Increasing and Decreasing by a Percentage

Given the original value and a percentage, find the new value: find the change and add it or take it away, or multiply the original value by (100 + x)% or (100 − x)% in one step.

Before you start

To find a percentage of a quantity, write the percentage as a fraction or as a decimal and multiply: $35\%$ of $240 is $\frac{35}{100} \times 240 = 84$ , or $0.35 \times 240 = 84$ , so it is $84.

The quantity that a percentage is taken of is the base, and the base is $100\%$ . A percentage can be more than $100\%$ : $115\%$ of a quantity is more than the whole quantity, and $115\% = \frac{115}{100} = 1.15$ .

In a bar model a bar is cut into equal units. If a bar of 10 units stands for $100\%$ , then 1 unit is $10\%$ ; if a bar of 20 units stands for $100\%$ , then 1 unit is $5\%$ .

Increasing a Quantity by a Percentage

When a price of $80 is increased by $15\%$ , the price goes up by $15\%$ of $80. The $80 is the original value. It is the base, so it is $100\%$ . The amount that the price goes up by is the increase, and the price after the increase is the new value.

The increase is always a percentage of the original value: "increased by 15%" means increased by 15% of the value before the change. This stays true even when it is the new value that is known, as in reverse percentages.

Rule: increasing a quantity by a percentage

To increase a quantity by $x\%$ :

Increase $= x\%$ of the original value

New value $=$ original value $+$ increase

Example 1: Increasing a price

A school bag costs $80. Its price is increased by $15\%$ . Find the new price of the school bag.

Solution

The original price, $80, is the base: it is $100\%$ .

Find the increase. It is $15\%$ of the original price: $\frac{15}{100} \times 80 = 12$ , so the increase is $12.

Add the increase to the original price: $80 + 12 = 92$ .

Check: $10\%$ of $80 is $8, and $5\%$ is half of that, $4. So $15\%$ of $80 is $8 + 4 = 12$ dollars, as found.

The new price of the school bag is $92.

Example 2: A pay rise

Mr Tan's monthly salary is $3600. It is increased by $8\%$ . Find (a) the increase in his monthly salary, (b) his new monthly salary.

Solution

The original salary, $3600, is the base: it is $100\%$ .

(a) Increase $= 8\%$ of $3600 = \frac{8}{100} \times 3600 = 288$ , so the increase is $288.

(b) New salary $= 3600 + 288 = 3888$ , so the new salary is $3888.

Mr Tan's monthly salary increases by $288, to $3888.

Decreasing a Quantity by a Percentage

A decrease works in the same way. When a quantity is decreased by $x\%$ , it goes down by $x\%$ of the original value. The amount it goes down by is the decrease, and it is taken away from the original value.

Rule: decreasing a quantity by a percentage

To decrease a quantity by $x\%$ :

Decrease $= x\%$ of the original value

New value $=$ original value $-$ decrease

Example 3: A smaller population

The population of a town was $45\,000$ . Over five years it decreased by $5\%$ . Find the population of the town after the five years.

Solution

The original population, $45\,000$ , is the base: it is $100\%$ .

Find the decrease: $5\%$ of $45\,000 = \frac{5}{100} \times 45\,000 = 2250$ .

Take the decrease away from the original population: $45\,000 - 2250 = 42\,750$ .

Check: $10\%$ of $45\,000$ is 4500, and $5\%$ is half of that, 2250, as found.

After the five years the population of the town was $42\,750$ .

In One Step: the Multiplier

Look again at the school bag of Example 1. The original price is $100\%$ of itself. The new price is the original price and $15\%$ of it more, so the new price is $100\% + 15\% = 115\%$ of the original price. That gives the new price in one step.

The bar model shows why. Draw the original price as a bar of 20 units, so that each unit is $5\%$ of the original price. The increase, $15\%$ , is 3 more units, so the new price is a bar of 23 units: $115\%$ of the original price.

The original price is 20 units, so 1 unit is $80 \div 20 = 4$ dollars.

The new price is 23 units: $23 \times 4 = 92$ , so it is $92, as in Example 1.

In one step: $115\%$ of $80 = \frac{115}{100} \times 80 = 1.15 \times 80 = 92$ .

Why the two ways agree

Finding the increase and adding it works out $80 + 15\%$ of $80$ . But 80 is $100\%$ of 80, so this is $100\%$ of $80 + 15\%$ of $80$ , which is $115\%$ of 80. The two ways find the same $115\%$ of the original value; in one step the percentages are added first.

A decrease works in the same way. Taking $30\%$ of a quantity away leaves $100\% - 30\% = 70\%$ of it: the original value less $30\%$ of it is $70\%$ of it.

Rule: the new value in one step

After an increase of $x\%$ : new value $= (100 + x)\%$ of the original value $= \frac{100 + x}{100} \times$ original value

After a decrease of $x\%$ : new value $= (100 - x)\%$ of the original value $= \frac{100 - x}{100} \times$ original value

The number that the original value is multiplied by, $\frac{100 + x}{100}$ or $\frac{100 - x}{100}$ , is the multiplier. It is more than 1 for an increase and less than 1 for a decrease.

Some changes and their multipliers

Change	New value as a percentage of the original value	Multiplier
An increase of $15\%$	$100\% + 15\% = 115\%$	$1.15$
An increase of $8\%$	$100\% + 8\% = 108\%$	$1.08$
An increase of $2.5\%$	$100\% + 2.5\% = 102.5\%$	$1.025$
An increase of $150\%$	$100\% + 150\% = 250\%$	$2.5$
A decrease of $30\%$	$100\% - 30\% = 70\%$	$0.7$
A decrease of $16\%$	$100\% - 16\% = 84\%$	$0.84$
A decrease of $7.5\%$	$100\% - 7.5\% = 92.5\%$	$0.925$

Write the multiplier with care: an increase of $8\%$ gives $1.08$ , not $1.8$ , and a decrease of $7.5\%$ gives $0.925$ .

Both ways give the same answer, so use the one that suits the question. When the question asks for the increase or the decrease as well as the new value, find the change first. When it asks only for the new value, the multiplier is quicker, and the other way is a good check.

Example 4: Both ways, with a percentage that is not a whole number

Last month the sales of a fashion shop came to $ $12\,400$ . This month its sales increased by $2.5\%$ . Find the shop's sales this month.

Solution

Last month's sales, $ $12\,400$ , are the original value: they are $100\%$ .

Way 1: find the increase and add it. Increase $= 2.5\%$ of $12\,400 = \frac{2.5}{100} \times 12\,400 = 0.025 \times 12\,400 = 310$ . Sales this month $= 12\,400 + 310 = 12\,710$ .

Way 2: in one step. Sales this month are $(100 + 2.5)\% = 102.5\%$ of last month's sales, so the multiplier is $1.025$ , not $1.25$ : $1.025 \times 12\,400 = 12\,710$ .

The two ways agree. Check: $1\%$ of $12\,400$ is 124, so $2.5\%$ is $2.5 \times 124 = 310$ , as found.

The shop's sales this month came to $ $12\,710$ .

Example 5: A school's enrolment

A secondary school had 1250 students in 2025. In 2026 its enrolment increased by $4\%$ . How many students did the school have in 2026?

Solution

The enrolment in 2025, 1250 students, is the original value: it is $100\%$ .

After an increase of $4\%$ , the enrolment is $(100 + 4)\% = 104\%$ of the 2025 enrolment. The multiplier is $\frac{104}{100} = 1.04$ .

Enrolment in 2026 $= 1.04 \times 1250 = 1300$

Check by the other way: $4\%$ of $1250 = \frac{4}{100} \times 1250 = 50$ , and $1250 + 50 = 1300$ .

The school had 1300 students in 2026.

Example 6: A sale price

In a sale at a mall, the price of a pair of shoes, $65, is reduced by $30\%$ . Find the sale price of the shoes.

Solution

The original price, $65, is $100\%$ . Draw it as a bar of 10 units, so that each unit is $10\%$ of the original price. Taking away $30\%$ takes away 3 units and leaves 7 units: the sale price is $100\% - 30\% = 70\%$ of the original price.

In one step: sale price $= 70\%$ of $65 = \frac{70}{100} \times 65 = 0.7 \times 65 = 45.5$

Money is written with two decimal places: $45.50.

Check by the other way: the reduction is $30\%$ of $65 = 0.3 \times 65 = 19.5$ , which is $19.50, and $65 - 19.50 = 45.50$ .

The sale price of the shoes is $45.50.

Example 7: Water used from a tank

A tank holds 78 litres of water. $21\%$ of the water is used. How much water is left in the tank?

Solution

The water at the start, 78 litres, is the original value: it is $100\%$ .

Using $21\%$ of the water decreases it by $21\%$ , so the water left is $(100 - 21)\% = 79\%$ of 78 litres. The multiplier is $0.79$ .

Water left $= 0.79 \times 78 = 61.62$

Check by the other way: $21\%$ of $78 = 0.21 \times 78 = 16.38$ , and $78 - 16.38 = 61.62$ .

61.62 litres of water are left in the tank.

Increases of 100% or More

An increase can be more than $100\%$ of the original value. An increase of $100\%$ adds the whole of the original value again, so the new value is $100\% + 100\% = 200\%$ of the original value: the quantity is doubled. An increase of $150\%$ makes the new value $100\% + 150\% = 250\%$ of the original value, which is 2.5 times the original value.

A decrease is different. A decrease of $100\%$ takes away the whole of the original value and leaves nothing. So a price, a mass or a number of people cannot be decreased by more than $100\%$ .

Example 8: An increase of 150%

A new town had a population of $12\,000$ in 2010. By 2025 its population had increased by $150\%$ . Find the population of the town in 2025.

Solution

The population in 2010, $12\,000$ , is the original value: it is $100\%$ . Draw it as a bar of 2 units, so that each unit is $50\%$ of the 2010 population. The increase, $150\%$ , is 3 more units, so the population in 2025 is a bar of 5 units: $250\%$ of the 2010 population.

In one step: the multiplier is $\frac{100 + 150}{100} = \frac{250}{100} = 2.5$ .

Population in 2025 $= 2.5 \times 12\,000 = 30\,000$

Check by the other way: the increase is $150\%$ of $12\,000 = 1.5 \times 12\,000 = 18\,000$ , and $12\,000 + 18\,000 = 30\,000$ . On the bar model, 1 unit is $12\,000 \div 2 = 6000$ , and 5 units are $5 \times 6000 = 30\,000$ .

The population of the town in 2025 was $30\,000$ .

Increased by, and increased to

"Increased by $150\%$ " means that the increase is $150\%$ of the original value, so the new value is $250\%$ of the original value: multiply by $2.5$ .

"Increased to $150\%$ of the original value" means that the new value is $150\%$ of the original value, an increase of $50\%$ : multiply by $1.5$ .

Read the question for the small word.

Answers That Are Not Exact

Give an amount of money to the nearest cent, and any other answer that is not exact correct to 3 significant figures, unless the question says otherwise. Keep the full calculator value in the working and round only at the end, writing $\approx$ before the rounded value.

Example 9: Rounding the new value

(a) A mobile phone plan costs $28.90 a month. Its price is increased by $7.5\%$ . Find the new price, to the nearest cent.
(b) A puppy has a mass of 2.45 kg. In a month its mass increases by $38\%$ . Find its new mass, correct to 3 significant figures.

Solution

(a) The original price, $28.90, is $100\%$ . The new price is $(100 + 7.5)\% = 107.5\%$ of it, so the multiplier is $1.075$ .

New price $= 1.075 \times 28.90 = 31.0675$

To the nearest cent, look at the digit after the cents. It is 7, so round up: $31.0675 \approx 31.07$ .

(b) The original mass, 2.45 kg, is $100\%$ . The new mass is $(100 + 38)\% = 138\%$ of it, so the multiplier is $1.38$ .

New mass $= 1.38 \times 2.45 = 3.381$

Correct to 3 significant figures, look at the fourth figure. It is 1, so round down: $3.381 \approx 3.38$ .

(a) The new price is $31.07 a month, to the nearest cent. (b) The new mass of the puppy is 3.38 kg, correct to 3 significant figures.

Common Mistakes

Taking the percentage of the wrong value. The increase or the decrease is a percentage of the original value, the value before the change, never of the new value.
Adding the percentage as if it were an amount. Increasing $80 by $15\%$ does not give $80 + 15 = 95$ dollars. Find $15\%$ of $80 first, which is $12, so the new price is $92.
Stopping at the change. In Example 6, $0.3 \times 65 = 19.5$ gives the reduction, $19.50, not the sale price. The sale price is $0.7 \times 65 = 45.5$ , which is $45.50.
Writing the wrong multiplier. An increase of $5\%$ gives the multiplier $1.05$ , not $1.5$ (which is an increase of $50\%$ ). A decrease of $5\%$ gives $0.95$ , not $0.05$ (which gives the decrease, not the new value).
Mixing up "increased by" and "increased to". An increase of $150\%$ multiplies the original value by $2.5$ . A new value that is $150\%$ of the original value is the original value multiplied by $1.5$ .
Rounding in the middle. Keep the calculator value until the last line, and round money to the nearest cent or other answers to 3 significant figures only at the end.

Practice

Work each question on paper before you open its solution. Where a question asks for both ways, use both: they should agree.

Question 1

Increase 60 by $25\%$ , (a) by finding the increase and adding it, (b) in one step, with the multiplier.

ShowHide solution

The original value, 60, is $100\%$ . As a bar model: 60 is a bar of 4 units, so that each unit is $25\%$ of 60, and the new value is a bar of 5 units, $125\%$ of 60.

(a) Increase $= 25\%$ of $60 = \frac{25}{100} \times 60 = 15$ . New value $= 60 + 15 = 75$ .

(b) New value $= (100 + 25)\% = 125\%$ of 60: $1.25 \times 60 = 75$ .

On the bar model, 1 unit is $60 \div 4 = 15$ , and 5 units are $5 \times 15 = 75$ .

The new value is 75.

Question 2

Decrease 240 by $35\%$ , (a) by finding the decrease and taking it away, (b) in one step, with the multiplier.

ShowHide solution

The original value, 240, is $100\%$ .

(a) Decrease $= 35\%$ of $240 = \frac{35}{100} \times 240 = 84$ . New value $= 240 - 84 = 156$ .

(b) The new value is $(100 - 35)\% = 65\%$ of 240: $0.65 \times 240 = 156$ .

The new value is 156.

Question 3

Write down the multiplier for each change.
(a) An increase of $12\%$ (b) A decrease of $12\%$ (c) An increase of $4.5\%$ (d) A decrease of $0.5\%$ (e) An increase of $200\%$

ShowHide solution

(a) $100\% + 12\% = 112\%$ , so the multiplier is $\frac{112}{100} = 1.12$ .

(b) $100\% - 12\% = 88\%$ , so the multiplier is $\frac{88}{100} = 0.88$ .

(d) $100\% - 0.5\% = 99.5\%$ , so the multiplier is $\frac{99.5}{100} = 0.995$ .

(e) $100\% + 200\% = 300\%$ , so the multiplier is $\frac{300}{100} = 3$ : the new value is three times the original value.

(a) $1.12$ (b) $0.88$ (c) $1.045$ (d) $0.995$ (e) $3$

Question 4

Hui Min's monthly salary is $2850. She is given a pay rise of $6\%$ . Find (a) her pay rise in dollars, (b) her new monthly salary.

ShowHide solution

The original salary, $2850, is $100\%$ .

(a) Pay rise $= 6\%$ of $2850 = \frac{6}{100} \times 2850 = 171$ , so the pay rise is $171.

(b) New salary $= 2850 + 171 = 3021$ , so it is $3021.

Check in one step: $1.06 \times 2850 = 3021$ .

Hui Min's pay rise is $171 a month, and her new monthly salary is $3021.

Question 5

A school had 1480 students last year. This year its enrolment fell by $5\%$ . How many students does the school have this year?

ShowHide solution

Last year's enrolment, 1480 students, is the original value: it is $100\%$ .

After a decrease of $5\%$ , the enrolment is $(100 - 5)\% = 95\%$ of last year's: the multiplier is $0.95$ .

Enrolment this year $= 0.95 \times 1480 = 1406$

Check by the other way: $10\%$ of 1480 is 148, so $5\%$ is 74, and $1480 - 74 = 1406$ .

The school has 1406 students this year.

Question 6

In a sale, a jacket marked $89.90 is sold at $20\%$ off the marked price. Find the sale price of the jacket.

ShowHide solution

The marked price, $89.90, is the original value: it is $100\%$ .

The sale price is $(100 - 20)\% = 80\%$ of the marked price: $0.8 \times 89.90 = 71.92$ .

Check by the other way: $20\%$ of $89.90 = 0.2 \times 89.90 = 17.98$ , and $89.90 - 17.98 = 71.92$ .

The sale price of the jacket is $71.92.

Question 7

The price of a pair of earphones before GST is $18.80. GST of $9\%$ is added to the price. Find the price of the earphones with GST, to the nearest cent.

ShowHide solution

The price before GST, $18.80, is the original value: it is $100\%$ .

Adding $9\%$ makes the price $(100 + 9)\% = 109\%$ of the price before GST: the multiplier is $1.09$ .

Price with GST $= 1.09 \times 18.80 = 20.492$

To the nearest cent, look at the digit after the cents. It is 2, so round down: $20.492 \approx 20.49$ .

The price of the earphones with GST is $20.49, to the nearest cent.

Question 8

A plant is 36.5 cm tall. In a week its height increases by $12\%$ . Find its new height, correct to 3 significant figures.

ShowHide solution

The original height, 36.5 cm, is $100\%$ .

The new height is $(100 + 12)\% = 112\%$ of it: the multiplier is $1.12$ .

New height $= 1.12 \times 36.5 = 40.88$

Correct to 3 significant figures, look at the fourth figure. It is 8, so round up: $40.88 \approx 40.9$ .

The new height of the plant is 40.9 cm, correct to 3 significant figures.

Question 9

A museum had 3600 visitors a week. After a new gallery opened, the number of visitors a week increased by $175\%$ . How many visitors a week did the museum have after the gallery opened?

ShowHide solution

The number of visitors before, 3600, is the original value: it is $100\%$ . As a bar model: 3600 is a bar of 4 units, so that each unit is $25\%$ of it. The increase, $175\%$ , is 7 more units, so the number after is a bar of 11 units, $275\%$ of the number before.

In one step: the multiplier is $\frac{100 + 175}{100} = \frac{275}{100} = 2.75$ .

Visitors a week after $= 2.75 \times 3600 = 9900$

Check by the other way: the increase is $175\%$ of $3600 = 1.75 \times 3600 = 6300$ , and $3600 + 6300 = 9900$ . On the bar model, 1 unit is $3600 \div 4 = 900$ , and 11 units are $11 \times 900 = 9900$ .

After the gallery opened, the museum had 9900 visitors a week.

Question 10

Shop A sold a bicycle for $240, and Shop B sold the same bicycle for $250. Then Shop A raised its price by $15\%$ and Shop B raised its price by $10\%$ . Which shop now sells the bicycle for less, and by how much?

ShowHide solution

Each shop's own old price is its original value, $100\%$ .

Shop A: new price $= 115\%$ of $240 = 1.15 \times 240 = 276$ , so $276.

Shop B: new price $= 110\%$ of $250 = 1.1 \times 250 = 275$ , so $275.

Difference: $276 - 275 = 1$ , so $1.

Shop A started $10 cheaper, but its rise, $36, is $11 more than Shop B's rise of $25.

Shop B now sells the bicycle for less, by $1.

Question 11

A book costs $40. Its price is increased by $150\%$ .
(a) Siti says that the new price is $60. What mistake has she made? Find the new price.
(b) Ravi says that the price of a book could also be decreased by $120\%$ . Explain why it could not.

ShowHide solution

(a) $60 is $150\%$ of $40, since $1.5 \times 40 = 60$ . Siti has made the new price $150\%$ of the original price, which is an increase of only $50\%$ .

An increase of $150\%$ makes the new price $100\% + 150\% = 250\%$ of the original price: $2.5 \times 40 = 100$ .

Check: the increase is $150\%$ of $40 = 1.5 \times 40 = 60$ , and $40 + 60 = 100$ .

(b) A decrease of $120\%$ would take away $120\%$ of $40, which is $1.2 \times 40 = 48$ dollars. That is more than the whole price: $40 - 48 = -8$ , and a price cannot be less than nothing. A price can be decreased by $100\%$ at most, which makes it free.

(a) Siti found $150\%$ of the price instead of adding $150\%$ of it to the price. The new price is $100. (b) A decrease of $120\%$ would take away more than the whole price.

Question 12

A hawker sells 240 plates of chicken rice a day at $4.50 a plate. She raises the price by $10\%$ , and then sells $15\%$ fewer plates a day.
(a) Find the new price of a plate and the new number of plates she sells a day.
(b) Does she now take in more money or less money a day than before, and by how much?

ShowHide solution

The two percentages are of different quantities, one of the price and one of the number of plates, so each is worked on its own original value.

(a) The original price, $4.50, is $100\%$ . The new price is $110\%$ of it: $1.1 \times 4.50 = 4.95$ , so $4.95 a plate.

The original number of plates, 240, is $100\%$ . The new number is $(100 - 15)\% = 85\%$ of it: $0.85 \times 240 = 204$ plates.

(b) Before: $240 \times 4.50 = 1080$ , so $1080 a day.

Now: $204 \times 4.95 = 1009.8$ , so $1009.80 a day.

Difference: $1080 - 1009.80 = 70.20$ , so $70.20.

(a) The new price is $4.95 a plate, and she sells 204 plates a day. (b) She now takes in less money: $70.20 a day less than before.

Key Points

The original value is the base, $100\%$ . An increase or a decrease of $x\%$ is $x\%$ of the original value.
Increasing by $x\%$ : new value $=$ original value $+$ $x\%$ of the original value.
Decreasing by $x\%$ : new value $=$ original value $-$ $x\%$ of the original value.
In one step: new value $= (100 + x)\%$ of the original value after an increase, and $(100 - x)\%$ of it after a decrease. The multiplier is more than 1 for an increase and less than 1 for a decrease.
The two ways agree, because the original value and $x\%$ of it together are $(100 + x)\%$ of it.
An increase of $100\%$ doubles a quantity, and an increase of $150\%$ makes it 2.5 times as large. A price, a mass or a number of people cannot be decreased by more than $100\%$ .
Money to the nearest cent, other answers that are not exact to 3 significant figures; round only at the end.