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