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Rate and Work Problems: Core practice

10 practice problems for this lesson. Work on paper, use hints when you need them, and check the answer or the full solution when you are ready.

Difficulty: Core (core-course level)

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Problem 1 of 10

Problem 1 The print run

A printer finishes one print run in $40$ minutes at a steady rate. What is its rate in print runs per hour?
1. Hint 1
  A rate is the fraction of a print run finished in one unit of time, and the unit asked for here is an hour.
2. Hint 2
  Write $40$ minutes as a fraction of an hour, then take the reciprocal of that time.
Answer

$\frac32$ print runs per hour, or $1.5$ print runs per hour.

Full solution

Forty minutes is $\frac{40}{60}=\frac23$ of an hour, so the completion time is $T=\frac23$ hour.

A job finished in time $T$ is done at the rate $\frac1T$ per unit of time.
Here the rate is
$\frac{1}{2/3}=\frac32$
print runs per hour.

As a check, in one hour the printer finishes one run in the first $40$ minutes and half of another run in the remaining $20$ minutes, which is $1.5$ runs.

Answer

$\frac32$ print runs per hour, or $1.5$ print runs per hour.

Key idea

Before turning a completion time into the rate $\frac1T$ , write the time in the unit the rate is asked for.
Problem 1: Mark complete
Problem 2 The road markers

A cart moves at a steady speed from the $21$ mile marker to the $45$ mile marker in $1.5$ hours. What is its speed in miles per hour?
1. Hint 1
  The distance traveled is the change in marker reading.
2. Hint 2
  Divide that distance by the elapsed time.
Answer

$16$ miles per hour.

Full solution

The traveled distance is $45-21=24$ miles.
The speed is
$r=\frac{24}{1.5}$ $r=16$
in miles per hour.

At this speed, $16(1.5)=24$ miles, matching the change in marker reading.

Answer

$16$ miles per hour.

Key idea

For travel in one direction along a route, the distance covered is the difference between the position readings, not the final reading alone.
Problem 2: Mark complete
Problem 3 The replaced helper

Workers A and B complete $\frac9{20}$ of a job per hour together. A works at $\frac15$ of a job per hour. Worker C replaces A and works at $\frac3{20}$ of a job per hour. All rates are steady, with no duplicated work or interference. Find the combined rate of the new team, B and C.
1. Hint 1
  First recover the contribution of B from the original team.
2. Hint 2
  Subtract the rate of A from the original combined rate, then add the rate of C.
Answer

$\frac25$ of a job per hour, or $0.4$ of a job per hour.

Full solution

The rates of A and B add to the team rate, so the rate of B is what remains after removing A, whose rate is $\frac15=\frac4{20}$ .
$\frac9{20}-\frac4{20}=\frac5{20}$
So B works at $\frac5{20}=\frac14$ of a job per hour.

The new team is B and C, so its rate is the sum of their rates.
$\frac5{20}+\frac3{20}=\frac8{20}$
This is $\frac25$ of a job per hour.

Adding B back to A gives $\frac15+\frac14=\frac9{20}$ , confirming the recovered rate.

Answer

$\frac25$ of a job per hour, or $0.4$ of a job per hour.

Key idea

When every member's rate but one is known, a team rate reveals the missing rate, which can then be added into a new team.
Problem 3: Mark complete
Problem 4 The delayed helper

Worker A can finish a job alone in $9$ hours and worker B in $3$ hours. A works alone for $3$ hours, then B joins until the job is complete. Both work steadily without repeating work or interfering. Find the total time from A starting to completion.
1. Hint 1
  Find the completed fraction before the team starts.
2. Hint 2
  The remaining fraction is divided by the combined rate after B joins.
Answer

$\frac92$ hours, or $4.5$ hours, which is $4$ hours $30$ minutes.

Full solution

A finishes the job in $9$ hours, so A works at $\frac19$ of the job per hour and completes $3\cdot\frac19=\frac13$ of the job in the first three hours, leaving $\frac23$ .
B works at $\frac13=\frac39$ of the job per hour, so once B joins the rate is
$\frac19+\frac39=\frac49$
of the job per hour.

Dividing by $\frac49$ is multiplying by $\frac94$ , so the remainder takes
$\frac23\cdot\frac94=\frac32$
hours.
The total is
$3+\frac32=\frac92$
hours.

A works for $\frac92$ hours and completes $\frac12$ of the job.
B works for $\frac32$ hours and completes $\frac12$ , so the whole job is completed.

Answer

$\frac92$ hours, or $4.5$ hours, which is $4$ hours $30$ minutes.

Key idea

When a team changes during a job, separate the time intervals and track the remaining work.
Problem 4: Mark complete
Problem 5 The partly filled tank

A tank is initially one-quarter full. A pipe could fill it from empty in $4$ hours, and a drain could empty it from full in $10$ hours. Assume both rates are constant while water is present. With both open, how long until the tank becomes full?
1. Hint 1
  The tank needs only its missing fraction, not a whole tank of additional water.
2. Hint 2
  Subtract the drain rate from the pipe rate before dividing the missing fraction by the net rate.
Answer

$5$ hours.

Full solution

The pipe fills at $\frac14$ of a tank per hour and the drain empties at $\frac1{10}$ of a tank per hour.
Over the common denominator $20$ these rates are $\frac5{20}$ and $\frac2{20}$ , so the net filling rate is
$\frac5{20}-\frac2{20}=\frac3{20}$
tank per hour.
This is positive, so the level rises.
The missing fraction is $\frac34$ , and dividing by $\frac3{20}$ is multiplying by $\frac{20}3$ , so
$T=\frac34\cdot\frac{20}3$ $T=5$
hours.

In five hours the pipe adds $\frac54$ of a tank and the drain removes $\frac5{10}=\frac12$ of a tank.
Writing $\frac12$ as $\frac24$ , the net gain is $\frac54-\frac24=\frac34$ , and adding it to the initial one-quarter gives a full tank.

Answer

$5$ hours.

Key idea

A positive net filling rate acts on the fraction still missing from a tank.
Problem 5: Mark complete
Problem 6 The recorded stop

A cyclist rides $18$ miles at $12$ miles per hour, stops for half an hour, then rides $10$ miles uphill at $5$ miles per hour. Find the average speed for the whole outing, including the stop, and compare it with the average speed while moving.
1. Hint 1
  The distance total is the same for both averages, but the time total differs.
2. Hint 2
  Find each riding time from its distance and speed, then include or exclude the stop as requested.
Answer

Including the stop: $7$ miles per hour. While moving: $8$ miles per hour. The whole-outing average is $1$ mile per hour lower.

Full solution

The riding times are $\frac{18}{12}=1.5$ hours and $\frac{10}{5}=2$ hours.
The total distance is $18+10=28$ miles, and the full elapsed time is $1.5+0.5+2=4$ hours.

The average including the stop is
$\frac{28}{4}=7$
miles per hour.
Excluding the stop, the moving time is $3.5$ hours, so the average is $\frac{28}{3.5}=8$ miles per hour.

The whole-outing average is $1$ mile per hour lower because the same distance is divided by a larger time.
The moving average is below the mean of the two speeds, $8.5$ , because the cyclist spent more riding time at $5$ miles per hour than at $12$ ; the stop pulls the whole-outing average lower still.

Answer

Including the stop: $7$ miles per hour. While moving: $8$ miles per hour. The whole-outing average is $1$ mile per hour lower.

Key idea

The time interval named in an average-speed question determines whether stopped time belongs in the denominator.
Problem 6: Mark complete
Problem 7 The missing work record

Workers A and B work together for $3$ hours, completing three-fifths of a job. B then leaves, and A finishes the rest in $6$ more hours. Assume steady rates with no duplication or interference. Find how long each worker would take to complete the whole job alone.
1. Hint 1
  The solo interval directly reveals the rate of A.
2. Hint 2
  The first interval reveals the combined rate, from which the rate of B can be recovered.
Answer

A: $15$ hours. B: $\frac{15}{2}$ hours, or $7.5$ hours, which is $7$ hours $30$ minutes.

Full solution

After the first three hours, $1-\frac35=\frac25$ of the job is left.
A completes that $\frac25$ alone in six hours, so its rate is $\frac25\div6=\frac1{15}$ of a job per hour.
Its whole-job time is therefore fifteen hours.

The pair completes $\frac35$ in three hours, so the combined rate is $\frac35\div3=\frac15$ of a job per hour.
Writing $\frac15$ as $\frac3{15}$ and subtracting A gives
$\frac3{15}-\frac1{15}=\frac2{15}$
for B, whose solo time is the reciprocal, $\frac{15}{2}=7.5$ hours.

In the actual schedule A works nine hours and completes $\frac9{15}=\frac35$ of the job, and B works three hours and completes $\frac6{15}=\frac25$ , checking the total.

Answer

A: $15$ hours. B: $\frac{15}{2}$ hours, or $7.5$ hours, which is $7$ hours $30$ minutes.

Key idea

Work completed during different team arrangements can reveal individual rates.
Problem 7: Mark complete
Problem 8 The open outlets

A half-full tank has one inlet that could fill an empty tank in $4$ hours and two outlets that each could empty a full tank in $8$ hours. Assume constant rates while water is present. Kai claims the tank will eventually fill because the inlet is faster than either outlet separately. Is Kai correct? Describe what happens with all three open.
1. Hint 1
  Both outlets remove water at the same time, so their effects combine.
2. Hint 2
  Add the two outlet rates, then compare that total with the inlet rate.
Answer

No; the level stays half full.

Full solution

The total outlet rate is
$\frac18+\frac18=\frac14$
tank per hour, equal to the inlet rate.
The net rate is
$\frac14-\frac14=0$

Therefore the amount of water does not change.
The tank stays half full rather than filling, so Kai is incorrect.

Answer

No; the level stays half full.

Key idea

An inlet must exceed the combined outlet rate for the water level to rise.
Problem 8: Mark complete
Problem 9 The doubled schedule

A trip consists of $t$ hours at $20$ miles per hour and then $2t$ hours at $50$ miles per hour, without stops, where $t>0$ . Dana claims that doubling both travel times, at the same two speeds, leaves the average speed unchanged. Is this true? Find the average before and after the change.
1. Hint 1
  Doubling both times scales both the total distance and total time.
2. Hint 2
  Write the average as a ratio involving $t$ , then repeat with $2t$ and $4t$ .
Answer

Yes; both averages are $40$ miles per hour.

Full solution

The original distance is $20t+50(2t)=120t$ miles, and the original time is $3t$ hours.
Since $t>0$ ,
$\frac{120t}{3t}=40$
miles per hour.

After the change, the distance is $240t$ miles and the time is $6t$ hours, giving $240t/(6t)=40$ miles per hour again.
Both totals doubled, so the ratio stays the same.

Answer

Yes; both averages are $40$ miles per hour.

Key idea

Scaling every travel time by the same positive factor preserves the average speed when the speeds stay fixed.
Problem 9: Mark complete
Problem 10 The two assignments

Two workers each have positive constant rates and can divide a job freely without duplicating work or slowing each other down. Ari says that assigning each worker exactly half the job must give the fastest completion, even if their rates differ. Decide whether the claim is true. Use workers with solo times of $2$ hours and $6$ hours to justify your decision.
1. Hint 1
  Find how long each worker takes to finish the share assigned to it; the job ends only when both are done.
2. Hint 2
  With equal halves, one worker sits idle while the other finishes. Try a split that keeps both workers busy until the end.
Answer

False. Equal halves take $3$ hours; giving the $2$ -hour worker $\frac34$ and the $6$ -hour worker $\frac14$ takes $1.5$ hours.

Full solution

With equal halves, the faster worker takes one hour and the slower takes three hours, so completion waits until three hours.

If the faster worker receives three-quarters and the slower one-quarter, their times are
$\frac{3/4}{1/2}=1.5$ $\frac{1/4}{1/6}=1.5$
hours.
Both finish in $1.5$ hours, less than three.

The combined rate is $1/2+1/6=2/3$ of a job per hour, so $1.5$ hours also matches the time obtained when both remain productive throughout.
The equal-share claim is therefore false.

Answer

False. Equal halves take $3$ hours; giving the $2$ -hour worker $\frac34$ and the $6$ -hour worker $\frac14$ takes $1.5$ hours.

Key idea

To finish together at unequal rates, workers generally need unequal shares of a divisible job.
Problem 10: Mark complete

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