CSAT Solved Papers/ 2025/Q78

2025 CSAT — Q78

Quant Arithmetic & numeracy 2.5 marks Easy

The average of three numbers $p, q$ and $r$ is $k$ . $p$ is as much more than the average as $q$ is less than the average. What is the value of $r$ ?

A k Answer
B k - 1
C k + 1
D k/2

Worked rationale

The average condition gives $\dfrac{p + q + r}{3} = k$ , so

p + q + r = 3k.

” $p$ is as much more than the average as $q$ is less than the average” means the deviations cancel:

p - k = k - q \;\Longrightarrow\; p + q = 2k.

Substitute:

2k + r = 3k \;\Longrightarrow\; r = k.

Answer: (a) k.

Why the other options miss

B

the wrong formula: invents an offset, mis-reading “as much more … as less” as a unit difference rather than equal-and-opposite deviations.
C

the same wrong formula, opposite sign: the identical misread with the sign flipped.
D

solved the wrong question: divides by an extra factor, confusing $r$ with a per-term average.

Specialist insight

The phrase “as much more … as … less than the average” is a symmetric-deviation statement: $p$ and $q$ sit equidistant on either side of $k$ , so they average to $k$ and contribute $2k$ to the sum. Since all three sum to $3k$ , the third number $r$ must itself equal $k$ to keep the mean at $k$ . The fast mental model: if two of three numbers already average to the mean, the third is the mean. No heavy algebra needed — recognising the cancellation is the whole point.

The trap, in one line

$p$ and $q$ are equidistant from $k$ , so $p+q=2k$ ; with $p+q+r=3k$ , that forces $r = k$ .

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