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The final exam of a particular class makes up 40% of the

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Re: Tough airthmetic [#permalink] New post 27 Aug 2013, 14:52
VeritasPrepKarishma wrote:
anish319 wrote:
karishma

how did you derive this: Since the weights are in the ratio 3:2, the distance on the scale will be in the ratio 2:3.
So x = 82.5%

I followed the whole weighted average discussion. what is C2 for the above question



Weights are the number of boys and number of girls. Here we need to find the number of girls so we do not have the ratio of weights. We need to find it. What we do have is C1, CAvg and C2. As shown in the diagram above, ratio of distance between C2 and CAvg and distance between CAvg and C1 gives the ratio of weights.

C1 = 17 yrs, CAvg = 18 yrs and C2 = 20 yrs
Hence distance on the scale between 17 and 18 is 1 and distance on the scale between 18 and 20 is 2. This gives us a ratio of 2:1 for the weights i.e. for the number of girls:number of boys.

Remember C is what you want to find the average of. Here average age is 18 yrs. So age is C.
Weights is the number of boys/girls.



Is there a document which can explain the Mixture concept in detail. I seem to be having issues with it.
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Re: Tough airthmetic [#permalink] New post 27 Aug 2013, 20:22
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karannanda wrote:
VeritasPrepKarishma wrote:
anish319 wrote:
karishma

how did you derive this: Since the weights are in the ratio 3:2, the distance on the scale will be in the ratio 2:3.
So x = 82.5%

I followed the whole weighted average discussion. what is C2 for the above question



Weights are the number of boys and number of girls. Here we need to find the number of girls so we do not have the ratio of weights. We need to find it. What we do have is C1, CAvg and C2. As shown in the diagram above, ratio of distance between C2 and CAvg and distance between CAvg and C1 gives the ratio of weights.

C1 = 17 yrs, CAvg = 18 yrs and C2 = 20 yrs
Hence distance on the scale between 17 and 18 is 1 and distance on the scale between 18 and 20 is 2. This gives us a ratio of 2:1 for the weights i.e. for the number of girls:number of boys.

Remember C is what you want to find the average of. Here average age is 18 yrs. So age is C.
Weights is the number of boys/girls.



Is there a document which can explain the Mixture concept in detail. I seem to be having issues with it.


Here are some posts. The first one explains the weighted average concept and the second one builds on it. The third one tackles mixtures using the weighted average concept.

http://www.veritasprep.com/blog/2011/03 ... -averages/
http://www.veritasprep.com/blog/2011/04 ... ge-brutes/
http://www.veritasprep.com/blog/2011/04 ... -mixtures/
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Re: The final exam of a particular class makes up 40% of the [#permalink] New post 07 Sep 2013, 11:28
I attacked this in a completely different way:

We know that the exam makes up 40%, so it's 2/5 of the grade.

Moe has 45% for 60% of the grade. We are trying to find out what he needs for those last 2/5ths in order to get 60% as a final average.

So, [(45*3)+(2X)]/5 = 60 gives us the number needed for the final 2 'components' (that 40% or 2/5) to make a final average of 60

Solve for X and get 82.5
Re: The final exam of a particular class makes up 40% of the   [#permalink] 07 Sep 2013, 11:28
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