SSC CGL 2024 Tier-I · 9 September 2024, Shift 3 · Quantitative Aptitude · Question 5 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 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.

If x, y and z are positive numbers and x + y + z = 1, then the least value of 1⁄x + 1⁄y + 1⁄z is:

(a)11
(b)7
(c)5
(d)9

Answer

Why

Correct — D. The stem is an image: x, y and z are positive with x + y + z = 1, and the least value of 1⁄x + 1⁄y + 1⁄z is wanted.

Apply AM ≥ HM to the three positive numbers:
(x + y + z)⁄3 ≥ 3⁄(1⁄x + 1⁄y + 1⁄z)

Write S = 1⁄x + 1⁄y + 1⁄z and substitute x + y + z = 1:
1⁄3 ≥ 3⁄S
Cross-multiplying (S is positive): S ≥ 9

Equality in AM ≥ HM needs all three equal, so x = y = z = 1⁄3, and then S = 3 + 3 + 3 = 9.

The least value is 9 → option (d).

Why the others are wrong

(a)
11 — Larger than the minimum, so it cannot be the least value. The equal split x = y = z = 1⁄3 already delivers 9, and 9 is smaller than 11.
(b)
7 — Below the bound. AM ≥ HM forces 1⁄x + 1⁄y + 1⁄z ≥ 9 for every positive triple summing to 1, so no choice of x, y, z reaches 7.
(c)
5 — Also unreachable. The most balanced split gives exactly 9, and moving away from equality only pushes the sum of reciprocals higher, never lower.

Concept

For positive numbers the standard chain is AM ≥ GM ≥ HM, with equality only when all the numbers are equal.

The constraint fixes the arithmetic mean at 1⁄3, so the harmonic mean is at most 1⁄3, and 3 ÷ HM — which is exactly 1⁄x + 1⁄y + 1⁄z — is at least 9.

Cauchy–Schwarz gives the same result in one line: (x + y + z)(1⁄x + 1⁄y + 1⁄z) ≥ 3² = 9, and with x + y + z = 1 the second bracket alone is at least 9.

There is no upper bound: push any one variable towards 0 and its reciprocal grows without limit, so only a minimum exists.

The whole stem is delivered as an image in this paper, with the constraint and the expression on one line — read the condition x + y + z = 1 first, because the bound follows from it and not from the expression alone.

Key facts

For positive reals, AM ≥ HM, with equality exactly when every number is equal.
If x + y + z = 1 with all positive, the least value of 1⁄x + 1⁄y + 1⁄z is 9, attained at x = y = z = 1⁄3.
The general statement is (x + y + z)(1⁄x + 1⁄y + 1⁄z) ≥ 9 for positive x, y, z.
The same expression has no maximum, since letting one variable approach 0 sends it to infinity.

Study next

AM ≥ GM ≥ HM and the equality conditionThe (a + b + c)(1⁄a + 1⁄b + 1⁄c) ≥ 9 result and its two-variable versionMinimum of x + 1⁄x for positive x

Common traps

Hunting for the minimum by trying values instead of applying an inequality.
Assuming an extreme value must come from an unequal split of the variables.
Reading "least value" as the smallest number printed among the options.

The version SSC sets is always the symmetric one — a fixed sum, a reciprocal sum to be minimised — so testing the equal split first settles it in seconds, and the inequality is there only to confirm that nothing beats it.

Related PYQs

No directly related past PYQ was found.