Swiss Chocolates, Lollipops, GumdropsDate: 02/10/2003 at 23:53:47 From: Della DA Subject: Math You have $100 and must buy 100 pieces of candy. Imported Swiss chocolates cost $15 each. Lollipops are $1 each. Gumdrops are $.25 each. You must purchase at least one of each item of candy. Pieces of candy must be whole. How many of each candy do you buy to spend $100 and to get exactly 100 pieces? Date: 02/11/2003 at 01:14:24 From: Doctor Jeremiah Subject: Re: Math Hi Della, You should have tried figuring it out for 3 Swiss chocolates. Do that now, and if you still can't get an answer you like then read what follows. So that I don't have to type so much I am going to use C for the number of Swiss chocolates, L for the number of lollipops, and G for the number of gumdrops. If there is a total of 100 pieces, then the number of chocolates (C) plus the number of lollipops (L) plus the number of gumdrops (G) must equal 100, because the sum of the number of all three types must be 100. So: C + L + G = 100 If 1 chocolate costs $15 and C is the number of Swiss chocolates, then the cost of all the chocolates must be $15 times C. If 1 lollipop costs $1 and L is the number of lollipops, then the cost of all the lollipops must be $1 times L. If 1 gumdrop costs $0.25 and "G" is the number of gumdrops, then the cost of all the gumdrops must be $0.25 times G. The total cost of all three types is $100, so the cost of the Swiss chocolates ($15C) plus the cost of the lollipops ($1L) plus the cost of the gumdrops ($0.25G) must be $100. The equation for that is: $15C + $1L + $0.25G = $100 Now, we have two equations: C + L + G = 100 15C + L + 0.25G = 100 But we have three unknown values (C, L, and G), so we can't yet solve this. We either need one more piece of information or we need to assume a value for the number of one of the types. Swiss chocolate costs $15, so we need at least one but not more than six. This is true because $15 times 7 = $105. So let's start by assuming that the number of Swiss chocolates is one. Now we have three equations: C + L + G = 100 15C + L + 0.25G = 100 C = 1 Since the third equation says that C=1 we can change all the C in the other two equations to 1 and get rid of the dependence on C. That leaves us with two equations with two unknowns: 1 + L + G = 100 15 + L + 0.25G = 100 To solve this for L and G we must get one of the unknowns by itself. I will pick L purely at random. After a bit of rearrangement of the first equation we get: L = 99 - G Since L is equal to 99-G we can change all the L in the other equation to 99-G and get this equation: 15 + 99-G + 0.25G = 100 Which can be solved for G like this: 15 + L + 0.25G = 100 15 + 99-G + 0.25G = 100 15 + 99 - 100 = G - 0.25G 14 = 0.75G 14/0.75 = G 56/3 = G Notice that this isn't a whole number. Since you can't buy part of a gumdrop, the number of Swiss chocolates cannot be one. So let's start over and assume that the number of Swiss chocolates is two. Now we have these three equations: C + L + G = 100 15C + L + 0.25G = 100 C = 2 Since the third equation says that C=2 we can change all the C in the other two equations to 2 and get rid of the dependence on C. That leaves us with two equations with two unknowns: 2 + L + G = 100 30 + L + 0.25G = 100 To solve this for L and G we must get one of the unknowns by itself. I will pick L purely at random. After a bit of rearrangement of the first equation we get: L = 98 - G Since L is equal to 98-G we can change all the L in the other equation to 98-G and get this equation: 30 + L + 0.25G = 100 15 + 98-G + 0.25G = 100 15 + 98 - 100 = G - 0.25G 13 = 0.75G 13/0.75 = G 52/3 = G Notice that this isn't a whole number. Since you can't buy part of a gumdrop, the number of Swiss chocolates cannot be two. So let's go back and assume that the number of Swiss chocolates is three. Now we have three equations: C + L + G = 100 15C + L + 0.25G = 100 C = 3 Since the third equation says that C=3 we can change all the C in the other two equations to 3 and get rid of the dependence on C. That leaves us with two equations with two unknowns: 3 + L + G = 100 45 + L + 0.25G = 100 To solve this for L and G we must get one of the unknowns by itself. I will pick L purely at random. After a bit of rearrangement of the first equation we get: L = 97 - G Since L is equal to 97-G we can change all the L in the other equation to 97-G and get this equation: 45 + L + 0.25G = 100 45 + 97-G + 0.25G = 100 45 + 97 - 100 = G - 0.25G 42 = 0.75G 42/0.75 = G 56 = G This is a whole number but it might still not be the right answer. So we need to solve for L and then check the results. Remember that: L = 97 - G And since G=56 we can solve for L and get: L = 41 The question is, will the original two equations work with L = 41, G = 56, and C = 3? C + L + G = 100 15C + L + 0.25G = 100 Becomes: 3 + 41 + 56 = 100 45 + 41 + 14 = 100 And those two equations are definitely true. 3 + 41 + 56 does equal 100 45 + 41 + 14 does equal 100 This means that the number of each type is: L = 41, G = 56, and C = 3. Which means that there are 3 Swiss chocolates, 41 lollipops, and 56 gumdrops. - Doctor Jeremiah, The Math Forum http://mathforum.org/dr.math/ |
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