2024 CSAT — Q6

Quant Logical & quantitative reasoning 2.5 marks Medium

In the expression $5 * 4 * 3 * 2 * 1$ , $*$ is chosen from $+,\ -,\ \times$ each at most two times. What is the smallest non-negative value of the expression?

A 3
B 2
C 1
D 0 Answer

Worked rationale

There are four operator slots, and each of $+,-,\times$ may be used at most twice. We want the smallest value that is still $\ge 0$ . The natural floor to aim for is $0$ — if we can hit it, we are done, since nothing non-negative is smaller.

Construct it directly (respecting BODMAS — $\times$ binds first):

5 - 4 - 3 + 2 \times 1 \;=\; 5 - 4 - 3 + 2 \;=\; 0.

Operator tally: $-$ twice, $+$ once, $\times$ once — all within the “at most twice” cap. The value is exactly $0$ , which is the least possible non-negative result.

Answer: (d) 0.

Why the other options miss

A

missed a case: stops at an easy expression (e.g. $5-4+3-2+1$ mishandled) and reports $3$ without searching for a zero.
B

missed a case: finds a value of $2$ and quits, not testing whether $0$ is reachable under the “each at most twice” budget.
C

an arithmetic slip: targets parity-of- $1$ but slips on a sign or a $\times$ precedence, landing on $1$ instead of the achievable $0$ .

Specialist insight

The phrase “smallest non-negative” is the whole game: it sets the target at $0$ , not at the global minimum (which could be deeply negative). So the task is a feasibility check — can a legal operator assignment make the expression exactly $0$ ? — not an optimisation. One clean witness ( $5-4-3+2\times1=0$ ) settles it; you do not need to enumerate all $3^4$ assignments. Always read the extremum direction first: “non-negative” caps the search at $0$ and saves you a full sweep.

The trap, in one line

"Smallest non-negative" means hunt for $0$ , not the most-negative value — and $5-4-3+2\times1=0$ uses no operator more than twice.

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