Instead of handing out candy this Halloween, hand out this problem. I dare you. Assume the price of each candy remains constant. Daniel bought 1 pound of jelly beans and 2 pounds of chocolates for $2.00. A week later, he bought 4 pounds of caramels and 1 pound of jelly beans, paying $3.00. The next week, he bought 3 pounds of licorice, 1 pound of jelly beans, and 1 pound of caramels for $1.50. How much would he have to pay on his next trip to the candy store if he bought 1 pound of each of the 4 kinds of candy? John
There are 5 unknowns and 4 equations. How you can solve it? So you say that there is not trick but just mathematics.
jpanhalt, The trick to solving this problem is not to try to find the price per pound of all four different candies. That would be trying to solve 4 unknowns with only 3 equations. Instead, find what the sum of all 4 candies rates are. To do that, find out what each rate is in terms of a single reference rate. I used jelly beans as the reference rate. Without trying to post all the algebra involved, the results are: Using Cm for carmel, Ch for chocolate Jb for Jellybeans Lq for licorice Cm = -0.25*Jb + 0.75 Ch = -0.50*Jb + 1.00 Jb = Jb Lq = -.025*Jb + 0.25 ---------------------------- summing the left and right sides Cm + Ch + Jb + Lq = 2.00 Notice that 3 equations are sufficient if three unknowns are solved in terms of a single selected unknown. So the price paid for a pound of each type of candy is $2.00, and mik3 was correct. Congratulation to mik3 for solving it fast and first! Ratch
Well, I didn't really solved it like you. I omitted the last equation Cm + Ch + Jb + Lq = ? and solved for the first three equations (three unknowns and three equations). Then I said that he forgot to include the licorice in the prize and the prize of the others came out to be 2%. It was just luck.
mik3, Being lucky is often better than being clever. Mr. Gates did not get to be a multi-BILLionaire just by being smart and clever. He had a incredible streak of good luck at the beginning of his career. Ratch
My solution was a little different. Once it is realized that the jelly beans are common to each purchase, it doesn't matter what they cost -- it is math puzzle after all. I set a sensible cost of $0 for them, which makes getting the total cost trivial. If you are stimulated to answer a question that wasn't asked, namely the price of each candy, one can do that by assigning another value to the jelly beans. If they are $1 per pound, then licorice is free. I like licorice, so that is my favorite non-solution. There are many more. For interest, try jelly beans at $0.50 per pound. John
I just noticed this problem, so am too late. However, I'll mention my method to solve which is very simple minded. Total all 3 purchases first: 3 JB + 2 Choc + 5 Caram + 3 Lic = $6.50 Now add the third purchase: 4 JB + 2 Choc + 6 Caram + 6 Lic = $8.00 Now add 2 times the first purchase 6 JB + 6 Choc + 6 Caram + 6 Lic = $12.00 Now divide by 6 1 JB + 1 Choc + 1 Caram + 1 Lic = $2.00