Finding the Percentage Increase or Decrease

How much has a quantity gone up or down, compared with what it was before? The change as a percentage of the original value: increases, decreases, increases of more than 100%, answers that are not exact, and values in different units.

Before you start

To express one quantity as a percentage of another, divide it by the quantity it is compared with, called the base, and multiply by 100%. For example, 18 marks out of 80 marks is 1880×100%=22.5%\frac{18}{80} \times 100\% = 22.5\% of the full marks.

To increase a quantity by a given percentage, add that percentage of it: $50 increased by 20% is $50 + $10 = $60, which is 120% of $50. This lesson goes the other way: the value before the change and the value after it are known, and the percentage is to be found.

An answer that is not exact is given correct to 3 significant figures, with "≈" in front of it: 15.384…≈15.415.384\ldots \approx 15.4. Keep the full calculator value until the last line.

Before one quantity is divided by another, both must be in the same unit of measurement: 1 kg = 1000 g, 1 litre = 1000 ml, 1 min = 60 s and $1 = 100 cents.

Percentage Increase

When a quantity goes up, its original value is its value before the change and its new value is its value after the change. The increase is how much it has gone up by: the new value minus the original value.

An increase of $5 is a lot on a $10 meal and very little on a $1000 computer. So the size of an increase is judged by comparing it with the original value. The percentage increase is the increase expressed as a percentage of the original value.

Rule: percentage increase

Increase=New value−Original value\text{Increase} = \text{New value} - \text{Original value}

Percentage increase=IncreaseOriginal value×100%\text{Percentage increase} = \dfrac{\text{Increase}}{\text{Original value}} \times 100\%

The original value is the base: it stands for 100%.

Example 1: A rise in salary

Mr Tan's monthly salary rose from $3500 to $3710. Find the percentage increase in his salary.

Solution

The original value is his salary before the rise, $3500. The new value is $3710. In the drawing the original salary is the whole bar, 100%, and the increase is the extra piece at the end of the new bar.

Two bars drawn to the same scale. The original bar is 3500 dollars, 100 percent. The new bar is 3710 dollars: the original bar with a short extra piece at its end, marked increase 210 dollars, which is an unknown percentage of 3500 dollars.OriginalNew$3500 (100%)$3710increase $210 = ?% of $3500

Increase=3710−3500=210\text{Increase} = 3710 - 3500 = 210, so the increase is $210.

Express the increase as a percentage of the original salary: 2103500×100%=6%\frac{210}{3500} \times 100\% = 6\%

Check by working forward. After a 6% increase the new salary is 100%+6%=106%100\% + 6\% = 106\% of the original salary, and 1.06×3500=37101.06 \times 3500 = 3710. This is the new salary given in the question.

An increase of 6% means that the salary went up by $6 for every $100 of the original salary.

The percentage increase in Mr Tan's salary is 6%.

Percentage Decrease

When a quantity goes down, the decrease is the original value minus the new value. The percentage decrease is the decrease expressed as a percentage of the original value. The base is again the original value, the value before the change.

In this lesson a fall is always described by the words "percentage decrease" and a positive number: "the price decreased by 16%", or "a percentage decrease of 16%". The word "decrease" already says that the value went down, so no minus sign is written in front of the percentage.

Rule: percentage decrease

Decrease=Original value−New value\text{Decrease} = \text{Original value} - \text{New value}

Percentage decrease=DecreaseOriginal value×100%\text{Percentage decrease} = \dfrac{\text{Decrease}}{\text{Original value}} \times 100\%

Example 2: A lower price

The price of a laptop at a shop in a mall fell from $1250 to $1050. Find the percentage decrease in its price.

Solution

The original price is $1250 and the new price is $1050. In the drawing the new bar is shorter than the original bar, and the dashed piece is the decrease.

Two bars drawn to the same scale. The original bar is 1250 dollars, 100 percent. The new bar is 1050 dollars, shorter than the original bar; the missing piece at its end is drawn dashed and marked decrease 200 dollars, which is an unknown percentage of 1250 dollars.OriginalNew$1250 (100%)$1050decrease $200 = ?% of $1250

Decrease=1250−1050=200\text{Decrease} = 1250 - 1050 = 200, so the decrease is $200.

Express the decrease as a percentage of the original price: 2001250×100%=16%\frac{200}{1250} \times 100\% = 16\%

Check by working forward. After a 16% decrease the new price is 100%−16%=84%100\% - 16\% = 84\% of the original price, and 0.84×1250=10500.84 \times 1250 = 1050.

The percentage decrease in the price of the laptop is 16%.

Example 3: Fewer shows

A magician gave 250 shows last year and 100 shows this year. Find the percentage decrease in the number of his shows.

Solution

The original value is last year's number of shows, 250. The new value is this year's number, 100.

Decrease=250−100=150\text{Decrease} = 250 - 100 = 150 shows

Express the decrease as a percentage of last year's number: 150250×100%=60%\frac{150}{250} \times 100\% = 60\%

Check: after a 60% decrease, 40% of the shows are left, and 0.4×250=1000.4 \times 250 = 100. This is this year's number.

The number of the magician's shows decreased by 60%.

How large can a decrease be?

A price, a mass or a number of shows cannot fall by more than all of it, so the decrease is never more than the original value, and the percentage decrease is at most 100%. A decrease of 100% means that the quantity has fallen to nothing.

An increase has no such limit, as the section on increases of more than 100% shows.

Always Divide by the Original Value

A percentage change compares the change with the value before the change. Dividing by the new value is the commonest mistake in this topic. It gives a different answer, because the new value is a different base.

Example 4: The wrong base

At a hawker stall the price of a bowl of fishball noodles went up from $4 to $5. Ali says that the percentage increase is 20%. Find the correct percentage increase and explain Ali's mistake.

Solution

The original price is $4 and the new price is $5.

Two bars drawn to the same scale. The original bar is 4 dollars, 100 percent. The new bar is 5 dollars: the original bar with an extra piece at its end, marked increase 1 dollar, which is an unknown percentage of 4 dollars.OriginalNew$4 (100%)$5increase $1 = ?% of $4

Increase=5−4=1\text{Increase} = 5 - 4 = 1, so the increase is $1.

Express the increase as a percentage of the original price: 14×100%=25%\frac{1}{4} \times 100\% = 25\%

Ali divided by the new price: 15×100%=20%\frac{1}{5} \times 100\% = 20\%. That is the increase as a percentage of the new price, $5. A percentage increase is a percentage of the original price.

Working forward shows which answer is right. Increasing $4 by 20% gives 120% of $4, and 1.2×4=4.81.2 \times 4 = 4.8, which is $4.80, not $5. Increasing $4 by 25% gives 125% of $4, and 1.25×4=51.25 \times 4 = 5, which is $5.

The percentage increase is 25%. Ali found the increase as a percentage of the new price, $5, instead of the original price, $4.

Example 5: Find the original value first

After a rise of $9, a ticket to a school concert costs $54. Find the percentage increase in the price of the ticket.

Solution

The $54 is the new price, after the rise. The original price is not given, so find it first: 54−9=4554 - 9 = 45, so the original price was $45.

Express the increase as a percentage of the original price: 945×100%=20%\frac{9}{45} \times 100\% = 20\%

Check: 120% of $45 is 1.2×45=541.2 \times 45 = 54, which is $54.

Dividing by the price given in the question, 954×100%≈16.7%\frac{9}{54} \times 100\% \approx 16.7\%, would use the wrong base, because $54 is the new price.

The percentage increase in the price of the ticket is 20%.

Which value is the original?

The original value is the value at the start of the change that the question asks about. When a price rises from $4 to $5, the original value is $4. If the price later falls from $5 back to $4, the original value of that fall is $5, so the fall is a decrease of 15×100%=20%\frac{1}{5} \times 100\% = 20\%. The change is $1 both times, but the bases are different, so the percentages are different.

Answers That Are Not Exact

Often the division does not come out exactly. Then work out the percentage on a calculator, keep the full value, and give the answer correct to 3 significant figures, unless the question asks for another degree of accuracy.

Example 6: A school's enrolment

The enrolment of a secondary school rose from 1300 students to 1500 students. Find the percentage increase in the enrolment.

Solution

Increase=1500−1300=200\text{Increase} = 1500 - 1300 = 200 students

Express the increase as a percentage of the original enrolment: 2001300×100%=15.384615…%\frac{200}{1300} \times 100\% = 15.384615\ldots\%

Correct to 3 significant figures, 15.384615…%≈15.4%15.384615\ldots\% \approx 15.4\%.

Do not round the fraction first. 2001300=0.153846…\frac{200}{1300} = 0.153846\ldots, and rounding it to 0.15 before multiplying by 100% gives 15%, which is not correct to 3 significant figures.

The enrolment increased by about 15.4% (correct to 3 significant figures).

Example 7: Fewer visitors

A library had 1260 visitors on one Saturday and 1085 visitors on the next Saturday. Find the percentage decrease in the number of visitors.

Solution

The original value is the number of visitors on the first Saturday, 1260.

Decrease=1260−1085=175\text{Decrease} = 1260 - 1085 = 175 visitors

Express the decrease as a percentage of the first Saturday's number: 1751260×100%=13.888…%\frac{175}{1260} \times 100\% = 13.888\ldots\%

Correct to 3 significant figures, 13.888…%≈13.9%13.888\ldots\% \approx 13.9\%.

The number of visitors decreased by about 13.9% (correct to 3 significant figures).

Increases of More Than 100%

An increase can be larger than the original value. Then the increase is more than 100% of the original value, and the percentage increase is more than 100%.

When a quantity doubles, the increase is equal to the original value, so the percentage increase is 100%. When a quantity becomes three times as large, the increase is twice the original value, so the percentage increase is 200%.

Example 8: Savings that more than doubled

Ravi's savings grew from $120 to $300. Find the percentage increase in his savings.

Solution

The original value is $120. In the drawing the new bar is more than twice as long as the original bar, so expect an answer of more than 100%.

Two bars drawn to the same scale. The original bar is 120 dollars, 100 percent. The new bar is 300 dollars: the original bar with an extra piece at its end that is longer than the original bar, marked increase 180 dollars, which is an unknown percentage of 120 dollars.OriginalNew$120 (100%)$300increase $180 = ?% of $120

Increase=300−120=180\text{Increase} = 300 - 120 = 180, so the increase is $180.

Express the increase as a percentage of the original savings: 180120×100%=150%\frac{180}{120} \times 100\% = 150\%

Check by working forward. After an increase of 150% the savings are 100%+150%=250%100\% + 150\% = 250\% of the original savings, and 2.5×120=3002.5 \times 120 = 300.

Do not mix up the two percentages. The new savings are 250% of the original savings, because 300120×100%=250%\frac{300}{120} \times 100\% = 250\%. The increase is 150% of the original savings. The percentage increase is the second of these.

The percentage increase in Ravi's savings is 150%.

Values in Different Units of Measurement

When the original value and the new value are given in different units of measurement, write them in the same unit first. The smaller unit usually gives whole numbers. Then find the change and the percentage as before. The answer is a percentage of the original value, with no unit of measurement.

Example 9: Cents and dollars

The price of a cup of teh at a school canteen rose from 85 cents to $1.10. Find the percentage increase in the price.

Solution

Write both prices in cents: $1.10 = 110 cents.

Increase=110−85=25\text{Increase} = 110 - 85 = 25 cents

Express the increase as a percentage of the original price: 2585×100%=29.411…%\frac{25}{85} \times 100\% = 29.411\ldots\%

Correct to 3 significant figures, 29.411…%≈29.4%29.411\ldots\% \approx 29.4\%.

The price of a cup of teh increased by about 29.4% (correct to 3 significant figures).

Example 10: Minutes and seconds

At the start of the year Siti's best time for the 50 m freestyle was 1 min 5 s. By the end of the year it was 52 s. Find the percentage decrease in her time.

Solution

Write both times in seconds: 1 min 5 s = 60 s + 5 s = 65 s.

Decrease=65−52=13\text{Decrease} = 65 - 52 = 13 s

Express the decrease as a percentage of her time at the start of the year: 1365×100%=20%\frac{13}{65} \times 100\% = 20\%

Do not write 1 min 5 s as 1.05 min. A minute has 60 seconds, not 100, so 1 min 5 s is 1560=11121\frac{5}{60} = 1\frac{1}{12} min.

Siti's time decreased by 20%.

Comparing Changes by Percentage

Two changes on different original values cannot be compared fairly by their amounts alone. Turn each change into a percentage of its own original value, and compare the percentages.

Example 11: Which price rose more?

At a hawker centre the price of a plate of chicken rice rose from $3.50 to $4. The price of a cup of kopi rose from $1.20 to $1.40. Which price rose by the greater percentage?

Solution

Chicken rice: the increase is 4−3.5=0.54 - 3.5 = 0.5, so $0.50. Its original price is $3.50.

Percentage increase in the price of chicken rice: 0.53.5×100%=14.285…%≈14.3%\frac{0.5}{3.5} \times 100\% = 14.285\ldots\% \approx 14.3\%

Kopi: the increase is 1.4−1.2=0.21.4 - 1.2 = 0.2, so $0.20. Its original price is $1.20.

Percentage increase in the price of kopi: 0.21.2×100%=16.666…%≈16.7%\frac{0.2}{1.2} \times 100\% = 16.666\ldots\% \approx 16.7\%

The price of chicken rice rose by more money (50 cents against 20 cents), but the price of kopi rose by the greater percentage, because 20 cents is a larger part of $1.20 than 50 cents is of $3.50.

The price of kopi rose by the greater percentage: about 16.7%, against about 14.3% for the chicken rice.

Common Mistakes

  • Dividing by the new value. The base of a percentage change is the original value. From $4 to $5 is an increase of 14×100%=25%\frac{1}{4} \times 100\% = 25\%, not 15×100%=20%\frac{1}{5} \times 100\% = 20\%.
  • Taking the value in the question as the original without reading it. "After a rise of $9, a ticket costs $54" gives the new price. The original price is $54 − $9 = $45, and the increase is 945×100%=20%\frac{9}{45} \times 100\% = 20\%.
  • Giving the new value as a percentage of the original instead of the increase. From $120 to $300 the new value is 250% of the original value, but the percentage increase is 150%.
  • Leaving out the 100%. 2001250=0.16\frac{200}{1250} = 0.16, which is 16%, not 0.16%.
  • Rounding too early. 2001300=0.153846…\frac{200}{1300} = 0.153846\ldots; rounding it to 0.15 gives 15% instead of 15.4%. Keep the calculator value and round only the final answer.
  • Mixing units of measurement. From 85 cents to $1.10 is not "1.10−851.10 - 85". Write both in cents first: 110−85=25110 - 85 = 25 cents. And 1 min 5 s is 65 s, not 1.05 min.
  • Writing a minus sign as well as the word "decrease". "A decrease of 16%" already says that the price went down. Do not write "a decrease of −16%".

Practice

Work each question on paper before you open its solution. Give an answer that is not exact correct to 3 significant figures.

Question 1

A school canteen sold 850 packets of fried rice in May and 935 packets in June. Find the percentage increase in the number of packets sold.

ShowHide solution

The original value is May's number, 850.

Increase=935−850=85\text{Increase} = 935 - 850 = 85 packets

Percentage increase=85850×100%=10%\text{Percentage increase} = \frac{85}{850} \times 100\% = 10\%

Check: 110% of 850 is 1.1×850=9351.1 \times 850 = 935.

The number of packets sold increased by 10%.

Question 2

Over ten years the number of residents of a housing estate fell from 48 00048\,000 to 45 12045\,120. Find the percentage decrease in the number of residents.

ShowHide solution

The original value is 48 00048\,000.

Decrease=48 000−45 120=2880\text{Decrease} = 48\,000 - 45\,120 = 2880 residents

Percentage decrease=288048 000×100%=6%\text{Percentage decrease} = \frac{2880}{48\,000} \times 100\% = 6\%

Check: 94% of 48 00048\,000 is 0.94×48 000=45 1200.94 \times 48\,000 = 45\,120.

The number of residents decreased by 6%.

Question 3

In March, 40 students of a school were late for school at least once. In April, 34 students were. Find the percentage decrease in the number of students who were late.

ShowHide solution

The original value is March's number, 40.

Decrease=40−34=6\text{Decrease} = 40 - 34 = 6 students

Percentage decrease=640×100%=15%\text{Percentage decrease} = \frac{6}{40} \times 100\% = 15\%

The number of students who were late decreased by 15%.

Question 4

In 2014 the world population was estimated at 7.2 billion. At that time it was projected to reach 9.6 billion by 2050. Find the percentage increase that was projected.

ShowHide solution

Both values are in billions, so they can be used as they are. The original value is 7.2 billion.

Increase=9.6−7.2=2.4\text{Increase} = 9.6 - 7.2 = 2.4 billion

Percentage increase=2.47.2×100%=33.333…%\text{Percentage increase} = \frac{2.4}{7.2} \times 100\% = 33.333\ldots\%

Correct to 3 significant figures, 33.333…%≈33.3%33.333\ldots\% \approx 33.3\%. (Exactly, it is 3313%33\frac{1}{3}\%, because 2.47.2=13\frac{2.4}{7.2} = \frac{1}{3}.)

The projected percentage increase was about 33.3% (correct to 3 significant figures).

Question 5

Siti read 15 books in Term 1 and 12 books in Term 2. She says that the number of books she read fell by 25%. Is she right? If not, find the correct percentage decrease and explain her mistake.

ShowHide solution

The original value is the number of books in Term 1, 15.

Decrease=15−12=3\text{Decrease} = 15 - 12 = 3 books

Percentage decrease=315×100%=20%\text{Percentage decrease} = \frac{3}{15} \times 100\% = 20\%

Siti worked out 312×100%=25%\frac{3}{12} \times 100\% = 25\%. She divided by the new value, 12, instead of the original value, 15.

Check: 80% of 15 is 0.8×15=120.8 \times 15 = 12, but 75% of 15 is 0.75×15=11.250.75 \times 15 = 11.25, not 12.

No. The percentage decrease is 20%. Siti found the decrease as a percentage of the new number, 12, instead of the original number, 15.

Question 6

The enrolment of a school fell by 57 students to 1083 students. Find the percentage decrease in the enrolment.

ShowHide solution

1083 is the new enrolment, after the fall. The original enrolment was 1083+57=11401083 + 57 = 1140 students.

Percentage decrease=571140×100%=5%\text{Percentage decrease} = \frac{57}{1140} \times 100\% = 5\%

Check: 95% of 1140 is 0.95×1140=10830.95 \times 1140 = 1083.

Dividing by 1083 would give 571083×100%≈5.26%\frac{57}{1083} \times 100\% \approx 5.26\%, which uses the new enrolment as the base.

The enrolment decreased by 5%.

Question 7

Mrs Goh planted a seedling 12 cm tall. Six months later it was 45 cm tall.
(a) Find the percentage increase in its height.
(b) Express its new height as a percentage of its original height.

ShowHide solution

(a) The original height is 12 cm. Increase=45−12=33\text{Increase} = 45 - 12 = 33 cm

Two bars drawn to the same scale. The original bar is 12 centimetres, 100 percent. The new bar is 45 centimetres: the original bar with an extra piece at its end, nearly three times as long as the original bar, marked increase 33 centimetres, which is an unknown percentage of 12 centimetres.OriginalNew12 cm (100%)45 cmincrease 33 cm = ?% of 12 cm

Percentage increase=3312×100%=275%\text{Percentage increase} = \frac{33}{12} \times 100\% = 275\%

(b) 4512×100%=375%\frac{45}{12} \times 100\% = 375\%

The two answers differ by 100%, because the new height is the original height (100%) together with the increase (275%).

(a) The height increased by 275%.
(b) The new height is 375% of the original height.

Question 8

A bottle of shampoo used to hold 750 ml. The new bottle holds 1.2 litres. Find the percentage increase in the volume of shampoo in a bottle.

ShowHide solution

Write both volumes in millilitres: 1.2 litres = 1.2×1000=12001.2 \times 1000 = 1200 ml.

Increase=1200−750=450\text{Increase} = 1200 - 750 = 450 ml

Percentage increase=450750×100%=60%\text{Percentage increase} = \frac{450}{750} \times 100\% = 60\%

The volume of shampoo in a bottle increased by 60%.

Question 9

In January Ravi ran 2.4 km in 12 min 30 s. In May he ran the same distance in 11 min 15 s. Find the percentage decrease in his time.

ShowHide solution

Write both times in seconds. 12 min 30 s = 12×60+30=75012 \times 60 + 30 = 750 s, and 11 min 15 s = 11×60+15=67511 \times 60 + 15 = 675 s.

Decrease=750−675=75\text{Decrease} = 750 - 675 = 75 s

Percentage decrease=75750×100%=10%\text{Percentage decrease} = \frac{75}{750} \times 100\% = 10\%

The distance, 2.4 km, is the same both times, so it is not needed.

Ravi's time decreased by 10%.

Question 10

For a charity fair, Class 1A raised $480 last year and $600 this year. Class 1B raised $350 last year and $455 this year.
(a) Which class raised more extra money this year?
(b) Which class increased the money it raised by the greater percentage?

ShowHide solution

(a) Class 1A: 600−480=120600 - 480 = 120, so $120 more. Class 1B: 455−350=105455 - 350 = 105, so $105 more.

(b) Class 1A: 120480×100%=25%\frac{120}{480} \times 100\% = 25\%

Class 1B: 105350×100%=30%\frac{105}{350} \times 100\% = 30\%

Class 1B raised less extra money, but its increase is a larger part of what it raised last year.

(a) Class 1A raised more extra money ($120 against $105).
(b) Class 1B increased by the greater percentage (30% against 25%).

Question 11

A baby had a mass of 3.2 kg at birth. At six months old her mass was 7450 g. Find the percentage increase in her mass.

ShowHide solution

Write both masses in grams: 3.2 kg = 3.2×1000=32003.2 \times 1000 = 3200 g.

Increase=7450−3200=4250\text{Increase} = 7450 - 3200 = 4250 g

Percentage increase=42503200×100%=132.8125%\text{Percentage increase} = \frac{4250}{3200} \times 100\% = 132.8125\%

Correct to 3 significant figures, 132.8125%≈133%132.8125\% \approx 133\%. The increase is more than the birth mass, so the percentage increase is more than 100%.

The baby's mass increased by about 133% (correct to 3 significant figures).

Question 12

A shop sold 1250 umbrellas in November, 3100 umbrellas in December and 980 umbrellas in January.
(a) Find the percentage increase in the number sold from November to December.
(b) Find the percentage decrease in the number sold from December to January.

ShowHide solution

(a) The change asked about starts in November, so the original value is 1250.

Increase=3100−1250=1850\text{Increase} = 3100 - 1250 = 1850 umbrellas

Percentage increase=18501250×100%=148%\text{Percentage increase} = \frac{1850}{1250} \times 100\% = 148\%

(b) This change starts in December, so its original value is 3100, not 1250.

Decrease=3100−980=2120\text{Decrease} = 3100 - 980 = 2120 umbrellas

Percentage decrease=21203100×100%=68.387…%≈68.4%\text{Percentage decrease} = \frac{2120}{3100} \times 100\% = 68.387\ldots\% \approx 68.4\%

(a) The number sold increased by 148%.
(b) The number sold decreased by about 68.4% (correct to 3 significant figures).

Key Points

  • Increase = new value − original value. Decrease = original value − new value.
  • Percentage increase or decrease=Increase or decreaseOriginal value×100%\text{Percentage increase or decrease} = \dfrac{\text{Increase or decrease}}{\text{Original value}} \times 100\%
  • The base is always the original value, the value before the change. Dividing by the new value gives a wrong answer.
  • A percentage decrease is written as a positive number with the word "decrease". It is never more than 100%.
  • A percentage increase can be more than 100%. Doubling is an increase of 100%, and three times as large is an increase of 200%.
  • Put both values in the same unit of measurement before finding the change.
  • Give an answer that is not exact correct to 3 significant figures, and round only at the end.
  • Check by working forward: increase or decrease the original value by the percentage found, and see that you get the new value.

Exercises on this lesson in the Secondary 1 book

‹ 5.5 Increasing and Decreasing by a Percentage 5.7 Percentage Points ›

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