Question 1120853
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An informal solution by trial and error is of course valid.<br>
But let's look at a formal mathematical process that can be used, along with some logical reasoning, to solve this kind of problem.<br>
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.<br>
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:<br>
{{{(x)(y)+(12-x)(y-3) = 132}}}
{{{xy+12y-36-xy+3x = 132}}}
{{{12y+3x = 168}}}
{{{4y+x = 56}}}<br>
Now use some logical reasoning to find exactly the value x must be.<br>
{{{4y+x = 56}}}
{{{x = 56-4y}}}
{{{x = 4(14-y)}}}<br>
"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.<br>
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.<br>
So the purchase was of 8 apples and 4 oranges.<br>
The problem doesn't ask for the price of each kind of fruit; but we can go ahead and find them.<br>
{{{8(y)+4(y-3) = 132}}}
{{{8y+4y-12 = 132}}}
{{{12y = 144}}}
{{{y = 12}}}<br>
The cost of each apple is y=12 cents; the cost of each orange is y-3 = 9 cents.<br>
Check: 8 apples at 12 cents + 4 oranges at 9 cents = 96 cents + 36 cents = $1.32