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

Question

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

A. 85
B. 86
C. 88
D. 90
E. 92

Bunuel · Accepted Answer

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.

KarishmaB · Answer

Sum/N = X
Sum = NX

(Sum +80)/ (N+1) = X-1
(Sum + 80) = (N+1)*(X - 1)
NX + 80 = NX - N + X - 1
X - N = 81 ........ (i)

(Sum + 110) / (N+1) = X+4
Sum + 110 = (X + 4)(N + 1)
NX+ 110 = NX + 4N + X + 4
4N + X = 106 ......... (ii)

Solve (i) and (ii)
4X - 4N = 324
4N + X = 106

5X = 430
X = 86

mau5 · Answer

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.

skamal7 · Answer

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

Bunuel · Answer

(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.

gmatquant25 · Answer

could anyone please elaborate on how the above two equations were derived ?

thanks ~

mau5 · Answer

could anyone please elaborate on how the above two equations were derived ?

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

Ultimor · Answer

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.

kabilank87 · Answer

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 ..!

PrakharGMAT · Answer

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

chetan2u · Answer

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

PrakharGMAT · Answer

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

chetan2u · Answer

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..

fskilnik · Answer

(All weights are in pounds.)

? = x

Excellent opportunity to use the  homogeneity nature of the average :

\sum
olimits_n { = \,\,nx\,\,\,\,\,\left( {n\,\,{m{students}}} ight)}

\left\{ \matrix{
 80 + \sum
olimits_n {\, = \sum
olimits_{n + 1} {\, = \,\,\left( {n + 1} ight)\left( {x - 1} ight)} } \hfill \cr 
 110 + \sum
olimits_n {\, = \sum
olimits_{n + 1} {\, = \,\,\left( {n + 1} ight)\left( {x + 4} ight)} } \hfill \cr} ight.\,\,\,\,\,\, \Rightarrow \,\,\,\,\,110 - 80 = \left( {n + 1} ight)\left

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

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

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

Regards, 
Fabio.

GMATGuruNY · Answer

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. B

ShankSouljaBoi · Answer

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

bumpbot · Answer

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

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