Data Interpretation

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Growth rates and successive changes

⏱ 4 min read🧩 5 question types🎯 14 practice Q

The idea in one minute

Growth questions track a value rising year after year: one year's rate, the total rate over several years, or the starting value when only the end is known.

Every version still rests on one line: new value $=$ old value $\times$ growth multiplier.

The growth multiplier

Growth of $r\%$ multiplies by $\left(1 + \dfrac{r}{100}\right)$ .

Revenue 80 crore grows 20%: $80 \times \dfrac{120}{100} = 96$ crore. A 20% fall would multiply by $\dfrac{80}{100}$ .

Rule: Never add percentages of different bases. Convert each to a multiplier and multiply.

Two rises in a row

10% growth twice is not 20%.

\left(\frac{110}{100}\right)^2 = \frac{121}{100} = 1.21 \Rightarrow 21\%

The extra 1% comes from the first rise earning the second year's growth.

Watch: Successive changes of +10% and −10% end below the start: $1.1 \times 0.9 = 0.99$ .

Rate over several years

Over two years, the total growth multiplier is the ratio of end to start: $\dfrac{144}{80} = 1.8$ , so 80% in two years. If asked for the yearly rate assuming equal years, take the square root — but most exam questions want the total rate only.

Tip: End ÷ start gives the total multiplier instantly. 144/80 = 1.8 means +80%.

Working backwards

A value after 50% growth is 45,000. Before the growth it was $\dfrac{45000}{1.5} = 30,000$ .

Divide by the multiplier; never subtract 50% of 45,000.

Example: If 14,400 is 72% of a number, the number is $14400 \div 0.72 = 20,000$ . Same reversal, percentage base.

Which year grew fastest?

App users (thousands): 40, 60, 96, 120.

Year	2021	2022	2023
Growth	$\dfrac{20}{40} = 50\%$	$\dfrac{36}{60} = 60\%$	$\dfrac{24}{96} = 25\%$

2022 grew fastest even though 2023 added 24k users, more than 2021's 20k.

Watch: Rate divides by that year's own base; each year has its own.

Depreciation: growth in reverse

Falling values follow the same multiplier idea with a number below 1.

A machine worth ₹50,000 loses 10% each year. After two years: $50000 \times 0.9 \times 0.9 = ₹40,500$ . The loss is $19\%$ , not $20\%$ , because the second year's 10% acts on a smaller value.

Watch: Straight-line thinking (50,000 − 5,000 − 5,000 = 40,000) is the planted option. Percent decay always shrinks the base.

Chained different rates

A town of 8,000 grows 10% one year, then 5% the next: $8000 \times 1.1 = 8800$ , then $8800 \times 1.05 = 9240$ . The total growth is $\dfrac{9240}{8000} - 1 = 15.5\%$ , and $10\% + 5\% = 15\%$ is close but wrong.

Tip: Multiply step by step in the order the years come; each rate feeds on the previous result.

Comparing two growers

Plant A: 100 → 180, up 80%. Plant B: 50 → 100, up 100%. B grew faster by 20 percentage points, though A added more units.

Tip: "By how many percentage points" subtracts the two rates; "by how much percent more" divides the rates. The exam usually wants the points.

Question types you will see

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

Type 1very common3 practice Q

One year's growth rate

How to spot it:

A table of yearly values; one year's percent growth is asked.

\frac{V_2 - V_1}{V_1} \times 100\%

Method

Take the two values around that year.
Divide the rise by the earlier value.
Multiply by 100.

Why it works:

A rate needs its own base year.

Try this

Revenue of a company (in ₹ crore):

Year	2020	2021	2022	2023
Revenue	80	96	120	144

The growth rate in 2021 over 2020 is:

Show solution

Rise $= 96 - 80 = 16$ .
$\dfrac{16}{80} \times 100 = 20\%$ .

Answer

20%

Type 2common3 practice Q

Growth over two years

How to spot it:

The total percentage change after two successive rises is asked.

\left(1+\frac{a}{100}\right)\left(1+\frac{b}{100}\right)-1

Method

Write each rise as a multiplier.
Multiply the multipliers.
Subtract 1 for the total rate.

Why it works:

Each rise compounds on the one before it.

Try this

An item's price rises 10% in the first year and 10% again in the second year. The total percentage rise is:

Year	Rise
1st	10%
2nd	10%

Show solution

$1.10 \times 1.10 = 1.21$ .
Total rise $= 21\%$ .

Answer

21%

Type 3common2 practice Q

Value before the growth

How to spot it:

The end value and the growth rate are known; the start is asked.

\text{old} = \frac{\text{new}}{1+\frac{r}{100}}

Method

Turn the rate into a multiplier.
Divide the end value by it.
Check by growing the answer back up.

Why it works:

Growth multiplies, so reversal divides.

Try this

A company's revenue grew 50% this year and reached ₹45,000 crore. The revenue before the growth was:

Quantity	Value
Revenue after growth	₹45,000 crore
Growth during the year	+50%

Show solution

Multiplier $= 1.5$ .
$45000 \div 1.5 = 30000$ .

Answer

₹30,000 crore

Type 4common2 practice Q

Fastest growing year in a table

How to spot it:

A value table across years; which year grew at the highest rate.

\max_i \frac{V_{i+1}-V_i}{V_i}

Method

Compute each year's rise.
Divide each by its own base.
Name the year with the top rate.

Why it works:

Equal units added can hide very different rates.

Try this

Users of an app (in thousands):

Year	2020	2021	2022	2023
Users	40	60	96	120

In which year was the growth rate the highest?

Show solution

2021: $20/40 = 50\%$ .
2022: $36/60 = 60\%$ ; 2023: $24/96 = 25\%$ .
Highest: 2022.

Answer

2022

Type 5common2 practice Q

Two entities compared

How to spot it:

Two plants, shops or teams; whose growth was faster, and by how much.

\frac{N_2 - N_1}{N_1} \times 100\% \text{ for each}

Method

Compute each grower's rate.
Compare the two rates.
Points: subtract; percent more: divide.

Why it works:

A bigger absolute gain can still be a smaller rate.

Try this

Production of two plants (in tonnes):

Plant	2022	2023
A	100	180
B	50	100

B's growth rate exceeds A's by how many percentage points?

Show solution

A: $\dfrac{80}{100} = 80\%$ .
B: $\dfrac{50}{50} = 100\%$ .
$100 - 80 = 20$ points.

Answer

20 percentage points

Formula sheet

Growth multiplier

\text{new} = \text{old}\left(1+\frac{r}{100}\right)

Successive growth

r_{total} = \left(1+\frac{a}{100}\right)\left(1+\frac{b}{100}\right)-1

Reverse to original

\text{old} = \frac{\text{new}}{1+\frac{r}{100}}

Total multiplier

\frac{\text{end}}{\text{start}} = 1 + \frac{r_{total}}{100}

Shortcuts that save time

⚡ Multipliers beat repeated percenting

Chain growth as multipliers: +10% then +10% is 1.1 x 1.1 = 1.21. One multiplication, no compounding errors.

Example

A value rises 10% each year for 2 years. Total growth?

Show solution

$1.10 \times 1.10 = 1.21$ .
Total growth $= 21\%$ .

Answer

21%

⚡ End over start

The total growth multiplier over any span is just the last value divided by the first. 144/80 = 1.8 → +80%.

Example

Revenue went from 80 crore to 144 crore in two years. Total growth?

Show solution

$144 \div 80 = 1.8$ .
Total growth $= 80\%$ .

Answer

80%

Mistakes to avoid

Where most students lose marks on this subtopic.

Mistake 01

Adding two successive rates: 10% + 10% = 20%.

Multiply the multipliers: 1.1 x 1.1 = 1.21.

Mistake 02

Subtracting the percent of the final value to reverse a rise.

Divide by the multiplier: 45000 ÷ 1.5.

Mistake 03

Mixing 'percentage points' with 'percent more'.

Points subtract rates; percent more divides them.

Mistake 04

Judging the fastest year by units added.

Divide each rise by that year's own base.

Mistake 05

Averaging yearly rates for the span rate.

Use end ÷ start for the true span multiplier.

Quick revision

Read this the night before the exam.

new $=$ old $\times \left(1+\dfrac{r}{100}\right)$ .
Successive: multiply multipliers; $+10\%,+10\% \to +21\%$ .
$+10\%$ then $-10\%$ ends $1\%$ below the start.
Reverse: divide by the multiplier.
Span rate: end $\div$ start.
Fastest year: each rise over its own base.

Practice: 14 questions

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

Topic test · 10 questions

Suggested time 7 min · wrong answers go to your mistake notebook automatically.

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