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The final exam of a particular class makes up 40% of the
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02 Dec 2010, 02:29
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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%
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Re: Tough airthmetic
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06 Dec 2010, 21:38
gettinit wrote: Karishma can you please explain your method a bit more? I don't understand how you solved the problem using it? thanks
What is weighted average? It is average when each value has a different weight. e.g. a group of friends has 10 boys and 20 girls. Average age of boys is 20 years and average age of girls is 17 years. What is the average age of the group? Here, the average is weighted since we have different number of boys and girls. We calculate it as follows: \(W Avg = \frac{20*10 + 17* 20}{10 + 20}\) What we are doing instinctively here is using weighted average formula which as given below: \(C_{avg} = \frac{C_1*W_1 + C_2 * W_2}{W_1 + W_2}\) You need to find the average of C and W is the weight. In the example above, C is age and W is number of boys and girls. The alligation method, or the scale method as we call it, is based on the weighted averages formula itself: \(C_{avg} = \frac{C_1*W_1 + C_2 * W_2}{W_1 + W_2}\) If I rearrange the formula, I get \(\frac{W_1}{W_2} = \frac{C_2  C_{avg}}{C_{avg}  C_1}\) So I get that weights will be in the same ratio as difference between higher value of C and average value of C and difference between average value of C and lower value of C. How does this help? Knowing this, we can directly make a diagram and get the answer. e.g. A group of friends has 10 boys and some girls. Average age of boys is 20 years and average age of girls is 17 years. The average age of the group is 18 years. How many girls are there? Draw: Attachment:
Ques1.jpg [ 5.97 KiB  Viewed 22279 times ]
On a scale (number line), mark 17 years as age of girls, 18 years as average and 20 years as age of boys. Now, distance between 17 and 18 is 1 and distance between 18 and 20 is 2, The ratio of W1/W2 will be 2:1 (Note, the numbers 1 and 2, give a ratio of 2:1 for girls:boys as seen by the formula) Since there are 10 boys, there will be 20 girls. This method is especially useful when you have the average and need to find the ratio of weights. [/quote]
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Re: Tough airthmetic
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02 Dec 2010, 02:55
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% ANS: B Lets Assume that total marks are 100. Final exam weightage : 40 Rest fo the exams : 60 Percentage scored by Moe out of 60 = 45% Hence, Total marks obtained out of 60 = 45% of 60 = 27 Marks required in the final exam to get 60% in end = 60(marks)  27 = 33 marks i.e. 33 marks required out of 40 = 82.5 marks required out of 100 in final exam ANS: 82.5 % (B)
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Re: Tough airthmetic
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02 Dec 2010, 02:56
I'd go with b 82.5 oa?
You get that with weighted avg, unless there us a super tricky wrinkle which makes this a 700 level prob OA?
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Re: Tough airthmetic
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02 Dec 2010, 04:12
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 [ 5.28 KiB  Viewed 22431 times ]
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.
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Re: Tough airthmetic
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04 Dec 2010, 00:11
rite2deepti wrote: gmat1011 wrote: I'd go with b 82.5 oa?
You get that with weighted avg, unless there us a super tricky wrinkle which makes this a 700 level prob OA?
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Posted from my mobile device It would be helpful if you can show ,how did you solve it by weighted average ... Thanks in advance for detailed explanation The concept is same as that explained by VeritasKarishma.. Weighted average=60=(40*x+60*45)/(60+40) > 60*100=40x+2700 > 40x=60002700=3300 >x=3300/40=82.5



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Re: Tough airthmetic
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06 Dec 2010, 22:23
Karishma I knew what weighted average was and the formula but did not understand how you applied the scale method. wow, I had no clue about this Karishma. Great way to look at it.



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Re: Tough airthmetic
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07 Dec 2010, 04:45
This method is a real time saver. If you are already comfortable with the concept of weighted averages, try using the diagram for a few questions. Thereafter, when you see a weighted 'average' in the question and the smaller C and greater C, just find the difference between average and smaller C and difference between average and greater C and flip the ratios. e.g. A group of friends has 10 boys and some girls. Average age of boys is 20 years and average age of girls is 17 years. The average age of the group is 18 years. How many girls are there? When I see this question, I do not bother with the diagram. I say 18  17 (girls' age)= 1 so 1 goes to no of boys and 20 (boys' age)  18 = 2 so 2 goes to no of girls. Ratio girls:boys = 2:1
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Re: Tough airthmetic
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08 Dec 2010, 18:33
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.
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Re: Tough airthmetic
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08 Dec 2010, 18:49
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. 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%"



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Re: Tough airthmetic
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09 Dec 2010, 03:56
anish319 wrote: 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. 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
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Re: Tough airthmetic
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16 Jan 2011, 12:21
Karishma  amazing system. DO u have any more examples so we can try it out? thanks.
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18 Jan 2011, 20:10
144144 wrote: Karishma  amazing system. DO u have any more examples so we can try it out?
thanks. If you are comfortable with the method, try this question... http://gmatclub.com/forum/aproblemfromveritasarithmeticbook107983.htmland I have dozens of questions on this concept. Keep solving and I will keep putting more and more questions!
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Re: Tough airthmetic
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27 Apr 2011, 16:59
@subhashghosh Check out the very first post of mine in this topic. It shows you how you can apply the scale method here. You can also apply the formula as you wrote in your message. w1/w2 = (A2  Avg)/(Avg  A1) The problem in your solution is that you have switched the ratio. You took 40/60 which is weightage to final/weightage to tests But you are considering A2 to be the score in the finals. If final weightage is w1 (since 40 is numerator), final % will be A1. Remember, you can take anything as w1 and the other as w2 but you need to be careful with A1 and A2 after that. Also, I like to keep A2 the one which is greater than the average so that I don't have to mess around with negatives very much. So this is how I will make the equation: 60/40 = (x  60)/(60  45) x = 82.5%
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Re: Tough airthmetic
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27 Apr 2011, 22:01
total = 100, final = 40 out of 60 he has obtained 60 * (0.45) = 27
for passing he needs = 60  27 = 33
hence percentage required = (33/40) * 100 = 82.5%



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Re: Tough airthmetic
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28 Apr 2011, 04:19
Final Exam is 40% The rest of the exam is 60%
Solution: \(.4X + .6(45) = 1(60)\) \(.4X + 27 = 60\) \(.4X = 60  27\) \(.4X = 33\) \(X = 82.5 %\)



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Re: Tough airthmetic
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05 Jun 2012, 04:47
kuttingchai wrote: Hello Karishma,
I like this method, its a time saver, I solved the same problem with plugin method and took 9 mins to get the answer.
I have one question though, this method work fine with 2 weighted avgs, what happen if we have 3+ weights?
This method works only when there are 2 elements. For 3 elements, you need to use the formula Aavg = (A1*w1 + A2*w2 + A3*w3)/(w1 + w2 + w3)
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Re: Tough airthmetic
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10 Dec 2012, 23:40
VeritasPrepKarishma wrote: gettinit wrote: Karishma can you please explain your method a bit more? I don't understand how you solved the problem using it? thanks
What is weighted average? It is average when each value has a different weight. e.g. a group of friends has 10 boys and 20 girls. Average age of boys is 20 years and average age of girls is 17 years. What is the average age of the group? Here, the average is weighted since we have different number of boys and girls. We calculate it as follows: \(W Avg = \frac{20*10 + 17* 20}{10 + 20}\) What we are doing instinctively here is using weighted average formula which as given below: \(C_{avg} = \frac{C_1*W_1 + C_2 * W_2}{W_1 + W_2}\) You need to find the average of C and W is the weight. In the example above, C is age and W is number of boys and girls. The alligation method, or the scale method as we call it, is based on the weighted averages formula itself: \(C_{avg} = \frac{C_1*W_1 + C_2 * W_2}{W_1 + W_2}\) If I rearrange the formula, I get \(\frac{W_1}{W_2} = \frac{C_2  C_{avg}}{C_{avg}  C_1}\) So I get that weights will be in the same ratio as difference between higher value of C and average value of C and difference between average value of C and lower value of C. How does this help? Knowing this, we can directly make a diagram and get the answer. e.g. A group of friends has 10 boys and some girls. Average age of boys is 20 years and average age of girls is 17 years. The average age of the group is 18 years. How many girls are there? Draw: Attachment: Ques1.jpg On a scale (number line), mark 17 years as age of girls, 18 years as average and 20 years as age of boys. Now, distance between 17 and 18 is 1 and distance between 18 and 20 is 2, The ratio of W1/W2 will be 2:1 (Note, the numbers 1 and 2, give a ratio of 2:1 for girls:boys as seen by the formula) Since there are 10 boys, there will be 20 girls. This method is especially useful when you have the average and need to find the ratio of weights. [/quote] 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 rearrange the formula, I get \(\frac{W_1}{W_2} = \frac{C_2  C_{avg}}{C_{avg}  C_1}\) How did you rearrange the formula to get that? I don't see it. Thanks in advance!



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Re: Tough airthmetic
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11 Dec 2012, 23:39
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 rearrange the formula, I get \(\frac{W_1}{W_2} = \frac{C_2  C_{avg}}{C_{avg}  C_1}\) How did you rearrange 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
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