The final exam of a particular class makes up 40% of the : GMAT Problem Solving (PS) - Page 2
Check GMAT Club Decision Tracker for the Latest School Decision Releases http://gmatclub.com/AppTrack

 It is currently 19 Jan 2017, 02:30

### GMAT Club Daily Prep

#### Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.

Customized
for You

we will pick new questions that match your level based on your Timer History

Track

every week, we’ll send you an estimated GMAT score based on your performance

Practice
Pays

we will pick new questions that match your level based on your Timer History

# Events & Promotions

###### Events & Promotions in June
Open Detailed Calendar

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

Author Message
TAGS:

### Hide Tags

Intern
Joined: 29 Apr 2013
Posts: 8
Followers: 0

Kudos [?]: 0 [0], given: 11

### Show Tags

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.
Veritas Prep GMAT Instructor
Joined: 16 Oct 2010
Posts: 7125
Location: Pune, India
Followers: 2134

Kudos [?]: 13644 [2] , given: 222

### Show Tags

27 Aug 2013, 20:22
2
KUDOS
Expert's post
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/
_________________

Karishma
Veritas Prep | GMAT Instructor
My Blog

Get started with Veritas Prep GMAT On Demand for $199 Veritas Prep Reviews Intern Joined: 01 Aug 2013 Posts: 11 Location: United States Concentration: Operations, International Business GMAT 1: 710 Q47 V41 GPA: 3.65 Followers: 0 Kudos [?]: 7 [0], given: 0 Re: The final exam of a particular class makes up 40% of the [#permalink] ### Show Tags 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 Senior Manager Joined: 15 Aug 2013 Posts: 328 Followers: 0 Kudos [?]: 53 [0], given: 23 Re: The final exam of a particular class makes up 40% of the [#permalink] ### Show Tags 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. Veritas Prep GMAT Instructor Joined: 16 Oct 2010 Posts: 7125 Location: Pune, India Followers: 2134 Kudos [?]: 13644 [1] , given: 222 Re: The final exam of a particular class makes up 40% of the [#permalink] ### Show Tags 25 Aug 2014, 00:01 1 This post received KUDOS Expert's post 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 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. 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. _________________ Karishma Veritas Prep | GMAT Instructor My Blog Get started with Veritas Prep GMAT On Demand for$199

Veritas Prep Reviews

Manager
Joined: 26 Feb 2015
Posts: 127
Followers: 0

Kudos [?]: 9 [0], given: 43

Re: The final exam of a particular class makes up 40% of the [#permalink]

### Show Tags

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 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.

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?
e-GMAT Representative
Joined: 04 Jan 2015
Posts: 484
Followers: 138

Kudos [?]: 1126 [0], given: 90

Re: The final exam of a particular class makes up 40% of the [#permalink]

### Show Tags

27 May 2015, 22:11
The question can be interpreted in an another way

Total marks needed to pass
We 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 marks
Since 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 marks
The 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 equation
We 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
_________________

| '4 out of Top 5' Instructors on gmatclub | 70 point improvement guarantee | www.e-gmat.com

Manager
Joined: 21 Jul 2014
Posts: 71
Location: United States
WE: Project Management (Non-Profit and Government)
Followers: 0

Kudos [?]: 11 [0], given: 58

Re: The final exam of a particular class makes up 40% of the [#permalink]

### Show Tags

28 May 2015, 07:46
EgmatQuantExpert wrote:
The question can be interpreted in an another way

Total marks needed to pass
We 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 marks
Since 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 marks
The 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 equation
We 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.
Veritas Prep GMAT Instructor
Joined: 16 Oct 2010
Posts: 7125
Location: Pune, India
Followers: 2134

Kudos [?]: 13644 [1] , given: 222

Re: The final exam of a particular class makes up 40% of the [#permalink]

### Show Tags

28 May 2015, 20:40
1
KUDOS
Expert's post
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 599 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.
_________________

Karishma
Veritas Prep | GMAT Instructor
My Blog

Get started with Veritas Prep GMAT On Demand for $199 Veritas Prep Reviews Senior Manager Joined: 15 Oct 2015 Posts: 379 Concentration: Finance, Strategy GMAT 1: 690 Q36 V48 GPA: 3.93 WE: Account Management (Retail Banking) Followers: 4 Kudos [?]: 79 [0], given: 211 Re: The final exam of a particular class makes up 40% of the [#permalink] ### Show Tags 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. $$\frac{40x}{100} + \frac{(60*45)}{100} = 60$$ $$\frac{2x}{5} + 18 = 60$$ Manager Joined: 05 Jul 2016 Posts: 50 Location: China Concentration: Finance, Nonprofit GMAT 1: 680 Q49 V33 GMAT 2: 690 Q51 V31 GMAT 3: 710 Q50 V36 GPA: 3.4 Followers: 0 Kudos [?]: 8 [0], given: 129 The final exam of a particular class makes up 40% of the [#permalink] ### Show Tags 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*(X-60%)=0 X=82.5. Rgds, _________________ It's better to burn out than to fade away. Senior Manager Joined: 07 Dec 2014 Posts: 491 Followers: 3 Kudos [?]: 81 [0], given: 2 Re: The final exam of a particular class makes up 40% of the [#permalink] ### Show Tags 25 Jul 2016, 13:00 let x=final exam grade needed .4*x+.6*.45=.6 x=.825=82.5% Senior Manager Status: Exam scheduled!! Joined: 05 Sep 2016 Posts: 411 Location: United States (WI) Concentration: Marketing, Technology WE: Other (Law) Followers: 3 Kudos [?]: 13 [0], given: 241 Re: The final exam of a particular class makes up 40% of the [#permalink] ### Show Tags 01 Dec 2016, 17:47 .60(45)+.40x = 60 27+0.40x=60 0.40x=33 x=82.5 B. Re: The final exam of a particular class makes up 40% of the [#permalink] 01 Dec 2016, 17:47 Go to page Previous 1 2 [ 33 posts ] Similar topics Replies Last post Similar Topics: 3 In an engineering class that contains 50 students, the final exam 3 14 May 2016, 05:08 8 On a particular exam, the boys in a history class averaged 86 points a 10 30 Jun 2015, 04:05 3 A student is to take her final exams in three subjects. The probabilit 3 08 Dec 2014, 05:19 9 A student is to take her final exams in two subjects. The probability 19 05 Dec 2014, 06:43 4 A dress is marked up 16 2/3% to a final price of$140. What 6 28 Jun 2013, 06:11
Display posts from previous: Sort by