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Re: Tough airthmetic [#permalink]
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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]
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27 Aug 2013, 20:22
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 ... gebrutes/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]
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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]
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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 10040 = 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.



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Re: The final exam of a particular class makes up 40% of the [#permalink]
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25 Aug 2014, 00:01
russ9 wrote:
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 10040 = 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.
Scale method: w1/w2 = (C2  Cavg)/(Cavg  C1) Cavg is the thing we need to average  marks here w1 and w2 are the weights allotted to marks  40% to finals and 60% to rest of the tests Question: The final exam of a particular class makes up 40% of the final grade, and Moe is failing the class with an average (arithmetic mean) of 45% just before taking the final exam. What grade does Moe need on his final exam in order to receive the passing grade average of 60% for the class? "The final exam of a particular class makes up 40% of the final grade"  this tells us that weight allotted to final exam is 40% and hence 60% is allotted to exams before finals. So we have w1 and w2. This is the tricky part. "Moe is failing the class with an average (arithmetic mean) of 45%"  This means in exams before finals, Moe has 45% marks. "What grade does Moe need on his final exam in order to receive the passing grade average of 60% for the class"  To pass, Moe needs average 60% marks i.e. Cavg should be 60%. So what we need is his marks in finals i.e. C2 w1/w2 = (C2  Cavg)/(Cavg  C1) 60/40 = (C2  60)/(60  45) This will give you C2.
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Re: The final exam of a particular class makes up 40% of the [#permalink]
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27 May 2015, 12:24
VeritasPrepKarishma wrote: russ9 wrote:
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 10040 = 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.
Scale method: w1/w2 = (C2  Cavg)/(Cavg  C1) Cavg is the thing we need to average  marks here w1 and w2 are the weights allotted to marks  40% to finals and 60% to rest of the tests Question: The final exam of a particular class makes up 40% of the final grade, and Moe is failing the class with an average (arithmetic mean) of 45% just before taking the final exam. What grade does Moe need on his final exam in order to receive the passing grade average of 60% for the class? "The final exam of a particular class makes up 40% of the final grade"  this tells us that weight allotted to final exam is 40% and hence 60% is allotted to exams before finals. So we have w1 and w2. This is the tricky part. "Moe is failing the class with an average (arithmetic mean) of 45%"  This means in exams before finals, Moe has 45% marks. "What grade does Moe need on his final exam in order to receive the passing grade average of 60% for the class"  To pass, Moe needs average 60% marks i.e. Cavg should be 60%. So what we need is his marks in finals i.e. C2 w1/w2 = (C2  Cavg)/(Cavg  C1) 60/40 = (C2  60)/(60  45) This will give you C2. This is still quite unclear to me. I too understand the boys and girls problem just fine, but like the previous poster metioned, I don't understand how we get that upper limit. Any chance you can show it on a number line too?



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Re: The final exam of a particular class makes up 40% of the [#permalink]
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27 May 2015, 22:11
The question can be interpreted in an another way Total marks needed to passWe are given that out of total marks(let's assume it to be \(y\)) Moe should receive 60% to pass i.e. 60% of \(y\) = \(0.6y\) Final exam marksSince the final exam makes up for 40% of the total marks, it constitutes 40% of \(y\) \(= 0.4y\) marks. Let's assume he should get \(x\)% of marks in his final exam to pass the exam i.e. he should get \(x\)% of \(0.4y\) Rest of the marksThe rest 60% of total marks constitute of 60% of \(y\) \(= 0.6y\) marks Out of these \(0.6y\) marks, Moe has got only 45% i.e. 45% of \(0.6y = 0.45 * 0.6y\) Writing the equationWe can write the equation for marks of Moe as Total marks needed by Moe to pass = Rest of the marks + Final exam marks\(0.6y = 0.45 * 0.6y +\) \(x\)% \(* 0.4y\) Solving this would give us \(x = 82.5\)% of marks he needs in his final exam to pass the class Hope this helps Regards Harsh
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Re: The final exam of a particular class makes up 40% of the [#permalink]
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28 May 2015, 07:46
EgmatQuantExpert wrote: The question can be interpreted in an another way Total marks needed to passWe are given that out of total marks(let's assume it to be \(y\)) Moe should receive 60% to pass i.e. 60% of \(y\) = \(0.6y\) Final exam marksSince the final exam makes up for 40% of the total marks, it constitutes 40% of \(y\) \(= 0.4y\) marks. Let's assume he should get \(x\)% of marks in his final exam to pass the exam i.e. he should get \(x\)% of \(0.4y\) Rest of the marksThe rest 60% of total marks constitute of 60% of \(y\) \(= 0.6y\) marks Out of these \(0.6y\) marks, Moe has got only 45% i.e. 45% of \(0.6y = 0.45 * 0.6y\) Writing the equationWe can write the equation for marks of Moe as Total marks needed by Moe to pass = Rest of the marks + Final exam marks\(0.6y = 0.45 * 0.6y +\) \(x\)% \(* 0.4y\) Solving this would give us \(x = 82.5\)% of marks he needs in his final exam to pass the class Hope this helps Regards, Harsh Quite simple way to interpret & arrive at the final solution.



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Re: The final exam of a particular class makes up 40% of the [#permalink]
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28 May 2015, 20:40
erikvm wrote: This is still quite unclear to me. I too understand the boys and girls problem just fine, but like the previous poster metioned, I don't understand how we get that upper limit. Any chance you can show it on a number line too?
We have condensed the scale method into the formula for ease. They are the same. In the formula, C2 is unknown and you solve it with simple equation manipulation to get the value of C2 (cross multiply etc). On the scale, this is how it will look: Attachment:
Scale Method.jpg [ 13.16 KiB  Viewed 934 times ]
Weight given to 45% marks is 60% and weight given to final marks is 40%. So ratio of weights is 3:2. So 45% and final marks will be away from the average in the ratio 2:3 (inverse of 3:2). 45 is actually 15 away from 60 (the 2 of the ratio) so the final marks (C2) will be (15/2)* 3 = 22.5 away from 60. This will take us to 82.5 as C2.
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Re: The final exam of a particular class makes up 40% of the [#permalink]
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11 Mar 2016, 08:43
rxs0005 wrote: The final exam of a particular class makes up 40% of the final grade, and Moe is failing the class with an average (arithmetic mean) of 45% just before taking the final exam. What grade does Moe need on his final exam in order to receive the passing grade average of 60% for the class? A. 75% B. 82.5% C. 85% D. 90% E. 92.5% \(\frac{40}{100} + \frac{60}{100} = \frac{100}{100}\) when you weight each of these scores with what More actually scored in each, x and 45 respectively u get 60. \(\frac{40x}{100} + \frac{(60*45)}{100} = 60\) \(\frac{2x}{5} + 18 = 60\)



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The final exam of a particular class makes up 40% of the [#permalink]
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24 Jul 2016, 19:10
VeritasPrepKarishma wrote: rxs0005 wrote: The final exam of a particular class makes up 40% of the final grade, and Moe is failing the class with an average (arithmetic mean) of 45% just before taking the final exam. What grade does Moe need on his final exam in order to receive the passing grade average of 60% for the class? (A) 75% (B) 82.5% (C) 85% (D) 90% (E) 92.5% Using the scale method (called alligation by some), you can very neatly and quickly solve this question of weighted averages. Weightage of mid terms (or whatever)  60% Weightage of final exams  40% Marks obtained in mid term  45% Average required  60% So marks obtained in finals  x% Now make a diagram like this: Attachment: Ques2.jpg 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. Hi Karishma, thank you so much for your hint that reminded me to recall the weighted average method. Here's the general equation: 2*(45%60%)+3*(X60%)=0 X=82.5. Rgds,
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Re: The final exam of a particular class makes up 40% of the [#permalink]
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25 Jul 2016, 13:00
let x=final exam grade needed .4*x+.6*.45=.6 x=.825=82.5%



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Re: The final exam of a particular class makes up 40% of the [#permalink]
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01 Dec 2016, 17:47
.60(45)+.40x = 60 27+0.40x=60 0.40x=33 x=82.5
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