CSAT Solved Papers/ 2023/Q79

2023 CSAT — Q79

Quant Counting & combinatorics 2.5 marks Hard

A cuboid of dimensions 7 cm×5 cm×3 cm7\ \text{cm} \times 5\ \text{cm} \times 3\ \text{cm} is painted red, green and blue colour on each pair of opposite faces of dimensions 7 cm×5 cm7\ \text{cm} \times 5\ \text{cm}, 5 cm×3 cm5\ \text{cm} \times 3\ \text{cm}, 7 cm×3 cm7\ \text{cm} \times 3\ \text{cm} respectively. Then the cuboid is cut and separated into various cubes each of side length 1 cm1\ \text{cm}. Which of the following statements is/are correct?

  1. There are exactly 1515 small cubes with no paint on any face.

  2. There are exactly 66 small cubes with exactly two faces, one painted with blue and the other with green.

Which of the statements given above is/are correct?

  1. A 1 only Answer
  2. B 2 only
  3. C Both 1 and 2
  4. D Neither 1 nor 2

Worked rationale

The cuboid is 7×5×37\times 5\times 3, cut into 11-cm cubes. Colours: 7×57\times 5 faces == red, 5×35\times 3 faces == green, 7×37\times 3 faces == blue.

Statement 1 (no paint == interior): strip one layer off each dimension:

(7−2)(5−2)(3−2)=5×3×1=15.Statement 1 is true.(7-2)(5-2)(3-2) = 5 \times 3 \times 1 = 15. \quad\text{Statement 1 is true.}

Statement 2 (exactly two faces, one blue and one green): such cubes sit on an edge where a blue (7×37\times 3) face meets a green (5×35\times 3) face. Those two face-types share the vertical edges of length 33 — there are 44 such edges. On each, the cubes with exactly two painted faces number 3−2=13 - 2 = 1 (the two ends are 33-face corners). So

4 edges×1=4 cubes,not 6.Statement 2 is false.4 \text{ edges} \times 1 = 4 \text{ cubes}, \quad\text{not } 6. \quad\text{Statement 2 is false.}

Answer: (a) 1 only.

Visual solution

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

Hand-drawn worked solution for UPSC 2023 CSAT Q79 — Counting & combinatorics
Tap the drawing to open it full size for the fine detail.

Why the other options miss

  • B
    missed a case: miscounts the blue–green edges (e.g. uses all 44 short edges with ≥\ge one interior cube wrongly, or pairs the wrong faces) to reach 66, while flubbing the interior count.
  • C
    off by one: gets the interior 1515 right but accepts the blue–green count as 66, over-counting the length-33 edge cubes.
  • D
    misreads the geometry: mishandles the (n−2)(n-2) stripping (e.g. (7−2)(5−2)(3−2)(7-2)(5-2)(3-2) miscomputed) and the edge pairing, rejecting both.

Specialist insight

Two cuboid facts: interior (no-paint) cubes are (ℓ−2)(w−2)(h−2)=5⋅3⋅1=15(\ell-2)(w-2)(h-2) = 5\cdot 3\cdot 1 = 15; two-face cubes of a specific colour pair live on the edges where those two faces meet. Blue (7×37\times 3) and green (5×35\times 3) meet along the four height-33 edges, each contributing 3−2=13 - 2 = 1 exactly-two-face cube ⇒4\Rightarrow 4, not 66. The trap is Statement 2’s plausible-looking "66" — the scoring move is to identify which edges carry the blue–green pair and count (h−2)(h-2) along each.

The trap, in one line

Interior =(7−2)(5−2)(3−2)=15=(7{-}2)(5{-}2)(3{-}2)=15 (St-1 ✓); blue–green meet on 44 height-33 edges ×(3−2)=4\times (3{-}2) = 4 cubes, not 66 (St-2 ✗) ⇒\Rightarrow (a).

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