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 scale on the map is 5 centimeters to 15 kilometers. What is the real distance \( d \), in kilometers, between the two cities?

    Solution

    First, express the unit rate with known quantities in kilometers per centimeter: \[ \frac{15 \ \text{kilometers}}{5 \ \text{centimeters}} \] Let \( d \) be the real distance in kilometers corresponding to 15 centimeters. The unit rate using the unknown quantity \( d \) is: \[ \frac{d}{15 \ \text{centimeters}} \] Since the two rates are equal: \[ \frac{15 \ \text{kilometers}}{5 \ \text{centimeters}} = \frac{d}{15 \ \text{centimeters}} \] Cross multiply: \[ 15 \ \text{kilometers} \times 15 \ \text{centimeters} = d \times 5 \ \text{centimeters} \] Divide both sides by \( 5 \ \text{centimeters} \): \[ \frac{15 \ \text{kilometers} \times 15 \ \text{centimeters}}{5 \ \text{centimeters}} = \frac{d \times 5 \ \text{centimeters}}{5 \ \text{centimeters}} \] Simplify: \[ \frac{15 \ \text{kilometers} \times 15 \ \cancel{\text{centimeters}}}{5 \ \cancel{\text{centimeters}}} = \frac{d \times \cancel{5} \ \cancel{\text{centimeters}}}{\cancel{5} \ \cancel{\text{centimeters}}} \] \[ d = 45 \ \text{kilometers} \]

  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 the unit rate with known quantities in miles per gallon: \[ \frac{220 \ \text{miles}}{10 \ \text{gallons}} \] Let \( x \) be the number of gallons needed to travel 330 miles. Then the unit rate using \( x \) is: \[ \frac{330 \ \text{miles}}{x} \] Since the two rates are equal: \[ \frac{220 \ \text{miles}}{10 \ \text{gallons}} = \frac{330 \ \text{miles}}{x} \] Cross multiply: \[ 220 \ \text{miles} \times x = 330 \ \text{miles} \times 10 \ \text{gallons} \] Divide both sides by \( 220 \ \text{miles} \): \[ \frac{220 \ \text{miles} \times x}{220 \ \text{miles}} = \frac{330 \ \text{miles} \times 10 \ \text{gallons}}{220 \ \text{miles}} \] Simplify to find \( x \): \[ x = 15 \ \text{gallons} \]

  3. Ten tickets to a cinema theater cost \$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 (the unit rate), which is given by: \[ c = \frac{\$66}{10 \text{ tickets}} = 6.6 \text{ dollars per ticket} \] To find the cost \( C \) of 22 tickets, multiply the cost of one ticket by 22: \[ C = 22 \text{ tickets} \times c = 22 \times 6.6 = \$145.2 \]

  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): \[ n = \frac{36 \text{ cans}}{4 \text{ boxes}} = 9 \text{ cans per box} \] In 7 boxes, the number of cans \( N \) is: \[ N = 7 \times n = 7 \text{ boxes} \times 9 \text{ cans per box} = 63 \text{ cans} \] b) Let \( B \) be the number of boxes needed to package 99 cans. Using the unit rate, write the proportion: \[ \frac{9 \text{ cans}}{1 \text{ box}} = \frac{99 \text{ cans}}{B} \] Cross multiply: \[ 9 \text{ cans} \times B = 99 \text{ cans} \times 1 \text{ box} \] Divide both sides by \( 9 \text{ cans} \): \[ \frac{9 \text{ cans} \times B}{9 \text{ cans}} = \frac{99 \text{ cans} \times 1 \text{ box}}{9 \text{ cans}} \] Simplify: \[ B = 11 \text{ boxes} \]

  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 kilogram of apples (unit rate): \[ c = \frac{\$15}{4 \text{ kg}} = \$3.75 \text{ per kg} \] Knowing the cost of 1 kg, the cost \( C \) of 11 kg is: \[ C = 11 \text{ kg} \times c = 11 \text{ kg} \times \$3.75 \text{ per kg} = \$41.25 \]

  6. It takes a pump 10 minutes to move 55 gallons of water up a hill. Using the same pump under the same conditions:
    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 the number of gallons \( n \) moved in one minute (unit rate): \[ n = \frac{55 \text{ gallons}}{10 \text{ minutes}} = 5.5 \text{ gallons per minute} \] The number \( N \) of gallons moved in 22 minutes is: \[ N = 22 \text{ minutes} \times 5.5 \text{ gallons per minute} = 121 \text{ gallons} \] b) Let \( T \) be the number of minutes to move 165 gallons. Using the rate equality: \[ \frac{55 \text{ gallons}}{10 \text{ minutes}} = \frac{165 \text{ gallons}}{T} \] Cross multiply: \[ 55 \text{ gallons} \times T = 165 \text{ gallons} \times 10 \text{ minutes} \] Divide both sides by \( 55 \text{ gallons} \) and simplify: \[ T = \frac{165 \text{ gallons} \times 10 \text{ minutes}}{55 \text{ gallons}} = 30 \text{ minutes} \]

  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 leak. Using equality of rates: \[ \frac{3 \text{ liters}}{5 \text{ hours}} = \frac{324 \text{ liters}}{T} \] Cross multiply: \[ 3 \text{ liters} \times T = 5 \text{ hours} \times 324 \text{ liters} \] Divide both sides by \( 3 \text{ liters} \) and simplify: \[ T = \frac{5 \text{ hours} \times 324 \text{ liters}}{3 \text{ liters}} = 540 \text{ hours} \]

  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. Using equality of rates: \[ \frac{20 \text{ cans}}{7300 \text{ grams}} = \frac{5 \text{ cans}}{W} \] Cross multiply: \[ 20 \text{ cans} \times W = 5 \text{ cans} \times 7300 \text{ grams} \] Divide both sides by \( 20 \text{ cans} \) and simplify: \[ W = \frac{5 \text{ cans} \times 7300 \text{ grams}}{20 \text{ cans}} = 1825 \text{ grams} \]

  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 1 liter every 180 seconds. What is the quantity of water in the container after one hour?

    Solution

    In this problem, we have two rates:
    The filling rate \( R_f \)
    The leaking rate \( R_l \) Calculate the two rates: \[ R_f = \frac{5 \text{ liters}}{45 \text{ seconds}}, \quad R_l = \frac{1 \text{ liter}}{180 \text{ seconds}} \] We also know: \[ 1 \text{ hour} = 3600 \text{ seconds} \] The quantity \( Q_1 \) of water pumped after 1 hour is: \[ Q_1 = R_f \times 3600 = \frac{5}{45} \times 3600 = 400 \text{ liters} \] The quantity \( Q_2 \) of water lost through leakage after 1 hour is: \[ Q_2 = R_l \times 3600 = \frac{1}{180} \times 3600 = 20 \text{ liters} \] Therefore, the quantity \( Q \) of water in the container after 1 hour is: \[ Q = Q_1 - Q_2 = 400 - 20 = 380 \text{ liters} \]

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