# Trick or Treat?

Discussion in 'Off-Topic' started by jpanhalt, Oct 16, 2009.

1. ### jpanhalt Thread Starter Expert

Jan 18, 2008
6,798
1,398
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

2. ### mik3 Senior Member

Feb 4, 2008
4,846
69
Are you sure the problem statement is correct?

3. ### mik3 Senior Member

Feb 4, 2008
4,846
69
He paid 2\$ because the sales assistant forgot to include the licorice in the price.

4. ### jpanhalt Thread Starter Expert

Jan 18, 2008
6,798
1,398
The problem is solvable.

John

5. ### mik3 Senior Member

Feb 4, 2008
4,846
69
There are 5 unknowns and 4 equations. How you can solve it?

So you say that there is not trick but just mathematics.

6. ### jpanhalt Thread Starter Expert

Jan 18, 2008
6,798
1,398
There is no trick to the question. I believe there is only one answer.

John

7. ### mik3 Senior Member

Feb 4, 2008
4,846
69
Do you know it?

8. ### jpanhalt Thread Starter Expert

Jan 18, 2008
6,798
1,398
I just PM'd you the answer.

John

9. ### Ratch New Member

Mar 20, 2007
1,068
4
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

10. ### mik3 Senior Member

Feb 4, 2008
4,846
69
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.

11. ### Ratch New Member

Mar 20, 2007
1,068
4
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

Last edited: Oct 16, 2009
12. ### jpanhalt Thread Starter Expert

Jan 18, 2008
6,798
1,398
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

13. ### steveb Senior Member

Jul 3, 2008
2,432
469
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