Ratios, Simple and Subtle

Chapter 5 removed letters from a system one at a time. Chapter 6 opens by putting one in. Prealgebra chapter 7 settled what $a : b$ says. What is new here is algebraic. Naming the size of one part with a letter turns a ratio statement into an equation, and chapters 4 and 5 already solve equations. Start with a ratio and a total.

Problem

A bindery finishes 552 booklets, stapled to sewn in the ratio

7 : 5

. Doubling both parts of

7 : 5

gives

14 : 10

, which totals

24

, and tripling gives

21 : 15

, which totals

36

. Keep scaling both parts by the same number until the two counts total

552

, then enter the number of stapled booklets.

Show a hint

The scaled totals $24$ and $36$ are both multiples of $12$ , and $12$ is $7 + 5$ . Scaling both parts by the same number multiplies the total by that number as well.
$552 \div 12 = 46$ , so both parts of $7 : 5$ get multiplied by $46$ .

Show the full solution

Since

7 + 5 = 12

and

552 \div 12 = 46

, both parts get multiplied by

46

, and the stapled count is

7 (46) = 322

. The sewn count is

5 (46) = 230

, and

322 + 230 = 552

. Every scaled total is a multiple of

12

, so dividing

552

12

gives the multiplier in one step instead of testing

2

3

4

and on up.

Problem

A stadium has 493 seats, covered seats to open seats in the ratio

9 : 8

. Write the two counts as

9 k

and

8 k

for a single number

k

, form one equation from the total, and enter the number of covered seats.

Show a hint

Let $k$ be the size of one part, so the covered seats number $9 k$ , the open seats number $8 k$ , and the two counts together make $493$ .
$9 k + 8 k$ collects to $17 k$ , so the equation reads $17 k = 493$ , a one-letter equation of the kind 4.2 solves.

Show the full solution

9 k + 8 k = 493

collects to

17 k = 493

, so

k = 29

and the covered seats number

9 (29) = 261

. The open seats number

8 (29) = 232

, and

261 + 232 = 493

. Naming the part size

k

makes the whole problem the single division

493 \div 17

Problem

A workshop owns 378 hand tools, all of them hammers or chisels, with hammers to chisels in the ratio

5 : 9

. A student computes

\frac{5}{9}

378

and reports

210

hammers. Enter the correct number of hammers.

Show a hint

The ratio $5 : 9$ compares hammers with chisels, while $378$ is the total of both kinds. A fraction of that total is taken over both kinds together, not over the chisels alone.
The tools split into $5 + 9 = 14$ equal parts, and the hammers are $5$ of those parts.

Show the full solution

The tools split into

5 + 9 = 14

equal parts, so one part is

378 \div 14 = 27

tools and the number of hammers is

5 (27) = 135

. There are

9 (27) = 243

chisels, and

135 + 243 = 378

. Hammers are

\frac{5}{14}

of the tools, not

\frac{5}{9}

, and with

210

hammers only

168

chisels remain, more hammers than chisels, the reverse of

5 : 9

Problem

A nursery grows 984 saplings, oak, maple and pine, in the ratio

6 : 7 : 11

. Write the three counts as

6 k

7 k

and

11 k

, and enter the number of pine saplings.

Show a hint

All three counts are built from the same $k$ , so they add to $6 k + 7 k + 11 k$ , and that sum is $984$ .
$6 k + 7 k + 11 k$ collects to $24 k$ , so $24 k = 984$ .

Show the full solution

6 k + 7 k + 11 k = 984

collects to

24 k = 984

, so

k = 41

and the pine count is

11 (41) = 451

. The others are

6 (41) = 246

oak and

7 (41) = 287

maple, and

246 + 287 + 451 = 984

. With three parts the sum has one more term than with two, and the rest of the method is the same.

Every block is the same size, and

k

is what one block is worth. The ratio

5 : 2

says the first amount is five blocks and the second is two, so the amounts are

5 k

and

2 k

and the total is

7 k

, whatever

k

turns out to be.

Writing $x : y = a : b$ is writing $\frac{x}{y} = \frac{a}{b}$ , and that is an equation in two letters. Multiply both sides by $y$ and then by $b$ , which is 4.4's move done twice, and what is left is $b x = a y$ . Each numerator is multiplied whole, parentheses and all. Neither side is a fraction any more, so $b x = a y$ is linear in $x$ and $y$ .

Problem

The crates and pallets in a depot are in the ratio

12 : 7

, and the depot has 224 pallets. Let

c

be the number of crates, so that statement is the equation

\frac{c}{224} = \frac{12}{7}

. Cross-multiply, solve, and enter the number of crates.

Show a hint

Both sides are single fractions, so clear the denominators. Multiply both sides by $224$ and then by $7$ , and see what is left.
Clearing gives $7 c = 224 \times 12$ . Since $224 = 7 \times 32$ , divide by $7$ before multiplying by $12$ , which keeps the numbers small.

Show the full solution

Cross-multiplying

\frac{c}{224} = \frac{12}{7}

gives

7 c = 2688

, so

c = 384

. Check,

\frac{384}{224}

reduces to

\frac{12}{7}

. In the ratio, crates come first, so

12

is the numerator on the same side as

c

. With the

12

and

7

swapped the result would be

\frac{224 \times 7}{12}

, not a whole number of crates.

Problem

The ratio of

5 (x - 4) + 3 x

2 (x + 21)

3 : 1

. Collect like terms on each side first, then cross-multiply and enter the value of

x

Show a hint

A ratio of $3 : 1$ says the first quantity is $3$ times the second. Distribute the 2.1 way and combine like terms the 2.4 way on each side, then write that sentence as an equation.
The sides collect to $8 x - 20$ and $2 x + 42$ , so the statement reads $8 x - 20 = 3 (2 x + 42)$ .

Show the full solution

Distributing gives

5 x - 20 + 3 x = 8 x - 20

on the left and

2 x + 42

on the right, so

\frac{8 x - 20}{2 x + 42} = \frac{3}{1}

, and cross-multiplying gives

8 x - 20 = 6 x + 126

, then

2 x = 146

and

x = 73

. Check,

564

188

3

1

. Writing

5 (x - 4)

5 x - 4

instead gives

x = 65

, where the sides are

500

and

172

, not a

3 : 1

pair.

An amount can be added to one part, taken from one part, or moved between the two, and the problem states the new ratio. Write the starting amounts as $a k$ and $b k$ , adjust each by what moved, and that second statement is one equation in $k$ . The two ratios have separate multipliers, so $13 k$ and $4 k$ cannot also be $10 k$ and $3 k$ .

Problem

A club's juniors and seniors are in the ratio

11 : 8

. Then 42 juniors join, no one leaves, and the ratio of juniors to seniors is now

3 : 2

. Enter how many juniors the club had before those 42 joined.

Show a hint

The senior count never changes, so write both starting counts from the first ratio and adjust only the junior count.
With $11 k$ juniors and $8 k$ seniors, the counts afterwards are $11 k + 42$ and $8 k$ , so the second statement is $\frac{11 k + 42}{8 k} = \frac{3}{2}$ .

Show the full solution

With

11 k

juniors and

8 k

seniors, the ratio after the 42 join is

\frac{11 k + 42}{8 k} = \frac{3}{2}

. Cross-multiplying gives

22 k + 84 = 24 k

, so

2 k = 84

and

k = 42

, and the juniors before were

11 (42) = 462

. Check, the

336

seniors are unchanged and

504 : 336

reduces to

3 : 2

, so the

504

is the count after, not the one asked for. Naming both counts instead gives the system

8 j = 11 s

and

2 j + 84 = 3 s

, and elimination reaches the same

j = 462

The same quantity can appear in two ratios with a different number in each. In $x : y = 7 : 6$ the quantity $y$ is $6$ parts, and in $y : z = 9 : 2$ it is $9$ , and those parts are not the same size. Scale each ratio, both parts by the same factor, until the two numbers for $y$ agree. The three amounts are then in one ratio $x : y : z$ .

Problem

In a stockroom, the ratio of blue crates to green crates is

15 : 8

, and the ratio of green crates to red crates is

8 : 11

. There are

165

blue crates. Enter the number of red crates.

Show a hint

Both ratios use the same number, $8$ , for the green crates, so blue, green, and red fit into one three-part ratio without rescaling either ratio.
Write that three-part ratio, then compare its blue part with the $165$ actual blue crates to get the multiplier that turns parts into counts.

Show the full solution

Both ratios use

8

for green, so blue to green to red is

15 : 8 : 11

. The blue part gives

15 k = 165

, so

k = 11

, and the red count is

11 k = 121

. Green is

8 k = 88

, and the checks pass,

165 : 88 = 15 : 8

and

88 : 121 = 8 : 11

. The chain is direct only because the shared quantity has the same number in both ratios.

Problem

For three quantities

x

y

, and

z

x : y = 16 : 9

and

y : z = 6 : 13

. Enter

\frac{z}{x}

as a fraction in lowest terms.

Show a hint

In one statement $y$ is $9$ parts and in the other it is $6$ , and those parts are not the same size. Scale each ratio, multiplying both of its parts by the same factor, until the two numbers for $y$ match.
The least common multiple of $9$ and $6$ is $18$ , so double both parts of $16 : 9$ and triple both parts of $6 : 13$ .

Show the full solution

Doubling

16 : 9

gives

32 : 18

and tripling

6 : 13

gives

18 : 39

, so with

y

18

parts in both,

x : y : z = 32 : 18 : 39

and

\frac{z}{x} = \frac{39}{32}

. Stacking the printed numbers gives

\frac{13}{16}

instead, which is the trap, since the

9

and the

6

stand for different sizes of part.

A ratio is one equation in two letters, so neither amount follows from it alone. Adding a second fact about the same two amounts makes a system, and 5.1 counted two equations in two letters as usually enough. It can be a total, a difference, one amount, or the ratio after a change. Cross-multiply the ratio, then use substitution or elimination.

Problem

A caterer orders bread rolls and pastries, rolls to pastries in the ratio

11 : 6

. Rolls cost 2 dollars each, pastries cost 5 dollars each, and the order comes to 468 dollars. Enter the number of pastries.

Show a hint

The ratio is one fact about the two counts and the total cost is another, so there are two equations in two letters here.
With $r$ rolls and $p$ pastries, cross-multiplying the ratio gives $6 r = 11 p$ , and the cost gives $2 r + 5 p = 468$ . Tripling the cost equation makes its $r$ -term $6 r$ , which matches the ratio equation.

Show the full solution

Cross-multiplying the ratio gives

6 r = 11 p

, and the cost gives

2 r + 5 p = 468

. Tripling the cost equation makes it

6 r + 15 p = 1404

, so replacing

6 r

with

11 p

leaves

26 p = 1404

and

p = 54

. Then

r = 99

, and

2 (99) + 5 (54) = 468

checks, though the

99

is the roll count and not the ask. Writing the counts as

11 k

and

6 k

instead gives

52 k = 468

and the same pair.

Problem

A library shelves fiction and nonfiction in the ratio

11 : 7

. Then 45 books move from fiction to nonfiction, nothing is added or discarded, and the ratio of fiction to nonfiction reads

4 : 3

. Enter how many fiction books the library had before the move.

Show a hint

After the move there are $45$ fewer fiction books and $45$ more nonfiction books. Write both starting counts with one letter, using the first ratio.
With $11 k$ fiction and $7 k$ nonfiction, the second ratio gives $3 (11 k - 45) = 4 (7 k + 45)$ .

Show the full solution

With

11 k

fiction and

7 k

nonfiction the new ratio gives

\frac{11 k - 45}{7 k + 45} = \frac{4}{3}

, so

33 k - 135 = 28 k + 180

, then

5 k = 315

and

k = 63

. The fiction count before the move was

11 (63) = 693

. The nonfiction count was

441

, and after the move

648 : 486

reduces to

4 : 3

. The usual slip is adding the

45

to fiction and taking it off nonfiction, which gives

5 k = - 315

and a negative count, so it rules itself out.

Problem

A festival sells full-price, student, and child tickets. Of the tickets sold, full-price to student is

7 : 4

, and student to child is

6 : 11

. The festival then sells another 75 child tickets, with no change to the other two counts, and full-price to child is now

7 : 9

. Enter the number of student tickets sold.

Show a hint

In one ratio the student tickets are $4$ parts and in the other they are $6$ , and those parts are not the same size. Combine the two statements into a single three-part ratio before naming any letter.
Matching the student number at $12$ makes the three-part ratio $21 : 12 : 22$ , so the counts are $21 k$ , $12 k$ and $22 k$ , and the last fact cross-multiplies to $9 (21 k) = 7 (22 k + 75)$ .

Show the full solution

Matching the student number at

12

turns

7 : 4

into

21 : 12

and

6 : 11

into

12 : 22

, so the counts are

21 k

12 k

and

22 k

. The last fact gives

\frac{21 k}{22 k + 75} = \frac{7}{9}

, so

189 k = 154 k + 525

, then

35 k = 525

and

k = 15

. The student count is

12 (15) = 180

. Check, full-price

315

against child

405

7 : 9

. The letter solved for is

k

, so

15

is a part size and not a ticket count.

Every problem here put one letter on the size of a part or cross-multiplied a ratio into $b x = a y$ , and no count came without a second fact. Lesson 6.2, Conversion Factors, uses a ratio of another kind, where the two parts are one amount written in two units. The question there is what such a ratio is worth and what multiplying by it does.

Practice these ideas

Practice

A crate holds 340 tiles, each one either plain or patterned. The ratio of plain tiles to patterned tiles is

9 : 11

. Enter the number of plain tiles.

Show the solution

9 k + 11 k = 340

gives

20 k = 340

, so

k = 17

and the plain tiles number

9 (17) = 153

. The patterned count is

11 (17) = 187

, and

153 + 187 = 340

Practice

A crate holds pears and quinces and nothing else, in the ratio

6 : 11

. Enter the fraction of the fruit that is quinces, in lowest terms.

Show the solution

The counts are

6 k

and

11 k

, so the crate holds

17 k

pieces and the quinces are

\frac{11 k}{17 k}

of it, which is

\frac{11}{17}

. A common wrong answer is

\frac{11}{6}

, the quince count over the pear count rather than over the whole crate, and a share of the crate cannot be more than

1

Practice

In a choir, the ratio of altos to sopranos is

8 : 13

, and there are 296 altos. Enter the number of sopranos.

Show the solution

The altos are

8

parts, so

8 k = 296

and

k = 37

, which makes the sopranos

13 (37) = 481

. Check,

296 : 481

reduces to

8 : 13

. No total is stated here and none is needed, since one amount fixes

k

on its own.

Practice

A jar holds only amber, jade and onyx marbles, in the ratio

9 : 16 : 5

. Enter the fraction of the marbles in the jar that are jade, in lowest terms.

Show the solution

The counts are

9 k

16 k

and

5 k

, so the jar holds

30 k

marbles and the jade share is

\frac{16 k}{30 k} = \frac{8}{15}

. With three parts the share of the whole is still one part over the sum of all the parts, and

\frac{16}{9}

\frac{16}{5}

would each compare jade with one other colour instead.

Practice

A silo holds 494 sacks of wheat, rye and oats in the ratio

5 : 9 : 12

. Enter the number of sacks of oats.

Show the solution

Write the counts as

5 k

9 k

, and

12 k

. Then

26 k = 494

, so

k = 19

and the oat count is

12 (19) = 228

. The other two are

5 (19) = 95

wheat and

9 (19) = 171

rye, and

95 + 171 + 228 = 494

Practice

A workshop stocks screws, bolts and rivets in the ratio

11 : 6 : 4

, and there are 156 bolts. Enter the total number of screws, bolts and rivets.

Show the solution

Each part is

k

pieces, and the bolts are

6

parts, so

6 k = 156

and

k = 26

. The three counts together are

11 + 6 + 4 = 21

parts, so the total is

21 \times 26 = 546

. Counting by kind, that is

286

screws,

156

bolts and

104

rivets, and

286 + 156 + 104 = 546

Practice

The equation

\frac{x}{84} = \frac{17}{12}

holds. Cross-multiply and enter

x

Show the solution

Cross-multiplying gives

12 x = 1428

, so

x = 119

. Cancelling the

12

into the

84

first turns the whole thing into

17 \times 7

, which is faster than multiplying and then dividing.

Practice

Cross-multiply

\frac{7 x - 12}{2 x + 50} = \frac{5}{3}

, collect the

x

-terms on one side and the numbers on the other, and enter

x

Show the solution

Cross-multiply,

3 (7 x - 12) = 5 (2 x + 50)

, so

21 x - 36 = 10 x + 250

, then

11 x = 286

and

x = 26

. Check, at

x = 26

the fraction is

\frac{170}{102}

, which reduces to

\frac{5}{3}

Practice

Two accounts hold money in the ratio

17 : 8

, and the first holds 288 dollars more than the second. Enter the balance of the smaller account, in dollars.

Show the solution

With balances

17 k

and

8 k

, the gap is

17 k - 8 k = 9 k

, so

9 k = 288

and

k = 32

, and the smaller balance is

8 (32) = 256

. The other balance is

544

, and

544 - 256 = 288

, with

544 : 256

equal to

17 : 8

in lowest terms. A difference is enough to find

k

, the same as a total would be.

Practice

A stall sells notebooks and folders in the ratio

9 : 7

. Notebooks cost 6 dollars each, folders cost 11 dollars each, and the stall takes 1572 dollars for them. Enter the number of folders sold.

Show the solution

With

9 k

notebooks and

7 k

folders, the money gives

54 k + 77 k = 1572

, so

131 k = 1572

and

k = 12

, which makes the folder count

7 (12) = 84

. There are

108

notebooks, and

648 + 924 = 1572

checks. The ratio alone fixes neither count, and the takings are the second fact that does.

Practice

Two positive whole numbers are in the ratio

13 : 6

, and both are less than

200

. Enter how many such pairs there are.

Show the solution

Every such pair is

13 k

and

6 k

for a positive whole number

k

, and

6 k

is the smaller of the two, so the only condition is that

13 k

stays under

200

. Since

13 \times 15 = 195

and

13 \times 16 = 208

k

can be

1

through

15

, a count of

15

. Without the limit there would be one such pair for every

k

, endlessly many.

Practice

In a model kit, pegs and struts come in the ratio

10 : 3

, and struts and panels come in the ratio

3 : 4

. A batch of these parts has

76

panels. Enter the number of pegs in the batch.

Show the solution

The struts match at

3

in both ratios, so pegs to struts to panels is

10 : 3 : 4

. The panels give

4 k = 76

, so

k = 19

and the pegs number

10 (19) = 190

. If the two strut counts had differed, both ratios would need scaling to a common strut count first.

Practice

A field holds 682 plants, all of them corn, beans or squash. Corn to beans is

6 : 5

and beans to squash is

10 : 9

. Enter the number of squash plants.

Show the solution

Doubling

6 : 5

gives

12 : 10

, which matches the beans in

10 : 9

, so corn to beans to squash is

12 : 10 : 9

. The three parts add to

31

, one part is

\frac{682}{31} = 22

plants, and squash is

9 (22) = 198

. Corn and beans come out to

264

and

220

, and

264 + 220 + 198 = 682

Practice

A survey counts

m

walkers,

n

cyclists and

p

drivers, with

m

n

5 : 12

and

n

p

8 : 9

. Scale so the two numbers for

n

agree, then enter

\frac{p}{m}

as a fraction in lowest terms.

Show the solution

Doubling

5 : 12

gives

10 : 24

and tripling

8 : 9

gives

24 : 27

, so

m : n : p = 10 : 24 : 27

and

\frac{p}{m} = \frac{27}{10}

. Using the printed numbers straight gives

\frac{9}{5}

, which is wrong because

n

12

parts in the first ratio and

8

parts in the second.

Practice

A pantry stocks honey jars and jam jars in the ratio

10 : 7

. After 48 more honey jars arrive and no new jam jars, the ratio of honey to jam is

8 : 5

. Enter the number of jam jars.

Show the solution

Write the counts as

10 k

honey jars and

7 k

jam jars. Then

\frac{10 k + 48}{7 k} = \frac{8}{5}

, and cross-multiplying gives

50 k + 240 = 56 k

, so

6 k = 240

k = 40

and the jam count is

7 (40) = 280

. Check, the honey counts are

400

and

448

, and

448 : 280

reduces to

8 : 5

. The

400

is honey, not the count asked for.

Practice

A vault holds silver and copper coins, and the ratio of silver to copper is

9 : 4

. The owner removes 32 silver coins and puts in 32 copper coins, after which the ratio of silver to copper is

5 : 4

. Enter the number of silver coins the vault held at the start.

Show the solution

With

9 k

silver and

4 k

copper at the start, the new ratio gives

4 (9 k - 32) = 5 (4 k + 32)

, so

36 k - 128 = 20 k + 160

, then

16 k = 288

and

k = 18

. The starting silver count is

9 (18) = 162

. The vault began with

162

silver and

72

copper, and afterwards

130 : 104

reduces to

5 : 4

Practice

At a gallery, the ratio of prints to paintings is

16 : 7

. Then 26 prints are sold and 13 paintings are added, and the ratio of prints to paintings is now

5 : 3

. Enter the total number of prints and paintings after the change.

Show the solution

With

16 k

prints and

7 k

paintings, the new ratio gives

3 (16 k - 26) = 5 (7 k + 13)

, so

48 k - 78 = 35 k + 65

, then

13 k = 143

and

k = 11

. The gallery held

176

prints and

77

paintings, and afterwards

150

and

90

, a total of

240

. The check is that

150 : 90

reduces to

5 : 3

, and the total fell from

253

because 26 works left and only 13 came in.

Practice

A cabinet holds oak, ash and birch dowels and nothing else, with oak to ash

9 : 5

and ash to birch

15 : 8

. After 105 more birch dowels arrive, the birch count equals the ash count. Enter the total number of dowels the cabinet held before that delivery.

Show the solution

Triple

9 : 5

to get

27 : 15

, so both statements now use

15

for the ash and oak to ash to birch is

27 : 15 : 8

. With the counts

27 k

15 k

and

8 k

, equal counts after the delivery mean

8 k + 105 = 15 k

, so

7 k = 105

and

k = 15

. The parts total

50

, so the cabinet held

50 (15) = 750

. The counts are

405

oak,

225

ash and

120

birch, and

120 + 105 = 225

as required.