One Quantity as a Percentage of Another

Finding what percentage one quantity is of another: name the base, put both quantities in the same unit, divide and multiply by 100%, and use the fact that the parts of a whole add up to 100%.

Before you start

A percentage is a number of hundredths. "Per cent" means "out of 100", so 35%=35100=0.3535\% = \frac{35}{100} = 0.35, and 100%=100100=1100\% = \frac{100}{100} = 1 is the whole.

To change a fraction to a percentage, multiply it by 100%100\%. This does not change its value, because 100%=1100\% = 1: 38×100%=3008%=37.5%\frac{3}{8} \times 100\% = \frac{300}{8}\% = 37.5\%.

To round to 3 significant figures, count three figures from the first non-zero digit and look at the next digit: 40.5405…≈40.540.5405\ldots \approx 40.5 and 5.8333…≈5.835.8333\ldots \approx 5.83. Round only at the last step of a calculation.

The Part, the Whole and the Base

Hui Min scores 34 marks in a test marked out of 40. To say how well she did, we compare her 34 marks with the 40 marks of the whole test. As a fraction, she scored 3440\frac{34}{40} of the marks. A percentage says the same thing out of 100: 3440=85100=85%\frac{34}{40} = \frac{85}{100} = 85\%.

The quantity we compare with, here the 40 marks of the whole test, is called the base. In this lesson the base is always a whole, and the quantity is a part of it. The base stands for 100%100\%, and the question tells you what it is.

Rule: one quantity as a percentage of another

To express one quantity as a percentage of another quantity of the same kind, both in the same unit, write the first quantity as a fraction of the second (the base) and multiply by 100%100\%:

percentage=the quantitythe base×100%\text{percentage} = \dfrac{\text{the quantity}}{\text{the base}} \times 100\%

The base is the quantity you compare with. It goes at the bottom of the fraction, and it stands for 100%100\%.

When the quantity is a part of the base, the percentage is never more than 100%100\%.

Words that name the base

The base comes after "of" or "out of": "express 18 as a percentage of 80", "what percentage of the class", "34 marks out of 40". In these three the base is 80, the whole class and 40.

In "the percentage of the students who wear spectacles", the base is all the students, and the students who wear spectacles are the part.

Example 1: Finding the percentage

(a) What percentage of 200 is 50?
(b) Express 18 as a percentage of 80.

Solution

(a) "Of 200" names the base: 200 is the whole and stands for 100%100\%. The quantity is 50.

Write 50 as a fraction of 200 and multiply by 100%100\%: 50200×100%=14×100%=25%\frac{50}{200} \times 100\% = \frac{1}{4} \times 100\% = 25\%

The fraction was simplified first: dividing 50 and 200 by 50 gives 14\frac{1}{4}, and a quarter of 100%100\% is 25%25\%.

(b) "As a percentage of 80" names the base: 80 is the whole. The quantity is 18.

1880×100%=180080%=22.5%\frac{18}{80} \times 100\% = \frac{1800}{80}\% = 22.5\%

Check: 18 is a little less than a quarter of 80 (a quarter of 80 is 20), so the answer should be a little less than 25%25\%. It is.

(a) 50 is 25%25\% of 200.
(b) 18 is 22.5%22.5\% of 80.

Example 2: Marks in a test

Hui Min scored 34 marks in a Mathematics test marked out of 40. What percentage of the total marks did she score?

Solution

The base is the total marks of the test, 40 marks. It stands for 100%100\%. Her score, 34 marks, is a part of it.

Percentage scored =3440×100%= \frac{34}{40} \times 100\%

One way, multiplying first: 3440×100%=340040%=85%\frac{34}{40} \times 100\% = \frac{3400}{40}\% = 85\%

Another way, making the denominator 100: 3440=1720=85100\frac{34}{40} = \frac{17}{20} = \frac{85}{100} (dividing the top and the bottom by 2, then multiplying them by 5). That is 85 hundredths, which is 85%85\%.

Hui Min scored 85%85\% of the total marks.

Parts of a Whole Add Up to 100%

When a whole is split into parts with nothing left over, the parts together make the whole. So their percentages of the whole add up to 100%100\%. If the percentages of all the parts but one are known, the last part, often called "the rest", is 100%100\% minus the others. In the drawings below, the whole is one bar standing for 100%100\%, with a small mark at every 10%10\%.

Rule: the parts of a whole

When a whole is made up of parts, the percentages of the parts add up to 100%100\%. The percentage of the last part is 100%100\% minus the percentages of the other parts.

Example 3: The rest

A box holds 10 mangoes. 7 of them are ripe and the rest are unripe. What percentage of the mangoes in the box are unripe?

Solution

The base is all the mangoes in the box: 10 mangoes stand for 100%100\%. The ripe and the unripe mangoes are the two parts of this whole.

A bar standing for the whole box of 10 mangoes, marked 0% at its left end and 100% at its right end. The first 7 tenths of the bar are labelled Ripe: 7. The last 3 tenths are labelled Unripe, with a question mark for the number and a question mark for the percentage.Ripe: 7Unripe: ??%0%100%10 mangoes = 100%

Number of unripe mangoes =10−7=3= 10 - 7 = 3

Percentage of unripe mangoes =number of unripe mangoestotal number of mangoes×100%=310×100%=30%= \frac{\text{number of unripe mangoes}}{\text{total number of mangoes}} \times 100\% = \frac{3}{10} \times 100\% = 30\%

Another way, from 100%100\%: the ripe mangoes are 710×100%=70%\frac{7}{10} \times 100\% = 70\% of the box, so the unripe mangoes are 100%−70%=30%100\% - 70\% = 30\% of it.

30%30\% of the mangoes in the box are unripe.

Example 4: Girls and boys

There are 160 students in Secondary 1 at a school, and 90 of them are girls. Find the percentage of the Secondary 1 students who are
(a) girls,
(b) boys.

Solution

The base is all 160 students in Secondary 1. Every one of them is a girl or a boy, so the girls and the boys are the two parts of this whole.

A bar standing for all 160 Secondary 1 students, marked 0% at its left end and 100% at its right end. The first part is labelled Girls: 90 and the second part Boys, with a question mark for the number. Each part has a question mark for its percentage.Girls: 90?%Boys: ??%0%100%160 students = 100%

(a) Percentage of girls =90160×100%=9000160%=56.25%= \frac{90}{160} \times 100\% = \frac{9000}{160}\% = 56.25\%

(b) Method 1, from the number of boys. Number of boys =160−90=70= 160 - 90 = 70, so the percentage of boys =70160×100%=7000160%=43.75%= \frac{70}{160} \times 100\% = \frac{7000}{160}\% = 43.75\%

Method 2, from 100%100\%. The two percentages add up to 100%100\%, so the percentage of boys =100%−56.25%=43.75%= 100\% - 56.25\% = 43.75\%

Check: 56.25%+43.75%=100%56.25\% + 43.75\% = 100\%

(a) 56.25%56.25\% of the Secondary 1 students are girls.
(b) 43.75%43.75\% of the Secondary 1 students are boys.

Example 5: A budget

Ravi receives $80 of pocket money a month. He spends $36 on food and $20 on transport, and he saves the rest. What percentage of his pocket money does he
(a) spend on food,
(b) spend on transport,
(c) save?

Solution

The base is his whole pocket money, $80. Food, transport and savings are the three parts of it.

A bar standing for the whole 80 dollars of pocket money, marked 0% at its left end and 100% at its right end. It is cut into three parts labelled Food 36 dollars, Transport 20 dollars, and Savings with a question mark for the amount. Each part has a question mark for its percentage.Food$36?%Transport$20?%Savings??%0%100%$80 = 100%

(a) Food: 3680×100%=360080%=45%\frac{36}{80} \times 100\% = \frac{3600}{80}\% = 45\%

(b) Transport: 2080×100%=200080%=25%\frac{20}{80} \times 100\% = \frac{2000}{80}\% = 25\%

(c) His savings are the rest. The three percentages add up to 100%100\%, so savings =100%−45%−25%=30%= 100\% - 45\% - 25\% = 30\%

Check by the amount: he saves $80 − $36 − $20 = $24, and 2480×100%=30%\frac{24}{80} \times 100\% = 30\%.

Ravi spends 45%45\% of his pocket money on food and 25%25\% on transport, and he saves 30%30\% of it.

The Same Unit of Measurement First

A percentage compares two quantities of the same kind, and a fraction such as 2502000\frac{250}{2000} means something only when both of its numbers count the same unit. So, before dividing, put the quantity and the base in the same unit of measurement. It is usually easier to change the larger unit into the smaller one, because then both numbers are whole numbers.

Units to know
MeasureChanging units
Length1 km = 1000 m, 1 m = 100 cm, 1 cm = 10 mm
Mass1 kg = 1000 g
Volume1 litre = 1000 ml
Time1 hour = 60 minutes, 1 minute = 60 seconds
Money$1 = 100 cents

Example 6: Changing to the same unit

Express
(a) 250 g as a percentage of 2 kg,
(b) 45 minutes as a percentage of 2 hours,
(c) 95 cents as a percentage of $2.50.

Solution

(a) The base is 2 kg. Change it to grams: 2 kg =2×1000=2000= 2 \times 1000 = 2000 g.

2502000×100%=25 0002000%=12.5%\frac{250}{2000} \times 100\% = \frac{25\,000}{2000}\% = 12.5\%

(b) The base is 2 hours. Change it to minutes: 2 hours =2×60=120= 2 \times 60 = 120 minutes.

45120×100%=4500120%=37.5%\frac{45}{120} \times 100\% = \frac{4500}{120}\% = 37.5\%

(c) The base is $2.50. Change it to cents: $2.50 =2.50×100=250= 2.50 \times 100 = 250 cents.

95250×100%=9500250%=38%\frac{95}{250} \times 100\% = \frac{9500}{250}\% = 38\%

(a) 250 g is 12.5%12.5\% of 2 kg.
(b) 45 minutes is 37.5%37.5\% of 2 hours.
(c) 95 cents is 38%38\% of $2.50.

What goes wrong if the units are not changed

Without changing the units, (a) would give 2502×100%=12 500%\frac{250}{2} \times 100\% = 12\,500\%. That says 250 g is 125 times as heavy as 2 kg, when it is only one eighth of it.

A part of a whole is never more than 100%100\% of it, so an answer like this is a sign that the units were not made the same.

Example 7: A mixture

Siti makes a jug of rose syrup drink by mixing 180 ml of rose syrup with 1.02 litres of water. What percentage of the drink is rose syrup?

Solution

The question asks for a percentage of the drink, so the base is the whole drink, the syrup and the water together. It is not the water alone.

A bar standing for the whole drink, marked 0% at its left end and 100% at its right end. A short first part is labelled Syrup, 180 ml, with a question mark for its percentage. The long second part is labelled Water, 1.02 litres.Syrup180 ml?%Water1.02 litres0%100%the whole drink = 100%

Change to the same unit: 1.02 litres =1.02×1000=1020= 1.02 \times 1000 = 1020 ml.

The whole drink =180+1020=1200= 180 + 1020 = 1200 ml

Percentage of rose syrup =1801200×100%=18 0001200%=15%= \frac{180}{1200} \times 100\% = \frac{18\,000}{1200}\% = 15\%

The water is the rest of the drink: 100%−15%=85%100\% - 15\% = 85\%. Check: 10201200×100%=85%\frac{1020}{1200} \times 100\% = 85\%.

Rose syrup is 15%15\% of the drink.

Answers That Are Not Exact

Often the division does not come out exactly. Then give the percentage correct to 3 significant figures, write ≈\approx in front of it, and say how it was rounded. Keep the full value in the working and round only at the last step. An answer that is exact, such as 56.25%56.25\% in Example 4, is written in full.

Example 8: A family gathering

At a family gathering there are 15 children and 22 adults. What percentage of the people at the gathering are
(a) children,
(b) adults?
Give each answer correct to 3 significant figures.

Solution

The base is everyone at the gathering: 15+22=3715 + 22 = 37 people.

(a) Children: 1537×100%=40.5405…%\frac{15}{37} \times 100\% = 40.5405\ldots\%

To 3 significant figures, keep 4, 0 and 5. The next digit is 4, which is less than 5, so round down: 40.5405…%≈40.5%40.5405\ldots\% \approx 40.5\%

(b) Adults: 2237×100%=59.4594…%\frac{22}{37} \times 100\% = 59.4594\ldots\%. Keep 5, 9 and 4. The next digit is 5, so round up: 59.4594…%≈59.5%59.4594\ldots\% \approx 59.5\%

Or from 100%100\%, using the full value of (a): 100%−40.5405…%=59.4594…%≈59.5%100\% - 40.5405\ldots\% = 59.4594\ldots\% \approx 59.5\%

(a) About 40.5%40.5\% of the people are children.
(b) About 59.5%59.5\% of the people are adults.
(Both correct to 3 significant figures.)

Example 9: A survey with three parts

In a survey, 120 students said how they travel to school: 74 take a bus, 39 take the MRT and 7 walk. Find the percentage of the students who travel each way. Give answers that are not exact correct to 3 significant figures.

Solution

The base is all 120 students in the survey. Check that the parts make the whole: 74+39+7=12074 + 39 + 7 = 120.

Bus: 74120×100%=61.666…%≈61.7%\frac{74}{120} \times 100\% = 61.666\ldots\% \approx 61.7\%

MRT: 39120×100%=32.5%\frac{39}{120} \times 100\% = 32.5\%, which is exact.

Walk: 7120×100%=5.8333…%≈5.83%\frac{7}{120} \times 100\% = 5.8333\ldots\% \approx 5.83\%. The first significant figure is the 5, so the three figures kept are 5, 8 and 3.

The rounded percentages add up to 61.7%+32.5%+5.83%=100.03%61.7\% + 32.5\% + 5.83\% = 100.03\%, not exactly 100%100\%. That is the effect of rounding, not a mistake. The exact percentages, 6123%61\frac{2}{3}\%, 32.5%32.5\% and 556%5\frac{5}{6}\%, add up to exactly 100%100\%.

About 61.7%61.7\% of the students take a bus, 32.5%32.5\% take the MRT and about 5.83%5.83\% walk (correct to 3 significant figures).

Finding the rest from full values

If the percentage who walk in Example 9 were found from the rounded values, 100%−61.7%−32.5%=5.8%100\% - 61.7\% - 32.5\% = 5.8\%, it would be wrong in the third figure.

Subtract the full values, 100%−61.666…%−32.5%=5.8333…%≈5.83%100\% - 61.666\ldots\% - 32.5\% = 5.8333\ldots\% \approx 5.83\%, or work out the part on its own.

Common Mistakes

  • Dividing the wrong way round. The base goes at the bottom. If 20 of the 25 students in a group passed, the percentage who passed is 2025×100%=80%\frac{20}{25} \times 100\% = 80\%, not 2520×100%=125%\frac{25}{20} \times 100\% = 125\%. A part of a whole cannot be more than 100%100\% of it.
  • Leaving out the ×100%\times 100\%. 1825=0.72\frac{18}{25} = 0.72, which is 72%72\%. Writing "0.72%0.72\%" makes the answer a hundred times too small.
  • Not changing to the same unit. 250 g as a percentage of 2 kg is not 2502×100%=12 500%\frac{250}{2} \times 100\% = 12\,500\%. In grams it is 2502000×100%=12.5%\frac{250}{2000} \times 100\% = 12.5\%.
  • Using a part as the base. In Example 7, 1801020×100%≈17.6%\frac{180}{1020} \times 100\% \approx 17.6\% compares the syrup with the water. The question asks for a percentage of the drink, so the base is the whole drink, 1200 ml, and the answer is 15%15\%.
  • Finding the rest from rounded percentages. 100%−61.7%−32.5%=5.8%100\% - 61.7\% - 32.5\% = 5.8\% is not the percentage who walk in Example 9. Use the full values, which give 5.83%5.83\%.
  • Expecting rounded percentages to add up to exactly 100%. Rounded percentages may add up to a little more or a little less, such as 100.03%100.03\%. The exact percentages add up to 100%100\%.

Practice

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

Question 1

Express 36 as a percentage of 48.

ShowHide solution

The base is 48.

3648×100%=34×100%=75%\frac{36}{48} \times 100\% = \frac{3}{4} \times 100\% = 75\% (dividing 36 and 48 by 12 first)

36 is 75%75\% of 48.

Question 2

What percentage of 250 is 40?

ShowHide solution

The base is 250, the number after "of".

40250×100%=4000250%=16%\frac{40}{250} \times 100\% = \frac{4000}{250}\% = 16\%

40 is 16%16\% of 250.

Question 3

Siti scored 43 marks out of 50 in a Science test. What percentage of the total marks did she score?

ShowHide solution

The base is the total marks, 50.

4350=86100\frac{43}{50} = \frac{86}{100} (multiplying the top and the bottom by 2), which is 86%86\%.

Or: 4350×100%=430050%=86%\frac{43}{50} \times 100\% = \frac{4300}{50}\% = 86\%

Siti scored 86%86\% of the total marks.

Question 4

In a class of 40 students, 14 walk to school. What percentage of the class
(a) walk to school,
(b) do not walk to school?

ShowHide solution

The base is the whole class, 40 students.

(a) 1440×100%=140040%=35%\frac{14}{40} \times 100\% = \frac{1400}{40}\% = 35\%

(b) Every student either walks or does not, so the two percentages add up to 100%100\%: 100%−35%=65%100\% - 35\% = 65\%

Check: 40−14=2640 - 14 = 26 students do not walk, and 2640×100%=65%\frac{26}{40} \times 100\% = 65\%.

(a) 35%35\% of the class walk to school.
(b) 65%65\% of the class do not walk to school.

Question 5

Express
(a) 350 m as a percentage of 2 km,
(b) 36 seconds as a percentage of 4 minutes,
(c) 85 cents as a percentage of $4.

ShowHide solution

(a) Change the base to metres: 2 km =2×1000=2000= 2 \times 1000 = 2000 m.

3502000×100%=35 0002000%=17.5%\frac{350}{2000} \times 100\% = \frac{35\,000}{2000}\% = 17.5\%

(b) Change the base to seconds: 4 minutes =4×60=240= 4 \times 60 = 240 seconds.

36240×100%=3600240%=15%\frac{36}{240} \times 100\% = \frac{3600}{240}\% = 15\%

(c) Change the base to cents: $4 =4×100=400= 4 \times 100 = 400 cents.

85400×100%=8500400%=21.25%\frac{85}{400} \times 100\% = \frac{8500}{400}\% = 21.25\%, which is exact.

(a) 17.5%17.5\%
(b) 15%15\%
(c) 21.25%21.25\%

Question 6

Mrs Goh's phone has 128 GB of storage. Her photos use 56 GB, her apps use 40 GB and the rest is free. What percentage of the storage
(a) do her photos use,
(b) is free?

ShowHide solution

The base is the whole storage, 128 GB. Photos, apps and free space are its three parts.

(a) Photos: 56128×100%=5600128%=43.75%\frac{56}{128} \times 100\% = \frac{5600}{128}\% = 43.75\%

For (b), first the apps: 40128×100%=4000128%=31.25%\frac{40}{128} \times 100\% = \frac{4000}{128}\% = 31.25\%

Free space is the rest: 100%−43.75%−31.25%=25%100\% - 43.75\% - 31.25\% = 25\%

Check: the free space is 128−56−40=32128 - 56 - 40 = 32 GB, and 32128×100%=25%\frac{32}{128} \times 100\% = 25\%.

(a) Her photos use 43.75%43.75\% of the storage.
(b) 25%25\% of the storage is free.

Question 7

(a) Ali scored 52 marks out of 75 in a test. What percentage of the total marks did he score?
(b) Express 5 cm as a percentage of 1.2 m.

ShowHide solution

(a) The base is the total marks, 75.

5275×100%=69.333…%≈69.3%\frac{52}{75} \times 100\% = 69.333\ldots\% \approx 69.3\% (the fourth figure is 3, so round down)

(b) Change the base to centimetres: 1.2 m =1.2×100=120= 1.2 \times 100 = 120 cm.

5120×100%=4.1666…%\frac{5}{120} \times 100\% = 4.1666\ldots\%

The first significant figure is the 4, so keep 4, 1 and 6. The next digit is 6, so round up: 4.1666…%≈4.17%4.1666\ldots\% \approx 4.17\%

(a) Ali scored about 69.3%69.3\% of the total marks.
(b) 5 cm is about 4.17%4.17\% of 1.2 m.
(Both correct to 3 significant figures.)

Question 8

A choir has 48 students, and 18 of them are boys. Ali says that the boys are 60%60\% of the choir, because 1830×100%=60%\frac{18}{30} \times 100\% = 60\%.
(a) What has Ali done wrong?
(b) Find the percentage of the choir who are boys, and the percentage who are girls.

ShowHide solution

(a) The 30 in Ali's fraction is the number of girls: 48−18=3048 - 18 = 30. He has compared the boys with the girls. The question asks for a percentage of the choir, so the base is all 48 students.

(b) Boys: 1848×100%=180048%=37.5%\frac{18}{48} \times 100\% = \frac{1800}{48}\% = 37.5\%

Girls: 100%−37.5%=62.5%100\% - 37.5\% = 62.5\%

Check: 3048×100%=300048%=62.5%\frac{30}{48} \times 100\% = \frac{3000}{48}\% = 62.5\%

(a) Ali used the number of girls as the base instead of the whole choir.
(b) 37.5%37.5\% of the choir are boys and 62.5%62.5\% are girls.

Question 9

A drinks stall at a hawker centre sold 360 drinks on Saturday: 135 cups of coffee, 96 cups of tea, and the rest were fruit juices. What percentage of the drinks sold were
(a) coffee,
(b) tea,
(c) fruit juices?

ShowHide solution

The base is all 360 drinks.

(a) 135360×100%=13 500360%=37.5%\frac{135}{360} \times 100\% = \frac{13\,500}{360}\% = 37.5\%, which is exact.

(b) 96360×100%=26.666…%≈26.7%\frac{96}{360} \times 100\% = 26.666\ldots\% \approx 26.7\%

(c) Number of fruit juices =360−135−96=129= 360 - 135 - 96 = 129

129360×100%=35.833…%≈35.8%\frac{129}{360} \times 100\% = 35.833\ldots\% \approx 35.8\%

Check with the full values: 37.5%+26.666…%+35.833…%=100%37.5\% + 26.666\ldots\% + 35.833\ldots\% = 100\%.

(a) 37.5%37.5\% of the drinks were coffee.
(b) About 26.7%26.7\% were tea.
(c) About 35.8%35.8\% were fruit juices.

Question 10

A secondary school has an enrolment of 1260 students, and 287 of them are in Secondary 1. What percentage of the students are
(a) in Secondary 1,
(b) not in Secondary 1?

ShowHide solution

The base is all 1260 students of the school.

(a) 2871260×100%=22.777…%≈22.8%\frac{287}{1260} \times 100\% = 22.777\ldots\% \approx 22.8\%

(b) From the full value: 100%−22.777…%=77.222…%≈77.2%100\% - 22.777\ldots\% = 77.222\ldots\% \approx 77.2\%

Check: 1260−287=9731260 - 287 = 973 students are not in Secondary 1, and 9731260×100%=77.222…%\frac{973}{1260} \times 100\% = 77.222\ldots\%.

(a) About 22.8%22.8\% of the students are in Secondary 1.
(b) About 77.2%77.2\% of the students are not in Secondary 1.

Question 11

Mei Ling mixes 450 g of flour, 150 g of sugar and 0.3 kg of butter to make a cake mixture. What percentage of the mixture is
(a) flour,
(b) sugar,
(c) butter?

ShowHide solution

Change the butter to grams: 0.3 kg =0.3×1000=300= 0.3 \times 1000 = 300 g.

The base is the whole mixture: 450+150+300=900450 + 150 + 300 = 900 g.

(a) Flour: 450900×100%=50%\frac{450}{900} \times 100\% = 50\%

(b) Sugar: 150900×100%=16.666…%≈16.7%\frac{150}{900} \times 100\% = 16.666\ldots\% \approx 16.7\%

(c) Butter: 300900×100%=33.333…%≈33.3%\frac{300}{900} \times 100\% = 33.333\ldots\% \approx 33.3\%

Check with the full values: 50%+16.666…%+33.333…%=100%50\% + 16.666\ldots\% + 33.333\ldots\% = 100\%.

(a) 50%50\% of the mixture is flour.
(b) About 16.7%16.7\% is sugar.
(c) About 33.3%33.3\% is butter.

Question 12

Over a weekend (Saturday and Sunday, 48 hours), Ravi sleeps for 17 h 30 min, does homework for 5 h 15 min and spends 3 h 45 min at his CCA.
(a) What percentage of the weekend does he spend on homework?
(b) What percentage of the weekend is left for everything else?

ShowHide solution

Put every time in minutes. The base is the whole weekend: 48×60=288048 \times 60 = 2880 minutes.

Sleep: 17×60+30=105017 \times 60 + 30 = 1050 minutes. Homework: 5×60+15=3155 \times 60 + 15 = 315 minutes. CCA: 3×60+45=2253 \times 60 + 45 = 225 minutes.

(a) 3152880×100%=10.9375%≈10.9%\frac{315}{2880} \times 100\% = 10.9375\% \approx 10.9\%

(b) Time used: 1050+315+225=15901050 + 315 + 225 = 1590 minutes. Time left: 2880−1590=12902880 - 1590 = 1290 minutes.

12902880×100%=44.791…%≈44.8%\frac{1290}{2880} \times 100\% = 44.791\ldots\% \approx 44.8\%

Check: sleep is 10502880×100%=36.458…%\frac{1050}{2880} \times 100\% = 36.458\ldots\% and CCA is 2252880×100%=7.8125%\frac{225}{2880} \times 100\% = 7.8125\%. With homework and the time left, the full values add up to 100%100\%: 36.458…%+10.9375%+7.8125%+44.791…%=100%36.458\ldots\% + 10.9375\% + 7.8125\% + 44.791\ldots\% = 100\%.

(a) Ravi spends about 10.9%10.9\% of the weekend on homework.
(b) About 44.8%44.8\% of the weekend is left for everything else.

Key Points

  • To express one quantity as a percentage of another: the quantitythe base×100%\dfrac{\text{the quantity}}{\text{the base}} \times 100\%.
  • The base is the whole you compare with. It comes after "of" or "out of", it goes at the bottom of the fraction, and it stands for 100%100\%.
  • Put the quantity and the base in the same unit of measurement first, usually the smaller unit.
  • A part of a whole is never more than 100%100\% of it. An answer over 100%100\% for a part means a mistake: the fraction is upside down, the units differ, or the base is not the whole.
  • The percentages of the parts of a whole add up to 100%100\%, so the rest is 100%100\% minus the other parts.
  • An answer that is not exact is given correct to 3 significant figures, with ≈\approx. Keep full values until the last step; rounded parts may add up to a little more or less than 100%100\%.

In this lesson every quantity has been a part of its base, so no percentage has been more than 100%100\%. A quantity can also be compared with a base that it is not a part of, such as the price of a bowl of noodles now with its price five years ago, and then the percentage can be more than 100%100\%. That comes next, in the lessons on percentages greater than 100% and on comparing two quantities by percentage.

Exercises on this lesson in the Secondary 1 book

‹ 5.1 Percentages, Fractions and Decimals 5.3 Percentages Greater than 100% ›

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