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Question

A number is first decreased by one-fifth of its value and then increased by one-fourth of the resulting value. If the final result is 48, what was the original number ?

This question was previously asked in
UPSSSC PET 2026 Question Paper (01-Nov-2026) (Shift-2)
The correct answer is

48

A decrease by one-fifth leaves four-fifths of the number, and an increase by one-fourth multiplies the value by five-fourths. Successive fractional changes are applied one after the other as multipliers.

Step 1 — Let the original number be \(x\). After decreasing it by one-fifth, the value becomes \(x - \frac{x}{5} = \frac{4x}{5}\).

Step 2 — Increasing this by one-fourth of itself gives \(\frac{4x}{5} \times \frac{5}{4} = x\). The two changes cancel exactly, so the final value equals the original number.

Step 3 — The final result is 48, therefore \(x = 48\).

Taking 46, 54 or 50 as the starting number would also return the same number at the end, i.e. 46, 54 or 50, not 48, so none of these fits.

48 option is correct.

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    Improper fraction

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    \(\frac{7}{11}\)  is

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