CSAT Solved Papers/ 2024/Q27

2024 CSAT — Q27

Quant Arithmetic & numeracy 2.5 marks Medium

A number is mistakenly divided by $4$ instead of multiplying by $4$ . What is the percentage change in the result due to this mistake?

A $25\%$
B $50\%$
C $72{\cdot}75\%$
D $93{\cdot}75\%$ Answer

Worked rationale

Take the number to be $x$ . The intended result is $4x$ ; the mistaken result is $\dfrac{x}{4}$ . Percentage change is measured against the intended (correct) result as base:

\frac{\text{mistaken} - \text{intended}}{\text{intended}}\times 100 = \frac{\tfrac{x}{4} - 4x}{4x}\times 100 = \frac{\tfrac14 - 4}{4}\times 100 = \frac{-3.75}{4}\times 100 = -93.75\%.

A drop of $93.75\%$ .

Answer: (d) $93{\cdot}75\%$ .

Why the other options miss

A

solved a different question: reads “divided by $4$ ” as a flat $25\%$ of the number, ignoring that the intended result was $4x$ , not $x$ .
B

the wrong base: compares $\tfrac{x}{4}$ to $x$ (a $75\%$ drop mis-stated), using $x$ as base instead of the intended $4x$ .
C

an arithmetic slip: sets up $\frac{4x-\,x/4}{4x}$ correctly in spirit but fumbles the $\tfrac{15}{16}$ , landing near but not on $93.75\%$ .

Specialist insight

The base for “percentage change in the result” is the correct result $4x$ , not the original number $x$ . The ratio of results is $\dfrac{x/4}{4x} = \dfrac{1}{16}$ — the mistaken value is just $\tfrac1{16}$ of what it should be, a fall of $\tfrac{15}{16} = 93.75\%$ . Reading ” $\times 4$ vs $\div 4$ ” as a factor of $16$ between the two results is the one-line route; the deadly slip is benchmarking against $x$ instead of against the intended $4x$ .

The trap, in one line

The two results differ by a factor of $16$ ( $\div4$ vs $\times4$ ), so the mistaken value is $\tfrac1{16}$ of intended — a $\tfrac{15}{16}=93.75\%$ drop, measured against $4x$ , not $x$ .

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