SOLUTION: You are at the store, and are trying to remember how much bread, meat, and cheese you were supposed to buy. The bread costs $3 per loaf, the meat costs $5 per pound, and the cheese

Question 1192262: You are at the store, and are trying to remember how much bread, meat, and cheese you were supposed to buy. The bread costs $3 per loaf, the meat costs $5 per pound, and the cheese costs $2 per pound. You know that you are to buy fifteen items, that the total cost will be $51, and that you are supposed to purchase one more meat than cheese. How many meat do you need to purchase?

Found 2 solutions by ikleyn, greenestamps:
Answer by ikleyn(52878) (Show Source): You can put this solution on YOUR website!
.
You are at the store, and are trying to remember how much bread, meat, and cheese you were supposed to buy.
The bread costs $3 per loaf, the meat costs $5 per pound, and the cheese costs $2 per pound.
You know that you are to buy fifteen items, that the total cost will be $51,
and that you are supposed to purchase one more meat than cheese.
How many meat do you need to purchase?
~~~~~~~~~~~~~~

x pounds of meat, (x-1) pounds of cheese and (15 - x - (x-1)) = 16-2x loaves of bread. The total money equation is 3*(16-2x) + 5x + 2*(x-1) = 51. (1) Simplify and find x 48 - 6x + 5x + 2x - 2 = 51 46 + x = 51 x = 51 - 46 ANSWER. 5 pounds of meat. CHECK. Then the left side of equation (1) is 3*(16-2*5) + 5*5 + 2*(5-1) = 51, which is precisely correct.

Solved.

Answer by greenestamps(13209) (Show Source): You can put this solution on YOUR website!

First an informal solution using logical reasoning....

(1) take away the one "extra" meat, leaving 14 items for a total of $46, with equal numbers of meat and cheese.
(2) The numbers of meat and cheese are the same, and the number of items remaining is 14; that means the number of breads is even.
(3) Since the number of breads is even, make groups of 2 breads each, with a cost of 2($3) = $6 per group. And since the numbers of meats and cheeses are the same, make groups of one meat and one cheese, with a cost of $5+$2 = $7 per group.
(4) So we need to make a total of $46 using groups of $6 and $7. There is only one way to do this: 3 groups at $6 each and 4 groups at $7 each for a total of 3($6)+4($7) = $18+28 = $46.

So we have 4 groups of one meat and one cheese, plus the "extra" meat we accounted for at the beginning; so the number of meat you are to purchase is 5.

ANSWER: 5 meats

CHECK: 6($3)+5($5)+4($2) = $18 + $25 + $8 = $51

A formal algebraic solution is easier than the informal one.... But solving a problem like this using logical reasoning is good brain exercise.

A typical formal algebraic solution....

x = # of cheese
x+1 = # of meat (one more than the number of cheese)
15-(x+x+1) = 14-2x = # of bread (the total number of items, 15, minus the total number of meat and cheese)

The total cost is $51:

2x+5(x+1)+3(14-2x)=51
2x+5x+5+42-6x=51
x=4

ANSWER: the number of meat to purchase is x+1 = 5

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