Data Interpretation

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medium importance~2 Q in Tier 119 formulas⚡ 10 shortcuts5 subtopics

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Averages from data

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⏱ 4 min read🧩 5 question types🎯 13 practice Q

The idea in one minute

Table averages rebuild the total first: average = total ÷ count.

Rows can carry their own counts (workers per salary band, machines per output), so weighted averages, not plain ones, are the default here.

Simple row average

Six days of factory output:

Day	Mon	Tue	Wed	Thu	Fri	Sat
Units	480	520	460	540	500	500

Total $= 3000$ , average $= 3000 \div 6 = 500$ units a day.

Rule: Total first, divide second. An average of a table row is one sum and one division.

Weighted average: counts differ

A plain average works only when every value has the same count. Salary bands do not:

Band	Staff	Salary (₹)
Clerks	40	25,000
Typists	30	32,000
Guards	20	28,000
Managers	30	40,000

\bar{x} = \frac{\sum (\text{count} \times \text{value})}{\sum \text{count}}

Clerks and typists alone: $\dfrac{40 \times 25000 + 30 \times 32000}{70} = \dfrac{1960000}{70} = ₹28,000$ . All four bands: $\dfrac{3720000}{120} = ₹31,000$ .

Watch: The unweighted mean of 25000, 32000, 28000, 40000 is 31,250 — close, and wrong. Options love it.

Where the weighted average sits

The weighted average always sits between the smallest and largest band values, pulled toward the biggest band.

Clerks (40 at ₹25,000) with guards (20 at ₹28,000): $\dfrac{40 \times 25000 + 20 \times 28000}{60} = ₹26,000$ . It leans toward ₹25,000 because clerks are the bigger band.

Tip: Use this to reject options before computing: an answer outside the band range is always wrong.

Drop one entry, add one entry

Removing a day changes both the total and the count.

Drop Wednesday's 460: new total $2540$ , new count 5, average $508$ . A Sunday of 520-average over 7 days needs $7 \times 520 = 3640$ , so Sunday's output $= 3640 - 3000 = 640$ .

Tip: Any average question with a change is two totals and one subtraction.

The sixth value that fixes the average

Five electricity bills average ₹2,560 (total ₹12,800). A sixth bill drops the average to ₹2,500, so the new total is $6 \times 2500 = 15000$ . The sixth bill is $15000 - 12800 = ₹2,200$ .

Example: Bill below the new average pulls it down; the size of the pull shows in the difference of totals.

Machine tables

Machines with one output each are plain rows:

Machine	M1	M2	M3	M4	M5	M6
Units	12000	15000	14000	16000	13000	14000

Total $= 84,000$ units, so the average is $84000 \div 6 = 14,000$ units per machine.

Watch: Keep the unit name (units, rupees, thousands) straight from the header row; two unit columns in one table is a classic DI setup.

Average of averages fails

Two shifts, averages 400 and 500, do not average to 450 unless the shifts are equal. If 10 workers and 20 workers produced those averages, the combined average is $\dfrac{10 \times 400 + 20 \times 500}{30} = 466.67$ .

Tip: Whenever group sizes differ, multiply each average by its size before adding.

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

Average of a row

How to spot it:

A table of daily or yearly values; the overall average is asked.

\bar{x} = \frac{\sum x}{n}

Method

Add the row or column.
Count the entries.
Divide.

Why it works:

One row, one total, one count.

Try this

Units produced by a factory on six days of a week:

Day	Mon	Tue	Wed	Thu	Fri	Sat
Units	480	520	460	540	500	500

The average daily production is:

Show solution

Total $= 480+520+460+540+500+500 = 3000$ .
Average $= 3000 \div 6 = 500$ .

Answer

500 units

Type 2very common2 practice Q

Weighted average from bands

How to spot it:

A table pairs counts with values (staff per salary, students per mark); one average asked.

\bar{x} = \frac{\sum n_i x_i}{\sum n_i}

Method

Multiply each count by its value.
Add the products.
Divide by the total count.

Why it works:

Bigger bands must pull the average harder.

Try this

Salary structure of a company:

Band	Clerks	Typists	Guards	Managers
Staff	40	30	20	30
Salary (₹)	25,000	32,000	28,000	40,000

The average salary of all 120 employees is:

Show solution

$\sum n x = 40 \times 25000 + 30 \times 32000 + 20 \times 28000 + 30 \times 40000$ .
$= 1000000 + 960000 + 560000 + 1200000 = 3720000$ .
Average $= 3720000 \div 120 = 31,000$ .

Answer

₹31,000

Type 3common2 practice Q

Average of a machine row

How to spot it:

Machines or branches each give one output; the average output is asked.

\bar{x} = \frac{\sum x}{n}

Method

Add the outputs.
Divide by the number of machines.
Keep the unit from the header.

Why it works:

It is still a row average; the row is just machines.

Try this

Output of six machines in a month:

Machine	M1	M2	M3	M4	M5	M6
Units	12,000	15,000	14,000	16,000	13,000	14,000

The average output per machine is:

Show solution

Total $= 12000+15000+14000+16000+13000+14000 = 84000$ .
Average $= 84000 \div 6 = 14,000$ .

Answer

14,000 units

Type 4common2 practice Q

Drop an entry, extend the week

How to spot it:

An entry is removed (lowest day) or added (seventh day) and the new average is asked.

\bar{x}_{new} = \frac{\text{total} \mp \text{entry}}{n \mp 1}

Method

Write the current total.
Remove or add the entry.
Divide by the new count.

Why it works:

Both the total and the count move together.

Try this

Units produced by a factory on six days:

Day	Mon	Tue	Wed	Thu	Fri	Sat
Units	480	520	460	540	500	500

If Wednesday is excluded, the average of the remaining days is:

Show solution

Total without Wednesday $= 3000 - 460 = 2540$ .
Days left $= 5$ .
Average $= 2540 \div 5 = 508$ .

Answer

508 units

Type 5common2 practice Q

The value that sets a target average

How to spot it:

A new reading must bring the average to a stated level.

x = (n+1)\bar{x}_{target} - \text{old total}

Method

Old total: old average times old count.
New total: target average times new count.
Subtract.

Why it works:

The new total must cover every old value plus the newcomer.

Try this

The average of five electricity bills is ₹2,560. After a sixth bill arrives, the average of all six becomes ₹2,500.

Bills	Count	Average (₹)
First five	5	2,560
All six	6	2,500

The sixth bill is:

Show solution

Old total $= 5 \times 2560 = 12800$ .
New total $= 6 \times 2500 = 15000$ .
Sixth bill $= 15000 - 12800 = 2200$ .

Answer

₹2,200

Formula sheet

Simple average

\bar{x} = \frac{\sum x}{n}

Weighted average

\bar{x} = \frac{\sum n_i x_i}{\sum n_i}

n_i is the count in each band or row.

Entry to hit a target average

x = (n+1)\bar{x}_{new} - n\bar{x}_{old}

Shortcuts that save time

⚡ Deviations from a round base

Table values near a round number: average = base + (sum of differences ÷ count). 480, 520, 460, 540, 500, 500 around 500: deviations sum to 0, average exactly 500.

Example

Find the average of 480, 520, 460, 540, 500, 500.

Show solution

Base 500: deviations $-20, +20, -40, +40, 0, 0$ .
Sum of deviations $= 0$ .
Average $= 500$ .

Answer

500

⚡ Balance around the new average

If an added value lowers the average, each old entry gains (old avg − new avg); the new entry supplies all of it.

Example

Five bills average 2,560. A sixth brings the average to 2,500. Find the sixth bill.

Show solution

Old total $= 5 \times 2560 = 12800$ .
New total $= 6 \times 2500 = 15000$ .
Sixth bill $= 15000 - 12800 = 2200$ .

Answer

₹2,200

Mistakes to avoid

Where most students lose marks on this subtopic.

Mistake 01

Averaging band values without their counts.

Weight by the staff or count column first.

Mistake 02

Averaging two group averages when groups differ in size.

Rebuild both totals and both counts.

Mistake 03

Dropping an entry but dividing by the old count.

The count falls by one with the entry.

Mistake 04

Mixing units from two different header rows.

Read the header of the exact row you use.

Mistake 05

Rounding mid-way and carrying the rounded value.

Keep fractions until the final line.

Quick revision

Read this the night before the exam.

Simple: $\dfrac{\sum x}{n}$ ; weighted: $\dfrac{\sum n_i x_i}{\sum n_i}$ .
Band tables: multiply count by value before adding.
Removed entry: new total ÷ new count.
Added entry: $(n+1)\times$ new average $-$ old total.
Average of averages is valid only for equal group sizes.
Deviations from a round base beat long addition.

Practice: 13 questions

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

Topic test · 13 questions

Suggested time 9 min · wrong answers go to your mistake notebook automatically.

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