SSC CGL 2024 Tier-I · 12 September 2024, Shift 1 · Quantitative Aptitude · Question 10 of 99

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 6 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.

Let t = 2⁄5, then the value of the expression t³ + (3⁄5)³ + (9⁄5)t is:

Answer

Why

Correct — D. Substitute t = 2⁄5 and put all three terms over 125.

t³ = (2⁄5)³ = 8⁄125
(3⁄5)³ = 27⁄125
(9⁄5)t = (9⁄5)(2⁄5) = 18⁄25 = 90⁄125

Sum = (8 + 27 + 90)⁄125 = 125⁄125 = 1 → option (d).

The shortcut: with a = 2⁄5 and b = 3⁄5, a + b = 1, so 3ab(a + b) = 3(2⁄5)(3⁄5) = 18⁄25, which is exactly the printed (9⁄5)t.

The expression is therefore a³ + b³ + 3ab(a + b) = (a + b)³ = 1³.

Why the others are wrong

(a)
1⁄2 cannot arise here at all. Every term is a number of 125ths, and 125 is odd, so no sum of them is a half.
(b)
2 is double the true total. It would need the numerators over 125 to reach 250, and 8 + 27 + 90 = 125.
(c)
1⁄5 is the difference 3⁄5 − 2⁄5, not the value of the sum. The expression adds three positive terms; it does not subtract the two fifths.

Concept

This is the cube identity (a + b)³ = a³ + b³ + 3ab(a + b) written out with one piece already simplified.

The setter picks a = 2⁄5 and b = 3⁄5 so that a + b = 1. That makes 3ab(a + b) collapse to a plain multiple of t, which is why the third term appears as (9⁄5)t and not as a bracket.

Once you see that, no cubing is needed: the answer is just (a + b)³. Direct substitution reaches the same 1 in three short lines, so either route is safe under exam time.

The stem and all four options are printed as images in the paper, so a text-only view of this row shows an empty question. The options read 1⁄2, 2, 1⁄5 and 1.

Key facts

(a + b)³ = a³ + b³ + 3ab(a + b), the same statement as a³ + b³ = (a + b)³ − 3ab(a + b).
With a = 2⁄5 and b = 3⁄5, a + b = 1, so 3ab(a + b) = 18⁄25 — the same value as (9⁄5)t.
8⁄125 + 27⁄125 + 90⁄125 = 125⁄125 = 1.

Study next

Expansions of (a + b)³ and (a − b)³ and their rearranged formsa³ + b³ + c³ − 3abc and its factorisationSpotting substitutions where a + b evaluates to 1 or 0

Common traps

Stopping at a³ + b³ = 8⁄125 + 27⁄125 = 7⁄25 and never adding the third term.
Treating (9⁄5)t as an unrelated linear term instead of the 3ab(a + b) piece of a cube.
Cubing 2⁄5 as 6⁄15 or 8⁄15 instead of 8⁄125.

SSC builds these substitution items so the bracket collapses to a clean 1 or 0 — look for that collapse before you start cubing anything.

The same habit of reading an expression as an identity rather than solving it is rewarded at Quant Q.16 of this paper, where x + 1⁄x = 15 is handed to you and x itself is never needed.

Related PYQs

No directly related past PYQ was found.