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Word Problems - Part I

Even the phrase "word problem" makes people sweat. The reason people have concerns about word problems is because most have no set procedure to handle them. Hopefully, we can rectify that.

Consider the following procedure:

1. Data - write down all important numbers from the problem. Identify them as well.
2. Variable - define a variable to represent the unknown quantity.
3. Plan - write a word equation, or plan, to describe the relationship between the known and unknown values.
4. Equation - translate the plan into an equation.
5. Solution - solve the equation and answer the question.

From now on, this procedure will be denoted by D, V, P, E, S.

Here's an example:

A student has received 75, 82, 71, and 84 on tests. She wants an average of 80. What should she get on the fifth test to get this average?

The test scores and desired average are the important numbers in the problem.

D: 75, 82, 71, 84 test scores. 80 average.

The unknown value is the fifth test score.

V: x = 5th score.

How are these values related? Well, to average test scores, we add up all the scores and divide by the number of tests.

P: [(sum of 4 tests + 5th test)/ 5] = average

Now, we substitute the values in:

E: [(75 + 82 + 71 + 84 + x)/ 5] = 80

Solving, we get:

S: [(312 + x)/ 5] = 80

and x = 88.

Another example:

An employee receives $492 in net pay. His deductions are 40% of his gross pay. What is his gross pay?

D: $492 is net pay, 40% are deductions

V: Let x = gross pay.

P: Net pay = gross pay - deductions

Deductions are 40% of gross, or (.40)x, so,

E: 492 = x - .4x

S: 492 = .6x

x = 820.

Here it is important to note that the net pay is not 40% of gross; the deductions are 40% of gross pay.

Try this problem:

New insulation costs $1080 to install, but saves 10% on heating. Currently, the family spends $60 per month in heating bills. How many months will it take the family to recover the cost of the insulation with savings from the heating bills?

Try to set this up using the D, V, P, E, S system.

Answer:

D: $1080 insulation, 10% savings, $60 per month heating.

V: x = number of months until savings equals insulation.

P: Total savings = Cost of insulation.

Where savings per month = percent savings times heating bills, or (.1)(60) = 6.

(savings per month)(number of months) = cost of insulation.

E: 6x = 1080.

S: x = 180.

Here is an example which teaches us a valuable skill:

A man invests $100,000 in two accounts; one at 8% and the other at 6.4%. After one year, he receives $7500 in total interest. How much has been invested in both accounts?

D: $100,000 total investment, 8% in account one, 6.4% in account two, $7500 total interest.

V: x = amount invested in account one.

But then how much was invested in account two? Often, two unknown values are compared by their total. For example, if 350 children go to a museum and x are boys then, 350-x are girls.

The 350-x represents the "left-overs"; how many are remaining after x are taken out. In this case, if x is the amount invested in account one, then 100,000-x is the amount invested in account two.

V: x = amount invested in account one.

100,000-x = amount invested in account two.

P: Simple interest is computed by I = PRT, so

Interest one + Interest two = Total Interest

(Principal one)(Rate one)(Time one) + (Principal two)(Rate two)(Time two) = Total Interest

E: x(.08)(1) + (100,000-x)(.064)(1) = 7500

S: .8x + 6400 - .064x = 7500

.016x = 1100 or

x = 68,750. We are not done; this is only the amount in account one. We need to compute the amount in account two, or 100,000-68,750 = 31,250.

Here are two to try on your own:

1. A concert is held and 600 people attend. Adult tickets are $5 and children's tickets are $2. If $2400 is collected in ticket prices, how many adults and children came to the concert?

2. A chemist wants to make 15ml of a 2% acid solution. She has 10% and 1% solution on hand. How much of each solution should she mix to obtain the desired solution? (Hint: this is similar to the interest problem above.)

Here are the answers:

1. D: 600 people, $5 adults, $2 children, $2400 total money.

V: x = number of adults,

600 -x = number of children.

P: (number of adults)(adult price) + (number of children)(kids price) = total

E: (x)(5) + (600-x)(2) = 2400.

S: 5x + 1200 - 2x = 2400

3x = 1200

x = 400, 400 adults and 200 children.

2. D: 15ml 2% solution. 10% and 1% mixture.

V: x = amount of 10% solution,

15 - x = amount of 1% solution.

P: Acid from 10% + acid from 1% = acid from 2%

(.1)(amount of 10%) + (.01)(amount of 1%) = (.02)(amount of 2%)

E: .1x + (.01)(15-x) = (.02)(15)

S: .1x + .15 - .01x = .3

.09x = .15

x = 1.666... so 1.66ml of 10% solution and 13.33ml of 1% solution.

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