It is currently 25 Jun 2017, 16:08

### GMAT Club Daily Prep

#### Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.

Customized
for You

we will pick new questions that match your level based on your Timer History

Track

every week, we’ll send you an estimated GMAT score based on your performance

Practice
Pays

we will pick new questions that match your level based on your Timer History

# Events & Promotions

###### Events & Promotions in June
Open Detailed Calendar

# Sets - good one

Author Message
Manager
Status: Berkeley Haas 2013
Joined: 23 Jul 2009
Posts: 191

### Show Tags

19 Aug 2009, 13:15
00:00

Difficulty:

(N/A)

Question Stats:

50% (04:31) correct 50% (05:18) wrong based on 3 sessions

### HideShow timer Statistics

This topic is locked. If you want to discuss this question please re-post it in the respective forum.

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
Joined: 14 Aug 2009
Posts: 123
Re: Sets - good one [#permalink]

### Show Tags

19 Aug 2009, 21:55

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
_________________

Kudos me if my reply helps!

Manager
Joined: 10 Jul 2009
Posts: 126
Location: Ukraine, Kyiv
Re: Sets - good one [#permalink]

### Show Tags

20 Sep 2009, 05:14
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

y=5
x=20

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

D
_________________

Never, never, never give up

Manager
Joined: 15 Sep 2009
Posts: 134
Re: Sets - good one [#permalink]

### Show Tags

20 Sep 2009, 05:27
I wud go with option D)320
Manager
Joined: 04 Sep 2009
Posts: 53
WE 1: Real estate investment consulting
Re: Sets - good one [#permalink]

### Show Tags

20 Sep 2009, 06:27
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, 06:27
Display posts from previous: Sort by