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SSC CGL Maths · Averages

Parts of a Total

When a list is split into parts with their own averages, the total of all minus the totals of the known parts gives whatever is left, and parts that overlap share a member.

All Averages notes

Why this matters

Eight questions in the bank, and seven of them are MODERATE, half of all the MODERATE rows in this chapter. They are long to read but short to solve, since each one is three totals and a subtraction. Read the counts carefully; that is where the marks go.

Concept 1 of 3: The members left after the known parts

The total of the whole list is fixed. Take away the totals of the parts you know, and what remains belongs to the members you have not yet accounted for.

Definition

  • Total of all =n×A= n \times A; each known part's total == its count ×\times its average.
  • Leftover total == total of all −- the known part totals.
  • If one part's average is given as '25%25\% more than' another part's, find that average first, then multiply by its count.
  • If the leftover members are linked ('the 11th is 6 less than the 13th'), write them all in terms of one unknown, add, and set equal to the leftover total.
  • Fast check: deviations from the overall average across all the parts add to 0.

Leftover total

Leftover total=nA−(n1A1+n2A2+⋯ )\text{Leftover total} = nA - (n_1A_1 + n_2A_2 + \cdots)

Worked example

The average of 9 numbers is 40. The first four average 35 and the last four average 44. What is the fifth number?
Practice this conceptself-check · 4 quick reps

The same idea in a real exam question:

SSC CGL · 2025 · Tier I, 13 Sep 2025, Shift 2 · Q58Moderate

Example 1 · Averages · Parts of a Total

The average of 13 numbers is 70. The average of the first five numbers is 60. The average of the next five numbers is 25%25\% higher than the average of the first five numbers. The 11th number is 6 less than the 13th number, and the 12th number is 4 greater than the 13th number. What is the average of the 11th and 12th numbers?

Applying a percentage to the wrong quantity

When a part's average is '25% more' than another's, raise the average by 25% and then multiply by that part's count. Raising the other part's total only works when the two parts have the same count.
Drill 3 more on the members left after the known parts

Concept 2 of 3: Overlapping parts share a member

If 'the first six' and 'the last six' of eleven numbers are both given, the sixth number sits in both parts. Adding the two part totals counts it twice, and the extra over the grand total is that shared number.

Definition

  • Check the counts first: 6+6=126 + 6 = 12 but there are only 11 numbers, so one number is shared.
  • Shared member == part 1 total ++ part 2 total −- total of all.
  • Calendars: a 31-day month's first 16 days and last 16 days share day 16.
  • Fast route with deviations from the overall average AA: shared =A+n1(A1−A)+n2(A2−A)= A + n_1(A_1 - A) + n_2(A_2 - A).

Shared member of two overlapping parts

Shared=n1A1+n2A2−nA\text{Shared} = n_1A_1 + n_2A_2 - nA

Worked example

The average of 9 numbers is 50. The first five average 48 and the last five average 53. Find the fifth number.
Practice this conceptself-check · 3 quick reps

The same idea in a real exam question:

SSC CGL · 2025 · Tier I, 21 Sep 2025, Shift 1 · Q66Moderate

Example 2 · Averages · Parts of a Total

The average of a group of 11 numbers is 61. If the first six numbers have an average of 57 and the last six numbers have an average of 65, what is the sixth number?

Missing the shared member

January has 31 days, so 'the first sixteen days' and 'the last sixteen days' overlap at day 16. Treating them as two separate halves leaves the totals impossible to match.
Drill 1 more on overlapping parts share a member

Concept 3 of 3: A member tied to the average of all

If one member is k above the average of everyone, the others together must be k below that average. Shared among the other n − 1, each is k ÷ (n − 1) below, so the average of all is the others' average plus that small amount.

Definition

  • Full method: call the average of all aa. Total na=na = (others' total) +(a+k)+ (a + k). Solve for aa.
  • Fast route: a=a = others' average +kn−1+ \frac{k}{n - 1}, and the special member is a+ka + k.
  • If the member is kk less than the average of all, subtract instead: a=a = others' average −kn−1- \frac{k}{n - 1}.

Average of all when one member is tied to it

a=xˉothers+kn−1a = \bar{x}_{\text{others}} + \frac{k}{n - 1}
  • aaverage of all n members
  • khow far the special member is above the average of all
  • nnumber of members, the special one included

Worked example

Ten friends share a bill. Nine of them pay ₹240 each, and the tenth pays ₹45 more than the average of all ten. How much does the tenth pay?
Practice this conceptself-check · 3 quick reps

The same idea in a real exam question:

SSC CGL · 2024 · Tier I, 23 Sep 2024, Shift 3 · Q51Moderate

Example 3 · Averages · Parts of a Total

In a team of seven members, the average age of six of the members is 42 years while the age of the seventh member of the team is 36 years more than the average age of all the seven members taken together. Find the age (in years) of the seventh member of the team.

Adding k to the others' average

The special member is k above the average of all, not above the others' average. The average of all is the others' average plus k ÷ (n − 1), and the special member is k above that.
Drill 1 more on a member tied to the average of all

Summary — formulas & gotchas at a glance

A revision cheat-sheet for the formulas and gotchas above. Click any concept name to jump back to its full explanation.

Formulas (3)

Watch out for (3)

Test yourself on Averages

20 past SSC CGL questions from this chapter, timed at 12 minutes and marked the way the exam marks it. You see your score and every answer the moment you finish. Free to start.

Take the Averages chapter testDrill all 8 questions