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Sets - good one

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Manager
Manager
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Status: Berkeley Haas 2013
Joined: 23 Jul 2009
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Kudos [?]: 31 [0], given: 16

Sets - good one [#permalink] New post 19 Aug 2009, 12:15
00:00
A
B
C
D
E

Difficulty:

  5% (low)

Question Stats:

50% (04:31) correct 50% (05:18) wrong based on 2 sessions
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.
Manager
Manager
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Joined: 14 Aug 2009
Posts: 123
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Kudos [?]: 89 [0], given: 13

Re: Sets - good one [#permalink] New post 19 Aug 2009, 20:55
Answer is D. 320

suppose the bags contain only raisins and peanuts are x, therefore,
bags only have almonds are 20x,
bags only have peanuts are 4x,
bags only have raisins are 40x,

suppose the bags that have almonds in it and have more than one item are t,
therefore t+20x=210

then we have this: 40x+x+4x+20x+t=435

x=5

so: the bags have only one item in it are: 40x+4x+20x=64x=320
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Manager
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Joined: 10 Jul 2009
Posts: 131
Location: Ukraine, Kyiv
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Kudos [?]: 27 [0], given: 60

Re: Sets - good one [#permalink] New post 20 Sep 2009, 04:14
Did it using Venn diagram principle.

we have:
only raisins = 10x
only peanuts = x
both peanuts and raisins = y
only almonds = 20y
x=1/5*20y
almonds=210

find:
x and y

solution:
435-10x-y-x=210
-11x-y=-225
-45y=-225

answer:
y=5
x=20

almonds=100
peanuts=20
raisins=200
total=320

D
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Manager
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Re: Sets - good one [#permalink] New post 20 Sep 2009, 04:27
I wud go with option D)320
Manager
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Joined: 04 Sep 2009
Posts: 53
WE 1: Real estate investment consulting
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Kudos [?]: 16 [0], given: 7

Re: Sets - good one [#permalink] New post 20 Sep 2009, 05:27
Got there via algebra:

Raisins (R) = 10 Peanuts (P)
Almonds (A) = 20 Raisins&Peanuts (RP)
P = 1/5 A

1) R + P + A + AR + AP + RP + ARP = 435
2) A + AP + AR + ARP = 210

1) - 2)
R + P + RP = 225
Easiest to describe via almonds:
2A + 1/5A + A/20 = 225
45A/20 = 225
A=100
P = 20
R = 200

Total = 320
Re: Sets - good one   [#permalink] 20 Sep 2009, 05:27
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