SSC CGL 2025 Tier-I · 12 September 2025, Shift 2 · 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 13 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.

If x = √7, determine the value of x + 1⁄x.

(a)8√7⁄7
(b)7√7⁄8
(c)7√7⁄9
(d)9√7⁄7

Answer

Why

Correct — A. Rationalise 1⁄x, then add like surds.

1⁄x = 1⁄√7 = √7⁄7 (multiply top and bottom by √7)
x = √7 = 7√7⁄7 (same denominator)
x + 1⁄x = 7√7⁄7 + √7⁄7 = 8√7⁄7 → option (a)

Check: (x² + 1) ⁄ x = 8⁄√7 = 8√7⁄7 ✓

Why the others are wrong

(b)
7√7⁄8 — 7√7⁄8 ≈ 2.32 is smaller than x = √7 ≈ 2.65 on its own. Adding a positive 1⁄x must make the sum larger than √7.
(c)
7√7⁄9 — 7√7⁄9 ≈ 2.06 is also below √7 ≈ 2.65, yet x + 1⁄x must exceed x because 1⁄x is positive.
(d)
9√7⁄7 — 9√7⁄7 = 9⁄√7, which is (x² + 2) ⁄ x, that is x + 2⁄x. It adds 1⁄x twice.

Concept

A fraction with a surd in the denominator is rationalised by multiplying top and bottom by that surd: 1⁄√7 = √7⁄7.

Once both terms share the denominator 7 they are like surds, and they add by their coefficients: 7√7⁄7 + √7⁄7 = 8√7⁄7.

The shortcut x + 1⁄x = (x² + 1) ⁄ x skips the split: x² = 7, so the sum is 8⁄√7.

A size check clears (b) and (c) at once: for positive x, x + 1⁄x is bigger than x, and both of those values are below √7 ≈ 2.65.

Key facts

1⁄√n = √n⁄n: multiply top and bottom by √n.
x + 1⁄x = (x² + 1) ⁄ x, so for x = √7 it is 8⁄√7.
√7 ≈ 2.646, so x + 1⁄x ≈ 2.646 + 0.378 = 3.024.

Study next

Squaring x + 1⁄x to get x² + 1⁄x²Rationalising denominators of the form a + √bx³ + 1⁄x³ = (x + 1⁄x)³ − 3(x + 1⁄x)

Common traps

Writing 1⁄√7 as √7. Rationalising gives √7⁄7, a seventh of √7.
Marking 7√7⁄8 because it looks like 8√7⁄7. It is smaller than √7 itself, which adding a positive 1⁄x cannot produce.

Here x is given and x + 1⁄x is asked. SSC also sets it in reverse: on 13 Sep 2025, 12:30, Quant Q.25, x + 1⁄x = 5 gives x² + 1⁄x² = 5² − 2 = 23.

On 18 Sep 2025, 12:30, Quant Q.21 the given is √x + 1⁄√x = 4, and squaring gives x + 1⁄x = 16 − 2 = 14.

Related PYQs

No directly related past PYQ was found.

Topic: Square roots — all 48 SSC CGL questions on it, every solved shift.