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210 bags contain almonds

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210 bags contain almonds

by sanju09 » Wed Feb 25, 2009 5:46 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
C. 316
D. 320
E. It cannot be determined from the given information.
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by BuckeyeT » Wed Feb 25, 2009 6:04 am
D. 320
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by bluementor » Wed Feb 25, 2009 6:05 am
Raisins only = Ro
Peanuts only = Po
Almonds only = Ao
Raisins and peanuts only = RP

Lets say Ao = x, then from the info within the question:
Po = (1/5)*x
RP = x/20
Ro = 10*Po = 2x

The three-set equation for this problem is as follows:

Total = Total_almonds + Ro + Po + RP
435 = 210 + 2x + x/5 + x/20
9x/4 = 225
x = 100

Therefore, bags that contain only one item = Ao + Po + Ro = x + x/5 + 2x = 3.2*(100) = 320.

Choose D.

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by BuckeyeT » Wed Feb 25, 2009 6:13 am
Explained answer...

Let r = raison only, a = almond only, p = peanut only
ra = raison/almond only, rp = raison/peanut only, ap = almond/peanut only, rap = raison/almond/peanut

r + a + p + ra + rp + ap + rap = 435
a + ra + ap + rap = 210
r = 10p
a = 20(rp)
p = 1/5 a

Combining the first two statements above, we get
r + a + p + ra + rp + ap + rap = 435
a + ra + ap + rap = 210
=r + p + rp = 225
=r + p + rp = 225

These can all be solved in terms of a from the third, fourth, and fifth statements above.

10(1/5 a) + 1/5 a + 1/20 a = 225
2a + 1/5 a + 1/20 a = 225
40/20 a + 4/20 a + 1/20 a = 225
45/20 a = 225
a = 100.

Solve for r and p.
p = 1/5 a
p = 1/5 (100)
p = 20.

r = 10 p
r = 10 (20)
r = 200.

a + r + p = 100 + 200 + 20 = 320.
D.
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by deepoe » Wed Feb 25, 2009 6:13 am
bluementor wrote:Raisins only = Ro
Peanuts only = Po
Almonds only = Ao
Raisins and peanuts only = RP

Lets say Ao = x, then from the info within the question:
Po = (1/5)*x
RP = x/20
Ro = 10*Po = 2x

The three-set equation for this problem is as follows:

Total = Total_almonds + Ro + Po + RP
435 = 210 + 2x + x/5 + x/20
9x/4 = 225
x = 100

Therefore, bags that contain only one item = Ao + Po + Ro = x + x/5 + 2x = 3.2*(100) = 320.

Choose D.

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I don't understand what you did here :oops: could you explain me?
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by deepoe » Wed Feb 25, 2009 6:21 am
Ah nevermind the explanation of BuckeyeT is clear to me:D
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by bluementor » Wed Feb 25, 2009 6:52 am
deepoe wrote:I don't understand what you did here :oops: could you explain me?
First I define the variables that I will use in this problem:
Raisins only = Ro
Peanuts only = Po
Almonds only = Ao
Raisins and peanuts only = RP

From the question, the relationship between the variables are given. First I assume Ao to be x. Then, I use this as the basis for the relationship to the other variables.

Info 1: The number of bags that contain only peanuts is one-fifth the number of bags that contain only almonds
Po = (1/5)*Ao
Po = (1/5)*x

Info 2: The number of bags that contain only almonds is 20 times the number of bags that contain only raisins and peanuts

Ao = 20*RP
x = 20*RP
RP = x/20

Info 3: The number of bags that contain only raisins is 10 times the number of bags that contain only peanuts

Ro = 10*Po
Ro = 10*(1/5)*x … (we use Info 1 to substitute Po)
Ro = 2x

Info 4: 210 bags contain almonds

210 = All the bags that contain almonds
210 = Almonds_only + (Almonds_and_Peanuts) + (Almonds_and_Raisins) + (Almonds_Peanuts_Raisins)

The diagram attached is a three-set representation of this problem. The 210 bags containing almonds is marked by the green region.

Therefore, the three set equation that can be derived from this diagram is:

Total = Total_almonds_(green region) + Ro + Po + RP
435 = 210 + 2x + x/5 + x/20
9x/4 = 225
x = 100

Therefore, bags that contain only one item = Ao + Po + Ro = x + x/5 + 2x = 3.2*(100) = 320.

Did this help? If not, let me know which part bothers you.

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by sureshbala » Wed Feb 25, 2009 7:27 am
Folks, believe me I am really shocked to see many equations here....
Anyway, here is my solution which is quite quick enough I guess...


Image
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Re: 210 bags contain almonds

by x2suresh » Wed Feb 25, 2009 7:35 am
sanju09 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.



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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by deepoe » Wed Feb 25, 2009 1:42 pm
bluementor how did you calculate the 9x?
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by bluementor » Wed Feb 25, 2009 4:09 pm
deepoe wrote:bluementor how did you calculate the 9x?
Total = Total_almonds + Ro + Po + RP
435 = 210 + 2x + x/5 + x/20
435 - 210 = x(2 + 1/5 + 1/20)
225 = x(2 + 4/20 + 1/20)
225 = x(2 + 5/20)
225 =x(2 + 1/4)
225 = x(9/4)
x = 100

hth

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by deepoe » Thu Feb 26, 2009 12:39 am
bluementor wrote:
deepoe wrote:bluementor how did you calculate the 9x?
Total = Total_almonds + Ro + Po + RP
435 = 210 + 2x + x/5 + x/20
435 - 210 = x(2 + 1/5 + 1/20)
225 = x(2 + 4/20 + 1/20)
225 = x(2 + 5/20)
225 =x(2 + 1/4)
225 = x(9/4)
x = 100

hth

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Thanks mate!
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by kamu » Tue Mar 03, 2009 2:54 am
sureshbala wrote:Folks, believe me I am really shocked to see many equations here....
Anyway, here is my solution which is quite quick enough I guess...


Image
Excellent.
Thanks.

Please could you explain how to use Venn Diagrams in different circumstances.
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by ironsferri » Fri Jun 04, 2010 12:27 pm
what is the difficulty level of this question????
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