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DRY FRUITS [#permalink] New post 09 Feb 2009, 04:55
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.

PLEASE HELP
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Re: DRY FRUITS [#permalink] New post 09 Feb 2009, 05:45
LoyalWater wrote:
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.

PLEASE HELP


R+A+P+RA+PA+RP+RAP=435 -->(1)

R=10P
A=20RP
P=1/5*A
A+RA+PA+RAP=210 -->(2)

FROM 1 AND 2

R+P+RP =435-210 = 225

10P+P+1/4*P = 225
-->P=20

R+A+P = 200+100 +20 =320
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Re: DRY FRUITS [#permalink] New post 09 Feb 2009, 06:01
D

Same explanation as that of Suresh.
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Re: DRY FRUITS [#permalink] New post 09 Feb 2009, 13:55
Let R be the number of bags containing Raisins.
Let A be the number of bags containing Almonds.
Let P be the number of bags containing Peanuts.
Let PR be the number of bags containing Peanuts and Raisins
Let OR be the number of bags containing only Raisins.
Let OA be the number of bags containing only Almonds.
Let OP be the number of bags containing only Peanuts.

From the given information we can draw the following facts.

Total no of bags containing Almonds ie A = 210.
OP = 1/5 OA --> eq 1
OA = 20 PR --> Eq 2
OR = 10 OP --> Eq 3

From Eq 1 and 2,

OP = 1/5 * 20 * (PR) = 4 * (PR) --> Eq 4

From Eq 4 and Eq 3,

OR = 10 * 4 * (PR) --> Eq 5

The question is asking for " How bags contains only one of the items"

ie OA + OP + OR = ?

Using Venn diagram Notation, we can say that

A + OP+ PR + OR = 485. (Now Substituting the values of OP, OR, PR from 4 and 5 in the below)

210 + 4(PR) + PR + 40(PR) = 485.

Solving it, we get PR = 5.

OA = 20 PR = 20 * 5 = 100
OP = 4 * PR = 4 * 5 = 20
OR = 40* O r= 40 * 5 = 200

OA + OR + OP = 320.

Edited the post.

Last edited by mrsmarthi on 09 Feb 2009, 19:06, edited 1 time in total.
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Re: DRY FRUITS [#permalink] New post 09 Feb 2009, 15:23
x2suresh wrote:
R+A+P+RA+PA+RP+RAP=435 -->(1)

R=10P
A=20RP
P=1/5*A
A+RA+PA+RAP=210 -->(2)

FROM 1 AND 2

R+P+RP =435-210 = 225

10P+P+1/4*P = 225
-->P=20

R+A+P = 200+100 +20 =320


I think you misinterpretted the questions.

There can be bags that contain only one item, cobination of any two items, or combination all three items.

So when Number of bags containing almonds = 210 means, it can be number of bags containing ONLY ALMONDS, and number of bags having a mixture of 2 or 3 items where Almonds can be one of the items.

Can you interpret the problem like this?
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Re: DRY FRUITS [#permalink] New post 09 Feb 2009, 18:16
mrsmarthi wrote:
x2suresh wrote:
R+A+P+RA+PA+RP+RAP=435 -->(1)

R=10P
A=20RP
P=1/5*A
A+RA+PA+RAP=210 -->(2)

FROM 1 AND 2

R+P+RP =435-210 = 225

10P+P+1/4*P = 225
-->P=20

R+A+P = 200+100 +20 =320


I think you misinterpretted the questions.

There can be bags that contain only one item, cobination of any two items, or combination all three items.

So when Number of bags containing almonds = 210 means, it can be number of bags containing ONLY ALMONDS, and number of bags having a mixture of 2 or 3 items where Almonds can be one of the items.

Can you interpret the problem like this?


I believe I interpreted correctly
A+RA+PA+RAP=210 -->(2)
A = No. of Bags with only almonds
RA = No. of Bags with almonds and Raisins
PA = No. of Bags with almonds and peanuts
RAP = No. of Bags with almonds,raisins and peants
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Re: DRY FRUITS [#permalink] New post 09 Feb 2009, 19:06
Oops....I was confused with the calc. Sorry. :cry:
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Re: DRY FRUITS [#permalink] New post 10 Feb 2009, 02:35
Agree with suresh.
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Re: DRY FRUITS [#permalink] New post 10 Feb 2009, 13:43
x2suresh wrote:
mrsmarthi wrote:
x2suresh wrote:
R+A+P+RA+PA+RP+RAP=435 -->(1)

R=10P
A=20RP
P=1/5*A
A+RA+PA+RAP=210 -->(2)

FROM 1 AND 2

R+P+RP =435-210 = 225

10P+P+1/4*P = 225
-->P=20

R+A+P = 200+100 +20 =320


I think you misinterpretted the questions.

There can be bags that contain only one item, cobination of any two items, or combination all three items.

So when Number of bags containing almonds = 210 means, it can be number of bags containing ONLY ALMONDS, and number of bags having a mixture of 2 or 3 items where Almonds can be one of the items.

Can you interpret the problem like this?


I believe I interpreted correctly
A+RA+PA+RAP=210 -->(2)
A = No. of Bags with only almonds
RA = No. of Bags with almonds and Raisins
PA = No. of Bags with almonds and peanuts
RAP = No. of Bags with almonds,raisins and peants

Isn't it just A=20(R+P), thats the way I read it
Re: DRY FRUITS   [#permalink] 10 Feb 2009, 13:43
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