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The average weight of a class is x pounds. When a new stud [#permalink]

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27 Feb 2013, 03:15

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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?

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(x-1) --> x = 86 pounds.

Re: The average weight of a class is x pounds. When a new stud [#permalink]

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27 Feb 2013, 04:15

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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 :

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

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,

Re: The average weight of a class is x pounds. When a new stud [#permalink]

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21 Apr 2013, 09:02

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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 :

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

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.
_________________

........................................................................................ See it big and keep it simple.

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.

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.

Re: The average weight of a class is x pounds. When a new stud [#permalink]

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18 Apr 2013, 12:24

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 :

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

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 ?

Re: The average weight of a class is x pounds. When a new stud [#permalink]

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18 Apr 2013, 14:01

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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
_________________

Re: The average weight of a class is x pounds. When a new stud [#permalink]

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21 Apr 2013, 23:15

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 )

Re: The average weight of a class is x pounds. When a new stud [#permalink]

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22 Apr 2013, 01:15

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(x-1) --> x = 86 pounds.

Answer: B.

Hope it's clear.

This is the best way to solve this.
_________________

When you feel like giving up, remember why you held on for so long in the first place.

Re: The average weight of a class is x pounds. When a new stud [#permalink]

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21 Oct 2014, 14:53

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Re: The average weight of a class is x pounds. When a new stud [#permalink]

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12 Dec 2015, 23:22

Hello from the GMAT Club BumpBot!

Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).

Want to see all other topics I dig out? Follow me (click follow button on profile). You will receive a summary of all topics I bump in your profile area as well as via email.
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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.

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).

The average weight of a class is x pounds. When a new stud [#permalink]

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13 May 2016, 18:32

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 x-1 to x+4, or 5 pounds w/x=30/5=n=6 students equation 1: 6(x-1)=w+80 equation 2: 5x=w subtracting e2 from e1, x=86 pounds

gmatclubot

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