CSAT Solved Papers/ 2022/Q80

2022 CSAT — Q80

Quant Arithmetic & numeracy 2.5 marks Medium

The average weight of $A, B, C$ is $40$ kg, the average weight of $B, D, E$ is $42$ kg and the weight of $F$ is equal to that of $B$ . What is the average weight of $A, B, C, D, E$ and $F$ ?

A 40.5 kg
B 40.8 kg
C 41 kg Answer
D Cannot be determined as data is inadequate

Worked rationale

Convert averages to sums:

A + B + C = 3 \times 40 = 120,\qquad B + D + E = 3 \times 42 = 126,\qquad F = B.

The total of all six is

A + B + C + D + E + F = (A+B+C) + (D+E) + F = 120 + (126 - B) + B = 120 + 126 = 246.

The $-B$ from $D+E = 126 - B$ and the $+B$ from $F = B$ cancel, so the total is fixed at $246$ regardless of $B$ . The average over six people:

\frac{246}{6} = 41 \text{ kg}.

Answer: (c) 41 kg.

Why the other options miss

A

an arithmetic slip: averages the two given averages ( $\tfrac{40+42}{2}$ -style) instead of summing correctly.
B

an arithmetic slip: divides a wrong total (e.g. $244$ or $245$ ) by $6$ .
D

gave up too early: sees the unknown $B$ and stops, missing that $F = B$ exactly cancels the $B$ hidden in $D + E$ .

Specialist insight

The elegance is the cancellation: $B$ appears in both group sums ( $A{+}B{+}C$ and $B{+}D{+}E$ ), so naive addition double-counts it — but $F = B$ supplies exactly the missing copy, and the total resolves to $120 + 126 = 246$ with no $B$ left. The “cannot be determined” trap is for anyone who treats $B$ as a free unknown without spotting that $F = B$ closes it.

The trap, in one line

$F = B$ cancels the $B$ in $D+E = 126 - B$ : total $= 120 + 126 = 246$ , average $41$ $\Rightarrow$ (c).

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