Averages — SSC CGL Quantitative Aptitude

An average is just a total shared equally. The one equation you need is Sum = Average × Count — every SSC average question is that identity read forwards or backwards. The scoring skill is the change view: when a value enters, leaves or is corrected, work with how much the total shifts, then divide by the count once. Don't recompute the whole sum.

1. What SSC actually asks

Tier 1: 1–2 Q · Tier 2: 1–2 Q. The recurring types: direct average, the "new member changes the average" family (someone joins/leaves/replaces), weighted average of two groups, average of a consecutive/AP sequence, wrong-value correction, and average speed (which is not the arithmetic mean of speeds).

2. The one identity, both directions

$\text{Average}=\frac{\text{Sum}}{\text{Count}}\iff \text{Sum}=\text{Average}\times \text{Count}$

Most problems give you two of the three and ask the third. The moment a question mentions a total, convert it to Average × Count and keep everything in totals until the final divide.

3. Change in average — the scoring shortcut

When the count is $n$ and the total changes by $\Delta$, the average changes by $\frac{\Delta }{n}$.

• Someone joins: new average pulls toward the newcomer's value. If a class of 30 averaging 12 years gains a 43-year-old teacher, total rises by 43, count by 1 → new average $=\frac{360+43}{31}=13$.
• Replacement (count unchanged): average change $=\frac{\text{new}-\text{old}}{n}$. "Average weight of 8 rises by 2.5 kg when a newcomer replaces a 65 kg person" → total rose by $8\times 2.5=20$ → newcomer $=65+20=85$ kg.
• Wrong value corrected: total shifts by (correct − wrong); divide by $n$ once. Misreading 36 as 26 across 10 numbers raises the true average by $\frac{10}{10}=1$.

The habit: find the shift in the total first, divide by the count exactly once at the end.

4. Weighted average — combining groups

$\bar{x}=\frac{n_1{\bar{x}}_1+n_2{\bar{x}}_2}{n_1+n_2}$

25 boys averaging 60 with 15 girls averaging 68 give a class average of $\frac{25(60)+15(68)}{40}=\frac{2520}{40}=63$. The combined average always lies between the two group averages, closer to the larger group — a fast sanity check.

5. Averages of sequences

For any evenly spaced list (consecutive integers, an AP, consecutive even/odd numbers):

$\text{Average}=\frac{\text{first}+\text{last}}{2}=\text{the middle term}$

• First $n$ natural numbers: $\frac{n+1}{2}$.
• First $n$ odd numbers: average $=n$ (their sum is $n^2$).
• First $n$ even numbers: average $=n+1$.

Five consecutive even numbers averaging 40 have 40 as the middle term, so they are 36, 38, 40, 42, 44 — no algebra needed.

6. Average speed — the classic trap

Average speed is total distance ÷ total time, never the plain average of the speeds. For equal distances at speeds $x$ and $y$:

$\text{Average speed}=\frac{2xy}{x+y}(\text{harmonic mean})$

Going at 40 km/h and returning at 60 km/h gives $\frac{2\times 40\times 60}{100}=48$ km/h — not 50. (See Time, Speed & Distance for the full treatment.)

7. Solved PYQ-style examples

Q1. Average of the first five multiples of 7? Solution. 7, 14, 21, 28, 35 — evenly spaced, so the average is the middle term 21.

Q2. A batsman's average after 16 innings is $x$. In the 17th he scores 85 and his average rises by 3. His new average? Solution. $16x+85=17(x+3)\Rightarrow x=34$, new average $=$ 37.

Q3. The average of 10 numbers is 15; a value 36 was wrongly read as 26. The correct average is… Solution. Total was short by 10 → true total up by 10 → average up by $\frac{10}{10}=1$ → 16.

Q4. Average of 11 results is 50; average of the first six is 49 and of the last six is 52. The sixth result? Solution. First-six sum 294, last-six sum 312, total 550. The sixth is counted twice: $294+312-550=$ 56.

Q5. A man walks to a town at 40 km/h and cycles back at 60 km/h. His average speed? Solution. Equal distances → harmonic mean $\frac{2(40)(60)}{100}=$ 48 km/h (not the tempting 50).

8. Exam protocol

1. Convert every "total/sum" into Average × Count immediately; stay in totals.
2. For joins/leaves/replacements, compute the shift in the total, then divide by the count once.
3. Weighted average: the answer must sit between the two group averages — reject anything outside.
4. Evenly spaced list? Average = middle term = (first + last)/2; skip the summation.
5. "Average speed" over equal distances is the harmonic mean, never (x + y)/2.