CSAT Solved Papers/ 2024/Q67

2024 CSAT — Q67

Quant Data sufficiency 2.5 marks Hard

A Question is given followed by two Statements I and II. Consider the Question and the Statements.

There are three distinct prime numbers whose sum is a prime number.

Question: What are those three numbers?

Statement-I: Their sum is less than $23$ .

Statement-II: One of the numbers is $5$ .

Which one of the following is correct in respect of the above Question and the Statements?

A The Question can be answered by using one of the Statements alone, but cannot be answered using the other Statement alone Answer
B The Question can be answered by using either Statement alone
C The Question can be answered by using both the Statements together, but cannot be answered using either Statement alone
D The Question cannot be answered even by using both the Statements together

Worked rationale

First a parity lock: if $2$ were one of the three primes, the sum $= 2 + (\text{odd}) + (\text{odd})$ is even and $\ge 10$ , hence not prime. So all three primes must be odd, and the sum is odd (a necessary condition for it to be prime).

Statement-I alone (sum $< 23$ ): enumerate triples of distinct odd primes with prime sum below $23$ : $3+5+7=15$ (no), $3+5+11=19$ (prime ✓), $3+5+13=21$ (no), $3+7+11=21$ (no); $5+7+11=23$ and $3+7+13=23$ are not $<23$ . Only $\{3,5,11\}$ qualifies — unique. Sufficient.

Statement-II alone (one number is $5$ ): $5 + p + q$ prime with $p,q$ distinct odd primes: $\{3,5,11\}=19$ ✓ and $\{5,7,11\}=23$ ✓ both work — at least two answers. Not sufficient.

So one statement alone (I) answers it, the other (II) does not.

Answer: (a) The Question can be answered by using one of the Statements alone, but cannot be answered using the other Statement alone.

Why the other options miss

B

missed a second valid set: thinks “one number is $5$ ” pins the triple, missing $\{5,7,11\}$ as a second valid set.
C

thought both were needed when one alone suffices: assumes neither statement is individually enough and only their combination ( $\{3,5,11\}$ ) works, overlooking that Statement I alone is already unique.
D

skipped the bounded search: fails the bounded enumeration under I, so it never sees the unique $\{3,5,11\}$ .

Specialist insight

The parity observation — including $2$ forces an even, non-prime sum — restricts everything to odd primes and is the gateway move. After that, Statement I is a small bounded enumeration (sum $< 23$ ) that yields exactly one prime-sum triple, $\{3,5,11\}$ , so it is self-sufficient. Statement II is weaker because “contains $5$ ” leaves room for $\{5,7,11\}$ . The DS lesson: a numeric bound (sum $<23$ ) can be far more constraining than a membership clue (one element $=5$ ) — always test each statement for uniqueness separately before defaulting to (c).

The trap, in one line

All three primes must be odd (else even sum); under sum $<23$ only $\{3,5,11\}=19$ survives (I alone is unique), but "one is $5$ " also allows $\{5,7,11\}$ — so $(a)$ , not $(c)$ .

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