Question 1210589
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The problem asks for the number of oranges he started with.  So let that be our variable.<br>
x = original number of oranges<br>
We will need another variable.<br>
y = original number of apples<br>
The original number of apples is 1/4 of the total number of fruits, so the total number of fruits is 4y. This gives us<br>
4y-(x+y) = 3y-x = original number of pears<br>
To start, then, we have<br>
apples: y
pears: 3y-x
oranges: x<br>
He sells 80 apples and 1/3 of the pears, leaving...<br>
apples: y-80
pears: (2/3)(3y-x) = 2y-(2/3)x
oranges: x<br>
He buys more oranges, increasing the number of oranges by 60% -- i.e., multiplying the number of oranges by 160%, or 8/5.<br>
apples: y-80
pears: (2/3)(3y-x) = 2y-(2/3)x
oranges: (8/5)x<br>
After doing that, he has twice as many pears as apples<br>
{{{2y-(2/3)x=2(y-80)}}}
{{{2y-(2/3)x=2y-160}}}
{{{(2/3)x=160}}}
{{{x=240}}}<br>
That is what the problem asked us to find, so we are done.<br>
ANSWER: x = 240 oranges<br>
Although the problem doesn't require us to do any more work, it is curious to continue to find the original numbers of apples and pears.<br>
The total number of fruits at the end was 20% more than the number at the beginning.<br>
at the beginning: {{{4y}}}
at the end: {{{(y-80)+(2y-(2/3)x)+(8/5)x=3y-80-160+384=3y+144}}}<br>
The total number at the end was 20% greater than -- i.e. 6/5 as much -- as at the beginning:<br>
{{{3y+144=(6/5)(4y)=(24/5)y}}}
{{{144=(9/5)y}}}
{{{y=(5/9)144=80}}}<br>
The fruits he started with:
apples: y = 80
pears: 3y-x = 240-240 = 0
oranges: 240<br>