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Can you please explain?

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by winniethepooh » Wed Jul 06, 2011 10:48 am
A total of 435 bags contain 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?
1. 240
4. 275
3. 316
2. 320
5. Cannot be determined.
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by Frankenstein » Wed Jul 06, 2011 11:59 am
Hi,
This can be solved by using venn diagrams. I cannot draw and show it here. But, I will just show the equations.
If 'x' denotes the part of only raisins and peanuts,
only raisins will be 40x
only almonds will be 20x
only peanuts will be 4x
Consider Almonds -> 20x + C = 210
Consider the union of all -> 40x+20x+4x+x+C = 435
Subtracting one equation from the other gives 45x = 225 =>x = 5
So, number of bags of only one kind of item is 40x+20x+4x = 64x = 64*5 = 320

Hence, 2
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by GMATGuruNY » Wed Jul 06, 2011 2:18 pm
winniethepooh wrote:A total of 435 bags contain 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. 240
B. 275
C. 316
D. 320
E. Cannot be determined.
I solved this quickly by guessing and checking.

Let P = only peanuts, R = only raisins, A = only almonds, RP = both raisins and peanuts but not almonds.

We are given the following relationships:
R = 10P.
P = (1/5)A.
A = 20(RP)
Since Total A=210, we know that R+P+RP = 435-210 = 225.

Since A = 20(RP), we know that RP must be small.
Given all the numbers in the problem, RP is probably a multiple of 5.

Let RP = 10.
Then A = 20(10) = 200.
P = (1/5)200 = 40.
R = 10*40 = 400.
Way too big.

Let RP = 5.
Then A = 20*5 = 100.
P = (1/5)*100 = 20.
R = 10*20 = 200.
R+P+RP = 200+20+5 = 225.
Success!

Total bags with only 1 item = A+R+P = 100+20+200 = 320.

The correct answer is D.
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by winniethepooh » Wed Jul 06, 2011 2:42 pm
Mitch, what pushed you to assume that RP is a multiple of 5?
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by GMATGuruNY » Wed Jul 06, 2011 3:06 pm
winniethepooh wrote:Mitch, what pushed you to assume that RP is a multiple of 5?
The numbers in the problem (435, 10, 20, 1/5, 210) suggest that the correct values for A, R, P and RP will be multiples of 5.
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by nh_cryptic » Fri Nov 18, 2011 10:49 am
hey Frankenstein

can u explain please how do u get it....

If 'x' denotes the part of only raisins and peanuts,
only raisins will be 40x
only almonds will be 20x
only peanuts will be 4x
Consider Almonds -> 20x + C = 210
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by iwillsurvive101 » Fri Nov 18, 2011 12:21 pm
Hi, What Level of problem would this be?
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