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A farmer sells buckets of onions for $2, buckets of potatoes for $4, a

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A farmer sells buckets of onions for $2, buckets of potatoes for $4, a  [#permalink]

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New post 31 Mar 2019, 08:16
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A farmer sells buckets of onions for $2, buckets of potatoes for $4, and buckets of squash for $5. How many buckets of squash did the farmer sell?

(1) The farmer sold one bucket of onions.

(2) In total, the farmer sold 7 buckets of produce for $33.

Source: McGraw-Hill's GMAT

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Re: A farmer sells buckets of onions for $2, buckets of potatoes for $4, a  [#permalink]

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New post 31 Mar 2019, 09:21
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Using Statement 2 alone, if the farmer sells 7 buckets for $33, then the average price per bucket is nearly $5. So almost all of the buckets sold must be $5 ones. Since onions and potatoes both sell for an even amount, for the total cost to be odd, the number of $5 buckets must be odd. The farmer did not sell 7 of the $5 buckets (that would give us too high a total), and did not sell only 3 of them (that would net $15, and the remaining four buckets could have sold for at most $16, which does not get us to $33), so must have sold 5 of them, for $25. Then the other two buckets would both need to be $4 buckets. So Statement 2 is sufficient.

The problem with the question though is that the two statements can't both be true. If the farmer sells, as Statement 1 tells us, one $2 bucket, there is no way Statement 2 can also be true, since you'd need the six remaining buckets to cost $31 in total, and six things each costing at most $5 can, in total, cost no more than $30.
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Re: A farmer sells buckets of onions for $2, buckets of potatoes for $4, a  [#permalink]

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New post 31 Mar 2019, 10:10
SajjadAhmad wrote:
A farmer sells buckets of onions for $2, buckets of potatoes for $4, and buckets of squash for $5. How many buckets of squash did the farmer sell?

(1) The farmer sold one bucket of onions.

(2) In total, the farmer sold 7 buckets of produce for $33.

Source: McGraw-Hill's GMAT


#1
no info about other buckets ; insufficient

#2
2x+5y+4z=33
and x+y+z= 7
so since sale is of 33$ it means sum must of odd + even integers
first maximize 5$ sales ; we see y=5 ; 25$ ; left 8 $ so we can have 2 of z ; sufficient
for other values of 5,15 we do nt get value to match 33$
IMOB
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Re: A farmer sells buckets of onions for $2, buckets of potatoes for $4, a  [#permalink]

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New post 31 Mar 2019, 20:25
SajjadAhmad wrote:
A farmer sells buckets of onions for $2, buckets of potatoes for $4, and buckets of squash for $5. How many buckets of squash did the farmer sell?

(1) The farmer sold one bucket of onions.

(2) In total, the farmer sold 7 buckets of produce for $33.

Source: McGraw-Hill's GMAT


The equation would be \(2x+4y+5z\)

we have to find z

Statement 1:
Nothing about squash. Insufficient.

Statement 2:

\(2x+4y+5z=33\)
This is in form of even+odd=odd
Odd part is contributed by 5.
But for\(z=1\), \(x+2y=14\)
& \(z=3\),\(x+2y=9\).

z has various values, insufficient
(1)+(2)
\(x=1,\)
So, \(2x+4y+5z=33.\)
\(4y+5z=31\)
for \(z=1, y=\frac{26}{4}\), not an integer.
for \(z=3,y=4\)
\(z=5, y=\frac{6}{4}\) not an integer.

So, z=3. Sufficient. IMO, Option C.
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Re: A farmer sells buckets of onions for $2, buckets of potatoes for $4, a   [#permalink] 31 Mar 2019, 20:25
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