she received.

(v) A lending library has a fixed charge for the first three days and an additional charge for each day Saritha paid ₹ 27 for a book kept for seven days, while Susy paid ₹ 21 for the book she kept for five days. Find the fixed charge and the charge for each extra day.

**Solution:**

(i) If we add 1 to the numerator and subtract 1 from the denominator, a fraction reduces to 1. It becomes 1/2 if we only add 1 to the denominator. What is the fraction?

Fraction has two parts numerator and denominator so assume the numerator as x, and denominator as y, two linear equations can be formed for the known situation.

Let the numerator = x

And the denominator = y

Then the fraction = x/y

When 1 is added to the numerator and 1 is subtracted from the denominator;

(x + 1)/(y - 1) = 1

x + 1 = y - 1

x + 1 = y -1

x - y + 1 + 1 = 0

x - y + 2 = 0 ...(1)

When 1 is added to the denominator;

x/(y + 1) = 1/2

2x = y + 1

2x - y - 1 = 0 ...(2)

By subtracting equation (2) from equation (1)

( x - y + 2) - (2x - y - 1) = 0

x - y + 2 - 2x + y + 1 = 0

- x + 3 = 0

x = 3

Substitute

x = 3 in equation (1)

3 - y + 2 = 0

y = 5

Equations are x - y + 2 = 0 and 2x - y -1 = 0 where the numerator of the fraction is x, and denominator is y.

The fraction is 3/5.

(ii) Five years ago, Nuri was thrice as old as Ten years later, Nuri will be twice as old as Sonu. How old are Nuri and Sonu?

Assuming the present age of Nuri as x years and Sonu as y years, two linear equations can be formed for the Known Solutions.

Let the present age of Nuri = x years

And the present age of Sonu = y years

5 years ago,

Nuri’s age = (x - 5) years

Sonu’s age = (y - 5) years

x - 5 = 3( y - 5)

x - 5 = 3y - 15

x - 3y - 5 + 15 = 0

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

10 years later,

Nuri’s age = (x + 10) years

Sonu’s age = (y + 10) years

x + 10 = 2(y + 10)

x + 10 = 2 y + 20

x - 2 y + 10 - 20 = 0

x - 2 y - 10 = 0 ...(2)

By subtracting equation (2) from equation (1)

( x - 3y +10) - ( x - 2 y -10) = 0

x - 3y +10 - x + 2 y +10 = 0

- y + 20 = 0

y = 20

Figure Substitute y = 20 in equation (1)

x - 3 × 20 + 10 = 0

x - 60 + 10 = 0

x - 50 = 0

x = 50

Linear equations are x - 3y +10 = 0 and x - 2 y -10 = 0 where the present age of Nuri is x and Sonu is y

The age of Sonu is 20 years and age of Nuri is 50 years.

(iii) The sum of the digits of a two-digit number is 9. Also, nine times this number is twice the number obtained by reversing the order of the digits. Find the number.

A two-digit number’s form is 10 y + x where y and x are ten’s and one’s digit respectively.

Let the one’s place = x

And the ten’s place = y

Then the number = 10y + x

Sum of the digits of the number;

x + y = 9 ...(1)

By reversing the order of the digits, the number =10x + y

Hence,

9 (10 y + x) = 2 (10x + y )

90 y + 9x = 20x + 2 y

20x + 2 y - 90 y - 9x = 0

11x - 88 y = 0

11( x - 8 y ) = 0

x - 8 y = 0 ...(2)

By subtracting equation (2) from equation (1)

( x + y ) - ( x - 8y ) = 9 - 0

x + y - x + 8 y = 9

9 y = 9

y = 1

Substitute y = 1 in equation (1)

x +1 = 9

x = 9 - 1

x = 8

Equations are x + y = 9 and 8x - y = 0 where y and x are ten’s and one’s digit respectively.

The two-digit number is 18.

(iv) Meena went to a bank to withdraw ₹. 2000. She asked the cashier to give her ₹. 50 and ₹. 100 notes Meena got ₹. 25 notes in all. Find how many notes of ₹. 50 and ₹. 100 she received.

Assuming the number of notes of ₹ 50 as x and ₹ 100 as y, two linear equations can be formed for the known Solutions.

Let number of notes of ₹ 50 = x

and number of notes of ₹ 100 = y

Meena got 25 notes in all;

x + y = 25 ...(1)

Meena withdrew ₹ 2000;

50x +100 y = 2000

50( x + 2 y) = 2000

x + 2 y = 2000/50

x + 2 y = 40 ...(2)

By subtracting equation (1) from equation (2)

( x + 2 y ) - ( x + y ) = 40 - 25

x + 2 y - x - y = 15

y = 15

y = 15

Substituting, y = 15 in equation (1)

x + 15 = 25

x =10

Equations are x + y = 25 and x + 2y = 40 where number of ₹ 50 and ₹ 100 notes are x and y respectively.

The number of ₹ 50 notes is 10 Number of ₹ 100 notes is 15.

(v) A lending library has a fixed charge for the first three days and an additional charge for each day Saritha paid ₹. 27 for a book kept for seven days, while Susy paid ₹. 21 for the book she kept for five days. Find the fixed charge and the charge for each extra day.

Assuming fixed charges as ₹ x and the additional charges for each extra day as ₹ y, two linear equations can be formed for the known situation.

Let the fixed charge = x

And charge per extra day = y

Saritha paid ₹ 27 for a book kept for 7 days;

x + (7 - 3) y = 27

x + 4 y = 27 ...(1)

Susy paid ₹ 21 for a book kept for 5 days;

x + (5 - 3) y = 21

x + 2 y = 21 ...(2)

By subtracting equation (2) from equation (1)

( x + 4 y ) - ( x + 2 y ) = 27 - 21

x + 4 y - x - 2y = 6

2y = 6

y = 6/2

y = 3

Substituting y = 3 in equation (3)

x + 4 × 3 = 27

x + 12 = 27

x = 27 - 12

x = 15

Equations are x + 2 y = 21 and x + 4y = 27 where fixed charge is ₹ x and charge for each extra day is ₹ y.

The fixed charge is ₹ 15 Charge for each extra day is ₹ 3

**Video Solution:**

### NCERT Solutions for Class 10 Maths - Chapter 3 Exercise 3.4 Question 2:

**Summary:**

1) Equations are x - y + 2 = 0 and 2x - y -1 = 0 where the numerator of the fraction is x, and denominator is y.The fraction is 3/5., 2) Linear equations are . x - 3y +10 = 0 and x - 2 y -10 = 0 where the present age of Nuri is x and Sonu is y.The age of Sonu is 20 years and age of Nuri is 50 years., 3) Equations are x + y = 9 and 8x - y = 0 where y and x are ten’s and one’s digit respectively.The two-digit number is 18., 4) Equations are x + y = 25 and x + 2y = 40 where number of ₹ 50 and ₹ 100 notes are x and y respectively.The number of ₹ 50 notes is 10 Number of ₹ 100 notes is 15., 5) Equations are x + 2 y = 21 and x + 4y = 27 where fixed charge is ₹ x and charge for each extra day is ₹ y.The fixed charge is ₹ 15 Charge for each extra day is ₹ 3.