Percents

What an amount is compared against is called the base, the number a percent is taken of. Twelve percent of one base and twelve percent of another are different amounts, so a percent alone leaves the amount unknown. Prealgebra chapter 9 did the arithmetic. New here is naming an unknown with a letter, which turns a percent sentence into an equation.

Problem

A quality check clears

64 %

of a shipment. Percent means per hundred, so

64 %

is the ratio

\frac{64}{100}

. Reduce it to lowest terms and enter the fraction.

Show a hint

Both parts of a ratio can be divided by the same number without changing it, and $64$ and $100$ share a factor larger than $2$ .
That factor is $4$ , so divide above and below the bar by $4$ , then check whether the result reduces again.

Show the full solution

Dividing both parts by

4

gives

\frac{64}{100} = \frac{16}{25}

. Since

16

and

25

share no common factor above

1

, that is lowest terms. As a decimal it is

64 \div 100 = 0.64

, the same number with the point moved two places left.

Problem

A sensor logs

\frac{7}{20}

of its readings as out of range. A percent is a ratio whose second part is

100

, so rewrite

\frac{7}{20}

with

100

below the bar. Enter the percent, just the number.

Show a hint

Multiplying both parts of a ratio by the same number leaves it unchanged, so find the factor that turns the $20$ below the bar into $100$ .
$20 \times 5 = 100$ , so multiply above the bar by $5$ as well.

Show the full solution

Multiplying both parts by

5

gives

\frac{7}{20} = \frac{35}{100}

, so the percent is

35

. A second route gives the same number,

\frac{7}{20} \times 100 = 35

, and it still works when the denominator does not divide

100

Problem

A label states that a bottle of saline is

6.5 %

salt by mass, so the ratio of salt to total mass is

\frac{6.5}{100}

. Lowest terms means the top and the bottom are whole numbers with no common factor other than

1

. Enter

6.5 %

as a fraction in lowest terms.

Show a hint

Multiplying the top and the bottom by the same number does not change the value, and multiplying by $10$ removes the decimal point.
That gives $\frac{65}{1000}$ , and $65$ and $1000$ share a factor of $5$ .

Show the full solution

Multiplying the top and the bottom by

10

gives

\frac{6.5}{100} = \frac{65}{1000}

, and dividing both by

5

gives

\frac{13}{200}

. Since

13

is prime and does not divide

200

, that is lowest terms. As a decimal it is

0.065

Problem

A machinist finds that

\frac{5}{6}

of a batch passes inspection. That fraction is already in lowest terms and no whole number times

6

gives

100

, so it cannot be rescaled to a whole number over

100

. Multiply

\frac{5}{6}

100

anyway and enter the exact percent as a fraction in lowest terms.

Show a hint

Rescaling is blocked here, so the route left is plain arithmetic. Multiplying $\frac{5}{6}$ by $100$ multiplies the numerator by $100$ and leaves the $6$ underneath alone.
$\frac{5}{6} \times 100 = \frac{500}{6}$ , and $500$ and $6$ are both even.

Show the full solution

\frac{5}{6} \times 100 = \frac{500}{6}

, and dividing the numerator and the denominator by

2

gives

\frac{250}{3}

. The same percent is often written

83 \frac{1}{3} %

. Long division gives

250 \div 3 = 83.333 \dots

, which never terminates, so

83.3 %

and

83.33 %

are roundings while

\frac{250}{3} %

is exact.

Each small square is one hundredth of the grid, and

37

of them are shaded, so the shaded count and the percent are the same number. The same amount is

\frac{37}{100}

as a fraction and

0.37

as a decimal. The

100

squares are what the

37

is counted against, and renaming the shaded part does not change how much is shaded.

A percent question names three numbers, an amount, a percent, and the whole the percent is taken of. The sentence $a$ is $p %$ of $b$ says $a$ equals $\frac{p}{100}$ of $b$ , and of means multiply, so the sentence is $a = \frac{p}{100} b$ . Two of the three are given and one is the letter. The next two problems leave out a different one.

Problem

A vineyard has 875 vines and

24 %

of them are a new cultivar. Here

p = 24

and

b = 875

, and the amount

a

is missing. Substitute into

a = \frac{p}{100} b

and enter the number of new-cultivar vines.

Show a hint

$24 %$ is the number $\frac{24}{100}$ , and of means multiply, so the count is $\frac{24}{100}$ times $875$ .
$\frac{24}{100}$ reduces to $\frac{6}{25}$ , and $875 \div 25 = 35$ , so divide before multiplying.

Show the full solution

a = \frac{24}{100} (875) = \frac{6}{25} (875) = 6 (35) = 210

. Reducing

\frac{24}{100}

first turns the work into

875 \div 25

and

6 \times 35

, shorter than multiplying

875

24

and then dividing by

100

Problem

A hall has 650 seats and 273 of them are booked. This time

a = 273

and

b = 650

are known and

p

is missing. Solve

a = \frac{p}{100} b

for

p

and enter the percent of the seats booked, just the number.

Show a hint

Substituting gives $273 = \frac{p}{100} (650)$ , and multiplying both sides by $100$ clears the fraction.
That leaves $650 p = 27300$ , a one-letter equation of the kind 4.2 solves in one division.

Show the full solution

From

273 = \frac{p}{100} (650)

, multiplying both sides by

100

gives

650 p = 27300

, so

p = 42

. Dividing part by whole gives the same number, since

\frac{273}{650} = 0.42

and

0.42 \times 100 = 42

The third case goes wrong most often. In $a = \frac{p}{100} b$ the base is the quantity being multiplied, so when the base is what is missing the equation has to be undone rather than run forward. The next two problems are both of that kind, and each one names the number the wrong route produces.

Problem

At a theatre, 350 seats are sold, and that is

28 %

of the seats in the house. A student computes

28 %

350

and reports a house of

98

seats. Enter the correct number of seats in the house.

Show a hint

The $28 %$ is taken of the whole house, so the house is the whole $b$ and the $350$ is the amount $a$ .
$350 = \frac{28}{100} b$ , and multiplying both sides by $100$ gives $35000 = 28 b$ .

Show the full solution

From

350 = \frac{28}{100} b

, multiplying both sides by

100

gives

28 b = 35000

, so

b = 1250

. Since

350

is the part rather than the whole, the correct move is to divide by

\frac{28}{100}

instead of multiplying, and

98

is smaller than the

350

seats already sold.

Problem

An inspector passes

62 %

of a shipment and sets the other 209 items aside. The 209 is not the

62 %

part, so work out what percent of the shipment it is before writing the equation. Enter the number of items in the shipment.

Show a hint

The whole shipment is $100 %$ of itself, and the set-aside items are what is left once the passing $62 %$ is taken out.
The set-aside items are $38 %$ , so the equation is $209 = \frac{38}{100} b$ .

Show the full solution

The set-aside items are

100 - 62 = 38

percent of the shipment, so

209 = \frac{38}{100} b

gives

38 b = 20900

and

b = 550

. Dividing

209

0.62

instead pairs the count with the wrong percent and gives about

337

. The check is

\frac{62}{100} (550) = 341

, the number that passed, and

341 + 209 = 550

The same amount is a different percent of different bases, and one percent taken against different bases gives different amounts. Fifteen percent of $200$ is $30$ , and fifteen percent of $800$ is $120$ . When a story states two totals, each percent belongs to the total its own sentence names, and adding the two gives a percent of neither.

Problem

A cannery runs two filling lines. The morning line fills 720 tins and

45 %

of them hold soup. The afternoon line fills 450 tins and

16 %

of them hold soup. Each percent is taken of its own line's run. Enter the total number of tins of soup.

Show a hint

The $45 %$ is taken of the morning run only and the $16 %$ of the afternoon run only, so the two bases are different and the counts have to be worked out separately.
The morning count is $\frac{45}{100} (720)$ , and the afternoon count is the same form with $16$ as the percent and $450$ as the base.

Show the full solution

The morning count is

\frac{45}{100} (720) = 324

and the afternoon count is

\frac{16}{100} (450) = 72

, so the total is

324 + 72 = 396

. Adding the percents and taking

61 %

of all

1170

tins gives

713.7

, which is wrong because neither percent was measured against

1170

Problem

A car park holds only vans and cars, 96 vans and 384 cars. Enter the percent of all the vehicles in the car park that are vans, just the number.

Show a hint

The comparison is against every vehicle in the car park, and that total is not printed in the problem.
The whole is $96 + 384 = 480$ , so the equation is $96 = \frac{p}{100} (480)$ .

Show the full solution

The car park holds

96 + 384 = 480

vehicles, so

96 = \frac{p}{100} (480)

gives

480 p = 9600

and

p = 20

. Taking

384

as the whole instead gives

25

, which is 6.1's part-to-part comparison rather than a share of every vehicle.

The equation $a = \frac{p}{100} b$ holds for any $p$ , not only for $p$ between $1$ and $100$ . With $p$ above $100$ , $\frac{p}{100}$ is greater than $1$ , so the amount comes out larger than the base. With $p$ below $1$ , $\frac{p}{100}$ is smaller than $\frac{1}{100}$ , so the base is more than a hundred times the amount.

Problem

Depot A holds 2380 tonnes of grain, which is

175 %

of what Depot B holds. Here

\frac{175}{100}

is greater than

1

. Enter the number of tonnes Depot B holds.

Show a hint

The 2380 tonnes is compared against what Depot B holds, so Depot B is the base. Call that base $b$ .
$2380 = \frac{175}{100} b$ , and $\frac{175}{100}$ reduces to $\frac{7}{4}$ .

Show the full solution

Depot B is the base, so

2380 = \frac{175}{100} b = \frac{7}{4} b

. Multiplying both sides by

4

gives

7 b = 9520

, so

b = 1360

. With a percent above

100

the amount is larger than the base, so any answer above 2380 would be wrong.

Problem

A quarry's tailings are

0.15 %

recoverable metal by mass, and that metal weighs 42 tonnes. Of those same tailings, 5040 tonnes have already been shipped to a second site. Enter what percent of the total tailings mass has been shipped, just the number.

Show a hint

$0.15 %$ is $\frac{0.15}{100}$ , which is $0.0015$ , and the metal is that fraction of all the tailings, so the tailings mass is a missing base.
Multiplying $42 = \frac{0.15}{100} b$ by $100$ gives $0.15 b = 4200$ , so $b$ comes out by dividing by $0.15$ . With $b$ known, solve $5040 = \frac{p}{100} b$ for $p$ .

Show the full solution

From

42 = \frac{0.15}{100} b

, multiplying both sides by

100

gives

0.15 b = 4200

, so

b = 28000

tonnes, and then

5040 = \frac{p}{100} (28000)

gives

28000 p = 504000

and

p = 18

. Since the percent is below

1

, the base is more than a hundred times the amount, which is why 42 tonnes of metal comes from 28000 tonnes of tailings.

Every question here was $a = \frac{p}{100} b$ with a different one of the three letters unknown, and the work each time was deciding which number is the base. Lesson 6.4, Percent Problems, keeps that equation and takes up the case where a quantity rises or falls by a percent, and the base there is what the quantity was before the change.

Practice these ideas

Practice

Write

85 %

as a fraction in lowest terms.

Show the solution

Dividing numerator and denominator of

\frac{85}{100}

5

gives

\frac{17}{20}

. Since

17

is prime and

20

is not a multiple of

17

, no further reducing is possible.

Practice

Write

9 %

as a decimal.

Show the solution

9 %

\frac{9}{100}

, so move the decimal point in

9.0

two places left to divide by

100

, which gives

9 % = 0.09

. One place left would give

0.9

, and that is

90 %

Practice

A pressure gauge reads

0.58

of full scale. Enter that reading as a percent, just the number.

Show the solution

Multiplying by

100

gives

0.58 \times 100 = 58

. Two decimal places means hundredths, so

0.58 = \frac{58}{100}

, which is

58

per hundred.

Practice

A crate is

\frac{13}{25}

full. Rewrite that ratio with

100

below the bar and enter the percent, just the number.

Show the solution

Multiplying both parts by

4

gives

\frac{13}{25} = \frac{52}{100}

, so the percent is

52

Practice

A battery loses

2.4 %

of its charge overnight. Write

2.4 %

as a decimal.

Show the solution

Dividing by

100

slides the decimal point two places to the left, so

2.4 % = \frac{2.4}{100} = 0.024

. One hop left of

2.4

gives

0.24

, and

0.24

24 %

, so the second hop is the one that matters.

Practice

A service fee is

7.5 %

of an order. Write

7.5 %

as a fraction in lowest terms, in the form

a / b

Show the solution

Multiplying top and bottom of

\frac{7.5}{100}

10

gives

\frac{75}{1000}

, and dividing both by

25

gives

\frac{3}{40}

. Multiplying top and bottom by the same number never changes the value, so

\frac{7.5}{100}

and

\frac{3}{40}

are the same number.

Practice

A kiln fires

\frac{4}{9}

of a load at a time. Since

9

does not divide

100

, that share is not a whole number of percent. Enter it as an exact percent, written as a single fraction in lowest terms rather than a mixed number or a rounded decimal.

Show the solution

A percent is a count of hundredths, so the number of percent is

\frac{4}{9} \times 100 = \frac{400}{9}

. The numerator

400

and the denominator

9

have no common factor other than

1

, so that is lowest terms, and the decimal form

44.444 \dots

does not terminate, which makes

44.4 %

a rounding and not the exact share.

Practice

A depot stores 840 tyres and

65 %

of them are winter tyres. Enter the number of winter tyres.

Show the solution

a = \frac{65}{100} (840) = \frac{13}{20} (840) = 13 (42) = 546

Practice

A tank holds 1600 litres of brine, and

3.5 %

of it is salt. Enter the number of litres of salt.

Show the solution

a = \frac{3.5}{100} (1600) = 0.035 (1600) = 56

Practice

A polling office mails out 700 survey forms, and 189 of them come back. Enter what percent of the forms came back, just the number.

Show the solution

Let

p

be the percent, so

189 = \frac{p}{100} (700)

. Multiplying both sides by

100

gives

700 p = 18900

, so

p = 27

. Checking,

\frac{27}{100}

700

27 \times 7 = 189

Practice

A shelf holds 640 books, 80 of them reference books. Enter what percent of the books on the shelf are reference books, just the number.

Show the solution

From

80 = \frac{p}{100} (640)

, multiplying both sides by

100

gives

640 p = 8000

, so

p = 12.5

. A percent need not be a whole number.

Practice

A deposit of 144 dollars is

32 %

of a fee. Enter the fee in dollars.

Show the solution

From

144 = \frac{32}{100} b

, multiplying both sides by

100

gives

32 b = 14400

, so

b = 450

. Checking,

\frac{32}{100} (450) = 144

Practice

A club has 78 members who have paid their dues, and that is

26 %

of the membership. Multiplying

78

26 %

gives

20.28

, fewer people than have already paid. Enter the number of members in the club.

Show the solution

With

b

the number of members,

78 = \frac{26}{100} b

, and multiplying both sides by

100

gives

26 b = 7800

, so

b = 300

. The

20.28

26 %

of the part, and a club is never smaller than the number of members who have paid.

Practice

Depot X ships 250 crates and

24 %

of them are fragile. Depot Y ships 380 crates and

15 %

of them are fragile. Enter the number of fragile crates the two depots ship together.

Show the solution

The counts are

\frac{24}{100} (250) = 60

and

\frac{15}{100} (380) = 57

, so the total is

60 + 57 = 117

. The two percents cannot be added, because

24 %

is taken of

250

crates and

15 %

380

crates, and

39 %

of all

630

crates would be

245.7

Practice

A tank has a capacity of 3060 litres, which is

120 %

of a second tank's capacity. Enter the capacity of the second tank in litres.

Show the solution

From

3060 = \frac{120}{100} b = \frac{6}{5} b

, multiplying both sides by

5

gives

6 b = 15300

, so

b = 2550

. A percent above

100

means the first tank is the larger of the two, so the answer had to be less than

3060

, and

1.2 (2550) = 3060

checks.

Practice

In a batch of 45000 screws,

0.24 %

are out of tolerance. Enter the number of screws out of tolerance.

Show the solution

0.24 % = \frac{0.24}{100} = 0.0024

, so the count is

0.0024 (45000) = 108

. Using

0.24

instead gives

10800

, which is

24 %

of the batch and a hundred times too many.

Practice

In a depot,

15 %

of the vehicles are vans and the rest are trucks, and there are 630 more trucks than vans. Enter the number of vehicles in the depot.

Show the solution

Writing the total as

b

, the vans number

\frac{15}{100} b

and the trucks

\frac{85}{100} b

, so

\frac{70}{100} b = 630

and

b = 900

. The check is

135

vans and

765

trucks, a gap of

630