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Rates and Unit Rates: 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 Supply labels

Label A compares $3$ meters of ribbon with $2$ meters of fabric. Label B compares $3$ meters of ribbon with a cost of 2 dollars. Which label gives a rate?
1. Hint 1
  Look at the units, rather than the names of the materials.
2. Hint 2
  A rate keeps two different units in the comparison.
Answer

Label B.

Full solution

Label A compares meters with meters, so it compares two lengths in the same unit.
Label B compares meters with dollars, which are different units.

Label B therefore gives a rate.
As a unit rate, in meters per dollar, it is
$\frac{3}{2}=1.5$

Answer

Label B.

Key idea

A rate compares quantities measured in different units.
Problem 1: Mark complete
Problem 2 Thread feed

A machine feeds $5.4$ meters of thread in $1.2$ minutes at a steady rate. How many meters does it feed per minute?
1. Hint 1
  The requested amount belongs to one minute.
2. Hint 2
  Meters per minute puts meters on top, so divide the meters by the number of minutes.
Answer

$4.5$ meters per minute.

Full solution

Dividing the time by $1.2$ gives one minute, so divide the thread amount by the same number.
$5.4\div1.2=4.5$

The machine feeds $4.5$ meters per minute.
Check that $4.5\times1.2=5.4$ , the original thread amount.

Answer

$4.5$ meters per minute.

Key idea

Dividing the first quantity by the second gives the amount for one unit, even when the time is not a whole number.
Problem 2: Mark complete
Problem 3 Moving belt

A belt moves steadily at $1.25$ meters per second. How far does it move in $24$ seconds?
1. Hint 1
  The rate tells how much distance belongs to each second.
2. Hint 2
  In twenty-four seconds the belt covers twenty-four copies of that one-second distance.
Answer

$30$ meters.

Full solution

Multiply the distance for one second by the number of seconds.
$1.25\times24=30$

The belt moves $30$ meters.
Dividing the result by $24$ seconds returns $1.25$ meters per second.

Answer

$30$ meters.

Key idea

Multiply an amount per unit by the number of units to find the total.
Problem 3: Mark complete
Problem 4 Two cyclists

Ana and Ben each cycle at a steady speed. Ana rides $39$ miles in $3$ hours, and Ben rides $50$ miles in $4$ hours. Who is faster, and how far does the faster rider travel in $6$ hours?
1. Hint 1
  Compare the distance each rider covers in one hour.
2. Hint 2
  Scale the faster rider's miles per hour up to six hours.
Answer

Ana is faster; she rides $78$ miles in $6$ hours.

Full solution

In miles per hour, Ana's speed is
$39\div3=13$
In miles per hour, Ben's speed is
$50\div4=12.5$
For a speed, the higher number of miles per hour is faster, and $13$ is greater than $12.5$ , so Ana is faster.

Scaling Ana's unit rate up to six hours gives, in miles,
$13\times6=78$

Answer

Ana is faster; she rides $78$ miles in $6$ hours.

Key idea

Speeds in the same units compare directly, and a unit rate scales up to a total for any length of time.
Problem 4: Mark complete
Problem 5 Reading log

A reader reads $96$ words during the first minute, $144$ during the next minute, and $60$ during the third and last minute. What is the average number of words read per minute across the whole log, and what is that average restated as words per $30$ seconds?
1. Hint 1
  The whole log supplies one total word count and one total time.
2. Hint 2
  Divide the total words by the three minutes.
3. Hint 3
  Thirty seconds is half a minute, so divide both quantities of the per-minute rate by two.
Answer

$100$ words per minute, which is $50$ words per $30$ seconds.

Full solution

Combine the words read in all three minutes.
$96+144+60=300$

The log covers three minutes, so in words per minute the average rate is
$300\div3=100$
Three minutes at that average accounts for the same $300$ words.

Thirty seconds is half of sixty seconds, so dividing both quantities of the rate by $2$ keeps it equal.
In words per $30$ seconds, the average is
$100\div2=50$
Like the per-minute figure, this is an average, not the count in every half minute.

Answer

$100$ words per minute, which is $50$ words per $30$ seconds.

Key idea

A total divided by its duration gives an average rate even when the amount varies from minute to minute.
Problem 5: Mark complete
Problem 6 Rain barrel

A sprinkler draws water from a rain barrel at a steady rate of $21$ liters every $14$ minutes. The barrel holds $40$ liters when the sprinkler starts. How many liters remain in the barrel after the sprinkler runs for $18$ minutes?
1. Hint 1
  Find the water used per minute before extending the running time.
2. Hint 2
  Subtract the total water used from the starting amount.
Answer

$13$ liters.

Full solution

In liters per minute, the sprinkler draws
$21\div14=1.5$
Over eighteen minutes, in liters, it draws
$1.5\times18=27$

The barrel began with forty liters, so in liters what remains is
$40-27=13$
The used and remaining amounts add back to the starting forty liters.

Answer

$13$ liters.

Key idea

Use a unit rate to calculate consumption before subtracting it from a starting supply.
Problem 6: Mark complete
Problem 7 Fabric suppliers

Supplier A sells $3$ meters of fabric for 12.60 dollars. Supplier B sells $50$ centimeters of the same fabric for 2.30 dollars. There are $100$ centimeters in a meter, and both suppliers sell any requested length at their stated rate. Which supplier charges less for $2.5$ meters, and how much less?
1. Hint 1
  Put both prices on the same length unit.
2. Hint 2
  Find dollars per meter for each supplier, then price two and a half meters at each rate.
Answer

Supplier A, by 1.00 dollar.

Full solution

In dollars per meter, Supplier A charges
$12.60\div3=4.20$
Two lengths of fifty centimeters make one meter, so in dollars per meter Supplier B charges
$2.30\times2=4.60$

Supplier A has the lower rate.
For two and a half meters, in dollars, Supplier A charges
$4.20\times2.5=10.50$
For the same length, in dollars, Supplier B charges
$4.60\times2.5=11.50$
Supplier A charges less, and in dollars the difference is
$11.50-10.50=1.00$

Answer

Supplier A, by 1.00 dollar.

Key idea

Rates must use the same units before their numerical values can be compared.
Problem 7: Mark complete
Problem 8 Bead labels

A set of identical beads contains $24$ beads and weighs $6$ grams. One label says $4$ beads per gram, and another says $0.25$ gram per bead. Can both labels be correct? Explain.
1. Hint 1
  The first label is about one gram, and the second is about one bead.
2. Hint 2
  Calculate each rate in the order its units name.
Answer

Yes; both labels are correct.

Full solution

Dividing the bead count by the weight in grams gives, in beads per gram,
$24\div6=4$
Dividing the weight in grams by the bead count gives, in grams per bead,
$6\div24=0.25$

The two rates answer different questions and agree with one another: four beads, each weighing a quarter gram, weigh one gram in total.

Answer

Yes; both labels are correct.

Key idea

Reversing the units reverses the division and gives a second rate that describes the same beads.
Problem 8: Mark complete
Problem 9 Sorting report

A worker sorts $60$ envelopes during $10$ minutes, and the report shows only those two numbers. Rina says this proves the worker sorted exactly $6$ envelopes in every minute. Is that conclusion justified? Explain.
1. Hint 1
  Distinguish the average for the full interval from what happened in each minute.
2. Hint 2
  Look for unequal whole-number minute counts whose total is sixty.
Answer

No. One possible log is $4$ envelopes in each of five minutes and $8$ in each of the other five; any whole-number minute counts that total $60$ and are not all $6$ also work.

Full solution

In envelopes per minute, the overall average is
$60\div10=6$
This division does not establish that the work was steady.

For example, the worker could sort four envelopes in each of five minutes and eight in each of five other minutes.
$5\times4+5\times8=60$
This log has the given total and duration without having six envelopes in any minute.

Answer

No. One possible log is $4$ envelopes in each of five minutes and $8$ in each of the other five; any whole-number minute counts that total $60$ and are not all $6$ also work.

Key idea

An average rate alone does not determine the amount in each individual interval.
Problem 9: Mark complete
Problem 10 Part of a minute

A pump moves $6$ liters of water in $0.75$ minute at a steady rate. Jo says that in one full minute the pump must move less than $6$ liters, because dividing always makes a number smaller. Is Jo right? Explain.
1. Hint 1
  One full minute is longer than $0.75$ minute, so ask whether the pump moves more or less water in it.
2. Hint 2
  Turning $0.75$ minute into one full minute takes a division, so do that same division to the liters and the rate stays equal.
3. Hint 3
  Check your per-minute amount by multiplying it by $0.75$ minute.
Answer

No; the pump moves $8$ liters per minute.

Full solution

Dividing both quantities of a rate by the same number keeps them matched, the equivalent-ratio move.
Dividing the time by its own value turns it into one minute.
$0.75\div0.75=1$

Dividing the liters by the same number gives, in liters per minute,
$6\div0.75=8$

Since $0.75$ is less than one, dividing by it scales both quantities up: the time from $0.75$ to $1$ minute, and the water from $6$ to $8$ liters.
A full minute is longer than $0.75$ minute, so at a steady rate the pump moves more water in it, not less.
Jo is not right.

As a check, in liters,
$8\times0.75=6$

Answer

No; the pump moves $8$ liters per minute.

Key idea

Dividing by a number between zero and one scales both quantities up, so the amount for a whole unit is larger than the amount for part of one.
Problem 10: Mark complete

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