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How to proceed - Bag Problem

by Uva@90 » Tue Jul 16, 2013 1:36 am
Hi All,
Please help me in solving this problem,

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.

Answer : D

Thanks in advance.

Regards,
Uva.
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by GMATGuruNY » Tue Jul 16, 2013 3:58 am
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 C316 D320 E350
Let R = only raisins, P = only peanuts, , RP = raisins and peanuts, A = only almonds.

T = 435.
The total number of bags with almonds = 210.
Thus, all of the OTHER bags contain raisins, peanuts, or both:
R+P+RP = 435-210 = 225.

The number of bags that contain only raisins is 10 times the number of bags that contain only peanuts:
R : P = 10:1.

The number of bags that contain only peanuts is one-fifth the number of bags that contain only almonds:
P = (1/5)A, implying that P:A = 1:5.

Combining R : P = 10:1 and P:A = 1:5, we get:
R : P : A = 10:1:5.

The number of bags that contain only almonds is 20 times the number of bags that contain only raisins and peanuts:
Thus, the value of A must be a multiple of 20:
R : P : A = 4(10:1:5) = 40:4:20.

Since the sum of the parts of this ratio = 40+4+20 = 64, the value of R+P+A must be a MULTIPLE OF 64.

Of the answer choices, only A and D are multiples of 64.
Since all of the values in the problem (435, 210, 225) are multiples of 5, the correct answer is also almost certainly a multiple of 5.

Answer choice D: R+P+A = 320
Since 320/64 = 5, the multiplier for R : P : A = 40:4:20 is 5:
R : P : A = 5(40:4:20) = 200:20:100.
Here, R=200, P=20, A=100.

Since R+P+RP = 225, and in the ratio above R+P = 220, RP = 5.
If D is the correct answer, the value of A will be 20 times the value of RP:
A/RP = 100/5 = 20.
Success!

The correct answer is D.
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by Uva@90 » Tue Jul 16, 2013 7:35 pm
GMATGuruNY 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 C316 D320 E350
Let R = only raisins, P = only peanuts, , RP = raisins and peanuts, A = only almonds.

T = 435.
The total number of bags with almonds = 210.
Thus, all of the OTHER bags contain raisins, peanuts, or both:
R+P+RP = 435-210 = 225.

The number of bags that contain only raisins is 10 times the number of bags that contain only peanuts:
R : P = 10:1.

The number of bags that contain only peanuts is one-fifth the number of bags that contain only almonds:
P = (1/5)A, implying that P:A = 1:5.

Combining R : P = 10:1 and P:A = 1:5, we get:
R : P : A = 10:1:5.

The number of bags that contain only almonds is 20 times the number of bags that contain only raisins and peanuts:
Thus, the value of A must be a multiple of 20:
R : P : A = 4(10:1:5) = 40:4:20.

Since the sum of the parts of this ratio = 40+4+20 = 64, the value of R+P+A must be a MULTIPLE OF 64.

Of the answer choices, only A and D are multiples of 64.
Since all of the values in the problem (435, 210, 225) are multiples of 5, the correct answer is also almost certainly a multiple of 5.

Answer choice D: R+P+A = 320
Since 320/64 = 5, the multiplier for R : P : A = 40:4:20 is 5:
R : P : A = 5(40:4:20) = 200:20:100.
Here, R=200, P=20, A=100.

Since R+P+RP = 225, and in the ratio above R+P = 220, RP = 5.
If D is the correct answer, the value of A will be 20 times the value of RP:
A/RP = 100/5 = 20.
Success!

The correct answer is D.
GMATGuruNY: Great Explanation.
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by satish_iitg » Fri Jul 19, 2013 7:56 am
Is it possible to solve this by Pie chart method ??
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