mikeCoolBoy wrote:IMO D
I used a Venn diagram to solve the problem.
Let's say
A = number of bags that only contain almonds
P = number of bags that only contain peanuts
R = number of bags that only contain raisins
AP= number of bags that contain almonds and peanuts
....
and so on
A + P + R + AP + AR + PR + APR = 435 (1)
We are told that 210 bags contain almonds so
A + AP + AR + APR = 210 (2)
substituting (2) in (1)
210 + P + R + PR = 435 --> P + R + PR = 225 (3)
The number of bags that contain only raisins is 10 times the number of bags that contain only peanuts
R = 10P (4)
The number of bags that contain only almonds is 20 times the number of bags that contain only raisins and peanuts
A = 20 PR (5)
PR = A/20
The number of bags that contain only peanuts is one-fifth the number of bags that contain only almonds
P = 1/5 A (6) substituting (6) in (4)
R = 2A
P + R + PR = 225 --> 1/5A + 2A + A/20 = 225 --> A = 100
R = 200
P = 20
P + R + A = 320 answer D
This problem is intimidating at face value and would probably force one to guess on the gmat with nerve and all that stuff but you've made it look like cake walk. The real problem is figuring this relationship: A + AP + AR + APR = 210. Everything else is given away ie there are too many equations relating A to R P to A etc. You can now solve simple equations and do division. The trick is to see that the word ONLY is frequent in the statement of the passge but in the 210 language there is no ONLY. This means those bags can contain others and that thought might lead you to decipher all the possibilities leading to the missing equation above. Good job Mike.
