Applications of Linear...

LAST REVISED Nov., 2008

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Algebra Module A22

Applications of Linear Equations

Module A22 − Applications of Linear Equations

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Introduction to Applications of Linear Equations Statement of Prerequisite Skills Complete all previous TLM modules before beginning this module.

Required Supporting Materials Access to the World Wide Web. Internet Explorer 5.5 or greater. Macromedia Flash Player.

Rationale Why is it important for you to learn this material? Applying the techniques learned in previous modules to the real world is the bridge that makes learning mathematics essential to the pursuit of a career in the technologies.

Learning Outcome When you complete this module you will be able to… Solve applied problems involving linear equations.

Learning Objectives 1. Solve distance problems. 2. Solve mixture problems. 3. Solve measurement problems. 4. Solve word problems involving work.

Connection Activity A seismic crew takes 5 days to work an area that measures one square km. A second crew takes 6 days to do the same job. How long would it take both crews working together to do the job?

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Module A22 − Applications of Linear Equations 2

OBJECTIVE ONE When you complete this objective you will be able to… Solve distance problems. The basic formula used is: Distance = Rate × Time In symbols: D = RT

Example 1 Two cars, A and B, having speeds of 30 km/h and 40 km/h respectively are 280 km apart. They start moving toward each other. How long will they travel before they meet? SOLUTION: Since time is the variable to be solved, use t to represent it. Using the distance formula stated above we see that car A will travel 30 × t km, and that car B will travel 40 × t km. Note here, both cars travel for the same length of time. When they meet they will have traveled the full 280 km, so the sum of their distances is 280, Thus: 30t + 40t = 280 And t = 4. t represented time so we have our answer of 4 hours.


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Example 2 Two planes start from Calgary at the same time and fly in opposite directions. One plane travels 40 km/h faster than the other. If they are 2000 km apart after 5 hours, find their speeds. SOLUTION: Let x be one of the unknowns, in this case “the slower speed.” Now x = slower plane’s speed, so the faster plane’s speed will equal (x + 40). Using D = RT we get: distance the slower plane travels = 5x (since time is 5h) and the distance the faster plane travels = 5(x + 40) Now the sum of their distances is 2000 km, so: 5x + 5(x + 40) = 2000 10x = 1800 x = 180 So: speed of the slower plane = 180 km/h and the speed of the faster plane = (180 + 40) = 220 km/h.


Experiential Activity One Solve the following word problems 1. Two trains approach one another along (straight) parallel tracks. They start out at the

same time from stations 90 km apart. One train travels at 100 km per hour, the second train at 80 km per hour. How far has the faster train gone when they pass each other?

2. A SAAB and a Volkswagen leave a toll area at the same time traveling in the same

direction along a straight road. The Volkswagen travels at two-thirds the rate of the SAAB. At the end of 4 hours they are 100 km apart. How fast is the SAAB traveling?

3. A boat goes upstream at the rate of 4 km per hour. It returns downstream at the rate

of 8 km per hour. If the round trip takes 3 hours, how far upstream did it go? 4. A student begins to walk along a straight road at the rate of 3 km per hour to the next

town, which is 18 km away. After 10 minutes a car picks him up, and in 15 more minutes he is in town. How fast was the car traveling?

5. Two vehicles leave the Hollywood Bowl at the same time and head for San

Francisco. The first vehicle takes 7 hours along the shorter route, which measures 420 km. The other vehicle, which travels 4 km per hour faster, arrives a half hour later. How long is the second route? Show Me

6. A cross-country skier walks 3000 m up a mountain road, rests for 10 minutes, and

then skis down the road. He returns a half hour after he started. If he skis four times as fast as he walks, at what rate does he ski?

7. A jogger can average 10 km per hour over level ground and 6 km per hour over hilly

ground. Altogether it takes him an hour and forty minutes to cover 12 km. How many of these km are hilly?


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Experiential Activity One Answers

1. 50 km

2. 75 km/h

3. 8 km

4. 70 km/h

5. 480 km

6. 750 m/min

7. 7 km


OBJECTIVE TWO When you complete this objective you will be able to… Solve mixture problems. These problems fit a pattern in their solution, just like the earlier money problems did. One thing that appears in almost every term of the solution equation is that each term has two "parts". One of the parts is a percentage and the other part is a mass or a volume. See the following examples for more information on this pattern.

EXAMPLE 1 How much metal alloy containing 25% copper must be added to 20 kg of pure copper to obtain an alloy having 50% copper? SOLUTION: Let x = the amount of metal containing 25% copper. NOTE: Pure copper means 100% copper. Form an equation in English. Base the equation on the copper you start with and the copper you finish with. Cu in first metal Cu in second metal = Total Cu in final alloy 25% of x + 100% of 20 = 50% of (x + 20) 0.25x + 20 = 0.50(x + 20) x = 40 Therefore the amount of 25% alloy is 40 kg.

EXAMPLE 2 A mixture of 40 kg of candy worth 60 cents a kg is to be made up by taking some worth 45 cents/kg and some at 85 cents/kg. How much of each should be taken? SOLUTION: Let x = the weight of the 45 cent candy, then (40 – x) = the weight of the 85 cent candy. The value of the 45 cent candy is 45x, and the value of the 85 cent candy is 85(40 – x). Now form an equation based on value: Value of 45 cent candy + Value of 85 cent candy = Value of final candy 45x + 85(40 – x) = (60)(40) 45x + 3400 – 85x = 2400 –40x = –1000 x = 25 Therefore we have 25 kg of 45 cent candy and (40 – 25) = 15 kg of 85 cent candy.


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EXAMPLE 3 How much water must be evaporated from 80 ml, of a 15% salt solution so that the resulting solution is 40% salt? SOLUTION: When the water evaporates it leaves the salt behind, thus increasing the percentage of salt in the solution. Note, the amount of salt stays the same in both the 15% solution and the 40% solution. We will let x = the amount of water evaporated. And we will build an equation based on salt content in each solution, thus:

Salt in 15% Solution = Salt in 40% Solution 15% of 80 = 40% of (80 – x)

.15(80) = .40(80 – x) 12 = 32 – .4x

x = 50 Thus, 50 ml of water must be evaporated.


Experiential Activity Two MIXTURE PROBLEMS 1. How many litres of a 20% salt solution should be combined with 18 litres of a 28%

salt solution to obtain a 22% solution? 2. How many litres of a 15% salt solution should be combined with a 25% salt

solution to obtain 40 litres of a 21% solution? Show Me 3. How much pure acid should be added to 5 litres of 40% acid solution to obtain a

50% acid solution? 4. How many litres of a chemical that is 44% nitric acid must be combined with 12

litres of a chemical that is 56% nitric acid to obtain a 47% nitric acid mixture? 5. How many grams of an alloy containing 60% gold must be added to 50 grams of

an alloy containing 45% gold to obtain an alloy containing 51% gold? 6. How much cream that contains 33% butterfat must be blended with milk that

contains 3% butterfat to obtain 10 litres of half-and-half that contains 8% butterfat?

7. Ten litres of butterscotch ice cream that contains 18% butterfat is mixed with 8

litres of vanilla ice cream that contains 12% butterfat to make butterscotch ripple ice cream. What percent butterfat does the butterscotch ripple contain?

8. A grocer mixes 6 kg of coffee that is 60% Peruvian with one that is 80% Peruvian.

How much of the second type should be used to obtain a mixture that is 78% Peruvian?

9. How much water must be evaporated from 60 ml of a 12% salt solution to make a

30% salt solution?


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Experiential Activity Two Answers 1. 54 litres 2. 16 litres of 15%, and 24 litres of 25% 3. 1 litres 4. 36 litres 5. 33.3 grams 6. 1.67 litres 7. 15.3% 8. 54 kg 9. 36 ml


OBJECTIVE THREE When you complete this objective you will be able to… Solve measurement problems. The problems in this section deal mostly with perimeters and areas for rectangles, squares, and triangles. They are good practical questions.

EXAMPLE 1 When each side of a square is increased by 4 m the area is increased by 64 m2. Determine the dimensions of the original square. SOLUTION: A square is a rectangle with all sides equal. You may wish to draw a square and label it's sides as indicated below. Let x = the length of the side of the original square. This means it's area is x2, i.e., side squared. Because we increase the length of each side by 4, the increased (new) area = (x + 4)2. Now form an English word equation from the statement of the problem: original area + increase in area = new area original area + 64 = new area Translate this into algebra: x2 + 64 = (x + 4)2 Solving we get: x2 + 64 = x2 + 8x + 16 8x = 48 x = 6 Therefore: The dimensions of the original square are 6 m by 6 m.


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EXAMPLE 2 A rectangle whose length is 2 cm more than its width has a perimeter of 36 cm. Find the dimensions. SOLUTION: Let x = the width of the rectangle. This means it's length will be (x + 2). Now: perimeter = 21 + 2w for a rectangle, so: 36 = 2(x + 2) + 2x 4x = 32 and x = 8, so the width = 8 cm, and the length = 10 cm.


Experiential Activity Three Solve the following word problems 1. The side of one square plastic part is 3 mm more than the side of another square

part. The perimeter of the larger part is 36 mm. What is the perimeter of the smaller part?

2. If each side of a square is reduced by 2 cm, the perimeter will be 24 cm. What is

the side of the square? 3. The length of a rectangle is 1 cm less than twice the width. Given that the

perimeter is 40 cm, find the dimensions. 4. A rectangular field is bounded on one side by a river. The other side and the two

ends are to be fenced. The side parallel to the river is 50 m longer than either of the ends, and 950 m of fencing is used. What are the dimensions of the field?

5. The area of a square exceeds the area of a rectangle by 3 cm2. The width of the

rectangle is 3 cm shorter and the length 4 cm longer than the side of the square. Find the side of the square. Show Me

6. The length of a rectangular floor is 8 m greater than its width. If each dimension is

increased by 2 m, the area is increased by 60 square m. Find the dimensions of the floor.

7. The perimeter of a rectangle is 110 m. Find the dimensions if the length is 5 m less

than twice the width. 8. The length of a rectangular swimming pool is twice its width. The pool is

surrounded by a cement walk 4 m wide. If the area of the walk is 784 square metres, determine the dimensions of the pool.


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Experiential Activity Three Answers 1. 24 mm 2. 8 cm 3. 13 cm × 7 cm 4. 350 m × 300 m 5. 9 cm 6. 18 m × 10 m 7. 35m × 20m 8. 60m × 30m


OBJECTIVE FOUR When you complete this objective you will be able to… Solve problems involving work. The type of problem we deal with in this objective is both interesting and unique. Generally the problem deals with the situation where a single task or job is being performed by more than one worker or machine simultaneously and we are required to find out how long it takes to complete the task. In general, the approach to solving this type of problem is to set up an equation based on the fractional amounts of the task completed per unit of time. For instance, if it takes 3 hours for a pipe to drain a tank then in 1 hour it can do 1/3 of the task.

EXAMPLE 1 Joe and Ed are to dig a trench. If Joe could do the job alone in 2 hours and Ed could do the job alone in 3 hours, how long would it take for the two men to do the job together? SOLUTION: You say 5 hours? Wrong, because 5 hours is more time than it would take either man to dig the trench by himself. We will make an equation based on how much work is done in 1 hour, our unit of time. Let x = the time in hours required for the two men to do the job together. Therefore 1/x = amount of work done in 1 hour also, Joe can do 1/2 of the job in 1 hour, and Ed can do 1/3 of the job in 1 hour so, let's form an equation based on the fractional amounts of the job completed in 1 hour (our unit of time) fraction of job done by Joe in 1 hour + fraction of job done

by Ed in 1 hour = fraction of job completed in 1 hour by both men

1/2 + 1/3 = 1/x

3 2 16 x+

=

x = 1.2

It takes 1.2 hours for the two men to complete the job.

NOTE on this example: A task can be performed faster by 2 or more men or machines working together than by 1 of them alone. Notice in this example that the answer of 1.2 hours is less than the time either of the men required alone.

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EXAMPLE 2 A 10,000 litre tank is to be drained by using 3 pipes. One pipe can drain the tank by itself in 6 hours, and each of the other two pipes would require 8 hours to drain the tank. Find how long it would take for all 3 pipes to drain the tank at the same time? SOLUTION: Let x = time in hours required to drain the tank. Therefore 1/x = amount of tank drained in 1 hour. Also, the first pipe can drain the tank in 6 hours, therefore it can do 1/6 of the job in 1 hour, the second pipe can drain the tank in 8 hours, therefore it can do 1/8 of the job in 1 hour, and the third pipe can drain the tank in 8 hours, therefore it can do 1/8 of the job in 1 hour. So in 1 hour we have: 1/6 of the job done + 1/8 of the job done + 1/8 of the job done = 1/x of the job done 1 1 1 16 8 84 3 3 1

2410 124

2.4

x

x

xx

+ + =

+ +=

=

=

It takes 2.4 hours for the 3 pipes working together to drain the tank. In EXAMPLE 2, if there were a 4th pipe filling the tank, and it took 10 hours to fill it then with all pipes working simultaneously the equation would be: 1 1 1 1 16 8 8 10 x

+ + − =

i.e., if emptying is called positive then the opposite operation of filling is called negative.

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EXAMPLE 3 A reservoir contains 500,000 litres of a liquid. It is being filled by a pipe which requires 20 hours to do the job, and emptied by two pipes each of which requires 8 hours to do the job. Find the time required to drain the reservoir. SOLUTION: Let x = time in hours required to drain the reservoir. Therefore 1/x = amount of reservoir drained in 1 hour. Setting up our equation as in the previous example, we get: 1/8 of the task + 1/8 the task – 1/20 of the task = 1/x of the task 1 1 1 18 8 201 1 14 204 1

205

x

x

xx

+ − =

− =

=

=

It takes 5 hours to drain the reservoir.

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Experiential Activity Four 1. A seismic crew takes 5 days to work an area that measures one square km. A second

crew takes 6 days to do the same job. How long would it take both crews working together to do the job?

2. One pipe can fill a certain tank in 6 hours, while a second pipe can fill the tank in 15

hours. Working together how long does it take to fill the tank? 3. One pipe can fill a tank in 10 hours while a second pipe can fill the tank in 6 hours.

Find the time it would take a third pipe to fill the tank so that when all three pipes are open, the tank is filled in 2 hours. Show Me

4. A winter blizzard left Peter’s driveway blocked with snow. Peter could clean the

snow away in 5 hours. His sister Paula can do the job in 4 hours. How long would it take the two of them working together to clear the driveway of snow?

5. A water fountain is being filled by one pipe and emptied by another pipe at the same

time. The pipe filling it would take 6 hours if operating alone. If it requires 24 hours to fill the fountain, how long would it require the second pipe to empty the reservoir if working alone?


Experiential Activity Four Answers 1. 2.72 days 2. 4.2857 hours 3. 4.2857 hours 4. 2.22 hours, or 2 hours 13 minutes 5. 8 hours

Practical Application Activity Complete the Applications of Linear Equations assignment in TLM.

Summary This module presented the student with a variety of well-organized linear theory word problems.

Applications of Linear...

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