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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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Presentation on theme: "Algebra Problems… Solutions Algebra Problems… Solutions next © 2007 Herbert I. Gross Set 1 By Herb I. Gross and Richard A. Medeiros."— Presentation transcript:

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

2 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

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

4 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)

5 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

6 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

7 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)

8 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 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 next © 2007 Herbert I. Gross We could make the following chart:

9 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

10 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

11 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

12 We may compute C either by rewriting equation (2) as C = 84 × (5 + 2/3) = (84 × 5) + (84 × 2/3 ) = = 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

13 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

14 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

15 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

16 next Note 2(b) If you used a calculator to compute 616 ÷ 84, you obtained an answer in the form 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

17 next Problem #3a next Answer: $ 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

18 Answer: $ 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 $ next © 2007 Herbert I. Gross

19 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

20 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

21 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

22 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

23 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

24 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

25 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

26 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

27 next Note 3(b) If one did replace S by 391 in formula (1) the formula would have been 0.17 × $391 or $ 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

28 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

29 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

30 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

31 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

32 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

33 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

34 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

35 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

36 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 $ $105 equals $136.50; and 70% of $ is $95.55, not $105. next © 2007 Herbert I. Gross

37 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

38 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

39 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

40 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

41 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

42 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

43 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

44 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

45 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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