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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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next If there are 2.54 centimeters in one inch, how many centimeters are there in 10 inches? Problem #1 (a) next © 2007 Herbert I. Gross Answer: 25.4 cm

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Answer: 2.54 Solution: 10 inches = 10 × 1 inch = 10 × 2.54 cm = 25.4 cm. next

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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 i nches 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 ) next © 2007 Herbert I. Gross Note 1(a)

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next 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

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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 next © 2007 Herbert I. Gross

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Being able to use approximations can be helpful in trying to determine whether an answer is reasonable. next 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. next © 2007 Herbert I. Gross Note 1(b)

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When all else fails (or even if all else hasn t failed) it s okay to use trial and error. next 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. next Number of inchesNumber of centimeters 12.54 10.16 3 4 5.08 7.62 2 next 410.16 © 2007 Herbert I. Gross We could make the following chart:

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next If apples cost $0.84 per pound, how much will 5 3/4 pounds of apples cost? Problem #2(a) next © 2007 Herbert I. Gross Answer: $4.76

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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: next Total cost = Cost per item × Number of items purchased Note 2(a) Cost = Cost item × items More symbolically… next © 2007 Herbert I. Gross

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Answer: $4.76 Solution: 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 the number of pounds purchased, the formula C = 84 × N (1) becomes… C = 84 × 5 2/3 (2) next © 2007 Herbert I. Gross

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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 next or we could rewrite equation (2) as C = 84 × 17/3 and obtain the same result. next © 2007 Herbert I. Gross

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next If apples cost $0.84 per pound, how many pounds can you buy for $6.16? Problem #2(b) next © 2007 Herbert I. Gross Answer: 7 1/3 pounds

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Answer: 7 1/3 Solution: To avoid having to use decimals we might let N represent the number of pounds we purchase and C to represent the total cost in cents. The formula then becomes: C = 84 × N next © 2007 Herbert I. Gross

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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 × N which can be paraphrased into the direct computation N = 616 ÷ 84 = 7.3 = 7 1/3 next © 2007 Herbert I. Gross

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next 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 7 1/3. 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. © 2007 Herbert I. Gross

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next Problem #3a next 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? © 2007 Herbert I. Gross

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Answer: $104.55 Solution: 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 in terms of dollars her commission is 615 × $0.17 or $104.55 next © 2007 Herbert I. Gross

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next 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. © 2007 Herbert I. Gross

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next 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. © 2007 Herbert I. Gross

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next 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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next 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 © 2007 Herbert I. Gross

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next Problem #3b next 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? © 2007 Herbert I. Gross

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Answer: $2,300 Solution: We may begin with the formula C = 0.17 × S (1) Since $391 represents her commission, we replace C by 391 in equation (1) to obtain the indirect computation 391 = 0.17 × S (2) Equation (2) can be paraphrased into the direct computation: S = 391 ÷ 0.17 or 2,300 next © 2007 Herbert I. Gross

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next 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). © 2007 Herbert I. Gross

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next 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). © 2007 Herbert I. Gross

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next 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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next 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 #4a next Answer: $87.50 © 2007 Herbert I. Gross

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next 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. © 2007 Herbert I. Gross

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Answer: $87.50 Solution: 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 next © 2007 Herbert I. Gross

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next 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 © 2007 Herbert I. Gross

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next 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 next © 2007 Herbert I. Gross

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next A sale advertises 30% off the regular price. How much was the regular price of a jacket if the sale price was $105? Problem #4b next Answer: $150 © 2007 Herbert I. Gross

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Answer: $150 Solution: Since the sale price (S) is $105, we may replace S by 105 in equation (1) to obtain the indirect computation: 105 = 0.70 × M which can be paraphrased as the direct computation M = 105 ÷ 0.70 or 150 next © 2007 Herbert I. Gross

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next 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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next 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. next © 2007 Herbert I. Gross

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next 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 #5a next Answer: 20 miles © 2007 Herbert I. Gross

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Answer: 20 miles Solution: 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. next © 2007 Herbert I. Gross

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next 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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next 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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next 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. © 2007 Herbert I. Gross

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next 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 #5b next Answer: 69 minutes or 1 hour 9 minutes © 2007 Herbert I. Gross

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Answer: 69 minutes Solution: If we use formula (1) we replace D by 46 to obtain the indirect computation 46 = 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 next © 2007 Herbert I. Gross

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next Note 5(b) We could also have used formula (2) to obtain the indirect computation 46 = 40 × H which can be paraphrased as the direct computation H = 46 ÷ 40 = 23 ÷ 20 or 1 3/20 and since H is expressed in hours the answer is 1 3/20 hours. © 2007 Herbert I. Gross

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next 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 9 minutes; which agrees with our previous answer of 69 minutes. © 2007 Herbert I. Gross

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