Hot X: Algebra Exposed

Transcription

1 Hot X: Algebra Exposed Solution Guide for Chapter 16 Here are the solutions for the Doing the Math exercises in Hot X: Algebra Exposed! DTM from p Let s call the number of dimes d, and the number of nickels n. We ll deal in cents, since that s the unit the problem is already using. Then we know that d + n = 10, and 10d + 5n = 80. We want to solve for n, so let s use the substitution d = 10 n in the second equation, and solve! 10d + 5n = 80 10(10 n) + 5n = n + 5n = n = 80 5n = 20 n = 4 So she has 4 nickels. Does that work out? So, then she d have 6 dimes, and that s 4(5) + 6(10) = = 80, yep! Answer: 4 nickels 3. Let s call the number of high heels she sold h and the number of flats she sold f. Then we know that h + f = 20 and that 80h + 60f = We want to solve for h, so let s use the substitution f = 20 h and stick that into the second equation: 80h + 60f = 1440

2 80h + 60(20 h) = h h = h = h = 240 h = 12 So Debbie sold 12 pairs of high heels, which means she sold 8 pairs of flats. Does that work? That would mean the money she made would be: 80(12) + 60(8) = = Yep! Answer: 12 pairs of high heels 4. We have dollars and cents, so let s convert the $1.20 to 120 cents and the $9 to 900 cents. This ll get rid of our decimals while we re solving. Next, let s call the number of pounds of gumballs g, and the number of pounds of jelly beans j. A 15-pound mixture means that g + j = 15, right? And the cost equation would be: 30g + 120j = 900. Let s use the substitution g = 15 j and stick it into the cost equation: 30g + 120j = (15 j) + 120j = j + 120j = j = j = 450 j = 5 So if he bought 5 pounds of jellybeans, then he bought 10 pounds of gumballs. Does this work? Let s try it: 30(10) + 120(5) = = 900 cents = $9. Yep! Answer: 5 pounds of jelly beans 5. Let s call the number of dimes d, the number of quarters q, and the number of nickels n. And let s use cents instead of dollars, so then $2 becomes 200 cents. We know that 10d + 25q + 5n = 200, and we re also told some relationships between our variables. Namely, that d = 5 + q, and 2n = q. Since we have q in both these relationship

3 equations, let s write the big equation all in terms of q. So we ll substitute d = 5 + q and n = q 2 and get: 10d + 25q + 5n = (5 + q) + 25q + 5 q 2 = q + 25q + 5q 2 = q + 5q q + 5q = q = 300 q = = = 4 = 200 Let s multiply both sides by 2, and we get: If there are 4 quarters, then there are d = 5 + q d = = 9 dimes and n = q 2 n = 4 2 = 2 nickels. Does this work? Substituting our answers in, we get: 10d + 25q + 5n = 10(9) + 25(4) + 5(2) = = 200. Yep! And the problem asked for the number of dimes, so: Answer: 9 dimes 6. So our units are chores and dollars. Let s call the number of chores she does c and the number of chores she forgets to do f. Then we re told that c + f = 13, right? And what s our money equation? Well, the money she makes will be $2.00c, and the money that is deducted is $1.50f. Make sense? That means the total money she ends up with can be written as $2.00c $1.50f which we ve been told equals $19, So we have our money equation! 2c 1.5f =19 Let s substitute c = 13 f and get: 2(13 f) 1.5f =19

4 2(13 f) 1.5f = f 1.5f = f = f = 7 f = 2 If the number of chores she forgets to do is 2, then she remembers to do 11 of them. Does this work? 2(11) 1.5(2) = 22 3 = 19 Yep! Answer: She forgot 2 chores DTM from p Ok, so our special ingredient is guava juice. Let s call the amount of Brand A that we ll use a and the amount of Brand B, b. Since we ll have 2 gallons total, we know that a + b = 2. That s one equation. Great! For the other equation, we can write a true statement by writing down the amount of actual guava juice contributed from each Brand, (40%)a + (10%)b, and setting it equal to the total amount of actual guava juice in the final mixture, (30%)(2), and we get: (40%)a + (10%)b = (30%)(2) (0.40)a + (0.10)b = (0.30)(2) Now we ll use the substitution a = 2 b, and we get: (0.40)(2 b) + (0.10)b = (0.30)(2) b + 0.1b = b = 0.2 b = = = 2 3 And since a + b = 2, we know that a = Done! Answer: gal. of Brand A, and 2 3 gal. of Brand B

5 3. So our special ingredient is pure glitter, and all our units are gallons. Nice! Where do we start? Let s label! Let s call the amount of pure glitter that she adds, g. This means when she starts with 4 gallons and adds g gallons, that the final mixture will have 4 + g gallons in it total. Make sense? Now, for our percents equation, we will keep track of pure glitter our special ingredient. She starts with 4 gallons of 10% glitter, so the total amount of pure glitter contributed from this will be (10%)4 gallons. Then she adds g gallons of 100% pure glitter, so that s (100%)g of pure glitter, and she ll end up with 4 + g gallons of 70% glitter, which can be expressed (70%)(4 + g). Here s our true statement, which you ll notice only has one unknown in it, so we can go ahead and solve it! (10%)4 + (100%)g = (70%)(4 + g) (0.10)4 + (1)g = (0.70)(4 + g) g = g 0.3g = 2.4 g = = = 24 3 = 8 So she should add 8 gallons of pure glitter! Answer: 8 gallons of pure glitter 4. Let s call the amount (in ounces) of Chipmunk Charlie s nut mix c and the amount of Rabbit Rosy s nut mix r. Then since we know that the final mixture will be 20 oz, this is true: c + r = 20. It sounds like the special ingredient in these nut mixes is hazelnuts, so let s now look at the special ingredient equation: The number of ounces of pure hazelnuts contributed from Chipmunk Charlie will be (25%)c, and the number of ounces of pure hazelnuts contributed from Rabbit Rosy will be (45%)r, and our final mix will have this amount of pure hazelnuts in it: (40%)(20), so we can write this true statement: (25%)c + (45%)r = (40%)(20) And using the substitution from our first equation: c = 20 r, we get:

6 (25%)(20 r) + (45%)r = (40%)(20) (0.25)(20 r) + (0.45)r = (0.40)(20) r r = 8 0.2r = 3 r = = = 30 2 = 15 That s 15 oz of Rabbit Rosy s nut mix. And since c = 20 r, that means we ll use 5 oz of Chipmunk Charlie s nut mix. Done! Answer: 5 oz. of Chipmunk Charlie s nut mix, and 15 oz. of Rabbit Rosy s nut mix 5. So our special ingredient is salt! But first, let s see what s going on with the total volume. Let s label: we ll call the amount (in gallons) of salty water in the tub that we start with s. Then if 5 gallons of water evaporate overnight, the next day we ll only have s 5 gallons of salty water left in the tub, right? Now, let s see what s going on with the special ingredient: pure salt. To begin with, we have s gallons of salty water, and since it s 6% salt, that makes (6%)s gallons of pure salt. We re told that the next morning, there is 8% salt in the tub. Since the total volume the next morning is s 5, that means that the amount of pure salt in the tub is (8%)(s 5). But wait, since only water evaporates (the salt all gets left behind), then the amount of pure salt doesn t change from the night to the next morning, so we can set those above two bolded expressions equal to each other; we ll have a true statement! (6%)s = (8%)(s 5) Remember, although we have steps we can follow, all word problems come down to labeling unknowns and figuring out how to write a true statement to solve. Let s do it! (6%)s = (8%)(s 5) (0.06)s = (0.08)(s 5) 0.06s = 0.08s s = 0.4

7 s = = = 40 2 = 20 So the amount of saltwater in the tub the night before is 20 gallons. Done! Answer: 20 gallons of saltwater 6. The special ingredient is pure gold. Let s not be distracted by the numbers 14 and 18, because we don t need them! So I ll say fourteen and eighteen to help with that. Let s call the amount of pure gold that we ll use g. So if we start with 40 grams of fourteen carat gold, and we add g grams of pure gold to it, then we ll end up with a total of 40 + g grams of our mixture (eighteen carat gold), right? Now let s work on writing a true statement relating the amounts of pure gold. The pure gold contributed from the 40 grams of fourteen carat gold is (60%)(40), right? And the amount of pure gold contributed from g grams of pure gold is (100%)g. We know we ll end up with 40 + g grams of eighteen carat gold, which is 75% pure gold, so the pure gold in our final mixture can be expressed as: (75%)(40 + g). So we have: (60%)40 + (100%)g = (75%)(40 + g) (0.6)40 + (1)g = (0.75)(40 + g) g = g 0.25g = 6 g = = = = 24 This means we ll need to add 24 grams of pure gold. Done! Answer: 24 grams of pure gold