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WORD PROBLEMS

Objectives: To gain some practice in working word problems.

Remark Students have trouble with word problems for one good reason: A word problem embodies two problems in one--the first is to translate the words into a symbolic mathematical language, the second is to use the mathematical language to solve the problem. The toughest part of a word problem is the translation into mathematical language. To help you to develop your skills in translating a sentence into mathematical language, we list a few general methods.

Recall 1: (Starting a word problem.)

1. Read the problem quickly to get a rough idea of the nature of the problem.
2. Isolate the question in the problem. This usually comes    at the end of the word problem.
3. Each problem should begin with Let x = something. The x should equal the quantity that you wish to find.     That quantity is usually determined by the question.
4. Translate English phrases into mathematical symbols    and find any relations between the known quantities and    the unknown quantities.
5. Interpret your answer and check if it makes sense.

Recall 2: Learn how to translate typical English phrases into mathematical language. Practice interpreting the following phrases. The English phrases are on the left, their mathematical equivalents are on the right.

   Two more than x                     x + 2
   His age (x), twenty years ago     x - 20
   Twice as much as x                  2x
   Three times as much as x          3x

Recall 3: (Further interpretation of English phrases.)

A number decreased by 5 is 20.

x - 5 = 20.

You now have the equation x - 5 = 20. So the number in question is x = 25. It is easy to check that 25 decreased by 5 is 20.

         Her age    5 years ago    was    7 years more than   half   her age now.

            x        -       5             =                    7      +      1/2           x

If you wish to know her age, use the equation that you have   translated the statement into:

x - 5 = 7 + x/2.

We solve this equation for x to find that x = 24. This is checked by asking whether or not a person who is 24 years old now would have been, five years ago, seven more than half her present age.

Recall 4: Isolate the question to find out what to let the unknown equal. The question usually comes at the end of the problem. For example, investigate the problem below.

Two numbers added together becomes 93. One of these numbers is 37 more than the other one. What are the numbers?

In this case we let x and y represent the two numbers. Then the first sentence becomes x + y = 93, and the second sentence translates to x = y + 37. The problem is easily solved by substituting y + 37 for x in the first equation. We get

y + 37 + y = 93,
2y = 93 - 37,
2y = 56,
or y = 28.

Note
Granted the problems that we exhibited in the Recalls above have been easy. Many of them could have been solved by inspection. The examples below offer a more extended range of problems. The examples below offer several types of word problems. You may find that the procedures differ slightly from type to type; but, for the most part, they are similar.

Examples

Example 1. A 20 foot wire is cut into two pieces. One piece is twelve feet shorter than 3 times the length of the other. How long is each piece?

Solution:    Let x = length of first piece, in feet.
Find a phrase that relates several quantities (known     or unknown and translate it into mathematical     language. This is a direct translation of the    second sentence.
     x = 3y - 12

Look for phrases that may lead to additional     information that may relate the unknowns. This     line comes from translating the first sentence.
    x + y = 20

If there are more than one variable, solve for one     unknown in terms of the other. Then use this     equation to eliminate one of the variables.
    y = 20 - x

Eliminate one of the variables. In this case we have eliminated the y.
x = 3(20 -x) - 12

x = 60 - 3* - 12             If you have an equation involving one unknown, then
4x = 48                        solve for the unknown.
x = 12   feet

x =12 and x + y = 20    Check that you have answered the question. In this case
12 + y = 20                  x = 12 feet is the length of the first piece. The question was
y = 8 feet                     about the lengths of two pieces. The equation that you use to
                                     eliminate one of the variables can be used to find the length
                                     of the other piece.

12 feet + 8 feet = 20 feet Check that the solution(s) make sense. Put the
12 feet = 3(8 feet) - 12 feet unknowns back into the question.

x= 12 feet and y = 8 feet Display your answer in a prominent place.

Example 2. Coke Ann, A Colombian drug smuggler leaves Cartagena for San Juan Puerto Rico on a high powered boat called The Pounder. Her average speed is 40 miles per hour. One hour later a Colombian undercover agent named Single-O-seven leaves from the same dock on a sea plane that travels at 100 miles per hour. How long will it be before Single-O-seven overtakes Coke Ann?

Solution:

1. The first thing to notice is that the names and the espionage are totally irrelevant to the problem. This is a problem about two people travelling at different speeds and leaving at different times.

2. Start with the question: How long will it be before Single-O-seven overtakes Coke Ann ? Let x be the time it takes for Single-O-seven to overtake Coke Ann.

3. Interpret the phrase Single-O-seven overtakes Coke Ann. Single-O-seven overtakes Coke Ann when they both have traveled the same distance. So, the problem reduces to finding the time when they have both traveled the same distance. This means that we should be finding the distances that both have traveled.

4. Coke Ann's distance:
   Since she travels at 40 miles/hour for (x + 1) hours, her distance is 40(x + 1) miles.

    Single-O-seven's distance:
    Since he travels at 100 miles/hour for x hours, his distance is 100*x miles.

5. Set the distances equal to each other to get the mathematical equation:

40(x + 1) = 100x.

6. Solve the equation for x.

                    40x + 40 = 100x,
                            60x = 40,
                                x = 4/6 hours or 40 minutes.

7. Check that the answer, x = 4/6 hour is correct by asking whether or not the distance that the plane traveled in that time is equal to the distance that the boat traveled in that time.

The distance that the plane travels in 4/6 an hour is:

100(4/6) = 200/3 miles.

The distance that the boat travels in hour is:

40(4/6 + 1) = 40(10/6) = 200/3 miles.

Note
There are several things worth noting here. First, the distance from Cartagena to San Juan was not needed. Second, the distance from Cartagena to the rendezvous point is a free byproduct of the calculations.

Example 3. Hoody Wallan, the director of a Hollywood film, requires a set to be erected as soon as possible. The studio employs two teams of carpenters. The first team can complete the job in 2 days less than it takes the second team. If the two teams work together, they could do it in 12/5 of a day. What is the fastest time that it would take if just one team were to do the job?

Solution: 1. Let x = the time it takes one team. Then the second team takes x + 2 days to do the same job.

2. Find out how much of the job each can do in one day. The first team can complete 1/x part of the job in one day. The second can complete 1/(x + 2) part of the job in one day.

3. Note that the fraction of the job that the first can do in a day plus the fraction of the job that the second can do in a day must equal the total fraction of the job that both can do in one day. We know that both can do 1/ (12/5) of the job in one day. So,

1/ x + 1/ (x + 2) = 5/ 12 .

4. This is the equation that we must solve. We multiply by the Least common denominator 12x(x + 2) to get

12(x + 2) + 12x = 5x(x + 2).

We may reduce this last equation to the quadratic polynomial 5x^2 -14x - 24 = 0. Notice that this polynomial equation factors as

(5x + 6)(x - 4) = 0

Note that the only acceptable (positive) solution is x = 4. The other team would take x + 2 days or 6 days to do the job; so the fastest team would take 4 days to do the job.

x = 4 is the answer.

Example 4. Two swimming pools compare as follows. One is square, the other is rectangular. One side of the rectangular pool is the same as a side of the square pool. The other side of the rectangular pool is 6 feet longer than its width and the areas of both pools together measure 176 square feet. Find the area of the square pool.

Solution: 1. Draw the two pools, side by side, labeling the dimensions.

2. Let x = length of one side of the square.

   3. Find the areas.
               Area of pool 1 = x^2,
               Area of pool 2 = x(x + 6).

4. We know that the total area is 176. So

Area of pool 1 + Area of pool 2 = 176,

x^2 + x(x + 6) = 176.

5. Solve for x.

        We have x^2 + x(x + 6) = 176,
                    2x^2 + 6x - 176 = 0
                     x^2 + 3x - 88 = 0

This last quadratic equation factors as
(x - 8)(x + 11) = 0.

This tells us that x = 8 or x = -11. Since x must be a length, it must be positive; so we ignore the solution x = -11. The area of the square pool must therefore be x^2 = 64 square feet.

6. Check that this makes sense by finding the area of the rectangular pool and checking that the sum of the areas of the two pools indeed measure to be 176 square feet.

x^2 + x(x + 6) = 64 + 8(8 + 6) = 64 + 64 + 42 = 176.

Note
The next two examples are normally called mixing problems. These generally give students a great deal of difficulty. because students do not have clear ideas of what to equate with what. In these examples we shall explain certain rules for compiling equations. There are two types of mixing problems--those dealing with amounts and those dealing with percentages. We shall handle both in the same way, even though they may not appear to be handled in the same way.

Example 5. Claudius, the King of Denmark requested his court chemist to concoct a poison for Hamlet, his nephew. The poison was to be mixed with wine. Cyanide is tasteless; but the king did not want the mixture to contain more than 10% cyanide, in case an autopsy revealed that he was poisoned. A wine containing 8% cyanide was to be mixed with a pint of a wine that contained 15% cyanide in order to reduce the percentage of cyanide in the wine to 10%. How much of the wine containing 8% cyanide should the chemist have used?

Solution: 1. Let x = pints of wine containing 8% cyanide.

2. The amount of pure cyanide in one mixture plus the amount of pure cyanide in the second solution must equal the amount of cyanide in the final mixture.

3. To find the amount of pure cyanide in each mixture, multiply the amount (in pints) by the percentage of cyanide. Thus, the 8% mixture contains 8x/100 pints of pure cyanide. The other mixture contains 15/100 pints of cyanide. The amount of cyanide in the final mixture is (x + 1)10/100. We therefore have:

8x/100 + 15/100 =10(x + 1)/100,

which reduces to 8x + 15 = 10x + 10, or 2x = 5. So the answer is that the chemist should mix x = 5/2 or 2+1/2 pints of the 8% mixture with the one pint of the 15% mixture.

4. Check that pints of an 8% mixture plus one pint of 15% mixture gives a mixture containing 10% cyanide.

Check that: (8/100)(5/2) + (15/100) = (10/100) (5/2 + 1) .
35 = 35

Example 6. Jack Hook, Captain of the pirate ship Watchout landed in Kingston Jamaica sometime late in the Nineteenth Century. The ship's first mate met a little girl in Kingston town and decided to stay. He opened a stand in the market and made a good living selling coffee and chocolate. He invented a mixture that later became known as mocha. To make the mixture affordable to Jamaicans, each day he would mix 50 pounds of coffee worth 15 shillings per pound. He would mix the coffee with chocolate worth 10 shillings per pound to make a mixture of mocha worth 12 shillings per pound. How much of the chocolate did he mix each day?

Solution: 1. Let x = pounds of chocolate.

2. The value of the coffee plus the value of the chocolate must equal the value of the mocha.

3. To find the value of each mixture, multiply the amount (in pounds) by the price per pound.
Thus, the coffee has a value of 50*15 = 750 shillings. The chocolate has a value of 10x shillings. The value of mocha is 12(x + 50) shillings. We therefore have:

750 + 10x = 12(x + 50),

which reduces to x = 75 pounds. So the answer is that the first mate used 75 pounds of chocolate.

4. Check that 50 pounds of coffee at 15 shillings per pound plus 75 pounds of chocolate at 10 shillings per pound equals 125 pounds of mocha worth 12 shillings per pound.

Check that: (50)(15) + (75)(10) = (125)(12).
1500 = 1500.

Note
The first mate made 1500 shillings per day, which translates to 75 Jamaican pounds or about 120 dollars per day. In today's dollars, this is equivalent to about 1200 dollars per day, not a bad income for an ex-pirate. Free enterprise and obeying the law does pay.

EXERCISES

1. The sum of two numbers is 66. The larger number is three less than twice the smaller number. Find the numbers.

2. A supermarket had a one day sale on Pepperidge Farm cookies. It sold three times as many packages of chocolate-chip cookies as oatmeal, and twice as many fudge-almond as oatmeal. If the supermarket sold 600 packages in one day, how many packages of each did it sell?

3 Otis lives one mile away from Tina. They begin to bicycle toward each other at the same time. The road from Otis's house to Tina's is downhill; so Otis can travel twice as fast as Tina, who peddles her bike uphill at 4 miles/hour. For the entire time of Otis' and Tina's journey, a fly travels back and forth between them at 12 miles/hour. How far will the fly fly?

4 In an FBI sting operation a group of agents leave a warehouse at 234 Vassar Street and move at 4 miles/hour to a second warehouse at 432 Rassav Street, eight miles from the Vassar Street warehouse. At the same time several members of the mob also move from the Vassar Street warehouse to the Rassav Street warehouse; but they move at a slower speed of 3 miles/hour. How much time will the FBI have at the Rassav Street warehouse before the mob arrives?

5 A compulsive gambler doubles his money at roulette. Feeling lucky he decides to play blackjack and loses $400. He returns to the roulette table and, once again, doubles his money. He seems to always lose when he plays blackjack and always win when he plays roulette. No matter, he decides to push his luck and play crap once more, only to lose another $400. If he ends with $20.00, how much money did he start with?

6 Two different kinds of tea are to be mixed to produce 3 pounds of tea that is valued at $3.20 /pound. If the value of the first and second kinds of tea are $3.80 and $2.70, respectively, how many pounds of tea of each type should go into the mixture?

7 A hobo was one fourth across a railroad bridge when he heard the train coming from behind. There was not enough room on the bridge for a person and a train. If the train was exactly one bridge-length away, which way should he run and why?

8 John Quizwiz averaged 73%, 84%, 68%, 89% and 91% on the first 5 Workouts of this worktext, respectively. If he averaged 100% on Workouts 6, 7, 8 and 9, what should he average on this Workout, in order that his average over the first 10 Workouts be 90%?

9 A nursery mixes three different kinds of grass seed--clover, alfalfa and bluegrass. The clover sells for $0.42 per pound; but the nursery has only 20 pounds left, and must use it all. The alfalfa sells for $0.35 per pound and the bluegrass for $0.55 per pound; but there are ample supplies of alfalfa and bluegrass. How much of each should be in the mixture, if the nursery can mix 100 pounds to be sold at $0.45 per pound?

10. Professor Egghead can write an algebra textbook in 18 months. Professor Wizzard can write one in one year. How long would it take them if they worked together?

11 Le Capital, is the French National Railroad's high speed train that links Paris to Marseille. It travels at an average speed of 210 kilometers/hour when traveling from Paris to Marseille and 190 kilometers/hour on returning from Marseille to Paris. What is the average speed of the train in making the journey back and forth?

12 A jumbo jet is on route from Boston to San Francisco. During the first half of the trip it averages 580 miles/hour. During the second half, the pilot decides to increase the average speed by 20 miles/hour. What will the final average speed be?

13 It is 9:00 a.m. (Pacific Time) in San Francisco when it is 12:00 p.m. (EST) in Boston. In Problem 12, assume that the plane left Boston at 12:00 p.m. (EST) and is scheduled to arrive at 1:50 p.m. ; and assume, once again, that the plane traveled half the 3000 distance between Boston and San Francisco at an average speed of 580 miles per hour. If the pilot decided to increase the average velocity during the second half of the trip in order to arrive at San Francisco on time, what should the average velocity of the plane be for the second half of the trip?

ANSWERS

1. 23 and 43

2. 100 oatmeal, 200 fudge-almond, 300 chocolate-chip

3 1 mile

4 8/12 hour or 40 minutes.

5 $305

6 15/11 pounds and 18/11 pounds.

7 Toward the train.

8. 95

9. 20 pounds of clover, 37pounds of alfalfa and 43 pounds of bluegrass.

10 216/30 months or 7.2 months

11 199.5 kilometers/hour

12 6960/118 miles/hour or 589.83 miles/hour

13 approx: 667.52 miles/hour