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GRE Math Challenge #89-A health food store prepares
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09 May 2015, 11:36
09 May 2015, 11:36
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Question Stats:
98% (00:58) correct
1% (00:47) wrong based on 55 sessions
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A health food store prepares a breakfast food that consists of oats, raisins, and nuts mixed in the ratio 9:2:1, respectively, by weight. If the nuts in the mixture weigh 9.2 pounds, how many pounds does the total mixture weigh?
(A) 82.2
(B) 92.2
(C) 101. 2
(D) 110.4
(E) 165.6
(A) 82.2
(B) 92.2
(C) 101. 2
(D) 110.4
(E) 165.6
Re: GRE Math Challenge #89-A health food store prepares
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18 Dec 2017, 06:23
18 Dec 2017, 06:23
1
sandy wrote:
A health food store prepares a breakfast food that consists of oats, raisins, and nuts mixed in the ratio 9:2:1, respectively, by weight. If the nuts in the mixture weigh 9.2 pounds, how many pounds does the total mixture weigh?
(A) 82.2
(B) 92.2
(C) 101. 2
(D) 110.4
(E) 165.6
(A) 82.2
(B) 92.2
(C) 101. 2
(D) 110.4
(E) 165.6
Here the weight of Nuts = 9.2 pounds and we need to find out the total weight of the mixture.
Let the total weight of mixture be = T
since oats:raisins : nuts = 9:2:1
i.e 9 parts of Total (T)+ 2 parts of Total (T) + 1 parts of total (T) = 12 T
Part/ Whole = \(\frac{Nuts}{Total mixture}\) = \(\frac{9.2}{Total weight}\)
or \(\frac{T}{12T}\)= \(\frac{9.2}{Total weight}\)
or Total weight = 110.4 i.e total weight of the mixture is 110.4 pounds
Re: GRE Math Challenge #89-A health food store prepares
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11 Feb 2019, 05:22
11 Feb 2019, 05:22
pranab01 wrote:
sandy wrote:
A health food store prepares a breakfast food that consists of oats, raisins, and nuts mixed in the ratio 9:2:1, respectively, by weight. If the nuts in the mixture weigh 9.2 pounds, how many pounds does the total mixture weigh?
(A) 82.2
(B) 92.2
(C) 101. 2
(D) 110.4
(E) 165.6
(A) 82.2
(B) 92.2
(C) 101. 2
(D) 110.4
(E) 165.6
Here the weight of Nuts = 9.2 pounds and we need to find out the total weight of the mixture.
Let the total weight of mixture be = T
since oats:raisins : nuts = 9:2:1
i.e 9 parts of Total (T)+ 2 parts of Total (T) + 1 parts of total (T) = 12 T
Part/ Whole = \(\frac{Nuts}{Total mixture}\) = \(\frac{9.2}{Total weight}\)
or \(\frac{T}{12T}\)= \(\frac{9.2}{Total weight}\)
or Total weight = 110.4 i.e total weight of the mixture is 110.4 pounds
I think that Total weight = 110.4 is true answer
