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Mixture Problem (Trail Mix)

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Mixture Problem (Trail Mix) [#permalink] New post 20 Jul 2011, 21:53
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Difficulty:

  5% (low)

Question Stats:

70% (02:30) correct 30% (01:42) wrong based on 20 sessions
A bowl of nuts is prepared for a party. Brand P mixed nuts are 20% almonds and Brand Q's Deluxe nuts are 25% almonds. If a bowl contains a total of 64 ounces of nuts, representing a mixture of both brands, and 15 ounces of the mixture are almonds, how many ounces of Brand Q's Deluxe mixed nuts are used?

(A) 16
(B) 20
(C) 32
(D) 44
(E) 48




[Reveal] Spoiler:
D



Does anyone have a good way of solving for this using allegation or a ratio formula?
Thanks
[Reveal] Spoiler: OA
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Re: Mixture Problem (Trail Mix) [#permalink] New post 20 Jul 2011, 22:32
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milanproda wrote:
A bowl of nuts is prepared for a party. Brand P mixed nuts are 20% almonds and Brand Q's Deluxe nuts are 25% almonds. If a bowl contains a total of 64 ounces of nuts, representing a mixture of both brands, and 15 ounces of the mixture are almonds, how many ounces of Brand Q's Deluxe mixed nuts are used?

(A) 16
(B) 20
(C) 32
(D) 44
(E) 48




[Reveal] Spoiler:
D



Does anyone have a good way of solving for this using allegation or a ratio formula?
Thanks


let X = Brand P nuts
Y = Brand Q nuts

X + Y = 64.........................................1
0.2 X + 0.25 Y = 15...........................2

solve for X
X = 20
Y = 44
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Re: Mixture Problem (Trail Mix) [#permalink] New post 21 Jul 2011, 00:48
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sudhir18n wrote:
milanproda wrote:
A bowl of nuts is prepared for a party. Brand P mixed nuts are 20% almonds and Brand Q's Deluxe nuts are 25% almonds. If a bowl contains a total of 64 ounces of nuts, representing a mixture of both brands, and 15 ounces of the mixture are almonds, how many ounces of Brand Q's Deluxe mixed nuts are used?

(A) 16
(B) 20
(C) 32
(D) 44
(E) 48




[Reveal] Spoiler:
D



Does anyone have a good way of solving for this using allegation or a ratio formula?
Thanks


let X = Brand P nuts
Y = Brand Q nuts

X + Y = 64.........................................1
0.2 X + 0.25 Y = 15...........................2

solve for X
X = 20
Y = 44

Yup.. U can't get a better and easier way for solving this.. :)
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Varun


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Re: Mixture Problem (Trail Mix) [#permalink] New post 21 Jul 2011, 01:54
Yea you are right unfortunately I am not the sharpest tool in the shed =D.

It makes perfect sense now, but I think I would have trouble recognizing that during the test. There is no single formula that you can use for these mixture problems, it sucks.

Thanks for the help though
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Re: Mixture Problem (Trail Mix) [#permalink] New post 15 Sep 2011, 02:13
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almonds in P = 1/5 ---> 64/320
almonds in Q = 1/4 ---> 80/320
almonds in mixture = 15/64 ---> 75/320 [to make calculations simple, i've taken lcm of the denominator]

ratio = \frac{80-75}{75-64} = \frac{5}{11}

because Q 'pulled' the avg towards itself, so Q:P = 11:5
11x+5x = 64 ---> x=4

weight of Q = 11*4 = 44
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Re: Mixture Problem (Trail Mix) [#permalink] New post 15 Sep 2011, 19:46
MBAhereIcome wrote:
almonds in P = 1/5 ---> 64/320
almonds in Q = 1/4 ---> 80/320
almonds in mixture = 15/64 ---> 75/320 [to make calculations simple, i've taken lcm of the denominator]

ratio = \frac{80-75}{75-64} = \frac{5}{11}

because Q 'pulled' the avg towards itself, so Q:P = 11:5
11x+5x = 64 ---> x=4

weight of Q = 11*4 = 44


The explanation given by MBAhereIcome is great. But I have two queries.

what is that formula used above to calculate the ratio? Can the formula above tell that the ratio is either Q:P or P:Q?
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Re: Mixture Problem (Trail Mix) [#permalink] New post 16 Sep 2011, 07:26
chawlavinu wrote:
MBAhereIcome wrote:
almonds in P = 1/5 ---> 64/320
almonds in Q = 1/4 ---> 80/320
almonds in mixture = 15/64 ---> 75/320 [to make calculations simple, i've taken lcm of the denominator]

ratio = \frac{80-75}{75-64} = \frac{5}{11}

because Q 'pulled' the avg towards itself, so Q:P = 11:5
11x+5x = 64 ---> x=4

weight of Q = 11*4 = 44


The explanation given by MBAhereIcome is great. But I have two queries.

what is that formula used above to calculate the ratio? Can the formula above tell that the ratio is either Q:P or P:Q?


yes, the formula is to calculate the ratio of P and Q. look at the question for a bit logically. the end mixture is 75 almonds, a ratio closer to Q than to P. so the quantity of Q MUST be higher in the end mixture. the end fraction is 5/11, so 11 must be the ratio of Q, then. i hope it's clear.
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Re: Mixture Problem (Trail Mix) [#permalink] New post 16 Sep 2011, 21:31
lets say x ounces of P is mixed with Q.

=> 64-x ounces of Q is present in the mixture (as the total = 64 ounces)

given total almond weight = 15 ounces

(20x/100)+(25/100)(64-x) = 15

=> x = 20

=> 64-20 = 44 ounces of Q is present in the mixture.

Answer is D.
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Re: Mixture Problem (Trail Mix) [#permalink] New post 17 Sep 2011, 01:45
x+y=64
0.2x+0.25y=15

y=44
OA is D
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Re: Mixture Problem (Trail Mix) [#permalink] New post 17 Sep 2011, 01:53
MBAhereIcome wrote:
almonds in P = 1/5 ---> 64/320

because Q 'pulled' the avg towards itself, so Q:P = 11:5
11x+5x = 64 ---> x=4

weight of Q = 11*4 = 44


we can also think in this way

since we know Q:P = 11:5 then Q=11/16*total=11*64/16 =44 (16=11+5)
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Re: Mixture Problem (Trail Mix) [#permalink] New post 17 Sep 2011, 02:03
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one more method (not sure that it is the good one ;)) )
we have 15 ounces of almond in total 64 ,then -
15/64 is approx 23%

so,

25% 20%
23%
------------------------------
3%(23-20) 2%(25-23)

2%/5%= 40%
since we have rounded 23% ,so 40% is also approx of the right answ. since 40% is near 44, not 48%, then the right answ is 44%
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Re: Mixture Problem (Trail Mix) [#permalink] New post 17 Sep 2011, 04:17
milanproda wrote:
Yea you are right unfortunately I am not the sharpest tool in the shed =D.

It makes perfect sense now, but I think I would have trouble recognizing that during the test. There is no single formula that you can use for these mixture problems, it sucks.

Thanks for the help though



try picking no from the answer stem if you are good in it..
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Re: Mixture Problem (Trail Mix) [#permalink] New post 17 Sep 2011, 23:41
Its 44.
Almonds: Nuts
p 1 : 5
q 1 : 4

Mix 15: 64
Since nuts of mixture are divisible by q . lets suppose only q is present in the mixture in that case we have ratio
16:64. Now since there is one more almond than required we remove the lcm of (4,5) = 20 nuts of q from the mixture and add 20 nuts of p into the mixture.
q 11 : 44
p 4 : 20
=> Mix 15:64
Hence the number of nuts of q is 44.
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CR notes
http://gmatclub.com/forum/massive-collection-of-verbal-questions-sc-rc-and-cr-106195.html#p832142
http://gmatclub.com/forum/1001-ds-questions-file-106193.html#p832133
http://gmatclub.com/forum/gmat-prep-critical-reasoning-collection-106783.html
http://gmatclub.com/forum/how-to-get-6-0-awa-my-guide-64327.html
http://gmatclub.com/forum/how-to-get-6-0-awa-my-guide-64327.html?hilit=chineseburned

Re: Mixture Problem (Trail Mix)   [#permalink] 17 Sep 2011, 23:41
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