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The average weight of the students Updated on: 10 Aug 2013, 05:55
Originally posted by Gian on 10 Aug 2013, 05:14.
Last edited by Zarrolou on 10 Aug 2013, 05:55, edited 1 time in total.
Added OA.
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Difficulty:

65% (hard)

Question Stats:

52% (02:15) correct 48% (01:43) wrong based on 23 sessions

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The average weight of the students in a class is 48kg. What is the number of students in the class?

1) If weights of two students were 64 and 45 instead of 46 and 54, then the average would be 50kg.
2) If two more students of weights 44 and 52kg joined the class, the average would remain the same.

Not convinced with the OA.

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Zarrolou
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The average weight of the students 10 Aug 2013, 05:54
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The average weight of the students in a class is 48kg. What is the number of students in the class?

1) If weights of two students were 64 and 45 instead of 46 and 54, then the average would be 50kg.
\(\frac{48n-46-54+64+45}{n}=50\) from this we get n=4.5.
This is sufficient, however this number makes no sense (you can't have 4.5 students).

2) If two more students of weights 44 and 52kg joined the class, the average would remain the same.
\(\frac{48n+44+52}{n+2}=48\), \(48n+44+52=48n+96\). This adds no info.
Not sufficient
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The average weight of the students 10 Aug 2013, 06:00
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Thanks, just wanted to double check whether I was correct.

Btw, I loved the previous line in your signature that one only requires a library and a gf in life :)
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The average weight of the students 10 Aug 2013, 06:09
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Gian
The average weight of the students in a class is 48kg. What is the number of students in the class?

1) If weights of two students were 64 and 45 instead of 46 and 54, then the average would be 50kg.
2) If two more students of weights 44 and 52kg joined the class, the average would remain the same.

Not convinced with the OA.
Show more


Gian,

My 2 cents.

let n be the number of students in the class. We are asked to solve for n.

Let w1, w2, w3......wn respectively be the weight of individual students upto n

so we know that

n * 48 = w1 + w2 + w3 + w4 .....+ wn

Option A:

Let w1 = 46 & w2= 54, this weight was changed to w1=64 & w2 = 45 and avg is 50.

so

n * 48 = 46 + 54 + w3 + w4+....wn ----- equation 1

n * 50 = 64 + 45 + w3+ w4.......wn ------- equation 2

Now subtract eq 1 from eq2

50n - 48n = 64+45+w3+...wn - ( 46+54+w3.....wn)

2n = 9

n= 4.5 students, As wWe were able to solve for N. So lets think option A is correct

Option B:

2 more students 44 & 52 were added and avg remained the same.

if you average (44+52)/2 == the avg is 48....

so it doesn't matter if n is 50 or 100 or 2 etc.....

so we cant solve for n.


In all I think this is a poor question because students cannot be in decimals. i.e. in A, we solved for n = 4.5

This was my understanding of the problem. Hope it helped if my explanation is correct.

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