Average

high importance~1 Q in Tier 116 formulas⚡ 12 shortcuts4 subtopics

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Subtopic 4 of 4·← Weighted average & two-group problems

Batsman problems, overlapping sums & multi-step sets

Batsman/cricketer template: a score in the next innings raises the average by d. If the new average is A' after n innings, the score was $A' + (n-1)d$ . Rearranged: new average $= \frac{\text{old total} + \text{score}}{n}$ .

Overlapping sums (temperatures, marks): if the average of Mon–Wed is a and of Tue–Thu is b, subtracting gives Thu − Mon = 3(b − a). Give one endpoint to find the other.

Split sets: an average over n numbers with partial averages over overlapping or consecutive chunks is solved by converting every average to a sum and using one variable for the unknown chunk.

📖 Notes 🧩 5 question types ⚡ Formulas 🎯 16 practice Q

Detailed notes

Batsman and bowler questions

A batsman's average = total runs ÷ number of innings (here every innings counts as "out"). The standard question: "He scores x runs in his nth innings and his average rises by d." Let the new average be A. Then the old average was $A - d$ for $n - 1$ innings: $(n-1)(A - d) + x = nA \ \Rightarrow\ A = x - (n-1)d$ Example: 98 runs in the 20th innings raise the average by 2 → new average $= 98 - 19 \times 2 = 60$ . The same bridge answers "how many must he score in the next innings to raise his average from 42 to 45 after 10 innings?" → $11 \times 45 - 10 \times 42 = 75$ .

A bowler's average is runs given ÷ wickets taken, and a LOWER value is better. If his average is a, and in a match he takes w wickets for r runs so that the average falls by d, with W wickets before the match: $aW + r = (a - d)(W + w)$ Average 12.4, 5 wickets for 26 runs, improves by 0.4 → $12.4W + 26 = 12(W + 5)$ → $W = 85$ .

Overlapping windows

The average of Monday to Thursday and the average of Tuesday to Friday share Tuesday, Wednesday and Thursday. Subtract the sums and the shared days cancel: $\text{Fri} - \text{Mon} = 4 \times (\text{second average} - \text{first average})$ If one of the two end days is known (or their ratio), the other follows at once. The "first 7 and last 7 of 13 numbers" question works the same way: the 7th number is counted twice, so $\text{middle} = \text{first-part sum} + \text{last-part sum} - \text{total sum}$ 13 numbers average 30; first 7 average 27; last 7 average 35 → 7th $= 189 + 245 - 390 = 44$ .

Split sets with relations

"The average of 6 numbers is 30; the first two average 24, the next two 33; of the remaining two, one is 4 more than the other." Go to sums: $180 - 48 - 66 = 66$ for the last two → $x + x + 4 = 66$ → 31 and 35. Name the smallest unknown x and write every other unknown in terms of x.

Ages over time

Every member of a group ages 1 year each year, so the average age also rises by 1 per year (for the same members).

n years ago, a family's average was A; now it is $A + n$ for the same members.
A child born later adds its age to the total and 1 to the count.
"Average age of the family at the birth of the youngest": take the present total, subtract the youngest's age from every other member and the youngest itself: $\frac{\text{total} - (\text{members}) \times \text{youngest's age}}{\text{members} - 1}$ . Example: 5 members average 24, youngest 8 → at the youngest's birth: $\frac{120 - 5 \times 8}{4} = 20$ .

Checking your answer

Rebuild the story with totals: old total, change, new total, new count. If the new total ÷ new count matches the stated average, you are right.

Quick revision

Batsman: new average $= x - (n-1)d$ ; needed score $=$ new total $-$ old total.
Bowler: runs-per-wicket totals on both sides of the match.
Overlap: difference of ends $=$ window size × difference of averages; shared middle $=$ part sums − total.
Split set: sums first, one variable.
Ages: average age rises 1 per year for a fixed group.

Types of questions asked

Every way this subtopic shows up in exams — how to recognise it, the formula or logic to use, and a solved example.

Type 1: Batsman: score that changes the averagevery common2 practice Q

How to spot it:

'Scores x in the nth innings and raises his average by d' or 'how many runs needed next innings to reach an average'.

A_{\text{new}} = x - (n-1)d, \qquad \text{needed} = (n+1)A' - nA

Write old total $= (n-1)(A - d)$ and new total $= nA$ .
Their difference is the score in the nth innings; solve for A.
For 'runs needed', subtract old total from the target total.

Why: the new score must lift every earlier innings by d and still leave its own share A.

Example: A batsman scores 98 in his 20th innings and his average rises by 2. His new average?

$98 - 19 \times 2 = 60$ . Check: $19 \times 58 + 98 = 1200 = 20 \times 60$ .

Type 2: Overlapping windows (days, first-k and last-k)common2 practice Q

How to spot it:

Averages over two overlapping sets — Mon–Thu and Tue–Fri, or first 7 and last 7 of 13 numbers.

\text{end}_2 - \text{end}_1 = k(\bar{b} - \bar{a}), \qquad \text{shared} = S_1 + S_2 - S

Convert both averages to sums.
Subtract (days) — shared items cancel — or add and remove the total (first/last parts) — the shared item is left.
Use the extra fact (one end, or a ratio) to finish.

Why: shared values appear in both sums and vanish on subtraction.

Example: The average of 13 numbers is 30. The first 7 average 27 and the last 7 average 35. Find the 7th number.

$7 \times 27 + 7 \times 35 - 13 \times 30 = 189 + 245 - 390 = 44$ .

Type 3: Split set with relations between the unknownscommon2 practice Q

How to spot it:

Averages of parts of a set plus 'one is 4 more than the other' or 'first is twice the second'.

\text{unknown part} = nA - \sum(\text{known parts})

Change every average into a sum.
Get the leftover sum for the unknown values.
Write the unknowns with one variable x from the relation and solve.

Why: once everything is a sum, the question is a one-variable equation.

Example: Six numbers average 30; the first two average 24 and the next two 33. Of the last two, one is 4 more than the other. The larger is?

Leftover $= 180 - 48 - 66 = 66$ → $2x + 4 = 66$ → $x = 31$ → larger $= 35$ .

Type 4: Ages over time (family and groups)common3 practice Q

How to spot it:

Average ages 'n years ago' or 'at the time of marriage / birth of the youngest', with members added later.

\text{average after } t \text{ years} = A + t \ \ (\text{same members})

Move every average to the SAME year (add t for each year passed).
Convert to totals and include or remove the new/old members.
At the birth of a member: subtract his age from every member, and drop him from the count.

Why: each person ages one year per year, so a fixed group's total rises by n per year.

Example: A family of 5 averages 24 years. The youngest is 8. What was the average age of the family at the birth of the youngest?

Then: $120 - 5 \times 8 = 80$ shared by 4 people → $20$ years.

Type 5: Bowling average (runs per wicket)occasional2 practice Q

How to spot it:

'A bowler's average is 12.4 runs per wicket; he takes 5 wickets for 26 runs and his average improves by 0.4.'

aW + r = (a - d)(W + w)

Let W be the wickets before the match; runs given so far $= aW$ .
After the match: runs $aW + r$ , wickets $W + w$ , average $a - d$ (improvement means it falls).
Solve the linear equation for W.

Why: bowling average is a runs-per-wicket average, so the same total bridge applies.

Example: A bowler's average is 12.4 runs per wicket. He takes 5 wickets for 26 runs and his average improves by 0.4. Wickets taken before this match?

$12.4W + 26 = 12(W + 5)$ → $0.4W = 34$ → $W = 85$ .

Formulas

Batsman score

\text{score} = A' + (n-1)d

Batsman new average

A' = \frac{(n-1)A + \text{score}}{n}

Overlap subtraction

\text{(Thu)} - \text{(Mon)} = 3(b - a)

Split sums

n\bar{x} = \sum_{\text{chunks}} (\text{chunk total})

Shortcut tricks

⚡ Batsman: work in totals

Old total + new score = new count × new average.

Example: A batsman scores 87 in his 17th innings and thereby increases his average by 3. Find his average after the 17th innings.

$16(A - 3) + 87 = 17A$ ... solve: let new average be A: $17A = 16(A-3) + 87$ ⇒ $A = 39$ .

⚡ Overlap: subtract the sums

The shared days cancel, leaving the difference of the end days.

Example: The average temperature of Mon, Tue, Wed is 37°C and of Tue, Wed, Thu is 34°C. If Monday was 40°C, find Thursday.

Thu = 102 − (111 − 40) = 31°C.

⚡ One variable for the unknown chunk

Name the smallest unknown x, express the rest from the given relations, and let the leftover sum close the equation.

Example: 8 numbers average 20. First two average 15.5, next three average 21⅓. The 6th is 5 less than the 7th, and the 8th is 7 more than the 7th. Find the 8th.

Leftover total = 160 − 95 = 65. Let 6th = x: x + (x+5) + (x+12) = 65 ⇒ x = 16 ⇒ 8th = 28.

Where students lose marks

In batsman problems, multiplying by n instead of (n − 1) for the old total.
Subtracting the averages (37 − 34) and reporting 3 as Thu − Mon without multiplying by the overlap length rule.
Forgetting that the 'next' chunk overlaps the previous one (11 numbers: first 6 and last 6 share the 6th).
Setting the relations backwards (6th is 5 LESS than 7th ⇒ 7th = 6th + 5).

Practice sets — 16 questions

Sets of 10, mixed across the question types above. Each answer comes with a step-by-step explanation.

Topic test · 10 questions

Suggested time 8 min · wrong answers go to your mistake notebook automatically.