SSC CGL 2025 Tier-I · 19 September 2025, Shift 3 · Quantitative Aptitude · Question 21 of 100

We transcribed this text from the pictures on a candidate's response sheet for this shift; the answer is the key SSC colour-marked on that sheet. Solution last reviewed 11 September 2026.

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SSC does not publish CGL question papers or a public answer key. From 2025 the candidate response sheet prints every question and option as a picture, so the text here is transcribed from those pictures for this shift and checked against them independently; where the layout carries meaning, the picture is shown as well. Each answer is the key SSC colour-marked on that sheet. SSC released a tentative key for challenge and later its final one, both only to candidates behind a login, and the sheet does not say which of the two it carries. The solutions are written by UnlockIAS.

Simplify: 1⁄(√7−√3) + 1⁄(√7+√3)

(a)√7
(b)2√7
(c)√7⁄2
(d)5√7

Answer

Why

Correct — C. Add the two fractions over one common denominator.

Denominator: (√7 − √3)(√7 + √3) = (√7)² − (√3)²
= 7 − 3 = 4
Numerator: (√7 + √3) + (√7 − √3) = 2√7 (the √3 terms cancel)

Sum = 2√7 ⁄ 4 = √7⁄2 → option (c)

Why the others are wrong

(a)
√7 — √7 is twice the true value. It divides 2√7 by 2, but the denominator is (√7)² − (√3)² = 7 − 3 = 4.
(b)
2√7 — 2√7 is the numerator alone. It never divides by the product of the two denominators, which is 4.
(d)
5√7 — 5√7 is ten times the value. No step of the working produces a 5: it uses only 2 (from √7 + √7) and 4 (from 7 − 3).

Concept

The identity (a − b)(a + b) = a² − b² turns a pair of surd denominators into a whole number. When two fractions have conjugate denominators, such as √7 − √3 and √7 + √3, their common denominator is rational at once.

In general, 1⁄(√a − √b) + 1⁄(√a + √b) = 2√a ⁄ (a − b). With a = 7 and b = 3 that is 2√7 ⁄ 4 = √7⁄2.

The stem is printed as a picture. It reads: Simplify 1⁄(√7 − √3) + 1⁄(√7 + √3).

Key facts

(√a − √b)(√a + √b) = a − b.
1⁄(√a − √b) + 1⁄(√a + √b) = 2√a ⁄ (a − b).
1⁄(√a − √b) − 1⁄(√a + √b) = 2√b ⁄ (a − b).

Study next

Rationalising a denominator of the form a + √bSquaring surd sums: (√a + √b)² = a + b + 2√(ab)Comparing sums of surds by squaring both sides

Common traps

Writing the denominator product as 7 + 3 = 10 instead of 7 − 3 = 4
Stopping at 2√7 and forgetting the denominator

Surds also appear at 19 Sep 2025, 09:00, Quant Q.3, which compares √7 + √2 with √6 + √3 and is settled by squaring both sums rather than by a conjugate.

Related PYQs

No directly related past PYQ was found.