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The average weight of a class is x pounds. When a new stud
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The average weight of a class is x pounds. When a new student weighing 80 pounds joins the class, the average decreases by 1 pound. In a few months the student’s weight increases to 110 pounds and the average weight of the class becomes x + 4 pounds. None of the other students’ weights changed. What is the value of x? A. 85 B. 86 C. 88 D. 90 E. 92
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Originally posted by emmak on 27 Feb 2013, 04:15.
Last edited by Bunuel on 27 Feb 2013, 05:27, edited 1 time in total.
Edited the question.




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Re: The average weight of a class is x pounds. When a new stud
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27 Feb 2013, 05:49
emmak wrote: The average weight of a class is x pounds. When a new student weighing 80 pounds joins the class, the average decreases by 1 pound. In a few months the student’s weight increases to 110 pounds and the average weight of the class becomes x + 4 pounds. None of the other students’ weights changed. What is the value of x?
A. 85 B. 86 C. 88 D. 90 E. 92 When the student weighs 80 pounds the average weight is x  1 pounds; When the student weighs 110 pounds the average weight is x + 4 pounds. So, the increase in total weight of 110  80 = 30 pounds corresponds to the increase in average weight of (x + 4)  (x  1) =5 pounds, which means that there are 30/5 = 6 students (including the new one). So, initially there were 5 student. Total weight = 5x + 80 = 6(x1) > x = 86 pounds. Answer: B. Hope it's clear.
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Re: The average weight of a class is x pounds. When a new stud
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27 Feb 2013, 05:15
emmak wrote: The average weight of a class is x pounds. When a new student weighing 80 pounds joins the class, the average decreases by 1 pound. In a few months the student’s weight increases to 110 pounds and the average weight of the class becomes x + 4 pounds. None of the other students’ weights changed. What is the value of x?
85
86
88
90
92 Let the number of students by y. We know that their weight is fixed. Thus, initially, when a new student with weight 80 pounds joins,we have : \(80+y*1 = x1.\) Again, when the same student weighs 110 pounds, we have : \(1104*y = x+4.\) Thus, multiplying the first equation by 4 and adding both we get, 320+110 =\(5*x\) or x = \(\frac{430}{5}\)= 86. B.
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Re: The average weight of a class is x pounds. When a new stud
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09 Mar 2013, 11:56
bunnel, Can you explain how Total weight = 5x + 80 = 6(x1) > x = 86 pounds.??



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10 Mar 2013, 06:17
skamal7 wrote: bunnel, Can you explain how Total weight = 5x + 80 = 6(x1) > x = 86 pounds.?? (Total weight) = (The # of students) * (The average weight) = 6*(x1). Similarly, the total weight for 5 students is 5x, which we know that is 80 pounds less than the total weight for 6 students. Hope it's clear.
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Re: The average weight of a class is x pounds. When a new stud
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18 Apr 2013, 13:24
vinaymimani wrote: emmak wrote: The average weight of a class is x pounds. When a new student weighing 80 pounds joins the class, the average decreases by 1 pound. In a few months the student’s weight increases to 110 pounds and the average weight of the class becomes x + 4 pounds. None of the other students’ weights changed. What is the value of x?
85
86
88
90
92 Let the number of students by y. We know that their weight is fixed. Thus, initially, when a new student with weight 80 pounds joins,we have : \(80+y*1 = x1.\) Again, when the same student weighs 110 pounds, we have : \(1104*y = x+4.\) Thus, multiplying the first equation by 4 and adding both we get, 320+110 =\(5*x\) or x = \(\frac{430}{5}\)= 86. B. could anyone please elaborate on how the above two equations were derived ? thanks ~



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Re: The average weight of a class is x pounds. When a new stud
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18 Apr 2013, 15:01
Quote: Let the number of students by y. We know that their weight is fixed. Thus, initially, when a new student with weight 80 pounds joins,we have :
\(80+y*1 = x1.\)
Again, when the same student weighs 110 pounds, we have :
\(1104*y = x+4.\)
Thus, multiplying the first equation by 4 and adding both we get,
320+110 =\(5*x\)
or x = \(\frac{430}{5}\)= 86.
B.
could anyone please elaborate on how the above two equations were derived ? Quote: thanks ~ When a new student joins and this results in a drop by 1 in the average, it is as if each student present in the class gave 1 pound to him. Also, after getting 1 pound from each student, the new weight the student has must equal the new average. Thus, 80+y*1 = x1 Similarly, when his weight becomes 110 pounds, to increase the average , he must have contributed 4 pounds to each student .Just as above, this new weight must equal the average. Thus, 1104*y = x+4
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Re: The average weight of a class is x pounds. When a new stud
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21 Apr 2013, 10:02
gmatquant25 wrote: vinaymimani wrote: emmak wrote: The average weight of a class is x pounds. When a new student weighing 80 pounds joins the class, the average decreases by 1 pound. In a few months the student’s weight increases to 110 pounds and the average weight of the class becomes x + 4 pounds. None of the other students’ weights changed. What is the value of x?
85
86
88
90
92 Let the number of students by y. We know that their weight is fixed. Thus, initially, when a new student with weight 80 pounds joins,we have : \(80+y*1 = x1.\) Again, when the same student weighs 110 pounds, we have : \(1104*y = x+4.\) Thus, multiplying the first equation by 4 and adding both we get, 320+110 =\(5*x\) or x = \(\frac{430}{5}\)= 86. B. could anyone please elaborate on how the above two equations were derived ? thanks ~ Hi, Ao  old average An  new average X  weight of the new student n  number of students including the new guy c  any constant (in our case 1) This is the equation for calculating the average in that case \(\frac{X + (n1)*Ao}{n}=Ao + c\) \(X=(1n)*Ao+n*(Ao+c)\) \(X=Ao+n*c\) \(X=Ao+cc+n*c\) (Ao+c = An) \(X=An+c*(n1)\) So the weight of the new student equals the new average plus n1(each of the old students) times c. Btw. it is not necessary to use above trick to get the result for that problem. If anyone is interested I can post an alternative way.
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Re: The average weight of a class is x pounds. When a new stud
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21 Apr 2013, 20:59
emmak wrote: The average weight of a class is x pounds. When a new student weighing 80 pounds joins the class, the average decreases by 1 pound. In a few months the student’s weight increases to 110 pounds and the average weight of the class becomes x + 4 pounds. None of the other students’ weights changed. What is the value of x?
A. 85 B. 86 C. 88 D. 90 E. 92 Check out this post for a discussion on mean and how to solve such questions logically. http://www.veritasprep.com/blog/2012/04 ... eticmean/
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Re: The average weight of a class is x pounds. When a new stud
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21 Apr 2013, 21:46
(x+4)(x1)=5 11080=30 30/5=6 80+6=86
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Re: The average weight of a class is x pounds. When a new stud
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22 Apr 2013, 00:15
emmak wrote: The average weight of a class is x pounds. When a new student weighing 80 pounds joins the class, the average decreases by 1 pound. In a few months the student’s weight increases to 110 pounds and the average weight of the class becomes x + 4 pounds. None of the other students’ weights changed. What is the value of x?
A. 85 B. 86 C. 88 D. 90 E. 92 lets assume Total weight = T ; Total number of people = N and average = X (as given in question ) as per the average formula .. X=T / N Condition 1 : ( X1 ) = ( T+80 )/ ( N+1) ... => XN=81 Condition 2 : (X+4) = ( T+110) /(N+1)... => X+ 4N = 106 Solving we get X=86 ..!
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22 Apr 2013, 02:15
Bunuel wrote: emmak wrote: The average weight of a class is x pounds. When a new student weighing 80 pounds joins the class, the average decreases by 1 pound. In a few months the student’s weight increases to 110 pounds and the average weight of the class becomes x + 4 pounds. None of the other students’ weights changed. What is the value of x?
A. 85 B. 86 C. 88 D. 90 E. 92 When the student weighs 80 pounds the average weight is x  1 pounds; When the student weighs 110 pounds the average weight is x + 4 pounds. So, the increase in total weight of 110  80 = 30 pounds corresponds to the increase in average weight of (x + 4)  (x  1) =5 pounds, which means that there are 30/5 = 6 students (including the new one). So, initially there were 5 student. Total weight = 5x + 80 = 6(x1) > x = 86 pounds. Answer: B. Hope it's clear. This is the best way to solve this.
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Re: The average weight of a class is x pounds. When a new stud
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13 May 2016, 09:40
Hi Experts / chetan2u, Please have a look on my approach. I have created 3 equations 1) (Sum) / N = X 2) When a new student of weight 80 is added the equations become (Sum +80)/ (N+1) = X1 3) When the students gain weight by 110. In this case the students have put on more weight so we just need to add 110 in the sum. (Sum + 110) / N = X+4 After solving these equations I am getting X=102 which is not correct. I know, I am missing something. Can you please assist. Thanks and Regards, Prakhar
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Re: The average weight of a class is x pounds. When a new stud
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13 May 2016, 09:47
PrakharGMAT wrote: Hi Experts / chetan2u, Please have a look on my approach. I have created 3 equations 1) (Sum) / N = X 2) When a new student of weight 80 is added the equations become (Sum +80)/ (N+1) = X1 3) When the students gain weight by 110. In this case the students have put on more weight so we just need to add 110 in the sum. (Sum + 110) / N= X+4 After solving these equations I am getting X=102 which is not correct. I know, I am missing something. Can you please assist. Thanks and Regards, Prakhar hi, i find the approach to be OK BUT the highlighted portion has to be N+1 and not N.. Try putting that and check your answer
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Re: The average weight of a class is x pounds. When a new stud
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13 May 2016, 09:54
Hi chetan2u, I am getting the corect answer now. But I am not able to understand, the question says In a few months the student’s weight increases to 110 pounds.So, the students have gain weight, then why do we need to denominator as (N+1). Can you please assist..? Thanks
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Re: The average weight of a class is x pounds. When a new stud
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13 May 2016, 09:58
PrakharGMAT wrote: Hi chetan2u, I am getting the corect answer now. But I am not able to understand, the question says In a few months the student’s weight increases to 110 pounds.So, the students have gain weight, then why do we need to denominator as (N+1). Can you please assist..? Thanks Prakhar , if you read the Q,it says the person who joined at 80 kgs has increased to 110 kgs.. and Including him TOTAL is N+1.. hope it is clear.. Quote: When a new studentweighing 80 pounds joins the class, the average decreases by 1 pound. In a few months the student’s weight increases to 110 pounds and the average weight of the class becomes x + 4 pounds.
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Re: The average weight of a class is x pounds. When a new stud
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13 May 2016, 10:50
Hi chetan2u, Thanks a lot for clearing my doubt.
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The average weight of a class is x pounds. When a new stud
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13 May 2016, 19:32
let w=total original weight of students n=number of students x=w/n=average weight of original students as w increases from 80 to 110, or 30 pounds, x increases from x1 to x+4, or 5 pounds w/x=30/5=n=6 students equation 1: 6(x1)=w+80 equation 2: 5x=w subtracting e2 from e1, x=86 pounds



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The average weight of a class is x pounds. When a new stud
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18 Dec 2016, 11:14
Here is my solution to this a=question =>
Let the number of students be N Average = x
Using
\(Mean = \frac{Sum}{#}\)
Sum=Nx
Now As per Question
Nx+80/N+1 = x1 And Nx+110/N+1
Solving the Equations we get x=86
Hence B
Method 2=>
110=x+4(N+1) Hence 110=x+4N+4
Also 80=x1(N+1) 81=xN
Solving we again get =>x=86
Hence B
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Re: The average weight of a class is x pounds. When a new stud
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30 Dec 2016, 08:38
my take
4 1 80x110
x = 80 + [(11080)*1/5)] = 80+6 = 86
the new average is a weighted average of the 2 weights of the student , weights are the effect his weight had on the original average x .
when his weight was 80 he had an effect of 1 and when his weight was 110 he had an effect of 4.




Re: The average weight of a class is x pounds. When a new stud
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