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emmak
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 :

\(80+y*1 = x-1.\)

Again, when the same student weighs 110 pounds, we have :

\(110-4*y = x+4.\)

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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Can you explain how Total weight = 5x + 80 = 6(x-1) --> x = 86 pounds.??
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bunnel,
Can you explain how Total weight = 5x + 80 = 6(x-1) --> x = 86 pounds.??

(Total weight) = (The # of students) * (The average weight) = 6*(x-1).

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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emmak
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 :

\(80+y*1 = x-1.\)

Again, when the same student weighs 110 pounds, we have :

\(110-4*y = x+4.\)

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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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 :

\(80+y*1 = x-1.\)

Again, when the same student weighs 110 pounds, we have :

\(110-4*y = x+4.\)

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 = x-1

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, 110-4*y = x+4
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gmatquant25
vinaymimani
emmak
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 :

\(80+y*1 = x-1.\)

Again, when the same student weighs 110 pounds, we have :

\(110-4*y = x+4.\)

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

\(\frac{X + (n-1)*Ao}{n}=Ao + c\)

\(X=(1-n)*Ao+n*(Ao+c)\)

\(X=Ao+n*c\)

\(X=Ao+c-c+n*c\)

(Ao+c = An)

\(X=An+c*(n-1)\)



So the weight of the new student equals the new average plus n-1(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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emmak
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 : ( X-1 ) = ( T+80 )/ ( N+1) ... => X-N=81

Condition 2 : (X+4) = ( T+110) /(N+1)... => X+ 4N = 106

Solving we get X=86 ..!
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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) = X-1

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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PrakharGMAT
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) = X-1

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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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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PrakharGMAT
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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emmak
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
(All weights are in pounds.)

\(? = x\)

Excellent opportunity to use the homogeneity nature of the average:

\(\sum\nolimits_n { = \,\,nx\,\,\,\,\,\left( {n\,\,{\rm{students}}} \right)}\)

\(\left\{ \matrix{\\
80 + \sum\nolimits_n {\, = \sum\nolimits_{n + 1} {\, = \,\,\left( {n + 1} \right)\left( {x - 1} \right)} } \hfill \cr \\
110 + \sum\nolimits_n {\, = \sum\nolimits_{n + 1} {\, = \,\,\left( {n + 1} \right)\left( {x + 4} \right)} } \hfill \cr} \right.\,\,\,\,\,\, \Rightarrow \,\,\,\,\,110 - 80 = \left( {n + 1} \right)\left[ {\left( {x + 4} \right) - \left( {x - 1} \right)} \right]\)

\(30 = 5\left( {n + 1} \right)\,\,\,\,\, \Rightarrow \,\,\,\,\,n = 5\)

\(80 + 5x = \left( {5 + 1} \right)\left( {x - 1} \right)\,\,\,\,\, \Rightarrow \,\,\,\,\,? = x = 86\)


This solution follows the notations and rationale taught in the GMATH method.

Regards,
Fabio.
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emmak
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

We can PLUG IN THE ANSWERS and use alligation.
To learn about alligation, check my posts here:
https://gmatclub.com/forum/a-car-averag ... 43216.html
https://gmatclub.com/forum/in-a-certain ... 93667.html
https://gmatclub.com/forum/two-mixtures ... 62-20.html
https://gmatclub.com/forum/in-a-200-mem ... 75-20.html

Let O = the original number of students and N = the new student.
Whether the new student weighs 80 pounds or 110 pounds, alligation must yield the SAME VALUE for O/N -- the ratio of old students to new student -- in each case.

D: x = 90
When the new student weighs 80 pounds, the average for all the students = x-1 = 90-1 = 89.
The following alligation is yielded:
O 90-----1-----89-----9-----80 N
Here, O/N = 9/1.
When the new student weighs 110 pounds, the average for all the students = x+4 = 90+4 = 94.
The following alligation is yielded:
O 90-----4-----94-----16-----110 N
Here, O/N = 16/4 = 4/1.
Since each alligation yields a different value for O/N, eliminate D.

B: x = 86
When the new student weighs 80 pounds, the average for all the students = x-1 = 86-1 = 85.
The following alligation is yielded:
O 86-----1-----85-----5-----80 N
Here, O/N = 5/1.
When the new student weighs 110 pounds, the average for all the students = x+4 = 86+4 = 90.
The following alligation is yielded:
O 86-----4-----90-----20-----110 N
Here, O/N = 20/4 = 5/1.
Success!
Each alligation yields the same value for O/N.

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emmak
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

My approach ... dont even think of multiplying the equations .It gets time consuming . Refer the image.
m is the no of students.
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