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Question 1190211: A boy scout troop is selling popcorn. There are three different kinds of popcorn in three different arrangements. Arrangement I contains 1 bag of cheddar cheese popcorn, 2 bags of caramel popcorn, and 3 bags of microwave popcorn. Arrangement II contains 3 bags of cheddar cheese popcorn, 1 bag of caramel popcorn, and 2 bags of microwave popcorn. Arrangement III contains 2 bags of cheddar cheese popcorn, 3 bags of caramel popcorn, and 1 bag of microwave popcorn. Jim needs 28 bags of cheddar cheese popcorn, 22 bags of caramel popcorn, and 22 bags of microwave popcorn to give as stocking stuffers for Christmas. How many of each arrangement should he buy?
Found 2 solutions by ikleyn, greenestamps:
Answer by ikleyn(52781)   (Show Source):
You can put this solution on YOUR website! .
A boy scout troop is selling popcorn. There are three different kinds of popcorn in three different arrangements.
Arrangement I contains 1 bag of cheddar cheese popcorn, 2 bags of caramel popcorn, and 3 bags of microwave popcorn.
Arrangement II contains 3 bags of cheddar cheese popcorn, 1 bag of caramel popcorn, and 2 bags of microwave popcorn.
Arrangement III contains 2 bags of cheddar cheese popcorn, 3 bags of caramel popcorn, and 1 bag of microwave popcorn.
Jim needs 28 bags of cheddar cheese popcorn, 22 bags of caramel popcorn, and 22 bags of microwave popcorn to give
as stocking stuffers for Christmas. How many of each arrangement should he buy?
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Let x be the # of arrangements I;
y be the # of arrangements II;
z be the # of arrangements III.
Write the system of equations as you read the problem
x + 3y + 2z = 28 (bags of cheddar cheese popcorn)
2x + y + 3z = 22 (bags of caramel popcorn)
3x + 2y + z = 22. (bags of microwave popcorn)
Use your pocket calculator. Solve the system by any method and get the ANSWER:
2 Arrangements I; 6 Arrangements II and 4 Arrangements III.
Solved.
Answer by greenestamps(13200)  (Show Source):
You can put this solution on YOUR website!
Let's look at a couple of ways you might try to solve this problem.
x = # of arrangement I
y = # of arrangement II
z = # of arrangement III
x+3y+2z = 28 (bags of cheddar cheese popcorn)
2x+y+3z = 22 (bags of caramel popcorn)
3x+2y+z = 22 (bags of microwave popcorn)
There are many ways of solving a system of 3 equations in 3 unknowns.
If this question were on a timed competitive exam, you would want to look for a "trick" to make it easier to find the answer.
One such trick is to add the three equations to see if it gives you something useful. In this case, we get
6x+6y+6z=72
x+y+z=12
This tells us that the total number of arrangements is 12. Unfortunately, that doesn't get us any closer to the answer in this particular problem.
Note, however, that if the question had only asked for the total number of arrangements (as indeed might be the case on a "trick" question on a competitive exam!), then you would be done.
However, that didn't help us here. So what do we do? There are still many choices....
A standard algebraic procedure, if you don't spot any tricks to make the solution easier, is to choose one variable to eliminate, leaving a system of two equation in two unknowns, and then solve that system by ordinary methods.
We can choose to eliminate any of the variables; with the coefficients of the given equations, any one of them will be easy to eliminate. So I randomly choose to eliminate z.
Use the term "z" in the third equation to eliminate z in each of the first two equations:
x+3y+2z=28
6x+4y+2z=44
5x+y=16 [1]
2x+y+3z=22
9x+6y+3z=66
7x+5y=44 [2]
Eliminate y between [1] and [2]:
25x+5y=80
7x+5y=44
18x=36
x=2
14+5y=44
5y=30
y=6
2+3(6)+2z=28
20+2z=28
2z=8
z=4
ANSWER:
arrangement I: x=2
arrangement II: y=6
arrangement III: z=4
CHECK:
x+3y+2z=2+18+8=28
2x+y+3z=4+6+12=22
3x+2y+z=6+12+4=22
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