CSAT Solved Papers/ 2024/Q47

2024 CSAT — Q47

Quant Arithmetic & numeracy 2.5 marks Medium

A person buys three articles $P$ , $Q$ and $R$ for $3{,}330$ rupees. If $P$ costs $25\%$ more than $R$ and $R$ costs $20\%$ more than $Q$ , then what is the cost of $P$ ?

A 1,000 rupees
B 1,200 rupees
C 1,250 rupees
D 1,350 rupees Answer

Worked rationale

Express everything in terms of the cheapest anchor $Q = q$ .

R = 1.20\,q,\qquad P = 1.25 \times R = 1.25 \times 1.20\,q = 1.50\,q.

Sum the three:

P + Q + R = 1.50q + 1.00q + 1.20q = 3.70q = 3330 \;\Rightarrow\; q = 900.

Therefore $P = 1.50 \times 900 = 1350$ .

Answer: (d) 1,350 rupees.

Why the other options miss

A

answered the wrong article: splits $3330$ evenly-ish or solves for $Q$ / $R$ , then mislabels it as $P$ .
B

wrong base for the chain: chains the percentages off the wrong base (takes $25\%$ and $20\%$ on $Q$ rather than $P$ on $R$ on $Q$ ), so $P$ comes out as $1.2q$ instead of $1.5q$ .
C

skipped a link in the chain: applies ” $25\%$ more than $R$ ” by adding $25\%$ of the total or treats $P = 1.25q$ directly, skipping the $R$ -on- $Q$ step.

Specialist insight

The chain ” $P$ is $25\%$ more than $R$ , $R$ is $20\%$ more than $Q$ ” must be composed in order: $P = 1.25 \times 1.20 \times Q = 1.5Q$ — the multipliers stack, they do not add to a flat $45\%$ . Anchoring on the smallest quantity $Q$ keeps all coefficients $\ge 1$ and gives the clean total $3.7q = 3330$ . The $1.25\times1.20 = 1.5$ collapse is the line that separates the correct $1350$ from the “off-base” traps; candidates who anchor on $P$ or add the percentages land on (b) or (c).

The trap, in one line

Compose the multipliers in order: $P = 1.25\times1.20\,Q = 1.5Q$ , then $3.7Q = 3330 \Rightarrow Q=900$ , $P=1350$ — don't add $25\%+20\%$ .

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