How to Solve Rate Problems - Grade 7 Math Questions With Detailed Solutions

How to solve questions on rates in math? Grade 7 math questions are presented along with detailed Solutions and explanations included.

What are rates in math and where are they needed?

The rate is a ratio of two quantities having different units.

Where are they needed?

Example 1: Car A travels 150 kilometers in 3 hours. Car B travels 220 kilometers in 4 hours. We assume that both car travels at constant speeds. Which of the two cars travels faster?

Solution

Car A travels 150 kilometers in 3 hours. In one hour it travels

\( \dfrac{150 \,\, \text{kilometers}}{3 \,\, \text{hours}} = \dfrac{50 \,\, \text{km}}{1 \,\, \text{hour}} \) = 50 km / hour

Car B travels 220 kilometers in 4 hours. In one hour it travels

\( \dfrac{220 \,\, \text{kilometers}}{4 \,\, \text{hours}} = \dfrac{55 \,\, \text{km}}{1 \,\, \text{hour}} \) = 55 km / hour

The quantities 50 km / hour and 55 km / hour are called unit rates because the denominator is one unit of time: 1 hour. In this case the unit rates can be used to find out which car travels faster because we now know how many kilometers are traveled by each car in one hour and we can therefore compare the speed (or rates) and say that car B travels faster.


Example 2: A car travels 150 kilometers in 3 hours. We assume that the car travels at a constant speed. How many hours are needed for this car to travel 250 kilometers at the same speed?

Let t be the number of hours needed to travel 250 kilometers. Since the car travels at a constant rate (speed), we can write that the unit rate is the same whatever values for distance and time we use. Hence we write

\( \dfrac{150 \,\, \text{km}}{3\,\,\text{hour}} = \dfrac{250 \,\, \text{km}}{\text{t}} \) , t in hours

The above equation in t has the form.

\( \dfrac{a}{b} = \dfrac{c}{d} \)

Mulitply both terms of the above by the product of the denominators \(b \times d\).

\( b \times d \times \dfrac{a}{b} = b \times d \times \dfrac{c}{d} \)

Simplify

\( \cancel{b}\times d \times\dfrac{a}{\cancel{b}} = b \times \cancel{d} \times \dfrac{c}{\cancel{d}} \)

to obtain

\( a \times d = b \times c \)

Hence the equations \( \dfrac{a}{b} = \dfrac{c}{d} \) and \( a \times d = b \times c \) are equivalent and have the same solution. This method of changing an equation from fractions on each side to products on each side is called "cross muliply" method which we will use to solve our problems.

We now go back to our equation \( \dfrac{150 \,\, \text{km}}{3\,\,\text{hour}} = \dfrac{250 \,\, \text{km}}{\text{t}} \) and use the "cross multiply" method to write it as follows.

\( 150 \,\, \text{km} \times t = 250 \text{km}\times 3 \text{hours} \)

Since we need to find t, we then isolate it by dividing both sides of the above equation by \( 150 \,\, \text{km} \).

\( \dfrac{150 \,\, \text{km} \times t}{150 \,\, \text{km}} = \dfrac{250 \text{km}\times 3 \text{hours}}{150 \,\, \text{km}} \)

Simplify.

\( \dfrac{\cancel{150 \,\, \text{km}} \times t}{\cancel{150 \,\, \text{km}}} = \dfrac{250 \cancel{\text{km}}\times 3 \text{hours}}{150 \,\, \cancel{\text{km}}} \)

\( t = \dfrac{250 \times 3}{150} \, \, \text{hours} = 5 \,\, \text{hours}\)


The exercises below with solutions and explanations are all about solving rate problems.

Solve the following rate problems.

  1. The distance between two cities on the map is 15 centimeters. The scales on the map is 5 centimeters to 15 kilometers. What is the real distance, in kilometers, between the two cities?

  2. A car consumes 10 gallons of fuel to travel a distance of 220 miles. Assuming a constant rate of consumption, how many gallons are needed to travel 330 miles?

  3. Ten tickets to a cinema theater costs $66. Wha is the cost of 22 tickets to the same cinema theater?

  4. Cans of soda are packaged in boxes containing the same number of cans. There are 36 cans in 4 boxes.
    a) How many cans are there in 7 boxes?
    b) How many boxes are needed to package 99 cans of soda?

  5. Joe bought 4 kilograms of apples at the cost of $15. How much would he pay for 11 kilograms of the same apples in the same shop?

  6. It takes a pump 10 minutes to move 55 gallons of water up a hill. Using the same pump under the same condition;
    a) how much water is moved in 22 minutes?
    b) how long does it take to move 165 gallons of water?

  7. A container with 324 liters of water, leaks 3 liters every 5 hours. How long does it take for the container to become empty?

  8. Twenty one cans of tomato paste of the same size have a weight of 7300 grams. What is the weight of 5 cans?

  9. An empty container is being filled with water at the rate of 5 liters every 45 seconds and leaks water at the rate of one liter every 180 seconds. What is the quantity of water in the container after one hour?

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Updated: 20 January 2017 (A Dendane)

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