Detailed Solutions Problems and Questions on Rate

Detailed solutions to the questions on how to solve rate problems? are presented.

  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 d, in kilometers, between the two cities?
    Solution
    Express unit rate with known quantities in kilometer / centimeter.
    unit rate: \( \dfrac{15 \,\, \text{kilometers}}{5 \,\, \text{centimeters}} \)
    Since d be the real distance in kilometers corresponding to 15 centimeters, we express the unit rate using unknown quantity d as follows
    unit rate: \( \dfrac{d}{15 \,\, \text{centimeters}} \)
    The two rates are equal. Hence
    \( \dfrac{15 \,\, \text{kilometers}}{5 \,\, \text{centimeters}} \) = \( \dfrac{d}{15 \,\, \text{centimeters}} \)
    Cross multiply the numerators and denominators to obtain an equivalent equation without fractions.
    \( 15 \,\, \text{kilometers} \times 15 \,\, \text{centimeters} = d \times 5 \,\, \text{centimeters} \)
    Find d by dividing both sides of the equation by 5 cm.
    \( \dfrac{15 \,\, \text{kilometers} \times 15 \,\, \text{centimeters}}{5 \,\, \text{centimeters}} = \dfrac{d \times 5 \,\, \text{centimeters}}{5 \,\, \text{centimeters}} \)
    Simplify.
    \( \dfrac{15 \,\, \text{kilometers} \times 15 \,\, \cancel{\text{centimeters}}}{5 \,\, \cancel{\text{centimeters}}} = \dfrac{d \times \cancel{5} \,\, \cancel{\text{centimeters}}}{\cancel{5} \,\, \cancel{\text{centimeters}}} \)
    d = 45 km


  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?
    Solution
    Express unit rate with known quantities in miles/gallon.
    unit rate: \( \dfrac{220 \,\, \text{miles}}{10 \,\, \text{gallons}} \)
    Let x be the number of gallons to find and express unit rate using unknown quantity x
    unit rate: \( \dfrac{330 \,\, \text{miles}}{x} \)
    The two rates are equal. Hence
    \( \dfrac{220 \,\, \text{miles}}{10 \,\, \text{gallons}} \) = \( \dfrac{330 \,\, \text{miles}}{x} \)
    Cross multiply the two fractions in the above equation.
    \( 220 \,\, \text{miles} \times x = 330 \,\, \text{miles} \times 10 \,\, \text{gallons} \)
    Divide both sides of the equation by 220 miles.
    \( \dfrac{220 \,\, \text{miles} \times x}{220 \,\, \text{miles}} = \dfrac{330 \,\, \text{miles} \times 10 \,\, \text{gallons}}{220 \,\, \text{miles}} \)
    Simplify to find x.
    x = 15 gallons.


  3. Ten tickets to a cinema theater costs $66. What is the cost of 22 tickets to the same cinema theater?
    Solution
    It may be easier to find the cost c of one ticket (or unit rate) which is given by.
    c = $66 / 10 tickets = 6.6 dollar / ticket
    To find the cost C of 22 tickets, we just multiply the cost c of one ticket by 22
    C = 22 tickets c = 22 tickets 6.6 dollars / ticket = $145.2


  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?
    Solution
    a) First find the number n of cans per box (unit rate).
    n = 36 cans / 4 boxes = 9 cans / box
    In 7 boxes we have N cans given by
    N = 7 n = 7 boxes 9 cans / box = 63 cans
    b) Let B be the number of boxes to package 99 cans. Use the rate to write an equation
    9 cans / box = 99 cans / B
    Cross multiply the fractions making the equation to obtain
    9 cans B = 99 cans 1 box
    Divide both sides by 9 cans and simplify
    9 cans B / 9 cans = 99 cans 1 box / 9 cans
    B = 11 boxes


  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?
    Solution
    First find the cost c of one kilograms of apples (unit rate).
    c = $15 / 4 kg = $3.75 / kg
    Knowing the cost of 1 kg, the cost C of 11 kg is given by.
    C = 11 kg c = 11 kg $3.75 / kg = $41.25


  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?
    Solution
    a) First find number of gallons n moved in one minute (unit rate).
    n = 55 gallons / 10 minutes = 5.5 gallons / minute
    The number N of gallons moved in 22 minutes is given by.
    N = 22 minutes 5.5 gallons / minute = 121 gallons
    b) Let T is the number of minutes to move 165 gallons. An equality of the rate gives
    55 gallons / 10 minutes = 165 gallons / T
    Cross multiply the fractions in the above equation to obtain an equivalent equation.
    55 gallons T = 165 gallons 10 minutes
    Divide both sides by 55 gallons and simplify.
    T = 165 gallons 10 minutes / 55 gallons = 30 minutes


  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?
    Solution
    Let T be the number of hours for 324 liters to be leaked. An equality of the rates gives
    3 liters / 5 hours = 324 liters / T
    Cross multiply the fractions in the above equation to obtain an equivalent equation.
    3 liters T = 5 hours 324 liters
    Divide both sides by 3 liters and simplify.
    T = 5 hours 324 liters / 3 liters = 540 hours


  8. Twenty one cans of tomato paste of the same size have a weight of 7300 grams. What is the weight of 5 cans?
    Solution
    Let W be the weight of 5 cans. An equality of the rates gives
    20 cans / 7300 grams = 5 cans / W
    Cross multiply to obtain an equivalent equation.
    20 cans W = 5 cans 7300 grams
    Divide both sides by 20 cans and simplify.
    W = 5 cans 7300 grams / 20 cans = 1825 grams


  9. An empty container is being filled with a water pump at the rate of 5 liters every 45 seconds. But the container also leaks water at the rate of one liter every 180 seconds. What is the quantity of water in the container after one hour?
    Solution
    In this problem we have two rates: one rate Rf at which water is being filled and the rate Rl at which water is being leaked. Find the two rates.
    Rf = 5 liters / 45 seconds
    Rl = 1 liter / 180 seconds
    We also know that
    1 hour = 3600 seconds
    The quantity Q1 of water pumped after 1 hour is given by
    Q1 = (5 liters / 45 seconds) 3600 = 400 liters
    The quantity Q2 of water lost through leakage after 1 hour is given by
    Q2 = (1 liter / 180 seconds) = 20 liters
    The quantity Q of water in the container after 1 hour is given by
    Q = Q1 - Q2 = 400 liters - 20 liters = 380 liters.


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