Question

The numerator of a certain fraction is 6 less than its denominator. if the numerator is increased by 3 and the denominator is decreased by 1, the fraction becomes 7/8. find the original fraction. ​

Answer 1

Let the numerator be N and denominator be D.

The numerator of a certain fraction is 6 less than its denominator.

According to question,

⇒ N = D - 6

If the numerator is increased by 3 and the denominator is decreased by 1, the fraction becomes 7/8.

Numerator = N + 3

Denominator = D - 1

Fraction = 7/8

According to question,

⇒ (N + 3)/(D - 1) = 7/8

⇒ 8(N + 3) = 7(D - 1)

⇒ 8N + 24 = 7D - 7

⇒ 8N - 7D = - 7 - 24

Substitute value of N above

⇒ 8(D - 6) - 7D = - 31

⇒ 8D - 48 - 7D = -31

⇒ D = -31 + 48

⇒ D = 17

Substitute value of D in N

⇒ N = 17 - 6

⇒ N = 11

Therefore,

Original fraction = N/D = 11/17

Answer 2

Given :-

• The numerator of a certain fraction is 6 less than its denominator.
• if the numerator is increased by 3 and the denominator is decreased by 1, the fraction becomes 7/8.

Solution :-

Let the denominator be of the fraction is = x

→ Then, numerator = (x - 6)

A/q,

→ Numerator is increased by 3 = (x -6) +3 = (x - 3)

→ Denominator is Decreased by 1 = (x - 1)

So,

→ (x - 3)/(x -1) = 7/8

→ 8(x - 3) = 7(x - 1)

→ 8x - 24 = 7x - 7

→ 8x - 7x = 24 - 7

→ x = 17

So, Denominator = x = 17

numerator = (x - 6) = 17 - 6 = 11

Hence, The Required fraction is = (11/17) ..