Oct 18 08:00 AM PDT  09:00 AM PDT Learn an intuitive, systematic approach that will maximize your success on Fillintheblank GMAT CR Questions. Oct 19 07:00 AM PDT  09:00 AM PDT Does GMAT RC seem like an uphill battle? eGMAT is conducting a free webinar to help you learn reading strategies that can enable you to solve 700+ level RC questions with at least 90% accuracy in less than 10 days. Sat., Oct 19th at 7 am PDT Oct 20 07:00 AM PDT  09:00 AM PDT Get personalized insights on how to achieve your Target Quant Score. Oct 22 08:00 PM PDT  09:00 PM PDT On Demand for $79. For a score of 4951 (from current actual score of 40+) AllInOne Standard & 700+ Level Questions (150 questions) Oct 23 08:00 AM PDT  09:00 AM PDT Join an exclusive interview with the people behind the test. If you're taking the GMAT, this is a webinar you cannot afford to miss!
Author 
Message 
TAGS:

Hide Tags

Intern
Joined: 29 Apr 2013
Posts: 5

Re: Tough airthmetic
[#permalink]
Show Tags
27 Aug 2013, 15:52
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: 9706
Location: Pune, India

Re: Tough airthmetic
[#permalink]
Show Tags
27 Aug 2013, 21:22
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/
_________________
Karishma Veritas Prep GMAT Instructor
Learn more about how Veritas Prep can help you achieve a great GMAT score by checking out their GMAT Prep Options >



Intern
Joined: 01 Aug 2013
Posts: 10
Location: United States
Concentration: Operations, International Business
GPA: 3.65

Re: The final exam of a particular class makes up 40% of the
[#permalink]
Show Tags
07 Sep 2013, 12:28
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



Manager
Joined: 15 Aug 2013
Posts: 228

Re: The final exam of a particular class makes up 40% of the
[#permalink]
Show Tags
24 Aug 2014, 14:17
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.



Veritas Prep GMAT Instructor
Joined: 16 Oct 2010
Posts: 9706
Location: Pune, India

Re: The final exam of a particular class makes up 40% of the
[#permalink]
Show Tags
25 Aug 2014, 01:01
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.
_________________
Karishma Veritas Prep GMAT Instructor
Learn more about how Veritas Prep can help you achieve a great GMAT score by checking out their GMAT Prep Options >



Manager
Joined: 26 Feb 2015
Posts: 110

Re: The final exam of a particular class makes up 40% of the
[#permalink]
Show Tags
27 May 2015, 13:24
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?



eGMAT Representative
Joined: 04 Jan 2015
Posts: 3074

Re: The final exam of a particular class makes up 40% of the
[#permalink]
Show Tags
27 May 2015, 23:11
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
_________________



Manager
Joined: 21 Jul 2014
Posts: 80
Location: India
Schools: Booth '21 (D), Ross '21 (D), Tuck '21 (D), Johnson '21 (WL), Kelley '21 (D), KenanFlagler '21 (II), LBS '21 (D), Insead Sept19 Intake (D), ISB '20 (A), Goizueta '21 (D), Fisher '21 (I), Mendoza '21 (I), Carlson '21 (I)
GPA: 4
WE: Project Management (Energy and Utilities)

Re: The final exam of a particular class makes up 40% of the
[#permalink]
Show Tags
28 May 2015, 08:46
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.



Veritas Prep GMAT Instructor
Joined: 16 Oct 2010
Posts: 9706
Location: Pune, India

Re: The final exam of a particular class makes up 40% of the
[#permalink]
Show Tags
28 May 2015, 21:40
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 1742 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
Learn more about how Veritas Prep can help you achieve a great GMAT score by checking out their GMAT Prep Options >



Senior Manager
Joined: 15 Oct 2015
Posts: 299
Concentration: Finance, Strategy
GPA: 3.93
WE: Account Management (Education)

Re: The final exam of a particular class makes up 40% of the
[#permalink]
Show Tags
11 Mar 2016, 09:43
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. \(\frac{40x}{100} + \frac{(60*45)}{100} = 60\) \(\frac{2x}{5} + 18 = 60\)



Intern
Joined: 05 Jul 2016
Posts: 30
Location: China
Concentration: Finance, Nonprofit
GMAT 1: 680 Q49 V33 GMAT 2: 690 Q51 V31 GMAT 3: 710 Q50 V36
GPA: 3.4

The final exam of a particular class makes up 40% of the
[#permalink]
Show Tags
24 Jul 2016, 20:10
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,
_________________
It's better to burn out than to fade away.



VP
Joined: 07 Dec 2014
Posts: 1222

Re: The final exam of a particular class makes up 40% of the
[#permalink]
Show Tags
25 Jul 2016, 14:00
let x=final exam grade needed .4*x+.6*.45=.6 x=.825=82.5%



Current Student
Status: DONE!
Joined: 05 Sep 2016
Posts: 357

Re: The final exam of a particular class makes up 40% of the
[#permalink]
Show Tags
01 Dec 2016, 18:47
.60(45)+.40x = 60 27+0.40x=60 0.40x=33 x=82.5
B.



EMPOWERgmat Instructor
Status: GMAT Assassin/CoFounder
Affiliations: EMPOWERgmat
Joined: 19 Dec 2014
Posts: 15271
Location: United States (CA)

Re: The final exam of a particular class makes up 40% of the
[#permalink]
Show Tags
08 Feb 2018, 11:51
Hi All, This question can be solved in a variety of ways, depending on how you choose to set up the math (you could also TEST THE ANSWERS here, but you'd likely find the right algebraic approach to be pretty easy). Since we know that the final exam is 40% of the final grade, we know that it is 2/5 of the final grade. This means that Moe's work so far (the 45% grade) accounts for 3/5 of the final grade. Since we're looking for the minimal score needed on the final exam to raise his grade to 60%, we can set up the following equation: (3/5)(45%) + (2/5)(X%) = 60% 27 + 2X/5 = 60 2X/5 = 33 X = (5)(33)/2 X = 165/2 X = 82.5 Final Answer: GMAT assassins aren't born, they're made, Rich
_________________
Contact Rich at: Rich.C@empowergmat.comThe Course Used By GMAT Club Moderators To Earn 750+ souvik101990 Score: 760 Q50 V42 ★★★★★ ENGRTOMBA2018 Score: 750 Q49 V44 ★★★★★



Intern
Joined: 26 May 2018
Posts: 3

Re: The final exam of a particular class makes up 40% of the
[#permalink]
Show Tags
14 Aug 2018, 06:40
Hi, How'd you get the 2:3 ratio? Thanks! VeritasKarishma 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.



Manager
Joined: 23 Apr 2018
Posts: 138

Re: The final exam of a particular class makes up 40% of the
[#permalink]
Show Tags
28 Sep 2018, 16:45
VeritasKarishma 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. How do you calculate the percentage from ratio here ? Posted from my mobile device



Intern
Joined: 05 Jan 2017
Posts: 34

Re: The final exam of a particular class makes up 40% of the
[#permalink]
Show Tags
22 Oct 2018, 03:59
by weighted avg: 60=(40*x+60*45)/(60+40) => 60*100=40x+2700 => 40x=60002700=3300 =>x=3300/40=82.5 (B)
_________________
 Kudos if my post helps!



Manager
Joined: 22 Sep 2018
Posts: 240

The final exam of a particular class makes up 40% of the
[#permalink]
Show Tags
22 Jan 2019, 17:44
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% Question isn't very difficult but I still ended up taking 4 minutes on it... I need to be able to process information much faster. Simple weighted average question: Since the final exam is worth 40% of Moe's grade, all the other tests Moe took are worth 60% of his grade. 60*45% + 40x = 60 \(x = \frac{33}{40} = 82.5%\)



Intern
Joined: 18 Mar 2018
Posts: 27

Re: The final exam of a particular class makes up 40% of the
[#permalink]
Show Tags
08 Feb 2019, 16:59
There is a very simple method of doing this.
Student is currently at 45, and would like to reach 60. Difference is 15.
The final exam only accounts for 40%, or 0.4. So, he needs to get that difference of 15 but with a 0.4 weighting.
By doing \(\frac{15}{0.4} = 32.5\), we now know how much more more he needs for his his final exam.
45 + 37.5 = 82.5 which is our answer



Manager
Joined: 09 Mar 2018
Posts: 55
Location: India

Re: The final exam of a particular class makes up 40% of the
[#permalink]
Show Tags
31 Mar 2019, 00:10
solving this question in under 2 mins :
our current percentage is 45% we wish to take it to 60%
now to take 45% to 60% we must score 75% in our next exam, how did I calculate this?
simply using mean formula
45  60?
a distance between 45 and 60 is 15, so our next score should be 75, but the crunch is this distance of 15 that we'll cover, only 40% of that will be counted. i.e 0.4*15 =6
so, to simplify for every 15 marks scored only 6 will be counted.
therefore, 15/6=2.5 and 15*2.5=37.5, and finally 45+37.5=82.5




Re: The final exam of a particular class makes up 40% of the
[#permalink]
31 Mar 2019, 00:10



Go to page
Previous
1 2 3
Next
[ 41 posts ]



