Batsman problems, overlapping sums & multi-step sets — notes, question types & practice · Average

01

Batsman: the innings question

A batsman's average = total runs ÷ innings played. The standard question: he scores x runs in his nth innings and the average rises by d. Let the new average be A. Then:

(n-1)(A - d) + x = nA \quad\Rightarrow\quad A = x - (n-1)d

98 runs in the 20th innings lift the average by 2 → new average = $98 - 19 \times 2 = 60$ .

Rule: Score in the nth innings = new average + (n − 1) × rise. It must feed every past innings plus its own share.

The reverse question asks how many runs are needed. To lift an average from 42 to 45 in the 11th innings: needed = $11 \times 45 - 10 \times 42 = 75$ .

02

Bowler's average

A bowler's average = runs given ÷ wickets taken, and a LOWER value is better. Average a, then a match of w wickets for r runs makes it fall by d. With W wickets before the match:

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

Average 12.4; a match of 5 wickets for 26 runs improves it by 0.4 → $12.4W + 26 = 12(W + 5)$ , so $W = 85$ .

Watch: "Improves" means the bowling average falls. Subtract, never add.

03

Overlapping windows

The average of Mon–Wed is 37 and of Tue–Thu is 34. The two sums share Tue and Wed, so subtracting kills them:

\text{Thu} - \text{Mon} = 3 \times (34 - 37) = -9

Mon was 40°, so Thu = $40 - 9 = 31$ °. Multiply the difference of averages by the length of the overlap.

04

First-k and last-k share a middle

13 numbers average 30. The first 7 average 27 and the last 7 average 35. The 7th number sits in both halves, so it is counted twice:

\text{middle} = 7 \times 27 + 7 \times 35 - 13 \times 30 = 189 + 245 - 390 = 44

Tip: Add the two part-sums and subtract the whole. What survives is the shared item.

05

Split sets with relations

"The average of 6 numbers is 30. The first two average 24, the next two 33. Of the last two, one is 4 more than the other."

Sums first: total 180, first pair 48, second pair 66, so the last pair totals 66. Name the smaller x: $x + (x + 4) = 66$ , so 31 and 35.

Rule: Every average becomes a sum. Name ONE unknown x and write the rest in terms of it.

06

Ages move together

Every member of a fixed group gets one year older each year, so the average age also rises by 1 per year.

n years ago the group's average was A → today it is $A + n$ (same members).
At the birth of the youngest: subtract the youngest's age from every member, and drop the youngest from the count. Five members average 24, youngest 8 → $\dfrac{120 - 5 \times 8}{4} = 20$ .

Tip: Rebuild the story with totals — old total, change, new total, new count. If new total ÷ new count misses the stated average, a step is wrong.

07

Question types you will see

Each type: how to recognise it, the method step by step, and one question to try.

Type 1very common2 practice Q

Batsman: the innings that moves the average

How to spot it:

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

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

Method

Write old total = (n − 1)(A − d) and new total = nA for the new average A.
Solve for A, or subtract old total from target total when runs needed are asked.
Check with the totals.

Why it works:

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

Try this

A batsman scores 98 runs in his 20th innings and raises his average by 2. Find his new average.

Show solution

New average = $98 - 19 \times 2$ .
= $98 - 38 = 60$ .
Check: $19 \times 58 + 98 = 1200 = 20 \times 60$ .

Answer

60

Type 2common2 practice Q

Overlapping windows

How to spot it:

Averages over two overlapping stretches — Mon–Wed and Tue–Thu, or the 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

Method

Turn both averages into sums.
Subtract the sums: the shared items cancel, leaving the end difference. For first-k/last-k, ADD and subtract the whole: the shared middle survives.
Use the extra fact (one end value, or a ratio) to finish.

Why it works:

Shared values appear in both sums, so they vanish on subtraction and double up on addition.

Try this

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

Show solution

Part sums: $7 \times 27 = 189$ , $7 \times 35 = 245$ ; whole = $13 \times 30 = 390$ .
7th number = $189 + 245 - 390$ .
= $44$ .

Answer

44

Type 3common2 practice Q

Split set with relations between unknowns

How to spot it:

Averages of parts of a list, plus a tie such as 'one number is 4 more than the other'.

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

Method

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

Why it works:

Once everything is a sum, the question is one equation in one variable.

Try this

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. Find the larger of the last two.

Show solution

Total = $180$ ; first pair = $48$ ; second pair = $66$ .
Last pair total = $180 - 48 - 66 = 66$ .
$x + (x + 4) = 66 \Rightarrow x = 31$ ; larger = $35$ .

Answer

35

Type 4common3 practice Q

Ages over time

How to spot it:

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

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

Method

Bring every average to the SAME year: add t for each year passed.
Work in totals; include or remove members as the story demands.
At a member's birth: subtract that age from every member, and drop one from the count.

Why it works:

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

Try this

A family of 5 members has an average age of 24 years. The youngest is 8. What was the average age of the family at the birth of the youngest?

Show solution

Present total = $5 \times 24 = 120$ years.
8 years ago everyone was 8 younger: $120 - 5 \times 8 = 80$ years over 4 members.
Average = $80 \div 4 = 20$ years.

Answer

20 years

Type 5occasional2 practice Q

Bowling average

How to spot it:

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

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

Method

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.
Solve the linear equation for W.

Why it works:

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

Try this

A bowler's average is 12.4 runs per wicket. In a match he takes 5 wickets for 26 runs, and his average improves by 0.4. How many wickets did he have before this match?

Show solution

$12.4W + 26 = 12(W + 5)$ .
$12.4W + 26 = 12W + 60 \Rightarrow 0.4W = 34$ .
$W = 85$ .

Answer

85

08

Formula sheet

Batsman score

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

A' = new average, d = rise.

Batsman new average

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

Overlap subtraction

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

Three-day windows; multiply by the overlap length.

Shared middle item

\text{shared} = S_1 + S_2 - S_{\text{total}}

Split sums

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

One equation per chunk.

09

Shortcuts that save time

⚡ Batsman: stay in totals

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

Example

A batsman scores 87 in his 17th innings and raises his average by 3. Find his average after the 17th innings.

Show solution

Let the new average be $A$ : $17A = 16(A - 3) + 87$ .
$17A = 16A + 39$ , so $A = 39$ .

Answer

39

⚡ Overlaps: subtract the sums

Shared days cancel, leaving only the difference of the end days.

Example

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

Show solution

Sums: $3 \times 37 = 111$ and $3 \times 34 = 102$ .
Thu − Mon = $102 - 111 = -9$ .
Thu = $40 - 9 = 31$ °C.

Answer

31°C

⚡ One variable for the unknown chunk

Name the smallest unknown x, express the rest through it, and close the equation with the leftover sum.

Example

8 numbers average 20. The first two average 15.5 and the next three average $\dfrac{64}{3}$ . The 6th number is 5 less than the 7th, and the 8th is 7 more than the 7th. Find the 8th number.

Show solution

Total = $160$ ; known chunks = $31 + 64 = 95$ ; leftover = $65$ .
Let the 6th be $x$ : $x + (x+5) + (x+12) = 65$ .
$x = 16$ , so the 8th = $16 + 12 = 28$ .

Answer

28

10

Mistakes to avoid

Where most students lose marks on this subtopic.

Mistake 01

In batsman questions, multiplying by n for the old total.

Before the nth innings there are only n − 1 innings. Old total = (n − 1) × old average.

Mistake 02

Reporting the difference of two overlapping averages as the end-day difference.

Multiply the difference of averages by the overlap length (3 days → × 3).

Mistake 03

Forgetting the two halves of 11 numbers share the 6th number.

First 6 + last 6 = 12 counts for 11 numbers: the middle one is counted twice.

Mistake 04

Setting a relation backwards ('the 6th is 5 less than the 7th').

Translate carefully: 6th = 7th − 5, so 7th = 6th + 5.

Mistake 05

Treating a bowling average like a batsman average (higher = better).

A bowler improves when the average falls. Use $aW + r = (a-d)(W+w)$ .

11

Quick revision

Read this the night before the exam.

Batsman: new average = x − (n − 1)d; needed runs = new total − old total.
Bowler: aW + r = (a − d)(W + w); 'improves' means falls.
Overlap: end difference = overlap length × difference of averages.
First-k and last-k: shared middle = part sums − whole sum.
Split set: sums first, then one variable x.
Fixed group: average age +1 per year; at a birth, subtract that age from everyone.

12

Practice: 16 questions

Sets of 10, mixed across the question types above. Every answer has a step-by-step explanation.

Topic test · 10 questions

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