CSAT Solved Papers/ 2024/Q36

2024 CSAT — Q36

Quant Arithmetic & numeracy 2.5 marks Medium

Consider the following:

Weight of $6$ boys $=$ Weight of $7$ girls $=$ Weight of $3$ men $=$ Weight of $4$ women

If the average weight of the women is $63$ kg, then what is the average weight of the boys?

A 40 kg
B 42 kg Answer
C 45 kg
D 63 kg

Worked rationale

The equalities are about total weights of each group. Let the common total be $W$ :

\text{(total of 6 boys)} = \text{(total of 7 girls)} = \text{(total of 3 men)} = \text{(total of 4 women)} = W.

From the women: $4$ women weigh $W$ and average $63$ kg, so

W = 4 \times 63 = 252 \text{ kg}.

From the boys: $6$ boys weigh the same total $W = 252$ kg, so the average boy weighs

\frac{252}{6} = 42 \text{ kg}.

Answer: (b) 42 kg.

Why the other options miss

A

an arithmetic slip: divides a slightly wrong total (e.g. $W=240$ ) by $6$ , or mis-multiplies $4\times63$ .
C

the ratio the wrong way round: sets up a per-person ratio $7:6$ or $4:6$ inverted, scaling $63$ by a $\tfrac{6}{7}$ -type factor incorrectly.
D

solved a different question: copies the women’s average, missing that boys and women have the same total but different counts ( $6$ vs $4$ ).

Specialist insight

The trap is “average” versus “total”. The chain equates group totals, not per-person weights, so the clean route is: women’s total $= 4\times63 = 252$ , and that same $252$ is shared by $6$ boys $\Rightarrow$ $42$ . You never need the girls or the men — only the two groups linked by the question (women $\to$ boys). Anchoring everything on the common total $W$ and using the counts $4$ and $6$ keeps the unitary step clean and avoids the ratio-inversion that produces (c).

The trap, in one line

The equality is of group totals, not averages: women's total $4\times63=252$ is shared by $6$ boys $\Rightarrow 252/6=42$ kg.

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