Question

the numerator of fraction is 1 less than its denominator .If one is added to both numerator and denominator . The fraction becomes 7\8 . find the fraction ​

Answer 1

Given:

The numerator of a fraction is 1 less than the denominator

So,

Let the denominator be x

And the numerator will be (x-1) as given in the question

Now,

According to  the question,

If one is added to both numerator and denominator, the fraction becomes $\frac{7}{8}$.

So the equation will be

$\boxed{\frac{(x-1) + \red{1}}{x+\red{1}} =\pink{\frac{7}{8}}}$

$\implies \frac{x-\not{1} +\not{1}}{x+1} =\frac{7}{8}$

$\implies \frac{x}{x+1} =\frac{7}{8}$

[Now doing cross-multiplication, we get,]

⇒ 8x = 7(x+1)

⇒ 8x = 7x + 7

⇒ 8x - 7x = 7

[By transporting 7x to LHS]

⇒ x = 7

Thus,

The value of the denominator is x = 7

And the numerator is = (x-1) = 7 - 1 = 6

Hence,

The required fraction is

$\boxed{\red{\frac{6}{7}}}$

- - -

Verification,

When 1 is added to both the numerator and denominator

⇒  $\frac{6+\green{1}}{7+\green{1}}= \frac{7}{8}$

Verified.

Answer 2

Answer:

6/7

Step-by-step explanation:

Assume that the denominator is x.

The numerator of fraction is 1 less than its denominator.

Therefore,

The numerator is x - 1. ...........(1)

Also given that, if one is added to both numerator and denominator then the fraction becomes 7/8.

→ (x + 1 - 1)/(x + 1) = 7/8

→ (x )/(x + 1) = 7/8

→ 8x = 7(x + 1)

→ 8x = 7x + 7

→ 8x - 7x = 7

→ x = 7

Hence, the numerator = x - 1 = 7 - 1 = 6 and denominator = x = 7

Therefore, the fraction is 6/7.