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Question

If the numerator of a fraction is increased by 30% and its denominator is decreased by 35%, the value of the fraction becomes $\frac{3}{15}$. Find the original fraction.

This question was previously asked in
RRB NTPC 2019 CBT 1 Question Paper (8-Mar-2021) (Shift 2)
The correct answer is
$\frac{1}{10}$

Finding Original Fraction Based on Percentage Changes

Let the original fraction be represented algebraically as $\frac{x}{y}$.

Numerator Modification

The numerator $x$ is increased by 30%. The new numerator is calculated as:

$ x + (30\% \times x) = x + (0.30 \times x) = 1.30x $

Denominator Modification

The denominator $y$ is decreased by 35%. The new denominator is calculated as:

$ y - (35\% \times y) = y - (0.35 \times y) = 0.65y $

New Fraction Value

The resulting fraction after these modifications is $\frac{1.30x}{0.65y}$.

We are given that this new fraction equals $\frac{3}{15}$, which simplifies to $\frac{1}{5}$.

$ \frac{1.30x}{0.65y} = \frac{1}{5} $

Solving for the Original Fraction

First, simplify the numerical coefficients in the equation:

$ \frac{1.30}{0.65} = \frac{130}{65} = 2 $

Substitute this back into the equation:

$ 2 \times \frac{x}{y} = \frac{1}{5} $

To find the original fraction $\frac{x}{y}$, divide both sides by 2:

$ \frac{x}{y} = \frac{1}{5 \times 2} $

$ \frac{x}{y} = \frac{1}{10} $

Thus, the original fraction is $\frac{1}{10}$.

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Similar Questions

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  2. $1.236576576...$
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  3. What smallest fraction should be added to $3\frac{2}{3} + 6\frac{7}{12} + 4\frac{9}{36} + 5 + 7\frac{1}{12}$ to make the sum a whole number?
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  5. The sum of the numerator and denominator of a fraction is 11. If the numerator is decreased by 1, the fraction becomes $\frac{1}{4}$. Find the fraction.
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Important Questions from Fractions

  1. Which fraction among the following is the least ?

    \(\frac{5}{11}, \frac{7}{12}, \frac{8}{13}, \frac{9}{17}\)

  2. Find the value of the following expression:

    \(\frac{{3 \div 1 \times 2 + 5 - 2}}{{3 \times 3 - 2}}\)

  3. Simplify the expression 441 ÷  \(\left[270 \div \frac{3}{7}+\left(17\div \frac{1}{3}\right)-\left(8\frac{1}{2}-\frac{5}{2}\right)\right]\)

  4. If the sum of two positive numbers is 65 and the square root of their product is 26, then the sum of their reciprocals is:

  5. The value of \(9 \div [\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\frac{1}{6}\div(\frac{3}{4}-\frac{1}{3})\;of\;\frac{2}{9}]\)  is:

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