PREVIOUS

NEXT

## Class ten NCERT Solutions Maths Chapter 3

#### TEXTBOOK’S EXERCISE 3.4

2. 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 Sonu. 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. [2010]

(iv) Meena went to a bank to withdraw ` 2000. She asked the cashier to give her ` 50 and ` 100 notes only. 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 thereafter. 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.

Sol. (i) Let the fraction = x/y

x + 1 = y – 1

x y + 2 = 0———-(i)

2x = y + 1

⇒ 2x y – 1 = 0 …(2)

Subtracting equation (1) from equation (2), we have

Substituting the value of x in equation (2), we have

2 × 3 – y – 1 = 0

⇒ 6 – y – 1 = 0

⇒ 5 – y = 0

y = 5

Hence, the required fraction is 3/5

(ii) Let present age of Nuri = x years, and present age of Sonu = y years

5 years ago, age of Nuri = (x – 5) years

5 years ago, age of Sonu = (y – 5) years

10 years later, age of Nuri = (x + 10) years

10 years later, age of Sonu = (y + 10) years

As per condition :

x – 5 = 3y– 15

x –3y + 10 = 0 …(1)

And, x + 10 = 2 (y + 10)

x + 10 = 2 y + 20

x – 2y = 10

x – 2y – 10 = 0 …(2) Subtracting equation (2) from equation (1), we have,

Substituting the value of y in equation (1), we have

x – 3 × 20 + 10 = 0

x – 60 + 10 = 0

x – 50 = 0

x = 50

Hence, present age of Nuri = 50 years, and present age of Sonu = 20 years.

(iii) Let digit in units place = x and digit in tens place = y

∴ Two digit number = 10y + x

Reversing the order of the digits, number = 10x + y

As per condition :

x + y = 9 …(1)

And, 9 (10y + x) = 20x + 2y

⇒ 90y + 9x = 20x – 2y

⇒ 88y – 11x = 0

⇒ 8y x = 0 …(2)

Adding equation (1) and equation (2), we have

Substituting the value of y in equation (1), we have

x + 1 = 9 ⇒ x = 8

Thus, the required two digit number is 10y + x = 10 × 1 + 8 = 18.

(iv) Let number of ` 50 notes = x and number of ` 100 notes = y

As per condition :

x + y = 25 …(1)

And 50x + 100y = 2000

x + 2y = 40 …(2)

Subtracting equation (1) from equation (2), we have

Substituting the value of y in equation (1), we have, x + 15 = 25 ⇒ x = 10

Hence, number of Rs 50 notes = 10, and number of Rs 100 notes = 15.

(v) Let fixed charge for first 3 days = Rs x

And additional charge for each day = Rs y

As per condition :

x + 4y = 27 …(1)

x +2y = 21 …(2)

Subtracting equation (2) from equation (1), we have

y= = 6/2= 3

Substituting the value of y in equation (1), we have

⇒ x + 4 × 3 = 27

⇒ x + 12 = 27

x = 15

Hence, charge for first 3 days = Rs 15, and extra charge per day = Rs 3.

PREVIOUS

NEXT