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1.5: Problem Solving Using Rates and Dimensional Analysis .

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A rate is the ratio of two quantities. A unit rate is a rate with a denominator of one.

Example \(\PageIndex{1}\)

Your car can drive 300 miles on a tank of 15 gallons.

Express this as a rate and as a unit rate, in miles per gallon.

Expressed as a rate, \(\dfrac{300 \text { miles }}{15 \text { gallons }}\).

We clan divide to find a unit rate: \(\dfrac{20 \text { miles }}{1 \text { gallon }}\), which we could also write as \(20 \dfrac{\text { miles }}{\text { gallon }}\), or just 20 miles per gallon.

Exercise \(\PageIndex{1}\)

Find the unit rates. If necessary, round your answers to the nearest hundredth. 6 pounds for $5.29

______________ dollars per pound ______________ pounds per dollar

$0.88 per pound, 1.13 pounds per dollar

Example \(\PageIndex{2}\)

Compare the electricity consumption per capita in China to the rate in Japan.

To address this question, we will first need data. From the CIA \({ }^1\) website we can find the electricity consumption in 2011 for China was 4,693,000,000,000 KWH (kilowatt-hours), or 4.693 trillion KWH, while the consumption for Japan was \(859,700,000,000\), or 859.7 billion KWH. To find the rate per capita (per person), we will also need the population of the two countries. From the World \(\mathrm{Bank}^2\), we can find the population of China is \(1,344,130,000\), or 1.344 billion, and the population of Japan is \(127,817,277\), or 127.8 million.

Computing the consumption per capita for each country:

China: \(\dfrac{4,693,000,000,000 \mathrm{KWH}}{1,344,130,000 \text { people }} \approx 3491.5 \mathrm{KWH}\) per person

Japan: \(\dfrac{859,700,000,000 \mathrm{KWH}}{127,817,277 \text { people }} \approx 6726 \mathrm{KWH}\) per person

While China uses more than 5 times the electricity of Japan overall, because the population of Japan is so much smaller, it turns out Japan uses almost twice the electricity per person compared to China.

Many problems can also be solved by multiplying a quantity by rates to change the units. This is the foundation of the Factor-Label process that we have been using already in this chapter.

Example \(\PageIndex{3}\)

a. How far can it drive on 40 gallons? b. How many gallons are needed to drive 50 miles?

We earlier found that 300 miles on 15 gallons gives a rate of 20 miles per gallon.

a. If we multiply the given 40 gallon quantity by this rate, the gallons unit "cancels" and we're left with a number of miles:

40 gallons \(\cdot \dfrac{20 \text { miles }}{\text { gallon }}=\dfrac{40 \text { gallons }}{1} \cdot \dfrac{20 \text { miles }}{\text { gallon }}=800\) miles

Notice that this could also have been achieved using the rate given in the problem:

40 gallons \(\cdot \dfrac{300 \text { miles }}{15 \text { gallons }}=\dfrac{40 \text { gallons }}{1} \cdot \dfrac{300 \text { miles }}{15 \text { gallons }}=\dfrac{40}{1} \cdot \dfrac{300 \text { miles }}{15}=\dfrac{1200}{15}\) miles \(=800\) miles

b. Notice if instead we were asked "how many gallons are needed to drive 50 miles?" we could answer this question by inverting the 20 mile per gallon rate so that the miles unit cancels and we're left with gallons:

50 miles \(\cdot \dfrac{1 \text { gallon }}{20 \text { miles }}=\dfrac{50 \text { miles }}{1} \cdot \dfrac{1 \text { gallon }}{20 \text { miles }}=\dfrac{50 \text { gallons }}{20}=2.5\) gallons

Example \(\PageIndex{4}\)

A bicycle is traveling at 15 miles per hour. How many feet will it cover in 20 seconds?

To answer this question, we need to convert 20 seconds into feet. If we know the speed of the bicycle in feet per second, this question would be simpler. Since we don’t, we will need to do additional unit conversions. We will need to know that 5280 ft = 1 mile. We might start by converting the 20 seconds into hours:

20 seconds \(\cdot \dfrac{1 \text { minute }}{60 \text { seconds }} \cdot \dfrac{1 \text { hour }}{60 \text { minutes }}=\dfrac{1}{180}\) hour

Now we can multiply by the 15 miles/hr

\(\dfrac{1}{180}\) hour \(\cdot \dfrac{15 \text { miles }}{1 \text { hour }}=\dfrac{1}{12}\) mile

Now we can convert to feet

\(\dfrac{1}{12}\) mile \(\cdot \dfrac{5280 \text { feet }}{1 \text { mile }}=440\) feet

We could have also done this entire calculation in one long set of products: \[ 20 \text { seconds } \cdot \dfrac{1 \text { minute }}{60 \text { seconds }} \cdot \dfrac{1 \text { hour }}{60 \text { minutes }} \cdot \dfrac{15 \text { miles }}{1 \text { hour }} \cdot \dfrac{5280 \text { feet }}{1 \text { mile }}=440 \text { feet } \]

Example \(\PageIndex{5}\)

You are walking through a hardware store and notice two sales on tubing.

• 3 yards of Tubing A costs $5.49.
• Tubing B sells for $1.88 for 2 feet.

Either tubing is acceptable for your project. Which tubing is less expensive? Find

Find the unit price for each tubing. This will make it easier to compare.

To compare the prices, you need to have the same unit of measure. You can choose to use dollars per foot (like we have for Tubing B) or dollars per yard (like we have for Tubing A.

Either will work. For this example, we will go with dollars per yard.

Tubing A: \(\$ 1.83\) per yard

Tubing B: \(\dfrac{\$ 0.94}{1 \text { foot }} \cdot \dfrac{3 \text { feet }}{1 \text { yard }} \) Use the conversion factor \(\dfrac{3 \text { feet }}{1 \text { yard }}\)

\( \dfrac{\$ 0.94}{1 \text { fort }} \cdot \dfrac{3 \text { feet }}{1 \text { yard }}=\dfrac{\$ 2.82}{1 \text { yard }} \) Cancel and multiply.

$2.82 per yard

Compare prices for 1 yard of each tubing.

Tubing A: \(\$ 1.83\) per yard Tubing B: \(\$ 2.82\) per yard

Tubing A is less expensive than Tubing B.

Example \(\PageIndex{6}\)

The cost of gasoline in Arizona is about $2.05 per gallon. When you travel over the border into Mexico, gasoline costs 14.81 pesos per liter. Where is gasoline more expensive?

Note: This problem requires a currency conversion factor. Currency conversions are constantly changing, but at the time of print $1 = 18.68 pesos.

To answer this question, we need to convert from gallons to liters AND from U.S. dollars to Mexican pesos.

\[\begin{array}{c}\dfrac{\$ 2.05}{1 \text { gallon }} \cdot \dfrac{1 \text { gallon }}{3.8 \text { liters }} \cdot \dfrac{18.68 \text { pesos }}{\$ 1} \\ \dfrac{\$ 2.05}{1 \text { gallon }} \cdot \dfrac{1 \text { gallon }}{3.8 \text { liter } s} \cdot \dfrac{18.68 \text { pesos }}{\$ 1} \\ \dfrac{2.05}{1} \cdot \dfrac{1}{3.8 \text { liters }} \cdot \dfrac{18.68 \text { pesos }}{1}=\dfrac{2.05 \cdot 18.68}{3.8} \dfrac{\text { pesos }}{\text { liters }} \\ \dfrac{2.05 \cdot 18.68 \text { pesos }}{3.8 \text { liters }}=10.08 \text { pesos per liter }\end{array}\]

The price in Arizona of $2.05 per gallon is equivalent to 10.08 pesos per liter. Since the actual price in Mexico is 14.81 pesos per liter, gasoline is more expensive in Mexico.

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Chapter 9: Radicals

9.10 Rate Word Problems: Work and Time

If it takes Felicia 4 hours to paint a room and her daughter Katy 12 hours to paint the same room, then working together, they could paint the room in 3 hours. The equation used to solve problems of this type is one of reciprocals. It is derived as follows:

[latex]\text{rate}\times \text{time}=\text{work done}[/latex]

For this problem:

[latex]\begin{array}{rrrl} \text{Felicia's rate: }&F_{\text{rate}}\times 4 \text{ h}&=&1\text{ room} \\ \\ \text{Katy's rate: }&K_{\text{rate}}\times 12 \text{ h}&=&1\text{ room} \\ \\ \text{Isolating for their rates: }&F&=&\dfrac{1}{4}\text{ h and }K = \dfrac{1}{12}\text{ h} \end{array}[/latex]

To make this into a solvable equation, find the total time [latex](T)[/latex] needed for Felicia and Katy to paint the room. This time is the sum of the rates of Felicia and Katy, or:

[latex]\begin{array}{rcrl} \text{Total time: } &T \left(\dfrac{1}{4}\text{ h}+\dfrac{1}{12}\text{ h}\right)&=&1\text{ room} \\ \\ \text{This can also be written as: }&\dfrac{1}{4}\text{ h}+\dfrac{1}{12}\text{ h}&=&\dfrac{1 \text{ room}}{T} \\ \\ \text{Solving this yields:}&0.25+0.083&=&\dfrac{1 \text{ room}}{T} \\ \\ &0.333&=&\dfrac{1 \text{ room}}{T} \\ \\ &t&=&\dfrac{1}{0.333}\text{ or }\dfrac{3\text{ h}}{\text{room}} \end{array}[/latex]

Example 9.10.1

Karl can clean a room in 3 hours. If his little sister Kyra helps, they can clean it in 2.4 hours. How long would it take Kyra to do the job alone?

The equation to solve is:

[latex]\begin{array}{rrrrl} \dfrac{1}{3}\text{ h}&+&\dfrac{1}{K}&=&\dfrac{1}{2.4}\text{ h} \\ \\ &&\dfrac{1}{K}&=&\dfrac{1}{2.4}\text{ h}-\dfrac{1}{3}\text{ h}\\ \\ &&\dfrac{1}{K}&=&0.0833\text{ or }K=12\text{ h} \end{array}[/latex]

Example 9.10.2

Doug takes twice as long as Becky to complete a project. Together they can complete the project in 10 hours. How long will it take each of them to complete the project alone?

[latex]\begin{array}{rrl} \dfrac{1}{R}+\dfrac{1}{2R}&=&\dfrac{1}{10}\text{ h,} \\ \text{where Doug's rate (} \dfrac{1}{D}\text{)}& =& \dfrac{1}{2}\times \text{ Becky's (}\dfrac{1}{R}\text{) rate.} \\ \\ \text{Sum the rates: }\dfrac{1}{R}+\dfrac{1}{2R}&=&\dfrac{2}{2R} + \dfrac{1}{2R} = \dfrac{3}{2R} \\ \\ \text{Solve for R: }\dfrac{3}{2R}&=&\dfrac{1}{10}\text{ h} \\ \text{which means }\dfrac{1}{R}&=&\dfrac{1}{10}\times\dfrac{2}{3}\text{ h} \\ \text{so }\dfrac{1}{R}& =& \dfrac{2}{30} \\ \text{ or }R &= &\dfrac{30}{2} \end{array}[/latex]

This means that the time it takes Becky to complete the project alone is [latex]15\text{ h}[/latex].

Since it takes Doug twice as long as Becky, the time for Doug is [latex]30\text{ h}[/latex].

Example 9.10.3

Joey can build a large shed in 10 days less than Cosmo can. If they built it together, it would take them 12 days. How long would it take each of them working alone?

[latex]\begin{array}{rl} \text{The equation to solve:}& \dfrac{1}{(C-10)}+\dfrac{1}{C}=\dfrac{1}{12}, \text{ where }J=C-10 \\ \\ \text{Multiply each term by the LCD:}&(C-10)(C)(12) \\ \\ \text{This leaves}&12C+12(C-10)=C(C-10) \\ \\ \text{Multiplying this out:}&12C+12C-120=C^2-10C \\ \\ \text{Which simplifies to}&C^2-34C+120=0 \\ \\ \text{Which will factor to}& (C-30)(C-4) = 0 \end{array}[/latex]

Cosmo can build the large shed in either 30 days or 4 days. Joey, therefore, can build the shed in 20 days or −6 days (rejected).

The solution is Cosmo takes 30 days to build and Joey takes 20 days.

Example 9.10.4

Clark can complete a job in one hour less than his apprentice. Together, they do the job in 1 hour and 12 minutes. How long would it take each of them working alone?

[latex]\begin{array}{rl} \text{Convert everything to hours:} & 1\text{ h }12\text{ min}=\dfrac{72}{60} \text{ h}=\dfrac{6}{5}\text{ h}\\ \\ \text{The equation to solve is} & \dfrac{1}{A}+\dfrac{1}{A-1}=\dfrac{1}{\dfrac{6}{5}}=\dfrac{5}{6}\\ \\ \text{Therefore the equation is} & \dfrac{1}{A}+\dfrac{1}{A-1}=\dfrac{5}{6} \\ \\ \begin{array}{r} \text{To remove the fractions, } \\ \text{multiply each term by the LCD} \end{array} & (A)(A-1)(6)\\ \\ \text{This leaves} & 6(A)+6(A-1)=5(A)(A-1) \\ \\ \text{Multiplying this out gives} & 6A-6+6A=5A^2-5A \\ \\ \text{Which simplifies to} & 5A^2-17A +6=0 \\ \\ \text{This will factor to} & (5A-2)(A-3)=0 \end{array}[/latex]

The apprentice can do the job in either [latex]\dfrac{2}{5}[/latex] h (reject) or 3 h. Clark takes 2 h.

Example 9.10.5

A sink can be filled by a pipe in 5 minutes, but it takes 7 minutes to drain a full sink. If both the pipe and the drain are open, how long will it take to fill the sink?

The 7 minutes to drain will be subtracted.

[latex]\begin{array}{rl} \text{The equation to solve is} & \dfrac{1}{5}-\dfrac{1}{7}=\dfrac{1}{X} \\ \\ \begin{array}{r} \text{To remove the fractions,} \\ \text{multiply each term by the LCD}\end{array} & (5)(7)(X)\\ \\ \text{This leaves } & (7)(X)-(5)(X)=(5)(7)\\ \\ \text{Multiplying this out gives} & 7X-5X=35\\ \\ \text{Which simplifies to} & 2X=35\text{ or }X=\dfrac{35}{2}\text{ or }17.5 \end{array}[/latex]

17.5 min or 17 min 30 sec is the solution

For Questions 1 to 8, write the formula defining the relation. Do Not Solve!!

• Bill’s father can paint a room in 2 hours less than it would take Bill to paint it. Working together, they can complete the job in 2 hours and 24 minutes. How much time would each require working alone?
• Of two inlet pipes, the smaller pipe takes four hours longer than the larger pipe to fill a pool. When both pipes are open, the pool is filled in three hours and forty-five minutes. If only the larger pipe is open, how many hours are required to fill the pool?
• Jack can wash and wax the family car in one hour less than it would take Bob. The two working together can complete the job in 1.2 hours. How much time would each require if they worked alone?
• If Yousef can do a piece of work alone in 6 days, and Bridgit can do it alone in 4 days, how long will it take the two to complete the job working together?
• Working alone, it takes John 8 hours longer than Carlos to do a job. Working together, they can do the job in 3 hours. How long would it take each to do the job working alone?
• Working alone, Maryam can do a piece of work in 3 days that Noor can do in 4 days and Elana can do in 5 days. How long will it take them to do it working together?
• Raj can do a piece of work in 4 days and Rubi can do it in half the time. How long would it take them to do the work together?
• A cistern can be filled by one pipe in 20 minutes and by another in 30 minutes. How long would it take both pipes together to fill the tank?

For Questions 9 to 20, find and solve the equation describing the relationship.

• If an apprentice can do a piece of work in 24 days, and apprentice and instructor together can do it in 6 days, how long would it take the instructor to do the work alone?
• A carpenter and his assistant can do a piece of work in 3.75 days. If the carpenter himself could do the work alone in 5 days, how long would the assistant take to do the work alone?
• If Sam can do a certain job in 3 days, while it would take Fred 6 days to do the same job, how long would it take them, working together, to complete the job?
• Tim can finish a certain job in 10 hours. It takes his wife JoAnn only 8 hours to do the same job. If they work together, how long will it take them to complete the job?
• Two people working together can complete a job in 6 hours. If one of them works twice as fast as the other, how long would it take the slower person, working alone, to do the job?
• If two people working together can do a job in 3 hours, how long would it take the faster person to do the same job if one of them is 3 times as fast as the other?
• A water tank can be filled by an inlet pipe in 8 hours. It takes twice that long for the outlet pipe to empty the tank. How long would it take to fill the tank if both pipes were open?
• A sink can be filled from the faucet in 5 minutes. It takes only 3 minutes to empty the sink when the drain is open. If the sink is full and both the faucet and the drain are open, how long will it take to empty the sink?
• It takes 10 hours to fill a pool with the inlet pipe. It can be emptied in 15 hours with the outlet pipe. If the pool is half full to begin with, how long will it take to fill it from there if both pipes are open?
• A sink is ¼ full when both the faucet and the drain are opened. The faucet alone can fill the sink in 6 minutes, while it takes 8 minutes to empty it with the drain. How long will it take to fill the remaining ¾ of the sink?
• A sink has two faucets: one for hot water and one for cold water. The sink can be filled by a cold-water faucet in 3.5 minutes. If both faucets are open, the sink is filled in 2.1 minutes. How long does it take to fill the sink with just the hot-water faucet open?
• A water tank is being filled by two inlet pipes. Pipe A can fill the tank in 4.5 hours, while both pipes together can fill the tank in 2 hours. How long does it take to fill the tank using only pipe B?

Answer Key 9.10

Intermediate Algebra Copyright © 2020 by Terrance Berg is licensed under a Creative Commons Attribution-NonCommercial-ShareAlike 4.0 International License , except where otherwise noted.

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

• 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?
• 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?
• Ten tickets to a cinema theater costs $66. Wha is the cost of 22 tickets to the same cinema theater?
• 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?
• 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?
• 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?
• A container with 324 liters of water, leaks 3 liters every 5 hours. How long does it take for the container to become empty?
• Twenty one cans of tomato paste of the same size have a weight of 7300 grams. What is the weight of 5 cans?
• 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?

Middle School Maths (Grades 6, 7, 8, 9) - Free Questions and Problems With Answers High School Maths (Grades 10, 11 and 12) - Free Questions and Problems With Answers Primary Maths (Grades 4 and 5) with Free Questions and Problems With Answers Home Page

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