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A bowl of nuts is prepared for a party. Brand P mixed nuts are 20% alm [#permalink]
20 Jul 2011, 21:53

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Difficulty:

35% (medium)

Question Stats:

71% (02:24) correct
29% (01:42) wrong based on 21 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?

Re: A bowl of nuts is prepared for a party. Brand P mixed nuts are 20% alm [#permalink]
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?

Re: A bowl of nuts is prepared for a party. Brand P mixed nuts are 20% alm [#permalink]
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?

Re: A bowl of nuts is prepared for a party. Brand P mixed nuts are 20% alm [#permalink]
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.

Re: A bowl of nuts is prepared for a party. Brand P mixed nuts are 20% alm [#permalink]
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

Re: A bowl of nuts is prepared for a party. Brand P mixed nuts are 20% alm [#permalink]
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?

Re: A bowl of nuts is prepared for a party. Brand P mixed nuts are 20% alm [#permalink]
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. _________________

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% _________________

Happy are those who dream dreams and are ready to pay the price to make them come true

Re: A bowl of nuts is prepared for a party. Brand P mixed nuts are 20% alm [#permalink]
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..

Re: A bowl of nuts is prepared for a party. Brand P mixed nuts are 20% alm [#permalink]
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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