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Thursday, January 24, 2008

I Want Candy!

Donna bought one pound of jellybeans and two pounds of chocolate for $2. A week later, she bought four pounds of caramels and one pound of jellybeans, paying $3. The next week, she bought three pounds of licorice, one pound of jellybeans, and one pound of caramels for $1.50.

How much would she have to pay on her next trip if she bought a pound of each of the four kinds of candy?

7 comments:

Morris Tao said...

J = jellybeans
C = Chocolate
A = Caramel
L = Licorice

From the problem you can write these three equations:
1) J + 2C = $2
2) 4A + J = $3
3) 3L + J + A = $1.50

Note you don't need to solve for J,C,A,L just the sum of them.

Rewriting the above 3 equations:
3 x (J + 2C) = 3 x $2
4A + J = $3
2 x (3L + J + A) = 2 x $1.50

Add up all three and you get:
6J + 6C + 6A + 6L = $12
or
J + C + A + L = $2

Anonymous said...

Where is this place? I want some free jellybeans!

Anonymous said...

So what is the answer

Ken Stanley said...

There is more to this than meets the eye.
With four unknowns and only three equations there are several possibilities:
1) There could be no solution
2) There could be a set of solutions that forms a line in the 4-dimensional space.
3) There could be a set of solutions that forms a plane or hyper-plane.

There cannot be a set of solutions that includes only a single point. Hence, there is not enough information to compute the price of the four types of candy.

However, it turns out that all of solutions of these three equations satisfy the equation:
J+C+L+A = $2 as Morris Tau showed.

Anonymous said...

J = jellybeans
C = chocolate
A = Caramel
L = licorice

J + 2C = $2
4A + J = $3
3L + J + A = $1.50

AND HOW ON EARTH DO THEY DO THE NEXT BIT?????

Anonymous said...

TELL ME THE ANSWER PLEASE

Anonymous said...

I am 13 in algebra. We have to figure this problem out. But using substitution. I have no idea how to do it!!!