CSAT Solved Papers/ 2021/Q30

2021 CSAT — Q30

Quant Arithmetic & numeracy 2.5 marks Medium

A student appeared in $6$ papers. The maximum marks are the same for each paper. His marks in these papers are in the proportion $5:6:7:8:9:10$ . Overall he scored $60\%$ . In how many number of papers did he score less than $60\%$ of the maximum marks?

A 2
B 3 Answer
C 4
D 5

Worked rationale

Let each paper carry maximum $M$ and let the marks be $5k,6k,7k,8k,9k,10k$ . The total maximum is $6M$ and overall he scored $60\%$ :

5k+6k+\dots+10k = 45k = 0.60\times 6M = 3.6M \;\Rightarrow\; k = 0.08M.

A paper with ratio value $n$ gives a percentage $\dfrac{nk}{M} = n\times 0.08 = 0.08n$ . The threshold " $60\%$ " means $0.08n < 0.60$ , i.e.

n < 7.5.

The ratio values $n=5,6,7$ satisfy this (percentages $40\%,48\%,56\%$ ); $n=8,9,10$ give $64\%,72\%,80\%\ge 60\%$ . So $3$ papers fall below $60\%$ .

Answer: (b) 3.

Why the other options miss

A

one short: treats $n=7$ ( $56\%$ ) as $\ge 60\%$ , dropping it below the cut wrongly.
C

one too many: includes $n=8$ ( $64\%$ ) among the below- $60\%$ papers.
D

wrong base for the comparison: compares each $n$ against the average ratio value rather than converting to a percentage of $M$ .

Specialist insight

The key conversion is ” $60\%$ overall” $\to k = 0.08M$ , which turns each ratio value into a clean percentage $0.08n$ . Then the count is a single threshold test $n < 7.5$ . The lone trap is the boundary value $n=7$ : $0.08\times 7 = 56\% < 60\%$ , so it counts. Convert the ratio to percentages of $M$ once and the borderline paper resolves itself; never eyeball “the small ones.”

The trap, in one line

Each paper is $0.08n$ of max; $0.08n<0.6\Rightarrow n<7.5$ , so $n=5,6,7$ — three papers $\Rightarrow$ (b).

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