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raisins, apples and peanuts

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by joannabanana » Sun Sep 05, 2010 11:52 am
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) It cannot be determined from the given information.
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by Rahul@gurome » Sun Sep 05, 2010 5:48 pm
Let the number of bags that contain only peanuts = P
Then number of bags that contain only raisins = 10P
Let the number of bags that contain only raisins and peanuts = x
Then number of bags that contain only almonds = 20x
P = (1/5)(20x) or P = 4x
20x = 210
Now, 435 = Peanuts only + Raisins only + only raisins and peanuts + total almonds
435 = P + 10P + x + 210
435 = 4x + 10(4x) + x + 210
435 = 45x + 210
225 = 45x or x = 5
Almonds only = 20*5 = 100
Number of bags that contain only one kind of item = Peanuts only + Raisins only + almonds only = 20 + 10*20 + 100 = 320

The correct answer is [spoiler](D)[/spoiler].

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by limestone » Sun Sep 05, 2010 6:15 pm
Let's call No of bags that contain raisins : R, almonds : A , peanuts : P, a mixture of R&P : RP
From the given information, I set up some equations a below:
R = 10*P ( the no. of bags contains only raisins is ten times the no. of bags contains only peanuts)
A = 20*RP (RP is different from R*P, it's the no. of bags that contain only a mixture of raisins and peanuts)
P = (1/5)*A
A + Axx + Ax = 210 ( Axx is the no. of bags that contain the mixture of almonds, raisins and peanuts - if any. Ax is the no. of bags that contain the mixture of almonds with raisins or with peanuts - if any)
As we already have total no. of bags : 435 = R+P+A + RP + Axx ( or ARP) + Ax ( AR or AP)
So 435 = R+P+RP + 210
Hence R+P+RP = 225 (1)
Now we convert all elements in equation (1) into P
R = 10*P
A = 20*RP, it means RP = (1/20) * A (I)
while P = (1/5) * A ( as above), then A = 5*P (II)
From (I) & (II) RP = (1/20) * A = (1/20) * 5*P = (1/4)*P
Now equation (1) becomes : 10*P + P + (1/4)*P = 225 or 11.25 * P = 225
P = 20
R = 10*P = 200
A = P*5 = 100 (from equation (II))
Total no of bags contain only one kind of them: 200+100+20 = 320, or D
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