Percentage increase / decrease & successive change — notes, question types & practice · Percentage

01

Change is measured against the old value

Per cent change $= \dfrac{\text{new} - \text{old}}{\text{old}} \times 100$ . The old value sits below the line.

A shop's sales rose from 18,000 to 20,700. Change $= 2700$ , and $\dfrac{2700}{18000} \times 100 = 15\%$ .

Rule: Old value below, always. Ask yourself "which number came first?" before dividing.

Watch: 250 growing to 300 is $+20\%$ , but 200 growing to 250 is $+25\%$ . The same 50 gives different percents because the bases differ.

02

Multipliers: the fastest method

Do not add and subtract in steps. Multiply by one fraction per change.

Change	Multiplier	Change	Multiplier
$+10\%$	$11/10$	$-10\%$	$9/10$
$+20\%$	$6/5$	$-20\%$	$4/5$
$+25\%$	$5/4$	$-25\%$	$3/4$
$+50\%$	$3/2$	$-50\%$	$1/2$

$18000 \times \dfrac{23}{20} = 20700$ : a $15\%$ rise in one multiplication, and $+15\%$ or $-15\%$ both come from the chips $23/20$ and $17/20$ .

To undo a change, divide by the same multiplier. $-20\%$ turned 50,000 into 40,000, so $40000 \div \dfrac{4}{5} = 50000$ .

03

Two changes one after the other

Multiply the two multipliers, or use one formula.

\text{net change} = a + b + \frac{ab}{100}

Here $a$ and $b$ carry their own signs. $+15\%$ then $+12\%$ : $15 + 12 + \dfrac{15 \times 12}{100} = 28.8\%$ , a rise.

With multipliers: $\dfrac{23}{20} \times \dfrac{28}{25} = \dfrac{161}{125} = 1.288$ , the same $+28.8\%$ .

$+20\%$ then $-10\%$ : $\dfrac{6}{5} \times \dfrac{9}{10} = \dfrac{27}{25}$ , a net rise of $8\%$ .

Careful: Never just add the percents. $20\%$ up and $10\%$ down is not $10\%$ up.

04

The round trip: x% up then x% down

Raise a value by $x\%$ and then cut it by $x\%$ . It does not come back.

\text{net loss} = \left(\frac{x}{10}\right)^2 \%

$10\%$ up then $10\%$ down: $\dfrac{11}{10} \times \dfrac{9}{10} = \dfrac{99}{100}$ , a $1\%$ loss. $20\%$ up then $20\%$ down: $\dfrac{6}{5} \times \dfrac{4}{5} = \dfrac{24}{25}$ , a $4\%$ loss.

Note: The order does not matter. Up first or down first, the loss is the same $x^2/100$ per cent.

05

Getting the original back

If a value fell or rose and the final value is given, divide by the multiplier to undo the change.

A number became 8,100 after two yearly falls of $10\%$ . Two chips of $9/10$ were used, so the original $= 8100 \times \dfrac{10}{9} \times \dfrac{10}{9} = 10000$ .

A number is decreased by $10\%$ , then by $20\%$ , and becomes 720. Original $= 720 \times \dfrac{10}{9} \times \dfrac{5}{4} = 1000$ .

Watch: To undo $+20\%$ , divide by $1.2$ . Subtracting $20\%$ back is wrong.

06

When the top and bottom of a fraction change

A fraction's value changes by the top's chip divided by the bottom's chip.

The numerator rises $50\%$ (chip $3/2$ ) and the denominator falls $25\%$ (chip $3/4$ ):

\text{new value} = \text{old} \times \frac{3/2}{3/4} = \text{old} \times 2

The fraction doubles.

Example: Numerator $+20\%$ (chip $6/5$ ), denominator $-10\%$ (chip $9/10$ ): value $\times \dfrac{6/5}{9/10} = \dfrac{4}{3}$ , a $33\dfrac{1}{3}\%$ rise.

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

Percentage increase or decrease

How to spot it:

Two values of the same thing are given and the per cent change is asked.

\frac{\text{new} - \text{old}}{\text{old}} \times 100

Method

Find the difference between the new and old values.
Put the old value below.
Multiply by 100 and attach increase or decrease.

Why it works:

Per cent change states the difference as a part of the starting value.

Try this

A shop's monthly sales rose from Rs 18,000 to Rs 20,700. What was the percentage increase?

Show solution

Difference $= 20700 - 18000 = 2700$ .
$\dfrac{2700}{18000} \times 100 = 15$ .

Answer

15%

Type 2very common3 practice Q

Two successive changes

How to spot it:

A value changes twice in a row and the net change is asked: '+20% and then -10%'.

a + b + \frac{ab}{100}

Method

Write both percents with their signs.
Add them and add ab/100 as well.
State rise or fall from the sign.
Cross-check by multiplying the two multipliers.

Why it works:

The second change acts on the changed value, so a small cross term ab/100 appears.

Try this

A number is increased by 20% and then decreased by 10%. What is the net percentage change?

Show solution

$a = 20$ , $b = -10$ .
$20 - 10 + \dfrac{20 \times (-10)}{100} = 10 - 2 = 8$ .
Chips: $\dfrac{6}{5} \times \dfrac{9}{10} = \dfrac{27}{25}$ , so $+8\%$ .

Answer

8% increase

Type 3common2 practice Q

Same x% up and then down

How to spot it:

A value rises by x% and falls by the same x%; the net effect is asked.

\left(\frac{x}{10}\right)^2 \% \text{ loss}

Method

Square x.
Divide by 100.
State it as a net loss.

Why it works:

The two chips multiply to $1 - \dfrac{x^2}{10000}$ , always a shade below 1.

Try this

The price of an article is raised by 20% and then reduced by 20%. What is the net change in price?

Show solution

Loss $= \dfrac{20 \times 20}{100} = 4$ per cent.
Chips: $\dfrac{6}{5} \times \dfrac{4}{5} = \dfrac{24}{25}$ .

Answer

4% decrease

Type 4very common3 practice Q

Find the original value

How to spot it:

A value after some change is given and the value before the change is asked.

\text{original} = \text{final} \times \frac{100}{100 \pm x}

Method

Write the multiplier for each change that happened.
Divide the final value by the product of the multipliers.
Check by applying the changes to your answer.

Why it works:

Changes were multiplications, so undoing them is division by the same multipliers.

Try this

A number is first decreased by 10% and then decreased by 20%. The result is 720. What was the original number?

Show solution

Chips used: $\dfrac{9}{10}$ and $\dfrac{4}{5}$ .
$720 \times \dfrac{10}{9} = 800$ .
$800 \times \dfrac{5}{4} = 1000$ .

Answer

1000

Type 5occasional2 practice Q

Numerator and denominator both change

How to spot it:

A fraction's top and bottom are each changed by a percent and the effect on the fraction is asked.

\text{new} = \text{old} \times \frac{\text{top chip}}{\text{bottom chip}}

Method

Write the chip for the numerator's change.
Write the chip for the denominator's change.
Multiply the old fraction by top chip over bottom chip.

Why it works:

The top scales the fraction up and the bottom scales it down, so the chips divide.

Try this

In a fraction the numerator is increased by 50% and the denominator is decreased by 25%. The new fraction is how many times the old one?

Show solution

Top chip $= \dfrac{3}{2}$ , bottom chip $= \dfrac{3}{4}$ .
Factor $= \dfrac{3/2}{3/4} = 2$ .

Answer

2 times

Type 6occasional

Three or more successive changes

How to spot it:

Three changes are applied one after another and the net effect is asked.

Method

Write a multiplier for each change.
Multiply all the multipliers together.
Compare the product with 1 to get the net per cent.

Why it works:

Multipliers keep working for any number of changes; only the multiplication grows.

Try this

A value is increased by 10%, then increased by 20%, and then decreased by 25%. What is the net percentage change?

Show solution

Chips: $\dfrac{11}{10}$ , $\dfrac{6}{5}$ , $\dfrac{3}{4}$ .
Product $= \dfrac{11}{10} \times \dfrac{6}{5} \times \dfrac{3}{4} = \dfrac{99}{100}$ .
$0.99$ means a $1\%$ fall.

Answer

1% decrease

08

Formula sheet

Per cent change

\frac{\text{new} - \text{old}}{\text{old}} \times 100

Old value below the line.

Multiplier

\text{new} = \text{old} \times \frac{100 \pm x}{100}

Plus for a rise, minus for a fall.

Successive changes

a + b + \frac{ab}{100}

a and b carry their own signs.

Same x% up and down

\text{net loss} = \frac{x^2}{100}\%

x% rise followed by x% cut.

Fraction with both parts changed

\text{new} = \text{old} \times \frac{\text{top chip}}{\text{bottom chip}}

09

Shortcuts that save time

⚡ One multiplication per change

Replace every rise or fall with its multiplier fraction and multiply once. This replaces three lines of working.

Example

A price fell by 12.5% to Rs 3,500. What was the price before the fall?

Show solution

$-12.5\%$ means the chip $\dfrac{7}{8}$ .
Original $= 3500 \times \dfrac{8}{7} = 4000$ .

Answer

Rs 4,000

⚡ a + b + ab/100 for two changes

Two changes in a row combine by this one formula. Keep the signs of both percents.

Example

A value rises by 15% and then by 12%. What is the net percentage change?

Show solution

$15 + 12 + \dfrac{15 \times 12}{100}$
$= 27 + 1.8 = 28.8$

Answer

28.8% increase

⚡ Round trip loses x squared over 100

A rise of x% followed by a cut of x% always ends below the start, by exactly $\dfrac{x^2}{100}$ per cent.

Example

A price is raised by 20% and then reduced by 20%. What is the net change?

Show solution

Loss $= \dfrac{20 \times 20}{100} = 4$ per cent.
Chips check: $\dfrac{6}{5} \times \dfrac{4}{5} = \dfrac{24}{25}$ .

Answer

4% decrease

10

Mistakes to avoid

Where most students lose marks on this subtopic.

Mistake 01

Adding percents of different bases: $20\%$ up then $10\%$ down called $10\%$ up.

Combine with multipliers or $a + b + \dfrac{ab}{100}$ with signs.

Mistake 02

Measuring change against the new value.

The old value is the base. Divide the difference by the old value.

Mistake 03

Undoing a $20\%$ rise by subtracting $20\%$ .

Divide by the multiplier: $\text{new} \div 1.2$ .

Mistake 04

Undoing a fall with the wrong chip, e.g. $\times 0.9$ for a $10\%$ fall.

A fall used $9/10$ ; undo it with $\times \dfrac{10}{9}$ .

Mistake 05

Calling an up-then-down round trip 'no change'.

It always loses $\dfrac{x^2}{100}$ per cent.

Mistake 06

Applying $+10\%$ twice and calling it $+20\%$ .

$1.1 imes 1.1 = 1.21$ , a $21\%$ rise.

11

Quick revision

Read this the night before the exam.

Per cent change $=$ difference $\div$ old $\times 100$ .
Rise $x\%$ : multiply by $\dfrac{100+x}{100}$ ; fall: $\dfrac{100-x}{100}$ .
Two changes: multiply the multipliers, or $a + b + \dfrac{ab}{100}$ .
$x\%$ up then $x\%$ down $= \dfrac{x^2}{100}\%$ net loss.
Undo a change by dividing by the same multiplier.
Fraction value $\times$ top chip $\div$ bottom chip.

12

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 5 min · wrong answers go to your mistake notebook automatically.