SSC CGL 2025 Tier-I · 15 September 2025, Shift 2 · Quantitative Aptitude · Question 6 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.

Where this paper and its answers come from

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.

There are 5 consecutive even numbers X₁, X₂, X₃, X₄, x5 and 4 consecutive odd numbers Y₁, Y₂, Y₃, Y₄. The average of the even numbers is 5 more than the average of the odd numbers. If the sum of the even numbers is 30 more than the sum of the odd numbers, find the average of the odd numbers.

(a)3
(b)4
(c)5
(d)6

Answer

Why

Correct — C. Let the average of the odd numbers be m, and turn both conditions into sums.

Average of the even numbers: m + 5
Sum of the 5 even numbers: 5(m + 5) = 5m + 25
Sum of the 4 odd numbers: 4m

Sum condition: (5m + 25) − 4m = 30
Simplify: m + 25 = 30
Solve: m = 5 → option (c)

Why the others are wrong

(a)
3 — With m = 3 the sums are 40 and 12, a gap of 28, not 30. The gap is m + 25, so it reaches 30 only at m = 5.
(b)
4 — With m = 4 the sums are 45 and 16, a gap of 29, one short of 30. Each step of 1 in m moves the gap by 1.
(d)
6 — With m = 6 the sums are 55 and 24, a gap of 31, one more than 30. Since the gap is m + 25, m must be 5.

Concept

Sum = count × average. The stem gives one condition on averages and one on sums, and turning both into sums leaves a single unknown.

Five numbers with average m + 5 sum to 5m + 25, and four with average m sum to 4m. Their difference, m + 25, is set equal to 30.

The data do not fully cohere. Four consecutive odd numbers always average to an even number, midway between the two middle ones: 1, 3, 5, 7 average 4, and 3, 5, 7, 9 average 6. No such set averages 5.

The key's 5 comes from the two numerical conditions alone, and it is the only option that satisfies them. The even side does work: average 10 gives 6, 8, 10, 12, 14, with sum 50.

Key facts

Sum of a set = number of items × average.
For an odd count of consecutive even (or odd) numbers, the average is the middle number.
Four consecutive odd numbers average to the even number between the two middle ones, e.g. 3, 5, 7, 9 average 6.

Study next

Sum and average of consecutive integers, such as five consecutive numbers summing to 80Averages when items are added to or removed from a groupParity of sums and averages of even and odd numbers

Common traps

Writing the even sum as 5m + 5 instead of 5(m + 5) = 5m + 25. The extra 5 applies to each of the five numbers.
Hunting for four consecutive odd numbers that average 5. None exist, so solve with the two equations.

Sum = count × average also solves 18 Sep 2024, 09:00, Quant Q.14: five consecutive numbers summing to 80 average 16, so the largest is 18.

26 Sep 2024, 12:30, Quant Q.17 halves 174 the same way: two consecutive even numbers average 87, so they are 86 and 88.

Related PYQs

No directly related past PYQ was found.

Topic: Averages — all 53 SSC CGL questions on it, every solved shift.