← Data Analysis & Interpretation · RITES Senior Technical Assistant Civil

Chapter 4 of 8

Averages, Percentages & Ratios in Data

In the RITES Senior Technical Assistant Civil syllabus under Data Analysis & Interpretation · 4 parts

Data Interpretation · Averages · Percentage · Ratio, Proportion & Partnership

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📑 Contents (27 sections)

Part 1 of 4

Data Interpretation

Last reviewed 22 Sept 2026 · 5 min read

What a DI set really tests

Not arithmetic — reading. Nearly every mark lost in this section comes from one of four things: the unit in the heading (lakh, crore, thousand), the row or column read one line off, "percentage increase" mistaken for "percentage point", or an average taken over the wrong number of years.

Read in this order: title → unit → row and column headings → the question. Only then look at the numbers.

The four presentations

Type	What it is good at	Watch out for
Table	exact values, several variables	which way the years run
Bar chart	comparing quantities	the scale, and bars that do not start at zero
Line graph	trend over time	a broken axis exaggerating a rise
Pie chart	shares of one whole	two pies with different totals

A caselet presents the same data as a paragraph of prose; convert it into a small table before answering anything.

The five questions

FormulaThe recurring calculations

Total of a row or column — plain addition
Average $= \frac{total}{number of entries}$ (count the entries, do not assume five)
Ratio — reduce before dividing
Percentage share $= \frac{part}{whole} \times 100$
Percentage change $= \frac{new - old}{old} \times 100$ — always over the old value

For pie charts: $1% = 3.6°$ , so a sector of $54°$ is $15%$ . Two pie charts can only be compared in absolute terms if both totals are given.

Working quickly

Keep the unit out of the arithmetic: if everything is "₹ lakh", work in plain numbers and put the unit back at the end.
Use fraction equivalents rather than long division: $\frac{1}{8} = 12.5%$ , $\frac{1}{7} \approx 14.3%$ , $\frac{1}{6} \approx 16.7%$ , $\frac{1}{3} \approx 33.3%$ .
When options are far apart, approximate: $\frac{85}{180}$ is a little under half, so 47% is right and 43% is not.
Do not recompute a total you already worked out for the previous question — DI sets are built so each answer feeds the next.

A worked set

The table shows the sales of two products, in ₹ lakh.

Year	Product A	Product B	Total
2021	40	60	100
2022	55	65	120
2023	70	80	150
2024	60	90	150
2025	85	95	180

Worked ExampleQuestion 1 — total

What are the total sales of product A over the five years?

Solution. $40 + 55 + 70 + 60 + 85 =$ ₹ 310 lakh.

Worked ExampleQuestion 2 — average

Find the average annual sales of product B.

Solution. $\frac{60 + 65 + 80 + 90 + 95}{5} = \frac{390}{5} =$ ₹ 78 lakh.

Worked ExampleQuestion 3 — percentage change

By what percent did the total sales grow from 2021 to 2025?

Solution. $\frac{180 - 100}{100} \times 100 =$ 80%.

Worked ExampleQuestion 4 — percentage share

In which year was product A's share of the total the highest?

Solution. Shares: $40%$ , $45.8%$ , $46.7%$ , $40%$ , $47.2%$ → 2025.

Worked ExampleQuestion 5 — ratio

Find the ratio of the sales of A to B in 2023.

Solution. $70 : 80 =$ 7 : 8.

Worked ExampleQuestion 6 — year-on-year growth

Find the percentage increase in the sales of A from 2022 to 2023.

Solution. $\frac{70 - 55}{55} \times 100 = \frac{15}{55} \times 100 \approx$ 27.3%.

Worked ExampleQuestion 7 — a trap

Product A's share rose from 40% in 2024 to 47.2% in 2025. Is that a rise of 7.2% or of 7.2 percentage points?

Solution. 7.2 percentage points. As a percentage increase it is $\frac{7.2}{40} \times 100 = 18%$ . The two are different numbers and the options will contain both.

Worked ExampleQuestion 8 — as a pie chart

If the 2025 sales are drawn as a pie chart, what angle does product A occupy?

Solution. $\frac{85}{180} \times 360° =$ 170°.

Worked ExampleQuestion 9 — a year below the average

In how many years were product A's sales below their own five-year average?

Solution. Average $= 310/5 = 62$ → 2021 (40), 2022 (55) and 2024 (60) are below it: 3 years.

Worked ExampleQuestion 10 — combining rows

If the company earns a profit of 20% on A and 15% on B, what is the profit in 2024?

Solution. $0.20 \times 60 + 0.15 \times 90 = 12 + 13.5 =$ ₹ 25.5 lakh.

Frequently tested points

Read the unit in the heading before any calculation.
Percentage change is over the old value; a share in percent changes by percentage points.
Average needs the count of entries actually being averaged.
Pie chart: $1% = 3.6°$ ; comparing two pies needs both totals.
A bar chart whose axis does not start at zero exaggerates differences.
Reduce ratios before dividing, and reuse totals already computed.
Approximate when the options are far apart; compute exactly only when they are close.

Common MistakeCommon mistakes

Dividing by the new value in a percentage change.
Treating a rise from 40% to 47.2% as a 7.2% increase.
Averaging over five years when the question asks about four.
Comparing sectors of two pie charts as though the totals were equal.
Losing the unit — answering 310 when the answer is ₹ 310 lakh.

Revision SummaryChapter summary

Data interpretation rewards careful reading of the title, unit and headings before the numbers.
Tables, bar charts, line graphs, pie charts and caselets present the same kinds of data with different traps.
Five calculations — total, average, ratio, share and percentage change — cover almost every question asked.
Percentage change and percentage points are different, and both appear in the options.
Speed comes from fraction equivalents, approximation and reusing totals, not from faster long division.

Part 2 of 4

Averages

Last reviewed 22 Sept 2026 · 6 min read

Meaning

The average (arithmetic mean) of $n$ values is

Average = \frac{Sum of the values}{Number of values}, Sum = Average \times n

Almost every average question is solved by moving between these two forms: turn each average into a sum, do the arithmetic on sums, then turn the answer back into an average.

How an average behaves

Change to every value	Effect on the average
Add $k$	Average increases by $k$
Subtract $k$	Average decreases by $k$
Multiply by $k$	Average is multiplied by $k$
Divide by $k$	Average is divided by $k$

The average always lies between the smallest and the largest value.

Averages of standard series

FormulaStandard results

First $n$ natural numbers: $\frac{n + 1}{2}$
First $n$ odd numbers ( $1, 3, 5, \dots$ ): $n$
First $n$ even numbers ( $2, 4, 6, \dots$ ): $n + 1$
Squares of the first $n$ naturals: $\frac{( n + 1 ) ( 2 n + 1 )}{6}$
Cubes of the first $n$ naturals: $\frac{n ( n + 1 ) ^{2}}{4}$
Any set of consecutive numbers (or numbers in AP): $\frac{first + last}{2}$ , which is also the middle term when the count is odd

Combined and weighted average

FormulaGroups put together

For two groups of sizes $n_{1}, n_{2}$ with averages $A_{1}, A_{2}$ :

A = \frac{n _{1} A _{1} + n _{2} A _{2}}{n _{1} + n _{2}}

The same formula with $n_{i}$ read as "weights" is the weighted average. Note that it is not $\frac{A _{1} + A _{2}}{2}$ unless the groups are equal in size.

Replacement, inclusion and correction

FormulaThe three standard moves

Someone joins: new average $= \frac{n A + x}{n + 1}$
Someone is replaced: if the average of $n$ values changes by $d$ when a value $x$ is replaced by $y$ , then $y = x + n d$
A value was read wrongly: correct average $=$ wrong average $+ \frac{correct value - wrong value}{n}$

Average speed

FormulaTwo legs of a journey

Equal distances at speeds $x$ and $y$ : $Average speed = \frac{2 x y}{x + y}$ (the harmonic mean)
Equal times at speeds $x$ and $y$ : $Average speed = \frac{x + y}{2}$
In general, average speed $= \frac{total distance}{total time}$ — never the plain average of the speeds when the distances differ

Worked examples

Worked ExampleExample 1 — sum from average

The average of 11 results is 50. The average of the first six is 49 and of the last six is 52. Find the sixth result.

Solution. The sixth is counted in both groups: $6 \times 49 + 6 \times 52 - 11 \times 50 = 294 + 312 - 550 =$ 56.

Worked ExampleExample 2 — one more person

The average age of 30 students is 14 years. Including the teacher, the average becomes 15. Find the teacher's age.

Solution. $30 \times 14 + T = 31 \times 15 \Rightarrow 420 + T = 465 \Rightarrow T =$ 45 years.

Worked ExampleExample 3 — replacement

The average weight of 8 people increases by 2.5 kg when a new person replaces one weighing 65 kg. Find the new person's weight.

Solution. $y = x + n d = 65 + 8 \times 2.5 =$ 85 kg.

Worked ExampleExample 4 — a wrong reading

The average of 10 numbers is 23. One number, 32, was read as 23. Find the correct average.

Solution. Correct average $= 23 + \frac{32 - 23}{10} = 23 + 0.9 =$ 23.9.

Worked ExampleExample 5 — two classes

Class A has 30 students averaging 60 marks and class B has 20 averaging 70. Find the combined average.

Solution. $\frac{30 \times 60 + 20 \times 70}{50} = \frac{1800 + 1400}{50} =$ 64 marks.

Worked ExampleExample 6 — a standard series

Find the average of the first 50 natural numbers.

Solution. $\frac{n + 1}{2} = \frac{51}{2} =$ 25.5.

Worked ExampleExample 7 — consecutive numbers

The average of five consecutive even numbers is 26. Find the largest.

Solution. The middle number is 26, so the numbers are 22, 24, 26, 28, 30 and the largest is 30.

Worked ExampleExample 8 — a batsman's average

A batsman averages 40 runs in 20 innings. He scores 82 in the 21st. Find the new average.

Solution. $\frac{20 \times 40 + 82}{21} = \frac{882}{21} =$ 42.

Worked ExampleExample 9 — average speed, equal distances

A man travels a distance at 40 km/h and returns at 60 km/h. Find the average speed.

Solution. $\frac{2 \times 40 \times 60}{100} =$ 48 km/h — not 50 km/h, because he spends longer on the slower leg.

Worked ExampleExample 10 — average speed in general

Someone covers 120 km at 60 km/h and the next 120 km at 40 km/h, then rests for 1 hour. Find the average speed for the whole outing.

Solution. Time $= 2 + 3 + 1 = 6$ h for 240 km, so $\frac{240}{6} =$ 40 km/h. Resting time counts.

Worked ExampleExample 11 — average of a group after a change

The average salary of 20 workers is ₹ 15,000. Two workers earning ₹ 12,000 and ₹ 18,000 leave. Find the new average.

Solution. Total $= 300, 000$ ; the two removed total 30,000, so 18 workers share ₹ 270,000 → ₹ 15,000. Removing values whose own average equals the group average leaves the average unchanged.

Frequently tested points

Sum $=$ average $\times$ count; convert every average into a sum first.
Adding $k$ to every value adds $k$ to the average; multiplying by $k$ multiplies it.
First $n$ naturals: $(n + 1) /2$ ; first $n$ odd: $n$ ; first $n$ even: $n + 1$ .
Consecutive numbers or an AP: average $=$ (first + last)/2 $=$ middle term.
Combined average is weighted by group size, never the plain mean of two averages.
Replacement rule: $y = x + n d$ .
Wrong reading: correct the average by (correct − wrong)/ $n$ .
Equal distances: average speed $= 2 x y / (x + y)$ ; equal times: $(x + y) /2$ .

Common MistakeCommon mistakes

Averaging two averages when the groups are of different sizes.
Using $(x + y) /2$ for a there-and-back journey; equal distance needs $2 x y / (x + y)$ .
Forgetting that the overlapping value is counted twice in "first six / last six" problems.
Ignoring halts when asked for average speed over a whole journey.

Revision SummaryChapter summary

An average is a sum divided by a count, and most problems are solved on the sums.
Shifting or scaling every value shifts or scales the average by the same amount.
Standard series have standard averages, and any set of consecutive numbers averages to its middle term.
Groups combine by a weighted average, and joining, replacing or correcting a value each has a one-line rule.
Average speed is total distance over total time — the harmonic mean for equal distances, the arithmetic mean for equal times.

Part 3 of 4

Percentage

Last reviewed 16 Sept 2026 · 5 min read

Meaning

Per cent means "per hundred": $x % = \frac{x}{100}$ .

Conversions

Fraction → %: multiply by 100 ( $3/8 = 37.5%$ ).
% → fraction: divide by 100 and simplify ( $12.5% = 1/8$ ).
Decimal ↔ %: move the decimal point two places ( $0.075 = 7.5%$ ).

Key fraction equivalents

%	Fraction	%	Fraction
50	1/2	12.5	1/8
33.33	1/3	11.11	1/9
25	1/4	10	1/10
20	1/5	9.09	1/11
16.67	1/6	8.33	1/12
14.28	1/7	6.25	1/16
66.67	2/3	37.5	3/8
75	3/4	62.5	5/8
40	2/5	87.5	7/8

Basic formulas

FormulaPercentage relations

x% of N $= \frac{x}{100} \times N$
What % is A of B? $= \frac{A}{B} \times 100$
Number whose x% is A $= \frac{A \times 100}{x}$
Percentage change $= \frac{New - Old}{Old} \times 100$
After increase of x%: New = Old × $(1 + \frac{x}{100})$ ; after decrease: Old × $(1 - \frac{x}{100})$

Percentage point — absolute difference between percentages (from 20% to 25% is a 5 percentage point increase, but a 25% increase).

Comparison formulas

Formula"More than / less than"

If A is x% more than B, then B is less than A by $\frac{x}{100 + x} \times 100%$
If A is x% less than B, then B is more than A by $\frac{x}{100 - x} \times 100%$

Successive percentage changes

FormulaNet effect

Two successive changes of a% and b% (use negative for decrease):

Net change = a + b + \frac{ab}{100} %

Increase x% then decrease x% → net loss of $x^{2} /100$ %
Apply repeatedly for three changes (combine two, then the third)

Product constancy (price and consumption)

FormulaExpenditure = price × consumption

If price increases by r%, consumption must decrease by

\frac{r}{100 + r} \times 100%

to keep expenditure constant; if price decreases by r%, consumption can increase by $\frac{r}{100 - r} \times 100%$ .

Area of rectangle with length and breadth changed by a% and b%: net change $= a + b + ab /100$ %; square/circle with side/radius changed by x%: area changes by $2 x + x^{2} /100$ %.

Population and depreciation

FormulaGrowth and depreciation

Population after n years (growth r% p.a.): $P_{n} = P (1 + \frac{r}{100})^{n}$
Population n years ago: $P / (1 + \frac{r}{100})^{n}$
Value after depreciation of r% p.a.: $V_{n} = V (1 - \frac{r}{100})^{n}$

Worked examples

Worked ExampleExample 1 — basic

(a) Find 37.5% of 640. (b) What percent of 250 is 45? (c) 18 is 12% of what number?

Solution. (a) $\frac{3}{8} \times 640 =$ 240; (b) $\frac{45}{250} \times 100 =$ 18%; (c) $\frac{18 \times 100}{12} =$ 150

Worked ExampleExample 2 — more than / less than

A's salary is 25% more than B's. By what percent is B's salary less than A's?

Solution. $\frac{25}{125} \times 100 =$ 20%

Worked ExampleExample 3 — successive change

The price of cement rises by 20% and then falls by 10%. Find the net change.

Solution. $20 - 10 + \frac{20 \times ( - 10 )}{100} =$ +8%

Worked ExampleExample 4 — equal increase and decrease

A number is increased by 10% and then decreased by 10%. Net change?

Solution. $- \frac{1 0 ^{2}}{100} =$ −1% (a decrease of 1%)

Worked ExampleExample 5 — consumption adjustment

The price of steel increases by 25%. By what percent must a builder reduce consumption to keep expenditure unchanged?

Solution. $\frac{25}{125} \times 100 =$ 20%

Worked ExampleExample 6 — area change

The length of a rectangular plot is increased by 20% and breadth decreased by 20%. Find the change in area. What if the side of a square increases by 10%?

Solution. Rectangle: $20 - 20 - 4 =$ −4%; square: $20 + 1 =$ +21%

Worked ExampleExample 7 — population

A town's population of 50 000 grows at 4% per year. Find the population after 2 years.

Solution. $50000 \times 1.0 4^{2} = 50000 \times 1.0816 =$ 54 080

Worked ExampleExample 8 — elections

In an election between two candidates, the winner got 58% of valid votes and won by 4800 votes. Find the total valid votes.

Solution. Margin = 58% − 42% = 16% → $0.16 V = 4800$ → V = 30 000

Worked ExampleExample 9 — examinations

A student needs 40% to pass. She scored 180 marks and failed by 20 marks. Find maximum marks.

Solution. Pass marks = 200 = 40% → maximum 500

Worked ExampleExample 10 — depreciation

A machine worth ₹ 5 lakh depreciates at 10% per year. Find its value after 3 years.

Solution. $5 \times 0. 9^{3} = 5 \times 0.729 =$ ₹ 3.645 lakh

Worked ExampleExample 11 — percentage points

Unemployment rose from 6% to 7.5%. Express the rise in percentage points and percent.

Solution. 1.5 percentage points; $\frac{1.5}{6} \times 100 =$ 25%

Frequently tested points

x% = x/100; know fraction equivalents (1/8 = 12.5%, 1/6 = 16.67%, 1/7 ≈ 14.28%).
% change = (new − old)/old × 100.
A is x% more than B → B is x/(100 + x) × 100% less than A; x% less → x/(100 − x) × 100% more.
Successive changes: a + b + ab/100; equal rise and fall → −x²/100 %.
Constant expenditure: price up r% → consumption down r/(100 + r) × 100%.
Square/circle area change for x% change in side/radius: 2x + x²/100.
Population $P (1 + r /100)^{n}$ ; depreciation $V (1 - r /100)^{n}$ .
Percentage point vs percent.

Common MistakeCommon mistakes

Calculating percentage change on the new value instead of the original.
Adding successive percentages directly (10% up and 10% down is not zero change).
Confusing percentage points with percent change.

Revision SummaryChapter summary

Percentages express parts per hundred and interconvert with fractions and decimals, with key equivalents enabling fast calculation.
Basic relations find a percentage of a quantity, a quantity from its percentage and percentage change.
Comparison formulas convert "more than" into "less than" statements, and successive changes combine as a + b + ab/100.
Product constancy links price and consumption, and area changes follow the same successive-change rule.
Growth and depreciation compound over time, with applications to elections, examinations and budgets.

Part 4 of 4

Ratio, Proportion & Partnership

Last reviewed 22 Sept 2026 · 6 min read

Meaning of a ratio

A ratio compares two quantities of the same kind by division: $a : b = \frac{a}{b}$ , where $a$ is the antecedent and $b$ the consequent. A ratio has no unit, so 4 kg : 800 g must first be written in one unit — 4000 g : 800 g = 5 : 1.

Multiplying or dividing both terms by the same non-zero number does not change a ratio: $6 : 9 = 2 : 3$ .
$a : b : c$ means the three quantities are in the proportion $a$ , $b$ , $c$ of one common part $k$ — write them as $ak$ , $bk$ , $c k$ and one equation finds $k$ .

Comparing two ratios

Cross-multiply: $a : b > c : d$ when $a d > b c$ . To compare $5 : 7$ and $7 : 10$ : $5 \times 10 = 50$ and $7 \times 7 = 49$ , so $5 : 7$ is greater.

Derived ratios

Name	For the ratio $a : b$
Duplicate	$a^{2} : b^{2}$
Sub-duplicate	$a : b$
Triplicate	$a^{3} : b^{3}$
Sub-triplicate	$3 a : 3 b$
Inverse	$b : a$
Compounded of $a : b$ and $c : d$	$a c : b d$

Proportion

FormulaProportion and its consequences

Four quantities are in proportion when $a : b = c : d$ , written $a : b :: c : d$ .

a \times d = b \times c (product of extremes = product of means)

Fourth proportional to $a$ , $b$ , $c$ is $\frac{b c}{a}$
Third proportional to $a$ and $b$ is $\frac{b ^{2}}{a}$
Mean proportional between $a$ and $b$ is $ab$
$a$ , $b$ , $c$ are in continued proportion when $b^{2} = a c$

FormulaComponendo and dividendo

If $\frac{a}{b} = \frac{c}{d}$ , then

\frac{a + b}{a - b} = \frac{c + d}{c - d}

This turns an equation such as $\frac{3 x + 2}{3 x - 2} = \frac{5}{3}$ into one step instead of cross-multiplication.

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