SSC CGL 2024 Tier-I · 25 September 2024, Shift 3 · Quantitative Aptitude · Question 6 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.

If a² + b² = 148 and ab = 54, then find the value of (a + b) ⁄ (a − b).

Answer

Why

Correct — D. The stem figure gives a² + b² = 148 and ab = 54, and asks for (a + b) ⁄ (a − b). Build both brackets from the two square identities.

(a + b)² = a² + b² + 2ab = 148 + 108 = 256, so a + b = 16
(a − b)² = a² + b² − 2ab = 148 − 108 = 40, so a − b = √40 = 2√10

Divide:
(a + b) ⁄ (a − b) = 16 ⁄ 2√10 = 8⁄√10 → option (d)

The numbers are real: a = 8 + √10 and b = 8 − √10 satisfy both conditions.

Why the others are wrong

(a)
2⁄√10 takes a + b as 4 instead of 16 — the square root applied twice to 256. (a + b)² = 256 gives a + b = 16, and 256 is the square of 16, not of 4.
(b)
8√7 needs a √7, which nothing here produces. The only surd in the working comes from (a − b)² = 40 = 4 × 10, giving √10, and 148 ± 108 yields 256 and 40 — neither is 7 times a square.
(c)
5√3 ≈ 8.66, while the true quotient 8⁄√10 ≈ 2.53. A √3 would need (a − b)² to be three times a perfect square, and it is 40.

Concept

Two identities do all the work, and they differ only in one sign:

(a + b)² = a² + b² + 2ab
(a − b)² = a² + b² − 2ab

Given a² + b² and ab, both brackets follow immediately, which is why SSC supplies exactly that pair rather than a and b themselves.

The design is deliberate: 148 + 108 = 256 is a perfect square so a + b is clean, while 148 − 108 = 40 is not, so the surd survives into the answer and separates the options.

Signs are technically open — a + b could be −16 and a − b could be −2√10, and the quotient of two negatives is the same positive value.

All four options are positive, so the intended reading is the positive roots throughout. Note also that 8⁄√10 rationalises to 4√10⁄5, which is the same number in a different dress.

Key facts

(a + b)² = a² + b² + 2ab and (a − b)² = a² + b² − 2ab.
With a² + b² = 148 and ab = 54: (a + b)² = 256 and (a − b)² = 40.
√40 = √(4 × 10) = 2√10, so the quotient 16 ⁄ 2√10 reduces to 8⁄√10.
The pair satisfying both conditions is a = 8 + √10 and b = 8 − √10.

Study next

Getting a³ + b³ and a³ − b³ from a + b, a − b and abThe a + 1⁄a family, where a² + 1⁄a² and a³ + 1⁄a³ are built the same wayRecovering a and b as the roots of t² − (a + b)t + ab = 0

Common traps

Rooting 256 twice and using a + b = 4.
Writing (a − b)² = a² + b² − ab, dropping the factor 2.
Rationalising to 4√10⁄5 and then not finding it, because the option list keeps the surd in the denominator.

SSC gives the two symmetric quantities and asks for a third, so the identities are the whole question. Algebraic identity work is also set at 09 Sep 2024, 16:00, Quant Q.8, on a + b = 2c.

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