SSC CGL 2024 Tier-I · 23 September 2024, Shift 2 · Quantitative Aptitude · Question 8 of 100

Text as printed on a candidate's response sheet for this shift; answer per SSC's tentative key as marked on that sheet. Solution last reviewed 10 September 2026.

Where this paper and its answers come from

SSC does not publish CGL question papers or a public answer key. The questions and options here are reproduced as printed on a candidate's own response sheet for this shift — the Commission's text, unchanged. Each answer is SSC's tentative key as marked on that sheet; SSC allowed challenges to it and never publishes a final key. The solutions are written by UnlockIAS.

The length, breadth and height of a room is 15 m, 9 m, and 5 m, respectively. From each can of paint 40 square metre of area is painted. How many cans of paint will be needed to paint only the walls of the room?

(a)6
(b)12
(c)8
(d)4

Answer

Why

Correct — A. Rule: the four walls of a room are its lateral surface area, 2(l + b) × h. The stem says only the walls, so floor and ceiling are excluded.

l + b = 15 + 9 = 24 m
2 × 24 = 48 m
48 × 5 = 240 m² of wall

Each can covers 40 m².
240 ⁄ 40 = 6 cans → option (a)

Why the others are wrong

(b)
12 — 12 needs 480 m², which is 4(l + b)h — every wall counted at both its length and its breadth. The lateral area is 2(15 + 9) × 5 = 240 m², half of that.
(c)
8 — 8 would need 320 m² to cover. That is not the walls (240 m²), not the walls plus floor (375 m²), and not the whole room (510 m²) — it matches no surface this room has.
(d)
4 — 4 buys only 160 m². The two longer walls alone come to 2 × 15 × 5 = 150 m², so four cans barely finish them and leave the two 9 m walls unpainted.

Concept

The phrase only the walls is the whole question. A cuboidal room has six faces, and that phrase selects the four vertical ones.

Those four are the lateral surface area, 2(l + b)h. The floor and the ceiling are the two l × b faces and stay out. Include them and you are computing the total surface area, 2(lb + bh + hl), which answers a different question.

The paint step is then a division: area ÷ coverage per can, rounded up whenever there is a remainder, because a part-used can is still a can. Here it divides exactly, so no rounding arises.

Note which dimension is the height. Length and breadth pair up inside the bracket, and only the height multiplies from outside.

Read the room as 15 by 5 with a height of 9 and the walls come to 2(15 + 5) × 9 = 360 m², which needs 9 cans instead of 6.

Key facts

Lateral surface area of a cuboid is 2(l + b)h, the four walls.
Total surface area of a cuboid is 2(lb + bh + hl), all six faces.
For this room the walls measure 2(15 + 9) × 5 = 240 square metres.
240 square metres at 40 square metres per can needs exactly 6 cans.

Study next

Total surface area against lateral surface area, and which wording selects whichCost of painting or whitewashing a room at a rate per square metreVolume of a cuboid and its longest diagonal, √(l² + b² + h²)

Common traps

Including the floor or the ceiling when the stem says only the walls
Rounding a fractional can down instead of up
Using 4(l + b)h, which counts every wall twice

Cuboid mensuration runs from the bare formula to a painting or fencing dressing that hides which faces are wanted.

Also asked 9 Sep 2024, 16:00, Quant Q.25 (volume from 8, 4 and 6 cm) and 18 Sep 2024, 09:00, Quant Q.5 (surface area of three cubes joined end to end).

Related PYQs

No directly related past PYQ was found.