CSAT Solved Papers/ 2024/Q20

2024 CSAT — Q20

Quant Statement validity 2.5 marks Medium

Consider the following statements in respect of the sum $S = x + y + z$ , where $x$ , $y$ and $z$ are distinct prime numbers each less than $10$ .

The unit digit of $S$ can be $0$ .
The unit digit of $S$ can be $9$ .
The unit digit of $S$ can be $5$ .

Which of the statements given above are correct?

A 1 and 2 only
B 2 and 3 only
C 1 and 3 only Answer
D 1, 2 and 3

Worked rationale

The primes below $10$ are $\{2, 3, 5, 7\}$ — exactly four of them. Choosing $3$ distinct primes gives only $\binom{4}{3} = 4$ possible sums; enumerate them all:

2+3+5 = 10,\quad 2+3+7 = 12,\quad 2+5+7 = 14,\quad 3+5+7 = 15.

The four possible values of $S$ are $\{10, 12, 14, 15\}$ , with unit digits $\{0, 2, 4, 5\}$ .

Statement 1 (unit digit $0$ ): yes — $S=10$ . True.
Statement 2 (unit digit $9$ ): no value in $\{10,12,14,15\}$ ends in $9$ . False.
Statement 3 (unit digit $5$ ): yes — $S=15$ . True.

So statements $1$ and $3$ are correct.

Answer: (c) 1 and 3 only.

Why the other options miss

A

assumes reachability instead of checking it: accepts statement 2 by loosely assuming “any unit digit is reachable,” never checking that $9 \notin \{0,2,4,5\}$ .
B

wrongly keeps the impossible statement 2 and drops the valid statement 1 (unit digit $0$ from $S=10$ ).
D

marks all three true without the four-line enumeration that rules out a unit digit of $9$ .

Specialist insight

“Distinct primes each less than $10$ ” is a tiny finite set — only $\{2,3,5,7\}$ , so only $4$ sums exist. The fastest path is to list all four sums rather than reason abstractly about achievable unit digits; the full list $\{10,12,14,15\}$ instantly confirms $0$ and $5$ and kills $9$ . Whenever the universe is this small, exhaustive enumeration is both faster and safer than parity/modular arguments — there is no case left to miss.

The trap, in one line

Only four sums exist ( $10,12,14,15$ ); unit digit $9$ is unreachable, so statements $1$ and $3$ hold, not all three.

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