CSAT Solved Papers/ 2025/Q15

2025 CSAT — Q15

Quant Counting & combinatorics 2.5 marks Hard

A solid cube is painted yellow on all its faces. The cube is then cut into $60$ smaller but equal pieces by making the minimum number of cuts. Which of the following statements is/are correct?

I. The minimum number of cuts is $9$ .

II. The number of smaller pieces which are not painted on any face is $6$ .

Select the correct answer using the code given below:

A I only
B II only
C Both I and II Answer
D Neither I nor II

Worked rationale

To split a cube into $a \times b \times c$ equal cuboidal pieces you slice along the three directions: $a-1$ , $b-1$ , $c-1$ cuts (parallel cuts in one direction can be made together, but the count of cuts is $(a-1)+(b-1)+(c-1)$ ). With $abc = 60$ , minimise

\text{cuts} = (a+b+c) - 3.

Sum $a+b+c$ is smallest when the three factors are as close together as possible. The factorisation of $60$ into three nearest factors is $3 \times 4 \times 5$ (sum $12$ ); any other, e.g. $2 \times 5 \times 6$ (sum $13$ ) or $2 \times 3 \times 10$ (sum $15$ ), is larger.

\text{minimum cuts} = (3+4+5) - 3 = 9. \quad \textbf{Statement I is correct.}

Unpainted pieces are the inner block stripped of the outer shell: $(a-2)(b-2)(c-2)$ . With $3\times4\times5$ :

(3-2)(4-2)(5-2) = 1 \times 2 \times 3 = 6. \quad \textbf{Statement II is correct.}

Answer: (c) Both I and II.

Visual solution

The same solve, worked by hand — read it, then trace it.

Hand-drawn worked solution for UPSC 2025 CSAT Q15 — Counting & combinatorics — Tap the drawing to open it full size for the fine detail.

Why the other options miss

A

an arithmetic slip on the inner block: gets the $3\times4\times5$ split and $9$ cuts but botches the inner block (e.g. computes $1\times1\times3 = 3$ , mis-subtracting), so rejects II.
B

wrong cut count: counts $6$ unpainted but uses cuts $= a+b+c = 12$ (or some other count), so rejects the " $9$ " in I.
D

uses a non-minimal split: picks a non-minimal factorisation (e.g. $2\times5\times6$ , $11$ cuts, inner $0\times3\times4 = 0$ ) and finds both statements false.

Specialist insight

Two facts carry this template. First, minimum cuts ⇔ factors closest together: for a fixed product, $a+b+c$ is minimised when $a,b,c$ are near-equal, so always reach for the “middle” factorisation ( $3,4,5$ for $60$ ). Second, the unpainted core is $(a-2)(b-2)(c-2)$ — peel one layer off each face. The dangerous slip is using a dimension that is too thin: a side of length $1$ contributes $(1-2) = -1$ , which is not a piece — but with $3,4,5$ the smallest side is $3$ , so $(3-2)=1$ is a genuine single inner layer. Here both numbers land cleanly, and checking the inner block confirms the chosen factorisation was the minimal one.

The trap, in one line

Minimum cuts use the closest factor triple ( $3\times4\times5$ , so $9$ cuts), and the unpainted core is $(a-2)(b-2)(c-2) = 1\cdot2\cdot3 = 6$ — both statements hold.

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