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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, 04:15

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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?

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

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27 Feb 2013, 05:49

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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?

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, 05:15

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Expert's post

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, 10:02

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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.

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

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21 Apr 2013, 20:59

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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?

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.

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

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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 :

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

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, 15:01

Expert's post

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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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 )

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

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

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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).

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