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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 11 Dec 2012, 23:39
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dcastan2 wrote:


Karishma, this was very helpful and your way of doing the scale method makes sense because MGMAT doesn't explain it clearly in the foundation book. My only question is your equation

C_{avg} = \frac{C_1*W_1 + C_2 * W_2}{W_1 + W_2}
If I re-arrange the formula, I get
\frac{W_1}{W_2} = \frac{C_2 - C_{avg}}{C_{avg} - C_1}

How did you re-arrange the formula to get that? I don't see it. Thanks in advance!


C_{avg} = \frac{C_1*W_1 + C_2 * W_2}{W_1 + W_2}

C_{avg}*(W_1 + W_2) = C_1*W_1 + C_2 * W_2 (Cross multiplying)

C_{avg}*W_1 + C_{avg}*W_2 = C_1*W_1 + C_2 * W_2

C_{avg}*W_1 - C_1*W_1 = C_2 * W_2 - C_{avg}*W_2

W_1(C_{avg} - C_1) = W_2(C_2 - C_{avg})

\frac{W_1}{W_2} = \frac{C_2 - C_{avg}}{C_{avg} - C_1}
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Re: The final exam of a particular class makes up 40% of the [#permalink] New post 13 Jun 2013, 02:07
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Bumping for review and further discussion*. Get a kudos point for an alternative solution!

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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: 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
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Re: The final exam of a particular class makes up 40% of the [#permalink] New post 24 Aug 2014, 13:17
VeritasPrepKarishma wrote:
anish319 wrote:
VeritasPrepKarishma wrote:
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%

If you have time, it would be a good idea to be comfortable with the scale method. It will save you a lot of time and energy.


I was actually about the above solution....sorry for the confusion. how did you get "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%"


Oh ok... I thought I was missing something!
In the question above, you have the average of marks. So your C is marks. Here, the difference is that weights are given to you and one of the C (i.e. C2) is missing.
We know that ratio of weights will be the distance on the number line.

So if you look at the diagram above, the ratio of weights is 3:2 (because weightage of mid terms is 60% and weightage of finals is 40%, so 60:40 = 3:2). This means that the distance on the number line should be in the ratio 2:3 (The ratio on the number line flips). So distance between 45 and 60 is 2 units. This means 1 unit is 15/2 = 7.5 on the number line.
Now we need to find what 3 units distance is because C2 (i.e. x in the diagram) will be 3 units away from 60. Since 1 unit is 7.5, 3 units will be 22.5. Adding 22.5 to 60, we get 82.5. So x, the missing extreme right value must be 82.5


Hi karishma,

I thought that I understood your line method until I read this statement. Whenever we look at the line method (You usually have an average weight and the two ends(weights) of the data given). So in the example with ages, there are 17yr old girls, 20yr old boys and the group is 18, you have a clear number that's above the average and below the average -- which makes complete sense.

Two questions:
1) Line method:
If I were to use the line method in this example -- we are already TOLD that the middle weight is 40. In addition to that, i'm a little confused because the "middle weight" is actually not in the middle? I'm not sure if I'm explaining this properly but if we are looking to score a 60% (that should be the average of the two ends, correct?) -- we know that we have scored 40% but we don't have the other end (a number higher than 60). How do we find a ratio without knowing that number, which incidentally, is the same number that the question is asking for.

2) If I use the equation:

Weight 1 / Weight 2 = Average 2 - Average. Avg / Average. Avg - A1

We know that the weight of the final is 40% which means that the weight of everything but the exam is 100-40 = 60%. So we have w1=40 and w2 = 60

Additionally, we know that we are looking to score a 60% in the class, which is different than the 60% I outlined above. So we have that the Average. Avg = 60.

We also know that currently, he has scored 45%, which is on the non final part, therefore, we are looking to find what he needs to score on the Final. Correct? Meaning, we need to find A1 in the equation above. Correct?

This seems to get me to 82.5 but it was extremely difficult for me to draw the gap between what weights correspond to what grades. I've read your "quarter wit" documents but it still eludes me a little. Any help would be appreciated.

Thanks.
Re: The final exam of a particular class makes up 40% of the   [#permalink] 24 Aug 2014, 13:17
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