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X grapes and Y peanuts were purchased. Grapes cost 5 dollars

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X grapes and Y peanuts were purchased. Grapes cost 5 dollars [#permalink] New post 13 Jan 2010, 01:02
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weird question from gmatfocus. I thought there are no bad official problems. Is this one of them or I'm I missing something here?

X grapes and Y peanuts were purchased. Grapes cost 5 dollars; peanuts cost 4 dollars. Total spent is 40. What is X?

1) ratio of x to y is 4 to 5
2) x+y = 9

You don't need either 1 or 2 to figure out the answer. Right?????
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Re: weighted average [#permalink] New post 13 Jan 2010, 02:28
I agree the question looks funny. The cost is for what: total grapes or per grapes??
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Re: weighted average [#permalink] New post 13 Jan 2010, 03:36
I guess its the total cost meaning 5x+4y = 40
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Re: weighted average [#permalink] New post 13 Jan 2010, 05:10
what is x ==> how many grapes were purchased.
that is not given in the question. only total cost is given.

x can be solved independently using either equation.
my answer is D
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Re: weighted average [#permalink] New post 13 Jan 2010, 06:00
my argument is u dont need either.
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Re: weighted average [#permalink] New post 13 Jan 2010, 06:22
thailandvc wrote:
my argument is u dont need either.


sorry didnt get u....what do u mean by " u dont ned either"?? I hope its a DS question with stmnt 1 and 2
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Re: weighted average [#permalink] New post 13 Jan 2010, 06:35
thailandvc wrote:
my argument is u dont need either.

hi i think where u r going wrong is that u r taking that both x and y need to be +ive int.. in that case there will be no other sol except x as 4 and y as 5....because rest values of y(1,2,3,5,6,7) doesnt give any int value of x...
but then it is possible that only grapes or only oranges were bought ie(x,y)can be(0,10)or(8,0)..
however both si and sii confirm ans to be (4,5)
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Re: weighted average [#permalink] New post 13 Jan 2010, 06:35
yes thats what i meant. you don't need statement 1 or 2 to solve for X. the question itself is sufficient to solve for X.
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Re: weighted average [#permalink] New post 13 Jan 2010, 06:40
chetan2u wrote:
thailandvc wrote:
my argument is u dont need either.

hi i think where u r going wrong is that u r taking that both x and y need to be +ive int.. in that case there will be no other sol except x as 4 and y as 5....because rest values of y(1,2,3,5,6,7) doesnt give any int value of x...
but then it is possible that only grapes or only oranges were bought ie(x,y)can be(0,10)or(8,0)..
however both si and sii confirm ans to be (4,5)


yup i think ur right. im so used to thinking about weighted average and the distance between x and y that i totally overlooked 0 as a possible answer.
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Re: weighted average [#permalink] New post 13 Jan 2010, 23:10
From the question
5x+4y=40

St 1: x/y=4/5
=> x=4y/5

Put this value in the above equation to find the value of x and y. So 1 is sufficient

St 2: x+y=9
=> x=9/y

Put this value in the equation to find the value of x and y(x and y will be only positive).. So 2 sufficient

Hence D is the answer
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Re: weighted average [#permalink] New post 14 Jan 2010, 00:54
chetan2u wrote:
thailandvc wrote:
my argument is u dont need either.

hi i think where u r going wrong is that u r taking that both x and y need to be +ive int.. in that case there will be no other sol except x as 4 and y as 5....because rest values of y(1,2,3,5,6,7) doesnt give any int value of x...
but then it is possible that only grapes or only oranges were bought ie(x,y)can be(0,10)or(8,0)..
however both si and sii confirm ans to be (4,5)




hi you are absolutly right. then can we "D" choice as right answer. or any thing ealse.
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Re: weighted average [#permalink] New post 14 Jan 2010, 04:50
hi sanjay the ans would be C in this case...
Re: weighted average   [#permalink] 14 Jan 2010, 04:50
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