Math, asked by atiqullahshiekh, 2 months ago

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

Answers

Answered by ImperialGladiator
10

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}

Similar questions