Question

if the numerator of a fraction increased by 10%and its denominator is decreased by 20%then the value of the fraction becomes 55/48 which is the original fraction

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Answer 1

Answer:

The fraction is $\sf \dfrac{5}{6}$

Explanation :

Let's say the original fraction is : $\sf \dfrac{N}{D}$

Now,

‘N’ increases by 10%

So,

→ N + (10% of N)

→ $\sf\dfrac{11N}{10}$

And also,

‘D’ decreases by 20%

So,

→ D - (20% of D)

→ $\sf \dfrac{4D}{5}$

According to the question :

The fraction now becomes $\sf\dfrac{55}{48}$

$\sf \implies \: \dfrac{ \: \: \frac{11N}{10} \: \: }{ \frac{4D}{5} } = \dfrac{55}{48}$

$\sf \implies \: \dfrac{ \: \: \frac{11N}{10} \: \: }{ \frac{4D}{5} } = \dfrac{55}{48}$

$\sf \implies \dfrac{11N}{10} \div \dfrac{4D}{5} = \dfrac{55}{48}$

$\sf \implies \dfrac{11N}{{10}} \times \dfrac{5}{4D} = \dfrac{55}{48}$

$\sf \implies \dfrac{55N}{40D} = \dfrac{55}{48}$

$\sf \implies \dfrac{N}{D} = \dfrac{55 \times 40}{55 \times 48}$

$\sf \implies \dfrac{N}{D} = \dfrac{5}{6}$

$\sf \implies \dfrac{N}{D} = \dfrac{40}{48}$

$\therefore \sf The \: original \: fraction \: is\: \dfrac{5}{6}$