Algebra Problems… Solutions Algebra Problems… Solutions next © 2007 Herbert I. Gross Set 1 By Herb I. Gross and Richard A. Medeiros.

Algebra Problems…Solutions

Algebra Problems…Solutions

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© 2007 Herbert I. Gross

Set 1By Herb I. Gross and Richard A. Medeiros

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If there are 2.54 centimeters in one inch, how many

centimeters are there in 10 inches?

Problem #1 (a)next


Answer: 25.4 cm

Answer: 2.54Solution:

10 inches = 10 × 1 inch =

10 × 2.54 cm = 25.4 cm.

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• The approach used here does not depend on the fact that we started with 10 inches.

Namely if each inch is equal to 2.54 centimeters, I inches will equal (I × 2.54)

centimeters.So if we let C denote the number of

centimeters in I inches, we “invent” the formula:

C = 2.54 × I (1)

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Note 1(a)

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If there are 2.54 centimeters in one inch, how many inches are

there in 10 centimeters? Round off your answer to the

nearest tenth of an inch.

Problem #1 (b)next

Answer: 3.9 in.© 2007 Herbert I. Gross

Answer: 3.9 in.Solution:We need to simply replace C by 10 in

formula (1) to obtain the indirect computation:10 = 2.54 × I (2)

And we solve equation (2) by rewriting

it as…

I = 10 ÷ 2.54 = 3.9

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• Being able to use approximations can be helpful in trying to determine whether an

answer is reasonable.

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10 ÷ 2.54 is a little less than 10 ÷ 2.5(= 10 ÷ 5/2 = 10 × 2/5 = 4)

For example:

This tells us that our answer (3.9 cm) is quite reasonable.

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Note 1(b)

• When all else fails (or even if all else hasn’t failed) it’s okay to use trial and error.

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Note 1(b)

For example

And since 10.16 is just a little more than 10, we would guess that the number of inches is

just a little less than 4.

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Number of inches Number of centimeters

1 2.54

10.16

3

4

5.08

7.62

2

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4 10.16


We could make the following chart:

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If apples cost $0.84 per pound, how much will 5 3/4 pounds of

apples cost?

Problem #2(a)next


Answer: $4.76

• The formula for this problem is the same type as the formula in the previous

problem. In other words, when items are being sold at a constant rate, the

relationship between cost and number of items bought is always:

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Total cost = Cost per item × Number of items purchased

Note 2(a)

Cost = Costitem

× items

More symbolically…

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Answer: $4.76Solution:In this case the cost per pound is 84 cents and the number of pounds purchased is 5 2/3. So if we let C represent the total cost

and N thenumber of pounds purchased, the formula C = 84 × N (1)

becomes… C = 84 × 5 2/3 (2)

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We may compute C either by rewriting equation (2) as

C = 84 × (5 + 2/3)= (84 × 5) + (84 × 2/3)

= 420 + 56= 476 cents or $4.76

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or we could rewrite equation (2) asC = 84 × 17/3‚ and obtain the same

result.

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If apples cost $0.84 per pound, how many pounds can you buy

for $6.16?

Problem #2(b)next


Answer: 7 1/3 pounds

Answer: 7 1/3Solution:

To avoid having to use decimals we might let N represent the numberof pounds we purchase and C to

represent the total cost in cents. Theformula then becomes:

C = 84 × N

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We know that the total cost is $6.16, but since C represents the total

cost in cents we replace C by 616 in (to obtain the indirect computation:

616 = 84 × Nwhich can be paraphrased into the

direct computationN = 616 ÷ 84 = 7.3 = 71/3

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Note 2(b)

• If you used a calculator to compute 616 ÷ 84, you obtained an answer

in the form 7.3333333. The exact answer is 71/3 . However since the decimal representation of 1/3 consists of an

endless cycle of 3's the calculator can only show the first few 3's.


nextProblem #3a

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Answer: $104.55

A salesperson earns a commission of 17% on all the

clothes she sells. How much is her commission if she sells

$615 worth of clothes?


Answer: $104.55Solution:This is another form of a constant rate

problem. Namely, the sales person earns $0.17 for each $1 worth of clothes that she sells. Hence if

she sells $615 worth of clothes, she earns 17 cents 615 times. Stated interms of dollars her commission is

615 × $0.17 or $104.55

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Note 3(a)

• 17% means 17 per 100. Thus we could also have said that she earns

$17 per $100 in sales or $170 per $1,000 in sales, etc.


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Note 3(a)

• In fact the previous note gives us a quick way to obtain a very rough

estimate of her commission. Namely because $616 is between $100 and $1,000, her commission must be between $17 and

$170. Moreover since 615 is closer in value to 1,000 than to 100, the commission

is closer to $170 than to $17.


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Note 3(a)

• Of course no matter what the amount of her sales were, the salesperson would

earn $0.17 per each dollar in sales. Thus we can write a formula that relates the

amount of her sales in dollars (S) and the commission she receives in dollars (C).

Namely:

C × 0.17 = S (1)© 2007 Herbert I. Gross

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Note 3(a)

• In terms of what we talked about in this lesson, this problem is an example of a

direct computation. Namely we replace S by 615 to obtain the direct computation…

C = 0.17 × 615


nextProblem #3b

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Answer: $2,300

A salesperson earns a commission of 17% on the

selling price of all the clothes she sells. What is the dollar

amount of her total sales if she earns a commission of $391?


Answer: $2,300Solution:

We may begin with the formulaC = 0.17 × S (1)

Since $391 represents her commission, we replace C by 391 in equation (1) to

obtain the indirect computation391 = 0.17 × S (2)

Equation (2) can be paraphrased into the direct computation:

S = 391 ÷ 0.17 or 2,300

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Note 3(b)

• If we preferred not to use decimals, we could paraphrase 391 ÷ 0.17

into the equivalent form 39,100 ÷ 17.

• It should be obvious that her commission cannot exceed the amount of her sales. In other words, if her commission was $391,

the amount of her sales had to exceed $391 (in fact, by quite a bit).


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Note 3(b)

• The above note helps us avoid a common error. Namely, a common

error is to replace the wrong letter by 391 (in other words, reading

comprehension is important and one should remember what noun a number

is modifying).


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Note 3(b)

• If one did replace S by 391 in formula (1) the formula would have been

0.17 × $391 or $66.47.

However, the fact that her commission was $391 tells us that the amount of her sales had to exceed $391 (which $66.47

certainly does not).© 2007 Herbert I. Gross

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A sale advertises “30% off the marked price”. How much will it cost to buy a jacket during this sale if the marked price of the

jacket is $125?

Problem #4anext

Answer: $87.50© 2007 Herbert I. Gross

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Note 4(a)

• It is easy to confuse “off” with “of”. 30% off the marked price means that the

marked price has been reduced by 30%. If 30% of the marked price is subtracted

from the marked price, what's left is 70% of the marked price. In other words “30% off” means the same thing as “70% of”.


Answer: $87.50Solution:This is another form of a constant rate

problem. Namely for every dollar of the marked price the customer will pay $.70. Since the marked price is

$125, the customer will pay 125 × $.70 or $87.50

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Note 4(a)

• We could also have taken 30% of $125 and subtracted it from $125.

In other words:

$125 – (30% of $125) = 70% of $125


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Note 4(a)

• To translate this problem into a formula, we may replace $125 by any marked price. If we then let M represent the marked price

in dollars and S denote the sale price in dollars, the formula becomes…

S = 0.70 × M (1)

• In terms of this problem we would replace M by 125 to obtain the direct computation…

S = 0.70 × 125

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A sale advertises “30% off the regular price”. How much was the regular price of a jacket if

the sale price was $105?

Problem #4bnext

Answer: $150© 2007 Herbert I. Gross

Answer: $150Solution:

Since the sale price (S) is $105, we may replace S by 105 in equation (1) to obtain the indirect computation:

105 = 0.70 × Mwhich can be paraphrased as the

direct computationM = 105 ÷ 0.70 or 150

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Note 4(b)

• Again notice how important reading comprehension is. For example,

in doing part (a) we saw that we could subtract 30% of the regular price from

the regular price to find 70% of the regular price. That is…

30% + 70% = 100%© 2007 Herbert I. Gross

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Note 4(b)

• However, it would be wrong to take 30% of $105 and add it to $105 to find the regular price. Namely $105 is the

sale price while the “30% off” applies to the regular price (not the sale price).

• As a check notice that 30% of $105 is $31.50 and $31.50 + $105 equals $136.50; and 70% of $136.50 is $95.55, not $105.

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A car travels at a constant rate of a 2/3 of a mile per minute. How many miles will the car

travel in 1/2 an hour?

Problem #5anext

Answer: 20 miles© 2007 Herbert I. Gross

Answer: 20 milesSolution:

This is yet another example of a problem involving constant rates. In particular in this problem we know that the car travels

2/3 of a mile during each minute that it travels. A half hour is 30 minutes.

Therefore in 1/2 of an hour the car travels 2/3 of a mile 30 times; or (30 × 2/3) miles.

30 × 2/3 = 20. Hence the car travels 20 miles in a 1/2 hour.

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Note 5(a)

• To develop the formula we need only notice that since the car travels 2/3 of a

mile each minute, in M minutes it will travel 2/3 of a mile M times. Thus if we let D

denote the number of miles the car travels in M minutes, the formula becomes…

D = 2/3 × M (1)© 2007 Herbert I. Gross

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Note 5(a)

• By changing the noun phrase “miles per minute” to the more familiar phrase “miles

per hour (mph)”, we notice that the rate of a 2/3 mile per minute can be rephrased in

many ways; among which is 40 miles per hour. That is in one hour the car travels 2/3

of a mile 60 times. Hence if we let H represent the number of hours the object traveled, formula (1) would be replaced by

D = 40 × H (2)© 2007 Herbert I. Gross

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Note 5(a)

• Thus in solving this problem if we used formula (1) we would replace M by 30, but if we used formula (2) we would replace H

by 1/2.


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A car travels at a constant rate of 2/3 of a mile per minute. How

long will it take for the car to travel 46 miles?

Problem #5bnext

Answer: 69 minutes or1 hour 9 minutes


Answer: 69 minutesSolution:If we use formula (1) we replace D by 46 to

obtain the indirect computation46 = M × 2/3

which we can paraphrase as the direct computation

M = 46 ÷ 2/3 or 46 × 3/2 or 69 and since M is expressed in minutes the

answer is 69 minutes

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Note 5(b)

• We could also have used formula (2) to obtain the indirect computation

46 = 40 × Hwhich can be paraphrased as the direct

computationH = 46 ÷ 40 = 23 ÷ 20 or 1 3/20

and since H is expressed in hours the answer is 1 3/20 hours.


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Note 5(b)

• Since 1 hour ÷ 20 = 60 minutes divided by 20 or 3 minutes, 1/20 of an hour is 3

minutes; therefore 3/20 of an hour is 3 × 3 minutes or 9 minutes. Hence the answer

can be expressed as 1 hour and 9minutes; which agrees with our previous

answer of 69 minutes.


Algebra Problems… Solutions Algebra Problems… Solutions next © 2007 Herbert I. Gross Set 1 By Herb I. Gross and Richard A. Medeiros.

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