Блог Анкара

In This Chapter

^ Figuring the distance traveled as the sum of two different distances ^ Equating two distances that have different rates of speed ^ Emptying or filling a tank with open intake and outflow valves ^ Divvying up the work with work problems

His chapter deals with a lot of moving around. People and cars and trains

Are moving around all the time. Liquids in containers are moving from one place to another. The first problems involve moving around and the distance traveled. The typical distance problems use the relationship that distance is equal to rate times time, D = Rt. By using this equation, you deal with trains meeting somewhere in the middle of nowhere and wives catching up to forgetful husbands.

Another type of moving around occurs with the classic work problems — how many hands it takes for a lighter workload — and how long it takes when everyone works together.

And, lastly, you’ll find a lot of moving around when fluids move from one container to an other. If you’re trying to fill a tub while the tub is leaking, you have to work quickly to find out how long it takes to actually make the tub overflow.

Summing Up the Distances

These first distance problems have a common theme: that the sum of the distances that two people travel, along the same straight line, equals the total distance. Sometimes, people are traveling toward one another, and the sum

Of the distances is how far each has traveled until they meet. Other times, people are traveling in opposite directions, and you sum up the distances traveled to compute how far apart they are.

Meeting somewhere in the middle

Two lovers spy each other on opposite sides of the room and run toward one another. They meet somewhere between where they started running — and where they meet depends on how fast each can run. You can assume that there are no chairs to dodge in their mad dashes. The setup used to solve this problem is pretty much the same as a problem involving two bulls rushing at one another from either side of the arena or two trains approaching one another from opposite directions along the same track.

Keep in mind the formula D = Rt Or Distance = Rate X Time. When two different distances are added together to equal a total distance, the individual rates and times are multiplied together first and then the products are added together:

D = d, + d2

Making good time

With distance problems, you can solve for the distance traveled or the speed at which objects are traveling or the amount of time spent. The two problems in this section involve solving for how much time it takes to reach a goal.

The Problem: Betsy and Bart see each other from opposite sides of a gymnasium that measures 440 feet across. They both start running toward one another at the same time. If Betsy can run at the rate of 4 feet per second and Bart runs at 7 feet per second, then how long does it take for them to meet?

The total distance to be covered is 440 feet. Set that amount equal to the sum of the distances that each runner covers. Betsy can run at the rate of 4 feet per second, and she runs for T Seconds. Bart can run at 7 feet per second and also runs for T Seconds. So your equation is 440 = 4t + 7t. Simplify on the right to get 440 = ,,t. Divide each side of the equation by,,, and you get that T = 40 seconds. It’ll take less than a minute for the two to reach one another, with Betsy running 4(40) = 160 feet and Bart running 7(40) = 280 feet.

In the preceding problem, the two people see each other and start running at the same time. What if one starts before the other, and they run for different amounts of time?

The Problem: Jon leaves Chicago at noon and heads south toward Bloomington traveling at 45 mph. Jane leaves Bloomington heading north for Chicago at 1 p. m., traveling at 55 mph. If Chicago and Bloomington are,45 miles apart, then what time will they meet?

^VLA* The total distance to be traveled is 145 miles. If Jon drives for T Hours, then Jane, who left an hour later, will drive for T - 1 hours. Take each rate in terms of miles per hour and multiply it by the respective amount of time traveled. Jon will drive for 45T Miles and Jane will travel for 55(t - 1) miles. Add the two distances together and set the sum equal to 145. Solve the equation for T.

45t + 55 (T – 1) = 145 45t + 55t – 55 = 145 100T -55 =145 100T = 200 T= 2

Because T Is the number of hours that Jon drives, he drives for two hours and Jane drives for one hour less or one hour. In any case, they meet up at 2 p. m.

Speeding things up

Distance is equal to rate times time. Participants in these distance problems may travel for the same amount of time or different amounts of time. They can travel at the same speed or different speeds. The next two problems let you determine how fast things are moving.

The Problem: Two trains leave the same station traveling in opposite directions. The first train leaves at 2 p. m. The second train leaves a half-hour later and travels at a speed averaging 15 miles per hour faster than the first train. By 8 o’clock that evening, they’re 600 miles apart. How fast are the two trains traveling?

Vjj. VLA/V Use the same D = Rt Format, setting the sum of the distances traveled equal to 600. The first train travels for 6 hours at rate r, and the second train, leaving a half-hour later, only travels for 5.5 hours but at a greater speed, R + 15. The equation and solution:

6r + 5.5 (R + 15) = 600 6r + 5.5r + 82.5 = 600 11.5r + 82.5 = 600

11.5r = 517.5 517.5 11.5

The first train is traveling at 45 mph. The faster, or 60 mph.

Second train is traveling 15 mph

Taxi driver

Imagine that you’re a taxi driver in Chicago and 30 mph, and the tank is half full of fuel. One mile

That you’re driving a 1994 yellow cab. Your pas- into the trip, the tank is down to two-fifths full of

Sengers are college students, and they want to fuel, and the trip is over 15 minutes later. What is

Travel 4 miles from home to school. You’re driving the name and age of the cab driver?

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The Problem: Alberto can bicycle 2 miles per hour less than twice as fast as Ollie, so Alberto didn’t leave for the rally until two hours after Ollie left. If the total distance they traveled was 504 miles and if Ollie traveled for 10 hours, then how fast can Alberto bicycle?

Let the rate at which Ollie can bicycle be r. Then Alberto’s rate is 2r – 2. If Ollie traveled for 10 hours, then Alberto traveled for 8 hours. The total distance is equal to the sum of the distances the two traveled. Ollie traveled 10R Miles, and Alberto traveled 8(2R - 2) miles.

10r + 8 (2r – 2) = 504 10r + 16r – 16 = 504 26R -16 =504 26R = 520

R 520

26

Ollie bicycles at 20 mph, so Alberto bicycles at 2(20) – 2 = 40 – 2 = 38 mph.

Making a beeline

You can solve for the time it takes to travel, and you can solve for how fast cars or trains travel. Determining the total distance that someone or something travels can also be very interesting. Consider the story of Super Bee Who flies at the speed of 90 mph. He leaves the engine of a westbound train that’s traveling at 60 mph and flies until he reaches an eastbound train that’s traveling at 75 mph along the same track. After barely touching the east-bound train, he flies back to the westbound train, touches it, and flies back

And forth and back and forth between the trains until the trains finally meet. Figure 16-1 shows you the two trains and Super Bee, Flying between them.

Figure 16-1:

Super Bee flies back and forth between the speeding trains

The Problem: How far does a bee travel if it flies back and forth between trains that are approaching each other on the same track — one train traveling at 60 mph and the other traveling 75 mph — if the bee flies at 90 mph and, when it started this journey, the trains were 648 miles apart?

You could determine how far the bee had to fly until it flew from the first train to the second train — while the trains are getting closer to one another — and then how far it had to fly back to the first train, and so on, until the trains meet. Or, a much simpler way to do this is to figure out how long it will take the trains to meet if they start 648 miles apart and are approaching each other at the rate of 135 mph (the sum of 60 mph and 75 mph). When you’ve determined how much time it takes, then you know how long the bee has been flying and can multiply the time and the rate of 90 mph to get the distance. Because D = Rt, In this case 648 = 135t. Dividing each side of the equation by 135, you get that T = 4.8 hours — the time it takes the trains to meet. Multiplying 90 x 4.8, you get a distance of 432 miles. That’s one tired bee.

Equating the Distances Traveled

Many distance problems have the scenario that one person catches up with another or one train or plane leaves later and finally passes the first one. The common thread or theme for these problems is that the distance traveled by the two participants is the same. You equate the distances. Some problems have you solve for time — how long it took to catch up. Or you may solve for how fast one or the other is traveling. And the question may even be about the distance — how far they traveled before arriving at the same place. In each case, though, the equation setup is the same.

7

The formula is D = Rt Or Distance = Rate X Time. And when two distances are equated, the individual rates and times are multiplied together first and then set equal to one another:

Making it a matter of time

When two people leave the same place at different times, one has to travel more quickly than the other to catch up — assuming that they’re both using the same route.

The Problem: Henry left for work at 7 a. m. and drove at an average speed of 45 mph. Unfortunately, he forgot to put some important papers in his briefcase. At 7:30, Betty found the papers, jumped in her car, and chased after Henry. She averaged 55 mph. How long did it take for Betty to catch up to Henry?

^VLA/V The distance that Henry drives and the distance that Betty drives is the

Same — at the moment Betty catches up to Henry. So multiply the rate that Henry drives times how long he drives and equate that to the rate that Betty drives times her amount of time. Let the amount of time that Henry drives be represented by T Hours. Then Betty drives for T - 0.5 hours. Multiplying rate times time, the equation and solution are:

45T=55(t-0.5) 45T=55T-27.5

27.5 = 10T

T=Nrr = 2.75

Henry drove for 2.75 hours, or 2 hours and 45 minutes. If Betty drove for half an hour less than that, she drove for 2 hours and 15 minutes.

You can solve directly for the amount of time that Betty traveled by letting Betty’s time be T And Henry’s time be T + 0.5. The equation is just a bit different, but the answer will still come out that Betty drove for 2.25 or 2 hours and 15 minutes.

In the classic story of the tortoise and the hare, the tortoise is slow and steady, and the hare is just too sure of himself. Consider the next problem as something that may have been the specific details of that story.

Trains in a tunnel

At 7 o’clock, a train enters a tunnel on the only to pass, go around, over or under one another

Track that runs through that tunnel. Another in the tunnel. However, it was possible for the

Train enters the exact same tunnel at 7 o’clock trains to make it to the other end of the tunnel

On the same day. There is no way for the trains untouched. How can that be?

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The Problem: The tortoise and the hare are both poised at the starting line of a 1-mile race. The tortoise can muster up a full 1-mile-per-hour speed, and the hare boasts a hopping speed of 8-miles-per-hour. The starter’s gun sounds, and both take off at their top speed. After one minute, the hare decides he has time to take a nap and snoozes for the next 54 minutes. He wakes up and makes a dash for the finish line. How long does it take him to catch up to the tortoise?

^VLAAf The distance that both the tortoise and the hare travel are the precise

Moment that the hare catches up to the tortoise. The rates are 1-mile-per-hour for the tortoise and 8-miles-per-hour for the hare. The tortoise moves along for T Hours, and the hare has T Minus 54 minutes (which has to be

54 9

Changed to miles per hour) so it’s T — = T — = T — 0.9 hours. Setting the two distances equal to one another and solving,

1t = 8 (T — 0.9) T = 8t — 7.2 7.2 = 7t

T = 7^ . L03

So the distances are equal and the hare catches up in a little more than one hour. But that’s too late. The tortoise crossed the finish line after one hour, and the hare hadn’t caught up to him yet, so the tortoise wins, yet again.

Speeding things up a bit

When you want to catch up with your older brother, you need to move faster than he’s moving, if he left before you.

The Problem: Don left home with his fishing gear and headed for his favorite spot at 6 a. m. Don bicycles at 6 mph. Don’s younger brother, Doug, left home at 6:30 and caught up with Don at 6:45 a. m.. How fast did Doug have to travel to catch up to Don at that time?

^VLA* The distance traveled by the brothers is the same, so set the products of the rates and respective times equal to one another. Don bicycled at 6 mph for 45 minutes. Doug traveled for just 15 minutes. Changing 45 minutes to 0.75 hour and 15 minutes to 0.25 hour, then the only unknown is Doug’s rate of speed, which is represented by r. The equation 6(0.75) = r(0.25) sets the two distances equal to one another. Simplifying on the left, you get 4.5 = 0.25r. Dividing each side by 0.25, you get that R = 18 mph. Doug must be riding a bike or driving.

The Problem: Carole left Tampa one hour after Warren. Carole caught up to Warren 200 miles from Tampa, because she drove at a rate of speed that’s 125 percent Warren’s speed. How fast did Carole and Warren drive?

^VLA/V Both Carole and Warren drove 200 miles. Let R Represent Warren’s speed and T Represent how long Warren drove to make that 200 miles. For Warren, you write the equation 200 = Rt. Carole drove the same 200 miles, but she drove faster and for one hour less than Warren. For Carole, you write the equation 200 = (1.25r)(t – 1). Going back to Warren’s equation, solve for T In terms of the 200 and r, and replace the T In Carole’s equation with that expression.

200 = Rt, t = 200 200 = (1.25r)(T — 1)

200 = (1.25r)(200 —

Next, distribute the 1.25r over the terms in the parentheses and solve for R.

200 = 1.25/X 200 — 1.25r X 1 200 = 250 — 1.25r

1.25R = 50

R = ^r°- = 40 1.25

Warren drove at 40 mph, so it took him 5 hours (200 40) to drive that 200 miles. Carole drove at 125 percent of Warren’s speed, so she drove at 50 mph. It took her four hours (200 50) to drive the same distance — one hour less than the time it took Warren.

Solving for the distance

The main element of the Distance = rate X Time Problems is the distance. You solve for rate or time when given the distance or given the fact that the distances traveled by the participants are equal. Just as important is finding out just how far people have traveled.

The Problem: Paula caught a bus to travel from school to home. Three hours later, her brother, Ken left school to go home driving his own car. If Ken drove 20 miles per hour faster than the bus, and arrived home at the same time as Paula, how far is it from school to home?

^VLA/V The distances are the same. You solve for the rate and time and then compute the distance. Letting R Represent the rate at which Paula is traveling on the bus, Ken is driving at the rate of R + 20 mph. The time that Paula traveled is T Hours, and Ken traveled T - 3 hours. Letting the two distances be the same, you get the equation Rt = (r + 20)(t – 3). Multiplying on the right, you get Rt = Rt -3r + 20t – 60. Subtracting Rt From each side and adding 60 to each side, the equation becomes 60 = 20t – 3r. It appears that there are many different times and rates that fit the solution as it’s given. Using the equation 60 = 20t – 3r you can try out some values for T And R. Some possibilities are shown in Table 16-1.

Several of the entries in the table are reasonable scenarios for Paula and Ken. To actually finish the problem, you need a little more information. Assume, for instance, that Ken drove 60 mph. That means that the bus Paula is riding on is driving at 40 mph. The equation now becomes 40t = 60(t – 3), which multiplies out to be 40t = 60t - 180. Simplifying, you get 180 = 20t Or T = 9. If Paula rode 9 hours at 40 mph, then the distance from school to home is 40 X 9 = 360 miles.

When you’re racing against someone bigger or older or faster, and she wants to make a contest of it, you can ask for a Head start. Applying this situation to a distance problem, you equate the distances by adding or subtracting the head start to make the finish come out the same. Getting a head start helps to even the playing field. It makes a race more of a contest and more interesting if the runners appear to have an equal chance of winning.

The Problem: The contestant from Kenya can run the 10,000-meter race at an average of 6 meters per second. The contestant from Ethiopia has a best time so far of 5.5 meters per second. How far back should the runner from Kenya start to have the expectation that they would cross the finish line at the same time if they both start at the same time?

In this problem, the distances aren’t really equal. You have to add on X Number of meters to represent the additional distance that the runner from Kenya must cover. The rates are also different, but the times will be the same, if they start at the same time and finish at thde same time. Take the distance formula, D = Rt And solve for T, Giving you T = —. Now set the time it takes the runner from Kenya equal to the time for the runner from Ethiopia. Then change the times to distances and rates.

TK = TE

~T~ = R

K e

Replace the denominators with the respective rates, and let the distances be 10,000 for the Ethiopian runner and 10,000 + X For the Kenyan runner. Then solve for X By cross-multiplying.

10,000 + X 10,000

- =

6 5.5 5.5 (10,000 + X) = 6 (10,000) 55,000 + 5.5x = 60,000

5.5X= 5,000

5,000 X = R r . 909.09

The runner from Kenya will have to start over 900 meters behind the other runner. And that’s assuming that she can run her usual pace for an extra 900 meters.

Working It Out with Work Problems

Work problems usually involve two or more people pitching in together to make a lighter load for everyone involved. Different people work at different rates of speed, so some people accomplish more of the total job than others.

Here are the overriding principals or procedures to use in doing these problems:

E* Let X Represent the amount of time needed to do the whole job.

E* Let 3, 4, ••• represent how much can be accomplished in One day By the person who can do the whole thing in 2, 3, 4, . . . days, respectively (other time units also apply).

E* Let X, ^3, xx, • • • represent how much of the Whole job Is accomplished by the person who can do the job in 2, 3, 4, . . . days, respectively.

E Add up all the fractions with X In the numerator and set them equal to the number 1 for the whole job (or a fraction for part of the job).

A traditional problem is that two or more people get together to paint a house. You assume that everyone has a paint brush, paint, ladder, and any other supplies needed to do the job. Working together, they reduce the amount of time necessary to complete the whole job.

The Problem: Tom, Dick, and Harry arrive early one morning at the job site and get ready to paint a huge, old, Victorian mansion. Tom, working by himself, could paint the whole house in 14 days. It would take Dick 10 days to do the job by himself. And Harry could do the job in 8 days. How long does it take for the three men to do the job working together?

Let X Represent the amount of time needed to do the whole job. Then Tom

Will do tt of the job, Dick will do ttt of the job, and Harry will do X Of the 14 10 8

Job. Add the fractions together and set them equal to 1.

Xc J Xc J Xc_ 1

14 10 8

Solve for X By first multiplying both sides of the equation by 280 and then simplifying.

X X 28020 + x X 28028 + X X 28035 = 1 X 280 yl 10 8

20x + 28x + 35x = 280

83X= 280

280 83

It will take the men not quite three and a half days to do the job by working together.

In some instances, though, the whole crew doesn’t show up at the same time.

The Problem: A three-man crew can harvest the field in six hours, while a four-man crew can harvest the field in four hours. If the three-man crew worked for one hour and then were joined by the fourth man, how long will it take the four-man crew to finish the job? How long does it take from start to finish to do the whole job?

^VLA/V First, determine how much is accomplished by the three-man crew before the fourth man shows up. Then determine how long it’ll take to finish up the job. If the three-man crew can do the harvesting in six hours, then in one hour,

They’ve done – i of the job and have 5 left to finish. Let X Represent how long it’ll take to finish the job, divide by 4 (the amount of time it takes the four-man crew to do the whole job) and set X Equal to the fraction of the job that’s left. Solve for X By cross-multiplying.

X = 5 4 = 6

6X= 20

20 O 1 63

It’ll take another 3 hours and 20 minutes to finish the job. Add that to the hour spent by the three-man crew, and the whole harvesting process took 4 hours and 20 minutes.

What if someone leaves before finishing the job? After all, people get tired or have other commitments. Have you ever been left to finish up a project that many other people had started? I’ll at least make this next job something fun to do.

The Problem: Sarah, Sue, and Sybil are making chocolate-chip cookies for the annual club bake sale. Working alone, it would take Sarah 8 hours to do all the baking. Sue could do the whole job in 10 hours, and it would take Sybil 12 hours by herself. They all started working early in the morning. But, after 2 hours, Sybil said that she had to leave for an appointment and wouldn’t be back. One hour after that, Sue got tired of listening to Sarah’s griping and left. How long did it take Sarah to finish up the job by herself, after the other two left?

£,?LAiV First, figure out how much of the job got accomplished during the first two hours, before Sybil left. If Sarah can do the job alone in 8 hours, then she did

2 (\ J = 2 = t in that two hours. Sue can do the job alone in 10 hours, so, in

That first two hours, she did 21 tq j = Tq = 5 of the job. And Sybil did

2 C 1j J = 12 = T in that first two hours. Add the three fractions together —

T + T + T = 15 + 12 + 10 = 37 — and you see that they’re over half finished

4 5 6 60 60 J J

With the job in the first two hours.

Next, determine how much more is accomplished in the next hour, with just

Sarah and Sue working. Add – i + ttt = 5 4 = -777 To get how much of the job 0 8 10 40 40 0 ‘

Is done by the two working together for an hour. Add the amount from the

First two hours to this amount, and you get the total amount of the job that’s

Been accomplished so far: 37 + -777 = 74 +a27 = t°7T. To determine how r 60 40 120 120

Much time it’ll take Sarah to finish the job, first subtract what’s been done

From 1, and then set that equal to X + 8.

1

101 = J9_ 120 120 X 19

- = -

8 120 X 19

= -~r—

8 T20i5

15X = 19

19 = i_4_ 15 t15

X

It’ll take Sarah another 1 hour and 16 minutes (changing the fractional remainder to minutes) to finish up the job by herself.

Incoming and Outgoing

Filling and draining pools and tanks are just other variations on work problems. When you have more than one water pipe pumping water into a tank, it takes less time than one pipe doing the filling by itself. An added twist to these problems is that the tanks can be emptying, too. If your pool has a leak in it, then water is going out one hole at the same time it’s coming in a pipe.

The Problem: A backyard swimming pool is being filled by two different hoses. One hose can fill the pool in 10 hours, and it takes the other hose 14 hours to fill the pool. How long will it take for the two hoses to fill the pool if they’re turned on at the same time?

DfQ^L Let X Represent the amount of time it’ll take the two hoses, working together, to fill the pool. Then write two fractions, one with X Divided by 10 and the other with X Divided by 14 to represent how much of the whole job each hose can accomplish. Add the two fractions together, set the sum equal to 1, and solve for X.

X J X _1

10 14

X # 707 + Tt x 705 _ 1 X 70

7x + 5x _ 70 12x _ 70

70 35 5

12 6 6

It’ll take 5 hours and 50 minutes to fill the pool with the two hoses.

The Problem: A pool takes six hours to fill and eight hours to drain. The drain was accidentally left open for the first three hours while the pool was filling and then closed. How long did it take for the pool to be filled?

^jlVLA* In the first three hours, the intake was 3 _ 4, which is obtained by dividing Ffr-zOs. 6 2 3

3 by the 6 hours it takes to fill the pool. At the same time, the outgo was 3,

8

Dividing 3 by the 8 hours it takes to drain. Subtract the two fractions to get a

Net filling of 4 — 3 _ 3 — 3 _ 4, which is the fractional amount that the pool

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Is full right now. Subtract that fraction from 1 to get the amount still needed to be filled: 1 — – g – _ Lg.

Let X Represent the number of hours it now takes to complete the filling of the pool. Write the fraction dividing X By 6, set it equal to the fraction of the pool needed to be filled, and solve for X.

X_ = 7

6 = 8 8x = 42

X = -482 = 5-^ =

It took another 5 hours and 15 minutes to fill the pool after the drain was closed, so the total time was 8 hours and 15 minutes.

The Problem: In the tank of a water tower, the intake valve closes automatically when the tank is full and opens up again when 5 of the water is drained

Off. The intake fills the tank in 4 hours, and the outlet drains the tank in 12 hours. If the outlet is open continuously, then how long a time is it between two different instances when the tank is completely full?

^VLA* Start with the first instance that the tank is full. The intake valve closes, and only the outlet valve is working, emptying the tank. You need to determine

How long it takes for the outlet valve to empty the tank by 5, leaving it 5 full.

Divide X, The amount of time it’ll take to lower the amount in the tank, by 12.

X 3

— = —

12 5

5X = 36

X = ~5~ = ^ 5

It takes 7 hours and 12 minutes for the amount of water in the tank to reach the level at which the intake valve switches on. Now you determine how long it takes to fill the tank again. Let X Represent the total amount of time needed. Divide X By 4 for the intake value and subtract the fraction formed by dividing

X By 12 for the outlet. Set the difference equal to 5, the amount of the tank

That needs to be filled. Solve for X By first multiplying each term by 60, the common denominator.

X

601

10X = 36

36 = 33 10 = 3 5

4

It’ll take another 3 hours and 36 minutes to fill the tank. Add the 7 hours and 12 minutes to get the tank to the level where the intake kicks in, and add the 3 hours and 36 minutes. It’s a total of 10 hours and 48 minutes between times that the tank is full.

Chapter 17