Key Stone Problem… Key Stone Problem… next Set 15 © 2007 Herbert I. Gross

Key Stone Problem…Key Stone Problem…

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Set 15© 2007 Herbert I. Gross

You will soon be assigned problems to test whether you have internalized the material

in Lesson 15 of our algebra course. The Keystone Illustration below are

prototypes of the problems you'll be doing. Work out the problem on your own.

Afterwards, study the detailed solutions we’ve provided. In particular, notice that several different ways are presented that could be used to solve each problem.

Instruction for the Keystone Problem

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© 2007 Herbert I. Gross

As a teacher/trainer, it is important for you to understand and be able to respond

in different ways to the different ways individual students learn. The more ways

you are ready to explain a problem, the better the chances are that the students

will come to understand.

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A salesperson is given two options for how her salary is to be calculated.

(1) base salary of $350 per week, plus a 5% commission on the amount of her sales:

(2) base salary of $200 per week, plus an 8% commission on the amount of her sales.

What must the weekly amount of her sales be in order for the two options to be equal; that is, to give her the same dollar amount?

Keystone Problem for Lesson 15

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Solution for the Problem

Suppose we let x denote the amount of her weekly sales, in dollars. (1) At 5% her commission is 0.05x, and her total take-home pay, in dollars, is given by…

0.05x + 350

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(2) At 8% her commission is 0.08x, and her total take-home pay, in dollars, is given by…

0.08x + 200

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Solution for the Problemnextnext

To know when the two options will be equal, we have to know the value of x for which0.05x + 350 will be equal to 0.08x + 200?

0.05x + 350 = 0.08x + 200–

0.05x = 0.05x

350 = 0.03x + 200

As a first step, we may subtract 0.05x from both sides of the equation to obtain…

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Solution for the Problemnextnext

Since 350 = 0.03x + 200, we next subtract 200 from both sides of our equation …

350 = 0.03x + 200– 200 = 200

150 = 0.03x

As the final step, we divide both sides of our equation by 0.03.

0.030.03 x = 5000

The answer, then, is: the two options are equal when her sales for the week amount to $5000.

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5,000

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As a check, we see that if we replace x by 5,000 …

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Check

0.05x + 350 =

600

While…0.08x + 200 =

600

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0.05(5,000) + 350 = 250 + 350, or

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0.08(5,000) + 200 = 400 + 200, or

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would be her total weekly pay in dollars

@ 5% commission

would be her total weekly pay in dollars

@ 8% commission

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That is: if her sales for the week total less than $5,000, she does better by taking the greater salary and the lower commission

rate, but if her sales total more than $5,000 for the week, she does better by taking the lower salary but the greater commission.

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In this problem, $5,000 is the equilibrium point.

When her sales total $5,000 for the week, she earns $600 with either option.

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Equilibrium Point

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If you are more comfortable working with whole numbers:

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Note

0.05x + 350 = 0.08x + 200100

to obtain…

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(

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) ( )100

100(0.05)x + 100(350) = 100(0.08x) + 100(200)

or…

5x + 35,000 = 8x + 20,000

just multiply both sides of the equation0.05x + 350 = 0.08x + 200

by 100 to obtain the equivalent equation…

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In solving the equation 5x + 35,000 = 8x + 20,000,we want all the x’s to be on one side ofthe equation. We may do this either by subtracting 5x from both sides or by subtracting 8x from both sides. Subtracting 5x from both sides gives us positive numbers to work with.However, many textbooks use the tradition that the x’s have to be on the left-hand side of the equation. In that case, it is 8x that we subtract from both sides, and the equation becomes…

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-3x + 35,000 = 20,000

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Continuing, we next subtract 35,000 from both sides of the equation and obtain…

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-3x + 35,000 = 20,000

and we can now divide both sides of this equation by -3, again obtaining the answer that x = 5,000.

x = 5,000-3x = -15,000-3 -3

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The point is: the underlying logic behind the arithmetic of signed numbers

allows us to proceed in whichever way makes us most comfortable.

Our personal preference would be to work with positive rather than negative

numbers, whenever we can.

SignedNumbers

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If all that’s wanted is to find a solution to a specific problem, there is nothing wrong

with using trial-and-error and totally circumventing the need to use algebra.

However, aside from the algebraic solution being the main theme of this course:

the fact is that as problems become more complicated (for example, when several variables are involved), trial-and-error

can become cumbersome and time-consuming.

Other Methods of Solution

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A very simple form of trial-and-error rrrrr involves obtaining two estimates, one of which is too big to be the correct answer

and the other of which is too small to be the correct

answer.

Over and Under Method

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For example, if there are no sales for the week, it is better for her to choose the option in which the salary is $350 rather than $200.

On the other hand, if her sales for the week were $10,000, the 8% commission would earn her $800 while the 5% commission would earn her only $500. In this case, the extra $300 in commission more than compensates for the $150 difference in the two salary options.

Thus, we see that the equilibrium point will be between $0 and $10,000 in sales.


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Now let’s see what happens if her sales for the week were, say, $3,000. At 8% she would earn $240 in commission; while at 5% she would earn $150. In this case, the 8% option would not be enough to offset the extra $150 in salary that the 5% option offers. So we see that the equilibrium point, while less than $10,000, must be more than $3,000.


By such trial and error, we can eventually obtain the exact answer, $5000. [By coincidence, here the exact answer comes out midway between our first lower estimate ($0) and first upper estimate ($10,000).]

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Sometimes, making a table or chart helps us discover a logical pattern. This is especially true when we deal with linear equations because their

rate of change is constant.

Making a Table Methodnext

For example with respect to our current problem, let's denote the

8% option as Option E (for “eight”) and

the 5% option as Option F (for “five”)

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It is often helpful to use suggestive symbols rather than the traditional x and y. When we see “x”

or “y” we may have to look back to remind ourselves which option they represent. However,

seeing E or F reminds us immediately whether we're talking about the 8% or the 5% option.

Making a Table Method

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Using Option E: for each dollar in sales, she earns (adds) $0.03 more in commission than with

Option F. However, having our table proceed in increments of $1 could be very tedious; so we might want to have the table proceed in, say,

increments of $1,000.

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Pictorially…

Making a Table Method

Sales 8% 8% + 200 Option E 5% + 3505% Option F

$1,000 $80 $80 + $200 $280 $50 + $350$50 $400

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$2,000 $160 $160 + $200 $360 $100 + $350$100 $450

$3,000 $240 $240 + $200 $440 $150 + $350$150 $500

$4,000 $320 $320 + $200 $520 $200 + $350$200 $550

$5,000 $400 $400 + $200 $600 $250 + $350$250 $600

$6,000 $480 $480 + $200 $680 $300 + $350$300 $650

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$5,000 $400 $400 + $200 $600 $250 + $350$250 $600

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In this example, using the chart was not too cumbersome because we arrived at the

correct answer quickly. The chart is also significant in helping us

realize that the difference between Option E and Option F changes by $30 every time the amount of her sales increases by $1,000.

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Sales Option E Option F Option F – Option E$1,000 $280 $400 $120$2,000 $360 $450 $90$3,000 $440 $500 $60$4,000 $520 $550 $30$5,000 $600 $600 $0$6,000 $680 $650 $-30$5,000 $600 $600 $0

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The meaning of the $-30 in the last line is: when the amount of her sales is $6,000,

Option F is $30 less than Option E.

This is important for understanding the significance of equilibrium points. Namely, that

if her sales are less than $5,000 for the week, Option F is the better deal: but if her sales are

more than $5,000 for the week, Option E is the better deal.

Sales Option E Option F Option F – Option E$1,000 $280 $400 $120$2,000 $360 $450 $90$3,000 $440 $500 $60$4,000 $520 $550 $30$5,000 $600 $600 $0$6,000 $680 $650 $-30

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If we are comfortable “thinking on our feet” we might observe right away that the

person who chooses the 8% option has to overcome a salary difference of $150.

Mental Math Method

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For every $1,000 in sales, the 8% option brings in $80 in commission; while the 5% option is only worth $50 in commission. That is: for

every $1,000 in sales, the 8% option exceeds the 5% option by $30.

Mental Math Method

Since 5 × $30 = $150 we see that, at the 8% option, we have to have 5 times

the commission on (added sales of) $1,000 in order to make up for the $150 difference in

base salary. Hence, the equilibrium point is 5 × $1,000 or $5,000.

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Key Stone Problem… Key Stone Problem… next Set 15 © 2007 Herbert I. Gross