Ratios, Percents, and Proportion: Chapter Test
20 multiple-choice questions and 10 core practice problems, drawn from across the chapter and mixed together.
Multiple choice
20 questions, 100 points in total, 5 points each. Answer in any order and change your mind as often as you like. When you submit, your answers lock and every question shows its worked solution.
Core practice
10 problems from across the chapter. 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 The sealed container
Three materials A, B, and C have masses in the ratio . Sealed together in a container, the three materials and the container have a combined mass of kg. The empty container has mass kg. Find the mass of each material.
- Hint 1
The stated ratio compares only the materials.
- Hint 2
Remove the container's own mass before finding the size of one ratio part.
Answer
A: kg. B: kg. C: kg.
Full solution
The materials together weigh kg.
Write their masses as , , and .
The masses are , , and kg.
They reduce to and total kg, which becomes the stated kg after the container is included.
Answer
A: kg. B: kg. C: kg.
Key idea
Remove amounts outside a stated ratio before dividing its total into shared parts.
- Hint 1
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Problem 2 The recycling load
In one week a recycling center collected kg of paper and kg of cardboard. Those two materials together are of the total mass the center collected that week. Find the total mass collected that week, in kilograms.
- Hint 1
The two named materials together form the part that the percent describes, and the total collected is the whole.
- Hint 2
Add the two masses to get the part, then write that part as the rate in decimal form times the unknown whole and solve for the whole.
Answer
kg.
Full solution
The paper and cardboard together weigh kg, and that combined mass is the part the percent describes.
Let be the total mass collected, in kilograms.
Divide both sides by .
Checking, of kg is kg, the combined mass of paper and cardboard.
The whole is larger than the part, as it must be for a rate below .
Answer
kg.
Key idea
Assemble the part a percent describes before dividing by the rate to recover the whole.
- Hint 1
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Problem 3 The paint coverage
A paint covers square meters per liter. Express its coverage in square centimeters per milliliter, using meter equal to centimeters and liter equal to milliliters.
- Hint 1
A rate has two units to convert, and one of them is a squared length.
- Hint 2
Apply the meter-to-centimeter factor once for each length in the area, and orient the volume factor so liters cancel from the denominator.
Answer
square centimeters per milliliter.
Full solution
Square meters sit on top of the rate, so the factor with centimeters above and meters below is applied twice, once for each length in the area.
One square meter is therefore
square centimeters, and the numerator becomes square centimeters.
Liters sit in the denominator, so the volume factor has liters on top and milliliters below: liter over milliliters.
The rate is
square centimeters per milliliter.
Checking, milliliters at square centimeters each cover square centimeters, which is square meters, the coverage of one liter.
Answer
square centimeters per milliliter.
Key idea
In a rate, convert each unit with its own factor, squaring the length factor for an area unit.
- Hint 1
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Problem 4 The fixed principal
An account has balance dollars after years at simple annual interest, with no deposits or withdrawals. Find the original principal and the interest it would earn in years at the same rate.
- Hint 1
The three-year balance contains both the principal and three years of interest.
- Hint 2
Undo the balance factor to reach the principal; with the principal and rate fixed, the interest grows in proportion to the time.
Answer
Principal: dollars. Interest in years: dollars.
Full solution
Let be the original principal in dollars.
The balance is with and .
Divide both sides by .
The seven-year interest is
dollars.
Checking, the three-year interest is dollars, and matches the stated balance.
With the principal and rate fixed, seven years earn of the three-year interest, and dollars again.
Answer
Principal: dollars. Interest in years: dollars.
Key idea
Reverse the balance multiplier to find a fixed principal before computing interest for another duration.
- Hint 1
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Problem 5 The three bins
Counts in bins A, B, and C satisfy and . Bins A and C together hold objects. Find the original count in each bin, and the ratio after objects are added to C only.
- Hint 1
The two ratios share B, so match its terms to get one three-term ratio.
- Hint 2
Write the original counts with one shared multiplier, and let the stated count of A and C together fix that multiplier.
- Hint 3
Add to the count in C, leave A as it is, and reduce the pair to lowest terms.
Answer
A: . B: . C: . After the addition, .
Full solution
Scaling by gives , which matches B in .
So , and the original counts are , , and , sharing one multiplier .
Bins A and C together hold objects.
The original counts are , , and .
After the addition C holds while A still holds .
Dividing both terms by gives
Checking, reduces to and reduces to , and matches the stated count.
Answer
A: . B: . C: . After the addition, .
Key idea
Merge two ratios through their shared term, then let a stated count fix the one shared multiplier before adjusting for a change.
- Hint 1
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Problem 6 The recorded price
A price is raised by and then reduced by of the raised price. It ends at dollars. Find the original price and the overall percentage change from that original.
- Hint 1
The final price includes two successive multipliers.
- Hint 2
Combine those factors, then reverse their product to recover the original.
Answer
Original: dollars. Overall change: an increase.
Full solution
Let the original price be dollars.
The combined factor is an increase.
Checking forward, dollars becomes dollars, then dollars, matching the final record.
Answer
Original: dollars. Overall change: an increase.
Key idea
A sequence of percentage changes is reversed by dividing by its combined nonzero factor.
- Hint 1
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Problem 7 The paired machines
One job is boxes. Machine A packs a job in hours at a steady rate, and machine B packs boxes per minute. Working together without duplicating work or interfering, how many minutes do they need to pack two jobs? Use hour equal to minutes.
- Hint 1
The two machines' rates must be in the same units before they are added.
- Hint 2
Convert the rate of A from boxes per hour to boxes per minute, add the rate of B, then divide the boxes in two jobs by the combined rate.
Answer
minutes.
Full solution
Machine A packs boxes per hour.
Hours sit in the denominator, so multiply by the factor hour over minutes.
So A packs boxes per minute.
Working together without interfering, the two rates add.
boxes per minute.
Two jobs are boxes, so the time is
minutes.
Checking, in minutes A packs boxes and B packs boxes, and , two jobs.
Answer
minutes.
Key idea
Convert rates to the same units before adding them, then divide the required work by the combined rate.
- Hint 1
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Problem 8 The matched settings
Positive output varies directly with input and inversely with input . At and , the output is . The input is then increased by and is decreased by . Find the new output, and the percent increase in alone, with unchanged at , that would give the same new output.
- Hint 1
Find the variation constant from the complete original record.
- Hint 2
Apply each percent multiplier to its own input, remembering that is in the denominator.
- Hint 3
For the second request, hold at , solve the variation equation for , and compare the result with .
Answer
New output: . Increase in alone: .
Full solution
The model is .
The original record gives , so .
The new inputs are and
The output is
With held at , the same output needs
Multiply both sides by and divide by .
The factor on is , a increase.
Checking,
The factor is also the output's own factor, , because with fixed the output varies directly with .
Answer
New output: . Increase in alone: .
Key idea
Percent changes in the direct and inverse inputs reach a combined-variation output through a quotient of multipliers, and with the inverse input fixed the direct input alone must supply that whole factor.
- Hint 1
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Problem 9 Kim's single record
A quantity is known to vary either directly or inversely with positive , and one record is . Kim says this record alone shows that the variation is direct. Is Kim right? Give the equation each kind of variation would have through this record, then use a second record, , to decide which kind holds.
- Hint 1
One point can be matched by a fixed ratio or by a fixed product.
- Hint 2
Find the ratio and the product of the first record, then test the second record against each.
Answer
No, not from one record alone. Direct: , or . Inverse: . The second record gives direct variation.
Full solution
Direct variation keeps the ratio fixed, so through its constant is and its equation is
Inverse variation keeps the product fixed, so its constant is and its equation is
Both equations pass through , so that record alone cannot decide between them.
Kim is not right to conclude direct variation from it.
For the second record, the ratio is , matching the direct constant.
The product is , not , so the inverse equation fails.
Only direct variation fits both records, with
Answer
No, not from one record alone. Direct: , or . Inverse: . The second record gives direct variation.
Key idea
One record fits both a direct and an inverse variation, so a second record is needed to tell a fixed ratio from a fixed product.
- Hint 1
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Problem 10 The trip report
A boat travels for hours at miles per hour, then covers another miles at a constant speed without stopping. Its average speed for the whole trip is miles per hour. Find the speed on the final leg, explain why the whole-trip average is not the plain mean of the two speeds, and find how long a final leg at that same speed would have to last, covering whatever distance that takes, for the whole-trip average to equal the plain mean.
- Hint 1
Average speed compares the total distance with the total time, so each speed counts for as long as it lasts.
- Hint 2
Use the stated average to find the total time, then the time and speed of the final leg.
- Hint 3
Compare the times spent at the two speeds; for the last request, let the final leg last hours and set the whole-trip average equal to the plain mean.
Answer
Final leg: miles per hour. The times are unequal, hours at and hours at , so the average is below the plain mean . The final leg would have to last hours.
Full solution
The first leg covers miles, so the whole trip is miles.
The stated average gives the total time.
hours.
The final leg therefore takes hours.
Its speed is
miles per hour.
The plain mean of the two speeds is miles per hour.
It would be the average only if the boat spent equal times at the two speeds.
Here it spends hours at the slower speed and only hours at the faster one, so the slower speed carries more weight and the average of falls below .
For the two to agree, let the final leg last hours at miles per hour.
Multiply both sides by the positive .
The final leg would have to last hours, the same time as the first leg.
Checking, miles in hours is miles per hour.
Answer
Final leg: miles per hour. The times are unequal, hours at and hours at , so the average is below the plain mean . The final leg would have to last hours.
Key idea
Average speed is total distance over total time, so it equals the plain mean of two different speeds only when equal times are spent at each.
- Hint 1