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# Each of 435 bags contains at least one of the following

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Manager
Joined: 14 Dec 2004
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Each of 435 bags contains at least one of the following [#permalink]

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04 Apr 2005, 11:01
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.

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Senior Manager
Joined: 19 Feb 2005
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Location: Milan Italy

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04 Apr 2005, 13:13
In my opinion, D=320
that was really tough
I reasoned with ratios
I came up with these ratios:
only P=4
only R=40
only A=20
R and P=1
435(total)-210(A+A/r+a/p)=225
225/(p+r+PandR)=5
now P=5*4; R=5*40; R and P=5*1
so A=5P(given by the stem) -> A=5*4*5=100
total A+r+p=100+200+20=320

sorry if my explanation sucks

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Manager
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04 Apr 2005, 15:35
This one took me a while to compute. I hope this is considered long for the actual gmat...

I approached the solution via venn diagrams.
Assume we have three circles, one for raisins, almonds, and peanuts.
Let x=amount for peanuts only region
Let 10x=amount for raisins only region
Let 'a' be the region common to raisins and peanuts only.
Let b be the region common to raisins and almonds only.
Let d be the region common to peanuts and almonds only.
Let c be the region common to all (The center of the diagram)
Then:
The region for almonds only = 20a ( From statement in problem)
Also, x=.2(20a) (From statement in problem)
Also, 5x+b+c+d=210 (From statement in problem)
Finally:

10x+210-5x+(x/4)+6x=435
Solve for x, x=20:

Bags with only 1 type:
10x+x+5x =

320

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SVP
Joined: 03 Jan 2005
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05 Apr 2005, 07:56
435-210=225 non almonds
10P+P+PR=225
A=20PR=5P
=>10P+P+1/4P=225
P=20
A=5P=100
R=10P=200
P+R+A=320
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keep on seeking, and you will find;
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Manager
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17 Aug 2005, 07:39
Honghu,

You make it look so simple.....When I put all my variables down, I totally got lost in them, how did you know you had to solve for P only first? How come you did not right away solve A=20RP, when A=210 then RP=10.5(which may not be right!)

Any alternate explanations?This totally stumped me!
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SVP
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17 Aug 2005, 20:05
Basically you just need to first translate all the words into algebra expressions and equations, then you try solve for them. When you see 10P+P+PR=225 you know if you have the relationship between P and PR you'd be able to solve for P, so you look at the equations you have written down and you find that you did have that relationship, so you use it. There may be many ways to solve for a problem, but the basic is the same and you just need to find the easiest route to go so you can save time.
_________________

Keep on asking, and it will be given you;
keep on seeking, and you will find;
keep on knocking, and it will be opened to you.

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Senior Manager
Joined: 29 Jun 2005
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24 Aug 2005, 10:41
cloaked_vessel wrote:
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.

Please, explain, how can it be?

oa = 20(or+op)
op = 1/5oa

oa=20or + 4a -> -3oa=20or.

Isn't it illogical?

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Manager
Joined: 09 Nov 2004
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03 Sep 2005, 06:44
HongHu wrote:
Basically you just need to first translate all the words into algebra expressions and equations, then you try solve for them. When you see 10P+P+PR=225 you know if you have the relationship between P and PR you'd be able to solve for P, so you look at the equations you have written down and you find that you did have that relationship, so you use it. There may be many ways to solve for a problem, but the basic is the same and you just need to find the easiest route to go so you can save time.

I'm back to this problem and confused again....

Is my starting =on going to be:

P+R+A+PR=435 right?
and at the point where I have A=20PR=5P , If I substiture A=210 then PR=210/20 which doesn't make sense.......How do you keep yourself from getting thrown off by equation results like these?
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03 Sep 2005, 06:44
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# Each of 435 bags contains at least one of the following

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