SOLUTION: a customer bought a dozen of pieces of fruits (apples and oranges) for 1.32 dollars. If an apple cost 3 cents more than an orange and more apples than oranges were purchased, how m

Algebra ->  Coordinate Systems and Linear Equations  -> Linear Equations and Systems Word Problems -> SOLUTION: a customer bought a dozen of pieces of fruits (apples and oranges) for 1.32 dollars. If an apple cost 3 cents more than an orange and more apples than oranges were purchased, how m      Log On


   

Question 1120853: a customer bought a dozen of pieces of fruits (apples and oranges) for 1.32 dollars. If an apple cost 3 cents more than an orange and more apples than oranges were purchased, how many pieces of each kind were bought?
Found 2 solutions by ankor@dixie-net.com, greenestamps:
Answer by ankor@dixie-net.com(22740) About Me  (Show Source):
You can put this solution on YOUR website!
a customer bought a dozen of pieces of fruits (apples and oranges) for 1.32 dollars.
If an apple cost 3 cents more than an orange and more apples than oranges were purchased,
:
let p = cost of the oranges
then
(p+3) = cost of the apples
:
No. of apples are at least 7. (apples are more than the oranges)
let's try 8 apples 4 oranges, friendly numbers
8(p+3) + 4p = 132
8p + 24 + 4p = 132
8p + 4p = 132 - 24
12p = 108
p = 108/12
p = 9 cents for oranges
then
9 + 3 = 12 cents for apples
"How many pieces of each kind were bought? "
8 apples, 4 oranges

Answer by greenestamps(13198) About Me  (Show Source):
You can put this solution on YOUR website!


An informal solution by trial and error is of course valid.

But let's look at a formal mathematical process that can be used, along with some logical reasoning, to solve this kind of problem.

Let x be the number of apples; then 12-x is the number of oranges.
Let y be the cost (in cents) of an apple; the y-3 is the cost of an orange.

The total cost of the fruit (x apples at y cents each, and (12-x) oranges at (y-3) cents each) was $1.32, or 132 cents:

%28x%29%28y%29%2B%2812-x%29%28y-3%29+=+132
xy%2B12y-36-xy%2B3x+=+132
12y%2B3x+=+168
4y%2Bx+=+56

Now use some logical reasoning to find exactly the value x must be.

4y%2Bx+=+56
x+=+56-4y
x+=+4%2814-y%29

"14" and "y" are whole numbers, so (14-y) is a whole number. This equation then tells us that x is a multiple of 4.

But we know x is less than 12; and we know it is greater than (12-x). The only value of x that satisfies all those conditions is x=8.

So the purchase was of 8 apples and 4 oranges.

The problem doesn't ask for the price of each kind of fruit; but we can go ahead and find them.

8%28y%29%2B4%28y-3%29+=+132
8y%2B4y-12+=+132
12y+=+144
y+=+12

The cost of each apple is y=12 cents; the cost of each orange is y-3 = 9 cents.

Check: 8 apples at 12 cents + 4 oranges at 9 cents = 96 cents + 36 cents = $1.32