CSAT Solved Papers/ 2024/Q26

2024 CSAT — Q26

Quant Arithmetic & numeracy 2.5 marks Medium

$P$ ’s salary is $20\%$ lower than $Q$ ‘s salary which is $20\%$ lower than $R$ ‘s salary. By how much percent is $R$ ‘s salary more than $P$ ‘s salary?

A $48{\cdot}75\%$
B $56{\cdot}25\%$ Answer
C $60{\cdot}50\%$
D $62{\cdot}25\%$

Worked rationale

Anchor on the top of the chain. Let $R = 100$ .

Q = R \times (1 - 0.20) = 80,\qquad P = Q \times (1 - 0.20) = 80 \times 0.8 = 64.

” $R$ more than $P$ ” takes $P$ as the base:

\frac{R - P}{P} \times 100 = \frac{100 - 64}{64}\times 100 = \frac{36}{64}\times 100 = 56.25\%.

Answer: (b) $56{\cdot}25\%$ .

Why the other options miss

A

wrong base for the percentage: computes $\frac{R-P}{R}=\frac{36}{100}=36\%$ … or averages the two $20\%$ cuts; takes the percentage on $R$ instead of on $P$ .
C

an arithmetic slip: a divide-by- $64$ slip, or treats $P = 64$ but reports a mis-rounded $\frac{36}{60}$ -type figure.
D

added the cuts instead of compounding: adds the two $20\%$ drops to $40\%$ then over-corrects the reverse direction, double-counting the chain.

Specialist insight

Two disciplines decide this. First, the cuts compound: $P = R\times0.8\times0.8 = 0.64R$ , not $R(1-0.40)$ — two $20\%$ reductions are a single $36\%$ reduction, not $40\%$ . Second, “how much percent is $R$ more than $P$ ” forces the base to be $P$ ( $64$ ), so the answer is $\frac{36}{64}=56.25\%$ , not $\frac{36}{100}=36\%$ . Setting $R=100$ makes both traps visible at once. The asymmetry — a $20\%$ drop needs a $25\%$ rise to reverse — is exactly why (a) and (b) differ.

The trap, in one line

Compound the two cuts ( $P=0.64R$ ) and put the " $R$ more than $P$ " percentage on base $P$ ( $64$ ): $\frac{36}{64}=56.25\%$ , not $\frac{36}{100}=36\%$ .

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