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hareesh860: Senior | Next Rank: 100 Posts; Posts: 31; Joined: Fri Oct 01, 2010 8:27 am; Thanked: 2 times

by hareesh860 » Tue Feb 07, 2012 2:38 am

Each of 435 bags contain at least one of the following 3 items: raisins,almonds and peanuts. The no of bags that contain only raisins is ten times the no of bags that contain only peanuts. The no of bags that contain only almonds is 20 times the no of bags that contain only raisins and peanuts.The no of bags that contain only peanuts is one fifth the no 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)350

Last edited by hareesh860 on Sat Feb 11, 2012 7:02 pm, edited 1 time in total.

Source: Beat The GMAT — Problem Solving | Tue Feb 07, 2012 2:38 am

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Anurag@Gurome: GMAT Instructor; Posts: 3835; Joined: Fri Apr 02, 2010 10:00 pmFollowed by: 523 members; Location: Milpitas, CA; Thanked: 1854 times; GMAT Score:770

by Anurag@Gurome » Tue Feb 07, 2012 4:11 am

hareesh860 wrote:Each of 435 bags contain at least one of the following 3 items: raisins,almonds and peanuts. The no of bags that contain only raisins is ten times the no of bags that contain only peanuts. The no of bags that contain only almonds is 20 times the no of bags that contain only raisins and peanuts.The no of bags that contain only peanuts is one fifth the no of bags that contain only almonds. 210 bags contain only almonds. how many bags contain only one kind of item?

a)256 b)260 c)316 d)320 e)350

Let the bags that contain only peanuts = P
Then # of bags that contain only raisins = 10P
Let us assume that the # of bags that contain only raisins and peanuts = N
Then # of bags that contain only almonds = 20N
# of bags that contain only peanuts = (1/5) * 20N = 4N
OR P = 4N implies P - 4N = 0 ... Equation (1)
210 bags contain only almonds.
Total # of bags = 435

From the Venn diagram,
Total bags = Bags with Almonds + 10P + N + P
435 = 210 + 11P + N
11P + N = 225 ... Equation (2)
Solving equations (1) and (2), P = 4(225 - 11P)
P = 900 - 44P
45P = 900
P = 20 and N = 225 - (11 * 20) = 5

Therefore, the bags that contain only one kind of item = P + 10P + 20N = 20 + 200 + 100 = 320

The correct answer is D.

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GMATGuruNY: GMAT Instructor; Posts: 15539; Joined: Tue May 25, 2010 12:04 pmFollowed by: 1906 members; Location: New York, NY; Thanked: 13060 times; GMAT Score:790

by GMATGuruNY » Tue Feb 07, 2012 4:36 am

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ronnie1985: Legendary Member; Posts: 626; Joined: Fri Dec 23, 2011 2:50 amFollowed by: 10 members; Location: Ahmedabad; Thanked: 31 times

by ronnie1985 » Tue Feb 07, 2012 7:07 am

using venn diagram one can easily arrive at the answer (D)

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mankey: Legendary Member; Posts: 627; Joined: Thu Jun 23, 2011 9:12 amFollowed by: 1 member; Thanked: 4 times

by mankey » Tue Feb 07, 2012 7:52 am

Dear Anurag

Please note problem in the language of the question. It says "210 bags contain ONLY almonds", whereas it should have read "210 bags contain almonds".

I am right in pointing out this error, please clarify.

Regards.

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Anurag@Gurome: GMAT Instructor; Posts: 3835; Joined: Fri Apr 02, 2010 10:00 pmFollowed by: 523 members; Location: Milpitas, CA; Thanked: 1854 times; GMAT Score:770

by Anurag@Gurome » Tue Feb 07, 2012 7:35 pm

mankey wrote:Dear Anurag

Please note problem in the language of the question. It says "210 bags contain ONLY almonds", whereas it should have read "210 bags contain almonds".

I am right in pointing out this error, please clarify.

Regards.

That's absolutely right, Mankey.

anjalimanas: Junior | Next Rank: 30 Posts; Posts: 18; Joined: Fri Apr 27, 2012 1:49 am

by anjalimanas » Thu Dec 06, 2012 2:10 am

Number of bags with peanuts only = p

Number of bags with raisin only = 10p

Let Number of bags with only raisin and peanut = n

Hence Number of bags with peanuts only = 4n

Number of bags with raisin only = 40n

Now 435 = 210 + 40n + n + 4n hence n = 5

Bags with only one item = 20n + 40n+ 4n = 64n = 320

This is easily visualized using venn diagram

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