If 1 is added to both of the numerator and denominator of a fraction,

1.	If 1 is added to both of the numerator and denominator of a fraction, it becomes 4/5 .If however, 5 is subtracted from both numerator and denominator, the fraction becomes 1/2 . Find the fraction.
Answer» Let the required fraction be x/y. Then, we have: x+1/y+1 = 4/5 ⇒ 5(x + 1) = 4(y + 1) ⇒ 5x + 5 = 4y + 4 ⇒ 5x – 4y = -1 ………(i) Again, we have: x−5/y−5 = 1/2 ⇒ 2(x – 5) = 1(y – 5) ⇒ 2x – 10 = y – 5 ⇒ 2x –y = 5 ………(ii) On multiplying (ii) by 4, we get: 8x – 4y = 20 ……..(iii) On subtracting (i) from (iii), we get: 3x = (20 – (-1) )=20 + 1 = 21 ⇒ 3x=21 ⇒ x=7 On substituting x = 7 in (i), we get 5 × 7 – 4y = -1 ⇒ 35 – 4y = -1 ⇒ 4y = 36 ⇒ y = 9 ∴ x = 7 and y = 9 Hence, the required fraction is 7/9.

If 1 is added to both of the numerator and denominator of a fraction, it becomes 4/5 .If however, 5 is subtracted from both numerator and denominator, the fraction becomes 1/2 . Find the fraction.

Answer»

Let the required fraction be x/y.

Then, we have:

x+1/y+1 = 4/5

⇒ 5(x + 1) = 4(y + 1)

⇒ 5x + 5 = 4y + 4

⇒ 5x – 4y = -1 ………(i)

Again, we have:

x−5/y−5 = 1/2

⇒ 2(x – 5) = 1(y – 5)

⇒ 2x – 10 = y – 5

⇒ 2x –y = 5 ………(ii)

On multiplying (ii) by 4, we get:

8x – 4y = 20 ……..(iii)

On subtracting (i) from (iii), we get:

3x = (20 – (-1) )=20 + 1 = 21

⇒ 3x=21

⇒ x=7

On substituting x = 7 in (i), we get

5 × 7 – 4y = -1

⇒ 35 – 4y = -1

⇒ 4y = 36

⇒ y = 9

∴ x = 7 and y = 9

Hence, the required fraction is 7/9.