SSC CGL 2024 Tier-I · 10 September 2024, Shift 1 · Quantitative Aptitude · Question 21 of 98

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.

If a right circular cone of height 24 cm has a volume of 1232 cm³, then the total surface area of the cone is (use π = 22⁄7):

(a)704 cm
(b)904 cm
(c)608 cm
(d)806 cm

Answer

Why

Correct — A. The stem is an image: a right circular cone of height 24 cm has volume 1232 cm³, find the total surface area with π = 22⁄7. The volume gives r, Pythagoras gives l.

V = ⅓πr²h → 1232 = ⅓ × 22⁄7 × r² × 24
1232 = 176⁄7 × r²
r² = 1232 × 7 ÷ 176 = 49, so r = 7 cm

l = √(r² + h²) = √(49 + 576) = √625 = 25 cm

TSA = πr(l + r) = 22⁄7 × 7 × (25 + 7)
= 22 × 32 = 704 → option (a).

Why the others are wrong

(b)
904 cm — 904 fails a one-second divisibility check. With r = 7 and π = 22⁄7 the total surface is 22 × (l + 7), always a multiple of 22, and 904 ÷ 22 = 41.09.
(c)
608 cm — 608 sits between the two pieces and matches neither. The curved surface πrl is 550 and the base πr² is 154, which add to 704 — no piece or pair of this cone comes to 608.
(d)
806 cm — 806 would need a slant height of about 29.6 cm. But r = 7 with h = 24 fixes l at √625 = 25 exactly, so the bracket (l + r) is 32 and nothing else.

Concept

A cone is pinned down by any two of r, h and l, locked together by l² = r² + h².

Volume uses the height: V = ⅓πr²h. Surface uses the slant: curved surface = πrl, base = πr², and total surface = πr(l + r).

So a volume-to-surface question is always three moves — r from the volume, l from Pythagoras, then substitute. Carrying h into a surface formula is the commonest slip, and it is silent, because the answer still looks plausible.

The numbers here are clean because (7, 24, 25) is a Pythagorean triple.

The options are printed as 704 cm, not 704 cm² — the paper drops the square on an area.

Read all four as square centimetres. The same dropped unit shows up at 12 Sep 2024, 12:30, Quant Q.5, where a sector area is offered as 1848 cm, and it never changes which option is right.

Key facts

For a cone, l² = r² + h², so the slant height is never the height.
Total surface area of a cone = πr(l + r), which is the curved surface πrl plus the base πr².
Volume of a cone = ⅓πr²h, so 1232 cm³ with h = 24 cm and π = 22⁄7 forces r = 7 cm.
(7, 24, 25) is a Pythagorean triple, alongside (3, 4, 5) and (5, 12, 13).

Study next

Curved surface area of a cone from radius and slant heightVolume of the cone generated by rotating a right triangle about a legThe frustum, and which of r, h and l survive when a cone is cut

Common traps

Substituting h = 24 into πr(l + r), which gives 22 × 31 = 682 and no option.
Stopping at the curved surface, 550, when the question says total.
Dropping the ⅓ from the volume formula, which makes r² = 49⁄3 and leaves no whole radius.

The cone is set from different pairs of its handles across shifts: curved surface from r and l at 18 Sep 2024, 12:30, Quant Q.23, and volume from r and h at 26 Sep 2024, 16:00, Quant Q.3.

A cone built by rotating a 3-4-5 triangle about one leg is asked at 10 Sep 2024, 12:30, Quant Q.12 — the same triple logic, run backwards.

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