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mixtures and ratio

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by rahul.s » Sun Jan 31, 2010 3:13 am
Trail mix P is 15 percent cashews and 85 percent almonds by weight; trail mix R is 40 percent cashews and 60 percent peanuts. If a mixture of P and R contains 20 percent cashews, what percent of the weight of the mixture is R?

A) 15 %
B) 20 %
C) 33 1/3%
D) 50 %
E) 80 %

OA: B
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by ajith » Sun Jan 31, 2010 3:35 am
rahul.s wrote:Trail mix P is 15 percent cashews and 85 percent almonds by weight; trail mix R is 40 percent cashews and 60 percent peanuts. If a mixture of P and R contains 20 percent cashews, what percent of the weight of the mixture is R?

A) 15 %
B) 20 %
C) 33 1/3%
D) 50 %
E) 80 %

OA: B
Say P and R are mixed in 1:x proportion

% of cashew in the mix =(1*15+x*40)/(1+x) = 20 => 20x = 5 => x =1/4

P&R are mixed 1:1/4 proportion

% of R in the mix = 1/4/(1+1/4) *100 = 1/5*100 = 20%
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by sars72 » Sun Jan 31, 2010 5:42 am
ajith wrote:
Say P and R are mixed in 1:x proportion

% of cashew in the mix =(1*15+x*40)/(1+x) = 20 => 20x = 5 => x =1/4

P&R are mixed 1:1/4 proportion

% of R in the mix = 1/4/(1+1/4) *100 = 1/5*100 = 20%
ajith, i've forgotten all that i studied about mixtures n alligations. I am not able to follow your calculations. Could you please elaborate a bit on them? Thanks.
Last edited by sars72 on Sun Jan 31, 2010 6:23 am, edited 1 time in total.
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by papgust » Sun Jan 31, 2010 5:59 am
I did it this way. I'm really horrible at setting up mixture equations. So, let me try by using a non-equation approach.

Set a line for distributing cashews. M is the mixture.

[5] [20]
P ------------- M -------------------- R
(15) (20) (40)

% of R in mixture = 5/(5+20) = 5/25 = 20%
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by rahul.s » Sun Jan 31, 2010 6:11 am
papgust wrote:% of R in mixture = 5/(5+20) = 5/25 = 20%
how'd you get that?
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by papgust » Sun Jan 31, 2010 6:21 am
I hope that this method could be solved for this question.

It's a weighted average concept. You could check out Stuart's method in this site. This method is the same as Stuart's.

If you are not able to search for stuart's method, check this out
https://www.beatthegmat.com/mba/2009/12/ ... art-2-of-2

------

Distance b/w P and M in the line is 5. Distance b/w M and R is 20. Total distance is 25

Weight of R is 5 [b/w P and M] and weight of P is 20 [b/w M and R]

So, R / (P+R) = 5/25
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by ajith » Sun Jan 31, 2010 6:29 am
sars72 wrote: ajith, i've forgotten all that i studied about mixtures n alligations. I am not able to follow your calculations. Could you please elaborate a bit on them? Thanks.
Any proportion can be written in 1: x form Say P & R are mixed in 1: x proportion

cashew coming from mixture P = 15 (15*1 - Since the proportion is 1: x)
cashew coming from mixture R = 40x (40*x - since the proportion is 1:x)

Total amount of mixture = 1+x

The cashew % in the mixture = 15+40x/1+x and it is given as 20

And it gives x =1/4

so P& R are mixed in 1:0.25 proportion so for every 4 units of P 1 unit of R is added hence % of R in the mixture = 1/5*100 =20%
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by thephoenix » Sun Jan 31, 2010 7:05 am
rahul.s wrote:
papgust wrote:% of R in mixture = 5/(5+20) = 5/25 = 20%
how'd you get that?
rahul its actually like this take any one component for which is common in bth and is also present in resultant
and write its % in following manner


P....................................R
15.................................40
then write the % of the same in resultant like this

15-----------------------------40
--------------20-----------------
cross diff
40-20...........................20-15
=20...........................5

now tot=20+5

the right side is for mix R
and left side is for mix P

% of R in mix is 5/25=1/4=20%

HTH
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by rahul.s » Sun Jan 31, 2010 8:11 am
stacey's post was amazing. made weighted average problems look simple and uncomplicated. i highly recommend everyone to go through the solution

follow the link.

https://www.beatthegmat.com/mba/2009/12/ ... art-2-of-2

thanks papgust :D
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by rahul.s » Sun Jan 31, 2010 8:15 am
thephoenix wrote:P....................................R
15.................................40
then write the % of the same in resultant like this

15-----------------------------40
--------------20-----------------
cross diff
40-20...........................20-15
=20...........................5
why is it 40 - 20 and 20 - 15?

i don't follow
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by papgust » Sun Jan 31, 2010 8:23 am
rahul.s wrote:
thephoenix wrote:P....................................R
15.................................40
then write the % of the same in resultant like this

15-----------------------------40
--------------20-----------------
cross diff
40-20...........................20-15
=20...........................5
why is it 40 - 20 and 20 - 15?

i don't follow
Imagine the line as a "tug of war".
40-20 = 20 is the weight of P.
20-15 = 5 is the weight of R.
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by thephoenix » Sun Jan 31, 2010 8:40 am
rahul.s wrote:
thephoenix wrote:P....................................R
15.................................40
then write the % of the same in resultant like this

15-----------------------------40
--------------20-----------------
cross diff
40-20...........................20-15
=20...........................5
why is it 40 - 20 and 20 - 15?

i don't follow
it a short cut trick
write the lower value in left most side and higher value in right most side
write the men in centre
below that write the diff of higher value and mean on left most side , and the diff of mean and lower value on right most side
this is your final ratio
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by bhumika.k.shah » Mon Feb 01, 2010 5:09 am
thephoenix wrote:
% of R in mix is 5/25=1/4=20%

HTH
% of R in mix is 5/25 = 1/5 = 20%
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