Joint Proportion and Rates

A quantity can depend on several others at the same time. The number of posters a print shop finishes depends on how many presses are running and on how long they run, and changing either one changes the count. The first two problems change the hours alone, then the presses and the hours together.

Problem

A print shop runs 6 presses for 4 hours and finishes 720 posters, every press printing at the same steady rate. The same 6 presses are then run for 7 hours. Enter how many posters they finish during those 7 hours.

Show a hint

Only the hours changed, since the press count stays at 6. That makes the poster count directly proportional to the hours, the direct proportion from 6.5.
The hours go from 4 to 7, so they are multiplied by $\frac{7}{4}$ , and the poster count is multiplied by that same factor.

Show the full solution

Multiplying the hours by

\frac{7}{4}

multiplies the posters by

\frac{7}{4}

, so the shop finishes

720 \times \frac{7}{4} = 1260

. Checking another way, the 6 presses together finish

720 \div 4 = 180

posters an hour, and

180 \times 7 = 1260

Problem

That same shop runs every press at the same steady rate, and 6 presses finish 720 posters in 4 hours. Enter how many posters 15 presses finish in 6 hours.

Show a hint

Change one quantity at a time. Take the presses from 6 to 15 with the hours held at 4, then take the hours from 4 to 6.
The two scale factors are $\frac{15}{6}$ and $\frac{6}{4}$ . Multiply $720$ by the first, then multiply that result by the second.

Show the full solution

Multiplying

720

\frac{15}{6}

for the presses gives

1800

, and multiplying

1800

\frac{6}{4}

for the hours gives

2700

. The hour factor multiplies the

1800

and not the original

720

, so the two factors multiply rather than add. Applying each one to

720

on its own and adding gives

1800 + 1080 = 2880

, which is wrong.

Problem

A camp's food store lasts a number of days proportional to the mass of food and inversely proportional to the number of campers. 240 kg feeds 15 campers for 32 days. Enter how many days 360 kg of food feeds 20 campers.

Show a hint

More food makes the store last longer and more campers make it run out sooner, so one change raises the number of days and the other lowers it.
The mass is multiplied by $\frac{360}{240}$ and the number of campers by $\frac{20}{15}$ , so the 32 days are multiplied by the first of those and divided by the second.

Show the full solution

Here

d = \frac{k m}{n}

, so the first case has

k = \frac{32 \times 15}{240} = 2

and then

d = \frac{2 \times 360}{20} = 36

. Scaling gives the same number, since with

\frac{3}{2}

times as much food and

\frac{4}{3}

times as many campers the count is

32 \times \frac{3}{2} \times \frac{3}{4} = 36

days.

Problem

A solar array's daily output

w

, in kilowatt-hours, is proportional to the number of panels

p

and to the hours of sun

h

, so

w = k p h

. An array of 12 panels produces 90 kilowatt-hours in 5 hours of sun. Find

k

first, then enter the output of 20 panels in 7 hours of sun, in kilowatt-hours.

Show a hint

One complete set of values is $p = 12$ , $h = 5$ and $w = 90$ , and putting all three into $w = k p h$ leaves $k$ as the only unknown.
With $k$ known, put $p = 20$ and $h = 7$ into the same equation and multiply.

Show the full solution

From

90 = k \times 12 \times 5

the constant is

k = \frac{3}{2}

, so the output of 20 panels in 7 hours is

\frac{3}{2} \times 20 \times 7 = 210

. Skipping

k

works too, since scaling by

\frac{20}{12}

for panels and

\frac{7}{5}

for hours gives

90 \times \frac{20}{12} \times \frac{7}{5} = 210

Dividing

x = k y z

y z

leaves

\frac{x}{y z} = k

, with every varying quantity on one side and the constant alone on the other. The letters take different values in the second set and the combination lands on the same

k

, so the two fractions equal each other and

k

never has to be computed.

A speed is a distance per unit of time, so a distance is a rate multiplied by a time. That is $d = r t$ , used directly in Prealgebra 7.3. At one steady speed the distance is proportional to the time, and over one fixed stretch of time it is proportional to the speed. The next two problems each leave one of the three quantities unknown.

Problem

A tram holds a steady 45 kilometers per hour for 40 minutes, then a steady 66 kilometers per hour for 30 minutes. Use

1 h = 60 min

. Enter the total distance covered, in kilometers.

Show a hint

Each stretch has its own rate and its own time, so $d = r t$ applies to each one separately and the two distances add.
The rates are per hour, so the times have to be in hours. 40 minutes is $\frac{2}{3}$ of an hour and 30 minutes is $\frac{1}{2}$ .

Show the full solution

The first stretch covers

45 \times \frac{2}{3} = 30

km and the second covers

66 \times \frac{1}{2} = 33

km, so the total in kilometers is

63

. Leaving the times in minutes gives

45 \times 40

, which is a distance in kilometer-minutes per hour and not in kilometers.

Problem

A delivery route is 288 km long. A truck covers the first 120 km at a steady 40 kilometers per hour, then covers the rest at one steady speed, and the whole route takes 6 hours. Enter the truck's speed on the second part, in kilometers per hour.

Show a hint

The first part has both a distance and a rate, so its time is fixed before anything about the second part is known.
The first part takes 3 hours, which leaves 3 of the 6 hours for the remaining 168 km.

Show the full solution

The first part takes

\frac{120}{40} = 3

hours, leaving

288 - 120 = 168

km for the remaining

6 - 3 = 3

hours, so the second speed in kilometers per hour is

56

. Dividing the whole route by the whole time gives

\frac{288}{6} = 48

, which is the trip's overall speed and not either part's.

Count a whole job as $1$ . Someone working at a steady pace who finishes in $n$ hours does the same amount in each of those hours, and $n$ of those equal amounts make the whole job, so the work done in one hour is $\frac{1}{n}$ of it. That fraction per hour is a rate, and work done is the rate times the time, the same shape as $d = r t$ .

Problem

A tiler lays a floor in 20 hours of steady work, laying the same amount in every hour. Enter the fraction of the floor laid in the first 9 hours, in lowest terms.

Show a hint

Call the whole floor $1$ , so the work done in all 20 hours totals $1$ .
A steady pace means equal work in equal hours, so the tiler lays $\frac{1}{20}$ of the floor in one hour.

Show the full solution

In one hour the tiler lays

\frac{1}{20}

of the floor, so nine hours of work amount to

9 \times \frac{1}{20} = \frac{9}{20}

. The per-hour fraction is the rate, and rate times time is the amount of the job done, the same shape as

d = r t

with the job standing in for the distance.

Problem

One crew repaves a road in 10 hours and a second crew repaves the same road in 15 hours, each at its own steady pace. Both crews work on the road at the same time. Enter how many hours the two crews working together take to repave the road.

Show a hint

In one hour the first crew repaves $\frac{1}{10}$ of the road and the second repaves $\frac{1}{15}$ , and both amounts are done during that same hour.
Add those two fractions over a common denominator to get the part of the road done in one hour, then the number of hours for one whole road is the reciprocal of that part.

Show the full solution

In one hour the crews repave

\frac{1}{10} + \frac{1}{15} = \frac{1}{6}

of the road, so the number of hours for one whole road is

6

. Adding the times gives 25 and averaging them gives 12.5, and both are longer than one crew takes alone, which cannot happen when a second crew helps.

When two objects move along the same line at the same time, the quantity to work with is how much the distance between them changes in one hour, not either speed on its own. That change is built from both speeds, and the way to build it depends on whether the two move toward each other or one follows the other.

Problem

Two ferries leave opposite ends of a 174 km channel at the same moment and travel toward each other, one at 24 kilometers per hour and the other at 34 kilometers per hour. Enter how many kilometers the faster ferry has covered when they meet.

Show a hint

What settles the meeting time is how fast the distance between the two ferries shrinks, not either speed on its own. In one hour the first covers 24 km of the channel and the second covers 34 km of it.
The 174 km gap therefore shrinks by 58 km each hour. Dividing gives the hours until they meet, and $d = r t$ on the faster ferry alone finishes it.

Show the full solution

The gap shrinks by

24 + 34 = 58

km each hour, so the ferries meet after

\frac{174}{58} = 3

hours and the faster one has covered

34 \times 3 = 102

. The slower ferry covers

24 \times 3 = 72

km, and

102 + 72 = 174

accounts for the whole channel.

Problem

A barge passes a lock at a steady 8 kilometers per hour. Three hours later a patrol boat leaves that lock along the same route in the same direction at a steady 20 kilometers per hour. Enter how far from the lock, in kilometers, the patrol boat catches the barge.

Show a hint

The patrol boat starts at the lock and the barge does not. Work out how far along the route the barge already is at the moment the patrol boat leaves.
Each hour the patrol boat covers 20 km and the barge covers 8 km, so after each hour the 24 km head start is 12 km smaller.

Show the full solution

The barge is

8 \times 3 = 24

km ahead when the patrol boat starts and the gap is

20 - 8 = 12

km smaller after each hour, so the catch takes

\frac{24}{12} = 2

hours and the distance from the lock in kilometers is

20 \times 2 = 40

. The barge has then been moving 5 hours, and

8 \times 5 = 40

agrees.

Every relationship in this chapter was one combination staying fixed while its parts changed. 6.1 fixed $\frac{a}{b}$ , 6.2 multiplied by a fraction worth $1$ , 6.3 and 6.4 fixed the second part of a comparison at 100, and 6.5 fixed $\frac{y}{x}$ or $x y$ . The combination fixed here is longer, and the last two problems use it.

Problem

The number of bricks a crew lays is proportional to the number of masons and to the number of days worked. Eight masons lay 2016 bricks in 6 days. A contractor needs 3780 bricks laid in 5 days. Enter how many masons that takes.

Show a hint

Write $b$ for bricks, $m$ for masons and $d$ for days. Joint proportion means $\frac{b}{m d}$ has the same value for both jobs, and here the unknown is in the denominator.
Eight masons working 6 days is 48 mason-days for 2016 bricks. Find the bricks one mason lays in one day, then ask how many masons put 3780 bricks into 5 days.

Show the full solution

With

b

bricks,

m

masons and

d

days, joint proportion gives

\frac{b}{m d} = k

, and from the first job

k = \frac{2016}{8 \times 6} = 42

bricks per mason per day. Then

3780 = 42 \times m \times 5

, so the number of masons is

18

. Scaling works without

k

too. The bricks are multiplied by

\frac{15}{8}

and the days by

\frac{5}{6}

, so the masons are multiplied by

\frac{15}{8} \div \frac{5}{6} = \frac{9}{4}

, and

8 \times \frac{9}{4} = 18

Problem

A workshop's looms are identical and each one weaves at the same steady rate. Six of them weaving for 5 hours produce 210 meters of cloth. Four of them are set to produce 252 meters of the same cloth. Enter the number of hours that takes.

Show a hint

How much cloth comes off depends on both how many looms run and how long they run, so work out first how many meters one loom weaves in one hour.
Six looms for 5 hours is 30 loom-hours of weaving in all, so $210 \div 30$ gives the meters per loom-hour. Four looms running together weave four times that much each hour.

Show the full solution

Six looms for 5 hours is 30 loom-hours, and

210 \div 30 = 7

meters per loom-hour. Four looms weave

4 \times 7 = 28

meters each hour, so the number of hours is

252 \div 28 = 9

. The cloth produced is jointly proportional to the number of looms and to the number of hours, so the single rate of 7 meters per loom-hour accounts for both changes at once.

Every relationship in this lesson had one constant and several quantities that change, and the work was always to gather the changing ones on one side of the bar and read the constant off the other. Distance, work rate and closing rate all have that shape. That closes chapter 6. Chapter 7, Graphing Lines, opens with 7.1, The Cartesian Plane.

Practice these ideas

Practice

The mass of a metal plate is proportional to its length and to its width. A plate 4 cm long and 9 cm wide has a mass of 54 grams. Enter the mass, in grams, of a plate 4 cm long and 14 cm wide.

Show the solution

Only the width changed, by a factor of

\frac{14}{9}

, so the mass in grams is

54 \times \frac{14}{9} = 84

. Nine divides 54 exactly six times, so the product reduces to

6 \times 14

Practice

W

is jointly proportional to

u

and

v

, so

W = k u v

for one constant

k

. When

u = 6

and

v = 4

W = 264

. Enter the value of

k

Show the solution

Substituting gives

264 = 24 k

, so

k = \frac{264}{24} = 11

. Checking it back,

11 \times 6 \times 4 = 264

Practice

A worker's pay is proportional to the number of shifts worked and to the hours in each shift. Four shifts of 6 hours pay 528 dollars in total. Enter the pay in dollars for 5 shifts of 9 hours.

Show the solution

The constant is

\frac{528}{4 \times 6} = 22

dollars per hour, so the pay for 5 shifts of 9 hours is

22 \times 5 \times 9 = 990

. Since

5

shifts is

\frac{5}{4}

4

and

9

hours is

\frac{9}{6}

6

, the pay is also

528 \times \frac{5}{4} \times \frac{9}{6}

, which is

990

again.

Practice

P

is jointly proportional to

q

and

r

, and

P = 189

when

q = 3

and

r = 9

. Enter

P

when

q = 5

and

r = 4

Show the solution

The constant is

k = \frac{189}{27} = 7

, so

P = 7 \times 5 \times 4 = 140

. Scaling agrees, since

189 \times \frac{5}{3} \times \frac{4}{9} = 140

Practice

y

is directly proportional to

m

and inversely proportional to

n

. When

m = 8

and

n = 6

y = 24

. Enter

y

when

m = 15

and

n = 10

Show the solution

Grouping gives

\frac{y n}{m} = k = \frac{24 \times 6}{8} = 18

, so

y = \frac{18 \times 15}{10} = 27

. Scaling agrees, since

24 \times \frac{15}{8} \times \frac{3}{5} = 27

Practice

The number of days a water store lasts is proportional to the volume stored and inversely proportional to the number of households drawing on it. 900 liters lasts 12 households 15 days. Enter the number of days 1200 liters lasts 20 households.

Show the solution

Substituting the first store's numbers into

d = \frac{k V}{H}

gives

k = \frac{15 \times 12}{900} = \frac{1}{5}

, and then

d = \frac{1}{5} \cdot \frac{1200}{20} = 12

. The households grew by the larger factor,

\frac{5}{3}

against

\frac{4}{3}

for the volume, so the number of days is below 15.

Practice

A tram covers 78 km in 65 minutes at a steady speed. Enter how many kilometers it covers in 100 minutes at that same speed.

Show the solution

The tram covers

\frac{78}{65} = 1.2

km each minute, so in 100 minutes it covers

1.2 \times 100 = 120

. Scaling the distance by the same factor as the time lands on the same number, since

78 \times \frac{100}{65} = 120

Practice

A survey drone covers 189 km in 3 hours 30 minutes at a steady speed. Use

1 h = 60 min

. Enter the drone's speed in kilometers per hour.

Show the solution

The time is

\frac{7}{2}

hours, so the speed in kilometers per hour is

189 \div \frac{7}{2} = 189 \times \frac{2}{7} = 54

Practice

A painter finishes a mural in 15 hours, covering the same amount of wall in every hour. Enter the fraction of the mural that is finished once 6 hours of work are done, in lowest terms.

Show the solution

One hour covers

\frac{1}{15}

of the mural, so six hours cover

\frac{6}{15}

, which reduces to

\frac{2}{5}

. Working at a steady rate makes the fraction finished the same as the fraction of the time spent.

Practice

One conveyor loads a barge in 14 hours and a second conveyor loads the same barge in 35 hours, each at its own steady rate. Both run at once on one barge. Enter how many hours they take to load it.

Show the solution

In one hour the conveyors load

\frac{1}{14} + \frac{1}{35} = \frac{5}{70} + \frac{2}{70} = \frac{7}{70} = \frac{1}{10}

of the barge, so the number of hours is

10

. Loading one tenth of the barge each hour means the whole barge takes ten hours, while adding the two times would give 49 hours, longer than either conveyor needs on its own.

Practice

Pumps A and B running together drain a tank in 12 hours, and pump A alone drains the tank in 20 hours. Enter how many hours pump B alone takes to drain the tank.

Show the solution

Pump B drains

\frac{1}{12} - \frac{1}{20} = \frac{5}{60} - \frac{3}{60} = \frac{2}{60} = \frac{1}{30}

of the tank per hour, so the number of hours is

30

. Checking,

\frac{1}{20} + \frac{1}{30} = \frac{1}{12}

Practice

Two drones 245 km apart set off at the same moment and fly toward each other along the same line, one at 42 kilometers per hour and the other at 28 kilometers per hour. Enter how many kilometers the faster drone has flown when they meet.

Show the solution

The gap closes at

42 + 28 = 70

km each hour, so they meet after

\frac{245}{70} = \frac{7}{2}

hours and the faster drone has flown

42 \times \frac{7}{2} = 147

. The slower drone flies

28 \times \frac{7}{2} = 98

km, and

147 + 98 = 245

accounts for the whole gap.

Practice

A hiker leaves a trailhead at a steady 7 kilometers per hour. Four hours later a cyclist leaves the same trailhead on the same path at a steady 21 kilometers per hour. Enter how far from the trailhead, in kilometers, the cyclist draws level with the hiker.

Show the solution

The hiker is

7 \times 4 = 28

km ahead when the cyclist starts, and the lead shrinks by

21 - 7 = 14

km each hour, so the catch takes 2 hours and the distance from the trailhead in kilometers is

21 \times 2 = 42

. The hiker has walked 6 hours by then, and

7 \times 6 = 42

agrees.

Practice

Sprinkler C waters a field in 10 hours and sprinkler D waters the same field in 40 hours, each at its own steady rate. Both run together for 4 hours, then sprinkler C is switched off and sprinkler D finishes alone. Enter how many further hours sprinkler D needs.

Show the solution

Together they water

\frac{1}{10} + \frac{1}{40} = \frac{5}{40} = \frac{1}{8}

of the field an hour, so four hours finish

\frac{4}{8} = \frac{1}{2}

of it. Sprinkler D covers the remaining half at

\frac{1}{40}

an hour, needing

\frac{1 / 2}{1 / 40} = 20

further hours. Adding the two solo times, or averaging them, answers a different question.

Practice

A packing plant running 8 machines for 5 hours a day takes 12 days to finish an order, every machine working at the same steady rate. The next order is twice as large and must be finished in 10 days using 12 machines. Enter how many hours a day those machines must run.

Show the solution

The first order is

8 \times 5 \times 12 = 480

machine-hours, so the second is

960

, and

960 \div (12 \times 10) = 8

. Forgetting to double the order gives

480 \div 120 = 4

instead.

Practice

A data hall's running cost is proportional to the number of racks and to the hours online, and inversely proportional to its efficiency rating. 20 racks online 6 hours at rating 4 cost 900 dollars. Enter the cost in dollars of 28 racks online 10 hours at rating 8.

Show the solution

Grouping gives

\frac{C e}{r h} = k

, and the first set fixes

k = \frac{900 \times 4}{120} = 30

. Then

C = \frac{30 \times 28 \times 10}{8} = 1050

Practice

Two cyclists ride toward each other along a 190 km road. The first sets out from one end at a steady 15 kilometers per hour. Two hours later the second sets out from the far end at a steady 25 kilometers per hour. Enter how far from the first cyclist's start, in kilometers, the two meet.

Show the solution

The first cyclist covers

15 \times 2 = 30

km before the second starts, leaving a 160 km gap, and the two then close on each other at

15 + 25 = 40

km per hour, so they meet 4 hours after the second sets out. The first has ridden 6 hours by then, so the distance in kilometers is

15 \times 6 = 90

. The second rides

25 \times 4 = 100

km, and

90 + 100 = 190