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Raisins, almonds, and peanuts

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by sam2304 » Sat Jan 07, 2012 11:07 pm
Each of 435 bags contains at least one of the following three items: raisins, almonds, and peanuts. The number of bags that contain only raisins is 10 times the number of bags that contain only peanuts. The number of bags that contain only almonds is 20 times the number of bags that contain only raisins and peanuts. The number of bags that contain only peanuts is one-fifth the number of bags that contain only almonds. 210 bags contain almonds. How many bags contain only one kind of item?
A.256
B.260
C.316
D.320
E.350
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by pemdas » Sun Jan 08, 2012 2:42 am
If 210 bags contain almonds, the rest contain only peanuts (p), only raisins (r), peanuts & raisins (pr). From the question, we deduce that 210 bags may contain only almonds (a), almonds & peanuts (ap), almonds & raisins (ar) and the total 435 makes 210+p+r+pr

Proportions provided by the question are
r=10p
a=20pr
p=a/5

Set the equation: 435=210+p+r+pr, 225=p+r+pr. Express all variables through a (only almonds) p=a/5, pr=a/20, r=10(a/5)

225=a/5+10(a/5)+a/20, 225=a/5 +2a +a/20, a=100
The bags with only almonds are 100
The bags with only peanuts are 100/5=20
The bags with only raisins are 10*(100/5)=200

Total makes 100+20+200=320

d
sam2304 wrote:Each of 435 bags contains at least one of the following three items: raisins, almonds, and peanuts. The number of bags that contain only raisins is 10 times the number of bags that contain only peanuts. The number of bags that contain only almonds is 20 times the number of bags that contain only raisins and peanuts. The number of bags that contain only peanuts is one-fifth the number of bags that contain only almonds. 210 bags contain almonds. How many bags contain only one kind of item?
A.256
B.260
C.316
D.320
E.350
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by GMATGuruNY » Sun Jan 08, 2012 3:45 am
I posted an explanation here:

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by sam2304 » Sun Jan 08, 2012 7:16 am
pemdas wrote: Set the equation: 435=210+p+r+pr
How did you arrive at this ? If they have given data for only almonds, only raisins and only peanuts then 2 items common and 3 items common should total to 435, but i don't see it in your equation only pr is there.
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by pemdas » Sun Jan 08, 2012 9:12 am
sam2304 wrote:
pemdas wrote: Set the equation: 435=210+p+r+pr
How did you arrive at this ? If they have given data for only almonds, only raisins and only peanuts then 2 items common and 3 items common should total to 435, but i don't see it in your equation only pr is there.
it's all included with my solution.
pemdas wrote:If 210 bags contain almonds, the rest contain only peanuts (p), only raisins (r), peanuts & raisins (pr). From the question, we deduce that 210 bags may contain only almonds (a), almonds & peanuts (ap), almonds & raisins (ar) and the total 435 makes 210+p+r+pr
...
Set the equation: 435=210+p+r+pr, 225=p+r+pr.
as per 3 items in common, there's no info in the question
do you have any idea about bag(s) with 3 items? the question specifies only 2 items in common
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by sam2304 » Sun Jan 08, 2012 9:55 am
pemdas wrote:210 bags may contain only almonds (a), almonds & peanuts (ap), almonds & raisins (ar)
Well thanks mate. I just overlooked the problem, thought the 210 represents only almonds and forgot that it includes the ap and ar as well. Now i got it right. :)

By the way, the OA is D.
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