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# A bowl of nuts is prepared for a party. Brand P mixed nuts are 20% alm

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

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20 Jul 2011, 22:53
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55% (hard)

Question Stats:

69% (03:03) correct 31% (02:12) wrong based on 269 sessions

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

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20 Jul 2011, 23:32
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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: A bowl of nuts is prepared for a party. Brand P mixed nuts are 20% alm [#permalink]

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21 Jul 2011, 01:48
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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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Re: A bowl of nuts is prepared for a party. Brand P mixed nuts are 20% alm [#permalink]

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21 Jul 2011, 02:54
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: A bowl of nuts is prepared for a party. Brand P mixed nuts are 20% alm [#permalink]

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15 Sep 2011, 03:13
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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: A bowl of nuts is prepared for a party. Brand P mixed nuts are 20% alm [#permalink]

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15 Sep 2011, 20:46
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: A bowl of nuts is prepared for a party. Brand P mixed nuts are 20% alm [#permalink]

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16 Sep 2011, 08:26
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: A bowl of nuts is prepared for a party. Brand P mixed nuts are 20% alm [#permalink]

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16 Sep 2011, 22:31
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.

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

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17 Sep 2011, 02:45
x+y=64
0.2x+0.25y=15

y=44
OA is D
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Re: A bowl of nuts is prepared for a party. Brand P mixed nuts are 20% alm [#permalink]

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17 Sep 2011, 02:53
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: A bowl of nuts is prepared for a party. Brand P mixed nuts are 20% alm [#permalink]

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17 Sep 2011, 03:03
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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: A bowl of nuts is prepared for a party. Brand P mixed nuts are 20% alm [#permalink]

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17 Sep 2011, 05:17
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: A bowl of nuts is prepared for a party. Brand P mixed nuts are 20% alm [#permalink]

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18 Sep 2011, 00:41
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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Re: A bowl of nuts is prepared for a party. Brand P mixed nuts are 20% alm [#permalink]

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

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30 Apr 2015, 22:37
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

Can somebody do the scale method to this question with the diagram?
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Re: A bowl of nuts is prepared for a party. Brand P mixed nuts are 20% alm [#permalink]

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03 Jul 2015, 06:41
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

why can we not use this formula here:

0,2P + 0,25Q = 0,23 (P+Q) ??? .... 15/64 = 0,23
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A bowl of nuts is prepared for a party. Brand P mixed nuts are 20% alm [#permalink]

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03 Jul 2015, 09:14
Hi LaxAvenger

I'm not sure if I understand your question. Could you kindly clarify? The ratio can be used, of course, once the % mix is calculated. I believe the difficulty of the question stems from calculating the % mix, in an efficient manner, i.e. without using long division. One work-around is finding the common denominator among the three ratios, and another way to arrive at the answer is through estimation.

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

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14 Sep 2015, 09:56
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

Let P be the amount of almonds in in brand P and Q be the amount of almonds in brand Q.
Then by allegations,ratio of amount of almonds from brand Q and P in the mixture will be

$$\frac{Q}{P}$$ = $$\frac{\frac{15}{64}-\frac{1}{5}}{\frac{1}{4}-\frac{15}{64}}$$ = $$\frac{\frac{11}{320}}{\frac{1}{64}$$ = $$\frac{11}{5}$$

Ratio of almonds from brand Q to the total number of almonds will be
$$\frac{Q}{P+Q}$$ = $$\frac{11}{16}$$

Then by proportion, total amount of brand Q nuts $$Q_t$$ in the mixture will be

$$\frac{11}{16}$$ = $$\frac{Q_t}{64}$$

or $$Q_t = 44$$

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

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28 Oct 2015, 19:43
.2(64-Q)+.25Q=15
.05Q=2.2
Q=44 ounces
Re: A bowl of nuts is prepared for a party. Brand P mixed nuts are 20% alm   [#permalink] 28 Oct 2015, 19:43
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