This page is designed to help students master the concept of rates and unit rates. Below, you will find introductory examples explaining how to set up rate equations, followed by carefully selected practice questions with step-by-step solutions.
A rate is a ratio comparing two quantities that have different units (e.g., kilometers and hours, or dollars and kilograms).
Car A travels $150$ kilometers in $3$ hours. Car B travels $220$ kilometers in $4$ hours. Assuming both cars travel at constant speeds, which of the two cars travels faster?
First, calculate how far Car A travels in exactly one hour:
$$\frac{150 \text{ km}}{3 \text{ hours}} = \frac{50 \text{ km}}{1 \text{ hour}} = 50 \text{ km/hr}$$
Next, calculate how far Car B travels in one hour:
$$\frac{220 \text{ km}}{4 \text{ hours}} = \frac{55 \text{ km}}{1 \text{ hour}} = 55 \text{ km/hr}$$
The quantities $50 \text{ km/hr}$ and $55 \text{ km/hr}$ are called unit rates because the denominator is exactly one unit of time. By comparing these unit rates, we can clearly see that Car B travels faster.
A car travels $150$ kilometers in $3$ hours. Assuming a constant speed, how many hours are needed for this car to travel $250$ kilometers?
Let $t$ be the number of hours needed to travel $250 \text{ km}$. Since the speed is constant, the rate remains the same. We can set up a proportion:
$$\frac{150 \text{ km}}{3 \text{ hours}} = \frac{250 \text{ km}}{t}$$
To solve for $t$, we can use the cross-multiply method. Multiply diagonally across the equals sign:
$$150 \text{ km} \times t = 250 \text{ km} \times 3 \text{ hours}$$
Isolate $t$ by dividing both sides by $150 \text{ km}$:
$$t = \frac{250 \times 3}{150} \text{ hours}$$
$$t = \frac{750}{150} \text{ hours} = \mathbf{5 \text{ hours}}$$
Note: If you get stuck, expand the solution block to see the step-by-step logic.
Let $d$ be the real distance in kilometers. Set up a proportion comparing the map scale to the unknown distance:
$$\frac{15 \text{ km}}{5 \text{ cm}} = \frac{d}{15 \text{ cm}}$$
Cross multiply:
$$15 \text{ km} \times 15 \text{ cm} = d \times 5 \text{ cm}$$
Divide both sides by $5 \text{ cm}$:
$$d = \frac{15 \times 15}{5} \text{ km} = \frac{225}{5} \text{ km} = \mathbf{45 \text{ km}}$$
Let $x$ be the number of gallons needed to travel $330 \text{ miles}$. Set up the proportion:
$$\frac{220 \text{ miles}}{10 \text{ gallons}} = \frac{330 \text{ miles}}{x}$$
Cross multiply:
$$220 \times x = 330 \times 10$$
Divide both sides by $220$ to isolate $x$:
$$x = \frac{3300}{220} = \mathbf{15 \text{ gallons}}$$
First, find the cost of exactly one ticket (the unit rate):
$$c = \frac{\$66}{10 \text{ tickets}} = \$6.60 \text{ per ticket}$$
To find the total cost ($C$) of $22$ tickets, multiply the unit rate by $22$:
$$C = 22 \text{ tickets} \times \$6.60 \text{ per ticket} = \mathbf{\$145.20}$$
Part a: First, find the number of cans per box (unit rate):
$$\text{Unit Rate} = \frac{36 \text{ cans}}{4 \text{ boxes}} = 9 \text{ cans per box}$$
In $7$ boxes, the number of cans is:
$$7 \text{ boxes} \times 9 \text{ cans per box} = \mathbf{63 \text{ cans}}$$
Part b: Let $B$ be the number of boxes needed to package $99 \text{ cans}$. Using the unit rate, write the proportion:
$$\frac{9 \text{ cans}}{1 \text{ box}} = \frac{99 \text{ cans}}{B}$$
Cross multiply and solve for $B$:
$$9 \times B = 99 \times 1 \implies B = \frac{99}{9} = \mathbf{11 \text{ boxes}}$$
First, find the cost of $1 \text{ kg}$ of apples (the unit rate):
$$c = \frac{\$15}{4 \text{ kg}} = \$3.75 \text{ per kg}$$
Knowing the cost of $1 \text{ kg}$, the cost ($C$) of $11 \text{ kg}$ is:
$$C = 11 \text{ kg} \times \$3.75 \text{ per kg} = \mathbf{\$41.25}$$
Part a: Find the number of gallons moved in one minute (unit rate):
$$\text{Unit Rate} = \frac{55 \text{ gallons}}{10 \text{ minutes}} = 5.5 \text{ gallons per minute}$$
The number of gallons moved in $22 \text{ minutes}$ is:
$$22 \text{ minutes} \times 5.5 \text{ gallons per minute} = \mathbf{121 \text{ gallons}}$$
Part b: Let $T$ be the number of minutes to move $165 \text{ gallons}$. Set up the proportion:
$$\frac{55 \text{ gallons}}{10 \text{ minutes}} = \frac{165 \text{ gallons}}{T}$$
Cross multiply and solve:
$$55 \times T = 165 \times 10 \implies T = \frac{1650}{55} = \mathbf{30 \text{ minutes}}$$
Let $T$ be the total number of hours it takes for all $324 \text{ liters}$ to leak. Using equality of rates:
$$\frac{3 \text{ liters}}{5 \text{ hours}} = \frac{324 \text{ liters}}{T}$$
Cross multiply:
$$3 \times T = 5 \times 324$$
Divide both sides by $3$ and simplify:
$$T = \frac{1620}{3} = \mathbf{540 \text{ hours}}$$
Let $W$ be the weight of $5 \text{ cans}$. Using equality of rates:
$$\frac{20 \text{ cans}}{7300 \text{ grams}} = \frac{5 \text{ cans}}{W}$$
Cross multiply:
$$20 \times W = 5 \times 7300$$
Divide both sides by $20$ and simplify:
$$W = \frac{36500}{20} = \mathbf{1825 \text{ grams}}$$
In this problem, we have two competing rates: the filling rate ($R_f$) and the leaking rate ($R_l$). Let's find the unit rate per second for both:
$$R_f = \frac{5 \text{ liters}}{45 \text{ seconds}} = \frac{1}{9} \text{ liters per second}$$
$$R_l = \frac{1 \text{ liter}}{180 \text{ seconds}}$$
We know that $1 \text{ hour} = 3600 \text{ seconds}$. Let's calculate the total volume for both actions over this time frame.
The quantity ($Q_1$) of water pumped in after $1 \text{ hour}$ is:
$$Q_1 = R_f \times 3600 = \frac{1}{9} \times 3600 = 400 \text{ liters}$$
The quantity ($Q_2$) of water lost through leakage after $1 \text{ hour}$ is:
$$Q_2 = R_l \times 3600 = \frac{1}{180} \times 3600 = 20 \text{ liters}$$
Therefore, the total quantity ($Q$) of water remaining in the container after $1 \text{ hour}$ is:
$$Q = Q_1 - Q_2 = 400 - 20 = \mathbf{380 \text{ liters}}$$