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Rate & Unit
Word Problems

Master the art of solving problems involving rates, unit conversions, and proportional reasoning for the SSAT Upper Level.

SSAT Upper Level: Quantitative

What Is a Rate?

A rate is a ratio that compares two quantities measured in different units. Whenever you see "per," "each," or "every," you are looking at a rate.

Speed
\( 60 \) miles per hour
Price
\( \$3.50 \) per pound
Output
\( 12 \) pages per minute

The Unit Rate

A unit rate is a rate with a denominator of 1. It tells you how much of one quantity corresponds to exactly one unit of the other.

General Rate
\( \frac{240 \text{ mi}}{4 \text{ hr}} \)
÷ 4 Unit Rate
\( \frac{60 \text{ mi}}{1 \text{ hr}} \)

Why Rates Matter on the SSAT

Rate problems appear in nearly every SSAT quantitative section. They test your ability to translate words into equations and manage units carefully.

Distance
"How far does the train travel in \( 5 \) hours?"
Work
"How long for two painters to finish the job together?"
Cost
"What is the total price for \( 3.5 \) pounds of apples?"

The Core Rate Formula

Amount Rate Quantity
\( A = R \times Q \)
\( Q = \frac{A}{R} \)
\( R = \frac{A}{Q} \)
 ×

Distance \( = \text{Rate} \times \text{Time} \)

The most common rate relationship on the SSAT. Memorize it as \( d = r \times t \), but understand that it is just the core formula applied to motion.

\( d = 300 \) miles
\( 60 \) mph
\( t = \frac{300}{60} = 5 \) hours

Rate vs. Ratio: Key Distinction

Rate Ratio
Units Different units (miles / hour) Same or no units (boys : girls)
Keyword "per," "each," "every" "to," "for every," "out of"
Example \( \$5 \) per lb \( 3 : 4 \)
Key Action Multiply / divide to convert Scale or set up proportion

Unit Conversion with Rates

Many SSAT problems require you to change units before or after applying a rate. Use conversion factors as fractions equal to \( 1 \).

GIVEN
\( 5 \frac{\text{ft}}{\text{sec}} \)
 × Factor
\( \frac{3600 \text{ sec}}{1 \text{ hr}} \)
RESULT
\( 18,000 \frac{\text{ft}}{\text{hr}} \)

Step-by-Step Approach

1
Identify the rate. Look for "per," "each," or "every" and write it as a fraction.
2
Identify what is given and what is asked. Label the known quantity and the unknown.
3
Check the units. Convert if necessary so that units align before calculating.
4
Apply the formula. Use \( \text{Amount} = \text{Rate} \times \text{Quantity} \) (or rearrange).
5
Verify units in your answer. Make sure the result carries the correct label.

Example 1: Simple Rate

Problem: A printer produces \( 8 \) pages per minute. How many pages does it produce in \( 45 \) minutes?

1
Rate: \( 8 \) pages per minute \( = \frac{8\;\text{pages}}{1\;\text{min}} \)
2
Given: \( t = 45 \) minutes. Find: total pages.
3
Units match: both in minutes. No conversion needed.
4
Calculate: \( 8 \times 45 = 360 \) pages.

Example 2: Unit Conversion

Problem: A cyclist rides at \( 12 \) miles per hour. How many feet does she travel in \( 10 \) minutes?

1
Rate: \( 12 \) mi/hr. Given: \( 10 \) minutes. Find: feet.
2
Convert time: \( 10\;\text{min} = \frac{10}{60} = \frac{1}{6}\;\text{hr} \).
3
Distance in miles: \( d = 12 \times \frac{1}{6} = 2\;\text{miles} \).
4
Convert to feet: \( 2 \times 5{,}280 = 10{,}560\;\text{feet} \).

Example 3: Combined Rates

Problem: Pipe A fills a tank in \( 6 \) hours. Pipe B fills the same tank in \( 3 \) hours. How long does it take both pipes together?

1
Rate A: \( \frac{1}{6} \) tank per hour. Rate B: \( \frac{1}{3} \) tank per hour.
2
Combined rate: \( \frac{1}{6} + \frac{1}{3} = \frac{1}{6} + \frac{2}{6} = \frac{3}{6} = \frac{1}{2} \) tank/hr.
3
Time: \( t = \frac{1\;\text{tank}}{\frac{1}{2}\;\text{tank/hr}} = 2\;\text{hours} \).

Example 4: Finding the Rate

Problem: A store sells \( 4.5 \) pounds of trail mix for \( \$11.25 \). What is the price per pound?

Total Cost
\( \$11.25 \)
 ÷ Quantity
\( 4.5 \) lbs

Common Mistakes

✕ Unit Mismatch
Mixing hours with minutes or miles with feet without converting first. Always check that units align.
✕ Adding Times
In combined-work problems, adding the times (e.g., \( 6 + 3 = 9 \)) is wrong. You must add the rates.
✕ Inverting the Rate
Writing \( \frac{\text{time}}{\text{distance}} \) instead of \( \frac{\text{distance}}{\text{time}} \). Keep the "per" unit in the denominator.

Summary Cheat Sheet

Problem Type Formula Key Move
Distance \( d = r \times t \) Multiply rate by time
Unit Price \( \text{price/unit} = \frac{\text{total cost}}{\text{quantity}} \) Divide total by count
Combined Work \( \frac{1}{t_A} + \frac{1}{t_B} = \frac{1}{t_{\text{total}}} \) Add rates, not times
Unit Conversion Multiply by \( \frac{\text{new unit}}{\text{old unit}} \) Cancel matching units

Question 1 of 5

A factory produces \( 240 \) widgets in \( 8 \) hours. At this rate, how many widgets are produced in \( 15 \) hours?

Question 2 of 5

A car travels \( 180 \) miles in \( 3 \) hours. How many feet does it travel in \( 1 \) minute?
(Use: \( 1\;\text{mile} = 5{,}280\;\text{feet} \))

Question 3 of 5

Machine A completes a job in \( 4 \) hours and Machine B completes the same job in \( 12 \) hours. Working together, how long will it take them?

Question 4 of 5

Oranges cost \( \$4.80 \) for \( 6 \) pounds. At this rate, how much do \( 10 \) pounds cost?

Question 5 of 5

A train leaves City X at \( 9{:}00 \) AM traveling at \( 80 \) mph. A second train leaves City X at \( 10{:}00 \) AM traveling at \( 100 \) mph in the same direction. At what time does the second train catch the first?