Each of 435 bags contains at least one of the following

The value of almonds from the venn diagram. 20y=5x. Also the explanation leading to the target. Thanks.

The number of bags that contain only almonds is 20 times the number of bags that contain only raisins and peanuts --> {only raisins and peanuts} = y --> {only almonds}=20y;

The number of bags that contain only peanuts is one-fifth the number of bags that contain only almonds --> {only peanuts}=x --> {only peanuts}={only almonds}/5 --> x={only almonds}/5 --> {only almonds}=5x.

{only almonds}=5x=20y --> y=x/4.

Total=435={Almonds}+10x+y+x --> 435=210+10x+x/4+x

What else?

Bunuel,

I would really appreciate your help. When I read this: "The number of bags that contain only almonds is 20 times the number of bags that contain only raisins and peanuts. ". I thought that the number of bags that only contain almonds is equal to the number of bags that contain raisins and the number of bags that only contain peanuts. So I said A=20*(10x+x)=20*(11x). Rather than what you had which is the 20y, where y is the number of bags that are a mixtures of both raisins and peanuts.

This is more of an english question but how is "only" used in word problems on the GMAT. When I get question with an "only A & B" is this equal to ("only" A) & ("only" B). Or should I interpret it as "only" (A&B)

Interpretation usually depends on the context.
'Bag containing only A and B' means the bag has A and B only, not say C.
If the question wants to imply only A and only B, it will clarify as such. The bag containing only A and the bag containing only B.

Note that here you have assumed that the statement implies this:
The number of bags that contain only almonds is 20 times the sum of the number of bags that contain only raisins and the number of bags that contain only peanuts.
If number of bags of almonds is 20 times the sum of 2 other numbers, they will specify it as such.
In the question they only give number of bags of almonds is 20 times the number of bags of only raisins and peanuts (which means that the intersection of all 3 in not included). So one number of bags is equal to another number of bags, not sum of two other numbers of bags.

Given:
R=10P
R/P=10/1

P=1/5A
A/P=5/1

R:P:A = 10:1:5

10x+x+5x=16x; answer should be divisible by 16.

A = 5x = 20(R+P) = 20y
5x=20y
y=1/4x

435 = Almonds + R + P + (R+P); 210 + 10x + x + 1/4x
225=45/4x
x=20

R=10x=200, A=5x=100,P=x=20
R+P+A=320 (ANSWER)

Answer: Option D

Please find attached the explanation

If you have no idea how to start setting up a solution for a problem, it's a worthwhile skill to develop to just brute force your way to a solution.

Let's try numbers. "Only peanuts" appears twice in our rules, so let's make something up for that.

Let's try 50. 50 only peanuts.
"The number of bags that contain only peanuts is one-fifth the number of bags that contain only almonds." So, 250 only almonds. Oops, that's too high: "210 bags contain almonds." We need to start with something smaller than 50 only peanuts.

Let's try 25. 25 only peanuts.
"The number of bags that contain only peanuts is one-fifth the number of bags that contain only almonds." So, 125 only almonds.
"The number of bags that contain only raisins is 10 times the number of bags that contain only peanuts." So, 250 only raisins.
"The number of bags that contain only almonds is 20 times the number of bags that contain only raisins and peanuts." So, 6.25 only raisins and peanuts.
We've accounted for 406.25 and there are 435 total, so we're missing 28.75.
We've yet to account for "only peanuts and almonds," "only raisins and almonds," and "all three." Each of those includes almonds. If we add 28.75 to 125, we get 153.75. We need 210. We needed more than 28.75, so we need lower the other things.

Let's try 20. 20 only peanuts.
"The number of bags that contain only peanuts is one-fifth the number of bags that contain only almonds." So, 100 only almonds.
"The number of bags that contain only raisins is 10 times the number of bags that contain only peanuts." So, 200 only raisins.
"The number of bags that contain only almonds is 20 times the number of bags that contain only raisins and peanuts." So, 5 only raisins and peanuts.
We've accounted for 325 and there are 435 total, so we're missing 110.
We've yet to account for "only peanuts and almonds," "only raisins and almonds," and "all three." Each of those includes almonds. If we add 110 to 100, we get 210. We need 210. Yay!

We are asked for the number of bags that contain only one item. 20 only peanuts + 100 only almonds + 200 only raisins = 320.

Answer choice D.

Hi Asifpirlo,

I like the approach, but there are 2 answer choices (256 and 320) that are divisible by 16. How do we determine which one among these is the answer?

Responding to a pm:

This is what the Venn diagram gives me:



I get that x = 4y

Now I write all x's in terms of y and add them up to 435.

435 = 40y + y + 4y + 210 (entire Almond circle)
y = 5

No of bags containing only 1 item = 40y + 4y + 20y = 64y = 64*5 = 320

Answer (D)

Check this post for discussion on how to use Venn diagrams in 3 overlapping sets: https://anaprep.com/sets-statistics-thr ... ping-sets/

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