Tom's sweet tooth
Tom bought 1 lb of jellybeans and 2 lb of chocolate for $16. A week later, he bought 4 lb of caramels and a pound of jellybeans for $24. The next week, he bought 3 lb of liquorice, 1 lb pound of jellybeans and 1 lb of caramels for $12.
How much would he have to pay for a pound of each of the four kinds of candy?
How much would he have to pay for a pound of each of the four kinds of candy?
Labels: mathschallenge
posted by Chris at
1:29 PM
11 Comments:
It looks like $16.
Some reason disallowes me to post using my GoogleAccount, that's why I use just name.
Hi quantense. The blogging software must be overloaded again.
$16 is right. No surprise there ;)
The layman hasn't got a hope of following what Roger Penrose has written in his "The Road To Reality".
Hi Chris. I heard it's very thick book. That's why I prefered Emperor.
Over a 1000 pages.
Sounds ominous :)
It also sounds voluminous ;)
The solution:
Let J = Jellybeans, C = Chocolate, A = cAramel and L = Licorice
Note you don't need to solve for J, C, A, L just the sum of them.
From the problem you can write the three equations:
1) 16 = J + 2C
2) 24 = 4A + J
3) 12 = A + J + 3L
Rewriting the above 3 equations:
1)*3 => 48 = 3J + 6C
2)*1 => 24 = 4A + J
3)*2 => 24 = 2A + 2J + 6L
Add up all three and you get:
6J + 6C + 6A + 6L = 96 => J + C + A + L = 16
Furthermore, given the assumption that the treats have a positive non-zero integer value, the individual values are solvable.
*1) 1J + 2C +0A +0L =16
*2) 1J + 0C +4A +0L =24
*3) 1J + 0C +1A +3L =12
based on *1 we see C <=7
based on *2 we see A <=5 and J>=4
based on *3 we see J<=8, L<=3
Given J<=8, based on *2 A=4 and J= 8 or A=5 and J=4
however A=4 and J=8 is incompatible with *3,
therefor J=4 , and A=5
accordingly, L=1, and C=6
Final solution:....
J=4, C=6, A=5, L=1
Anonmous. That was excellent. I hadn't even thought to try that. I more than forgive you for assuming integral costs. Well done.
I have to agree with them.. thats a lot of candy!!!!
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