SSC CGL 2024 Tier-I · 9 September 2024, Shift 3 · Quantitative Aptitude · Question 9 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 7845K854 is divisible by 11, then what is the value of K?

(a)9
(b)8
(c)6
(d)7

Answer

Why

Correct — A. Apply the divisibility rule for 11 to 7845K854.

Number the digits from the left:
7 (1st), 8 (2nd), 4 (3rd), 5 (4th), K (5th), 8 (6th), 5 (7th), 4 (8th)

Odd places: 7 + 4 + K + 5 = 16 + K
Even places: 8 + 5 + 8 + 4 = 25

Rule: the difference of the two sums must be 0 or a multiple of 11.
(16 + K) − 25 = K − 9

K − 9 = 0 gives K = 9. Setting K − 9 = ±11 would need K = 20 or K = −2, both outside the digits 0 to 9.

So K = 9, giving 78,459,854 = 11 × 7,132,714 → option (a).

Why the others are wrong

(b)
8 — K = 8 leaves a difference of 8 − 9 = −1, and −1 is not a multiple of 11, so 78,458,854 fails the test by 1.
(c)
6 — K = 6 leaves a difference of −3. Only 0, ±11, ±22 and so on pass the rule, and −3 is none of them.
(d)
7 — K = 7 leaves a difference of −2, again not a multiple of 11. Only K = 9 brings the two place-sums level at 25 each.

Concept

The test for 11 is the alternating sum of the digits: add those in the odd places, add those in the even places, and subtract one total from the other.

If that difference is 0 or a multiple of 11, the number is divisible by 11.

It does not matter whether you count places from the left or from the right — swapping the two groups only flips the sign, and a multiple of 11 stays a multiple of 11 when negated.

Here the fixed digits already sit 9 apart, so K has exactly one job: supply the missing 9.

Because K is confined to a single digit, the multiple-of-11 condition has only one solution here — a nine-digit version of the same question can have two, which is why the ±11 branch is worth writing down rather than assuming away.

Key facts

A number is divisible by 11 when (sum of digits in odd places) − (sum of digits in even places) is 0 or a multiple of 11.
In 7845K854 the odd-place digits are 7, 4, K and 5, and the even-place digits are 8, 5, 8 and 4.
Those sums are 16 + K and 25, so K = 9 and the number is 78,459,854.
78,459,854 divided by 11 is exactly 7,132,714.

Study next

Divisibility rules for 7, 11 and 13 taken togetherFinding a missing digit that satisfies two divisibility conditions at onceRemainder on division by 11 read straight off the alternating sum

Common traps

Miscounting the positions and swapping the odd and even groups.
Forgetting that K itself sits in an odd place and belongs in that sum.
Accepting a difference of 11 or −11 without checking that K stays a single digit.

SSC plants the unknown inside an eight- or nine-digit string and asks for the digit that satisfies one divisibility rule. Counting positions consistently from one end is the whole discipline of the question.

Related PYQs

No directly related past PYQ was found.