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M17-31

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M17-31  [#permalink]

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New post 16 Sep 2014, 00:02
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The average grade of students in group 1 is 4.0 while that of students in group 2 is 3.0. If the groups were merged into one big group, what would be the average grade in this group?


(1) There are twice as many students in group 1 as in group 2.

(2) There are 4 more students in group 1 than in group 2.

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Re M17-31  [#permalink]

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New post 16 Sep 2014, 00:02
Official Solution:


Let \(N1\) denote the number of students in group 1 and \(N2\) denote the number of students in group 2. The average grade in the new group would be \(\frac{N1*4 + N2*3}{N1 + N2}\).

Statement (1) by itself is sufficient. Using S1, we get \(\frac{2*N2*4 + N2*3}{2*N2 + N2} = \frac{11}{3}\).

Statement (2) by itself is insufficient. The variables do not cancel out from the fraction.


Answer: A
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M17-31  [#permalink]

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New post 23 May 2017, 07:36
I'm sorry I am not comprehending this. Isn't it possible to still calculate an average using statement two as long as the numbers used have four more students in group 1? If you get the time, could you please elaborate on why statement two is insufficient? Thanks you !!
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Re: M17-31  [#permalink]

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New post 23 May 2017, 07:44
NanaA wrote:
I'm sorry I am not comprehending this. Isn't it possible to still calculate an average using statement two as long as the numbers used have four more students in group 1? If you get the time, could you please elaborate on why statement two is insufficient? Thanks you !!


We don't have the ratio from the second statement.

(2) says - there are 4 more students in group 1 than in group 2: (group 1) = (group 2) - 4
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Re: M17-31  [#permalink]

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New post 23 May 2017, 07:52
NanaA wrote:
I'm sorry I am not comprehending this. Isn't it possible to still calculate an average using statement two as long as the numbers used have four more students in group 1? If you get the time, could you please elaborate on why statement two is insufficient? Thanks you !!


You can test statement 2 by pluggin in quick numbers

If class 1 has 5 students, then class 2 will have 1
The equation would be (5* 4.0 + 1 * 3.0) / ( 5 + 1) = 23 / 6 or 3 and 5/6

If class 1 has 6 students, then class 2 will have 2
The equation would be (6* 4.0 + 2 * 3.0) / ( 6 + 2) = 30 / 8 or 3 and 3/4

Since you get different answers, statement II is insufficient
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Re: M17-31  [#permalink]

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New post 24 May 2017, 05:40
The answer has to be option A. My take :

We need to find average of the entire lot. Avg of the first lot is 4 and Avg of the second lot is 3. To find the combined Avg we will need the ratio of the number of students in first lot to that in the second lot.

Option A - Sufficient
Option B - Says the number in first lot is 4 more than that in the second lot. Does not give us the ratio.

Hence option A must be the answer.
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Re: M17-31  [#permalink]

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New post 30 Aug 2018, 09:03
In this question, concept tested is weighted average.

Remember, if we know value of individual datapoint and ratio/weight of datapoint then we can easily calculate weighted average.

Hence, Statement A is clearly sufficient.

In Statement B, you can't find ratio/weight of datapoint. Hence, not sufficient.

Ans: A
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Re: M17-31 &nbs [#permalink] 30 Aug 2018, 09:03
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