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# 23 students appeared for the Geography paper. What is the su

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Manager
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23 students appeared for the Geography paper. What is the su  [#permalink]

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11 Aug 2014, 01:26
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62% (01:58) correct 38% (01:22) wrong based on 124 sessions

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23 students appeared for the Geography paper. What is the sum of marks obtained by them?

(1) The average mark obtained by the students equals the median mark obtained by the students

(2) The top 12 students have an average of 84 marks and the bottom 12 students have an average of 62 marks

Can anyone please explain me this problem.
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Joined: 31 Jul 2014
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Re: 23 students appeared for the Geography paper. What is the su  [#permalink]

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11 Aug 2014, 04:32
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desaichinmay22 wrote:
23 students appeared for the Geography paper. What is the sum of marks obtained by them?

(1) The average mark obtained by the students equals the median mark obtained by the students

(2) The top 12 students have an average of 84 marks and the bottom 12 students have an average of 62 marks

Can anyone please explain me this problem.

Let assume first 11 people marks be X, 12th person marks be y (which is median), last 11 people marks be z

A) It says median is average.....Y marks is average...... so its given (11x+y+11z)/23 = y..........Not sufficient -- Eq(1)

B) so average of first 12 people (11x + y)/12 = 84.........Eq(2)
and avergae of last 12 people (y+11z)/12 = 62 ..................Eq(3)

so solving for Eq 1, Eq2, Eq3....you get value for Y....so we cant determine sum
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Re: 23 students appeared for the Geography paper. What is the su  [#permalink]

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11 Aug 2014, 04:49
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I just a little bit correct nice idea in the previous post:)

Let assume that the average mark of first 11 people be $$x$$, 12th person mark be $$y$$ (which is median), and the average mar of last 11 people be $$z$$.
Wу need to find $$11x+y+11z$$.

1) It says median is average, so $$\frac{11x+y+11z}{23} = y$$ or $$11x+y+11z=23y$$. Since we don't know anything about $$y$$, the statement is insufficient.

2) We have two equations:
$$\frac{11x+ y}{12} = 84$$
$$\frac{y+11z}{12} = 62$$

or
$$11x=84*12-y$$
$$11z= 62*12-y$$

So, the question is $$11x+y+11z=84*12-y+y+62*12-y=84*12+62*12+y$$. We still don't know anything about y. Insufficient

(1)+(2) $$84*12+62*12+y=23y$$. So, we can find $$y$$ and answer the question. Sufficient

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23 students appeared for the Geography paper. What is the su  [#permalink]

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08 Oct 2015, 07:37
desaichinmay22 wrote:
23 students appeared for the Geography paper. What is the sum of marks obtained by them?

(1) The average mark obtained by the students equals the median mark obtained by the students

(2) The top 12 students have an average of 84 marks and the bottom 12 students have an average of 62 marks

Can anyone please explain me this problem.

Statement I} We do not know the individual marks , so, by itself, this information is of not much use to us.
<b>Insufficient</b>

Statement II} We have the averages of top 12 and Bottom 12 students. Each of these will include one common student, the median!. Let this student's marks be x. So we have :-

TopSum+BotSum+2x = (84+62)*12
From this, we can derive :-
TotSum + x = 146 * 12
But we cannot progress further from here, as we donot know TotSum or x. so <b> Insufficient</b>

Combining I and II]
Now we know that avg marks of 23 students equals the median marks!
So, TotSum = 23*x!

Substituting this in our earlier eq :-
23*x+x= 146*12

Therefore 24*x = 146*12.

No need to solve any further, we can get the median and then the avg and sum. <b>Sufficient</b>

Hence the correct Ans is C
Pls. point out if I missed anything
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Re: 23 students appeared for the Geography paper. What is the su  [#permalink]

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08 Oct 2015, 22:43
Forget conventional ways of solving math questions. In DS, Variable approach is the easiest and quickest way to find the answer without actually solving the problem. Remember equal number of variables and independent equations ensures a solution.

23 students appeared for the Geography paper. What is the sum of marks obtained by them?

(1) The average mark obtained by the students equals the median mark obtained by the students

(2) The top 12 students have an average of 84 marks and the bottom 12 students have an average of 62 marks

There are 23 variables (23 students) and we require 23 equations in order to match the numbers, but only 2 equations are given; this makes (E) likely to be the answer.
If we look at the conditions together, the total average =(12*84+12*62)/(12+12)=73=median, so the total sum becomes 12*84+12*62-73=1679. The conditions provide a unique answer, therefore they are sufficient and the answer becomes (C)

For cases where we need 3 more equation, such as original conditions with “3 variables”, or “4 variables and 1 equation”, or “5 variables and 2 equations”, we have 1 equation each in both 1) and 2). Therefore, there is 80% chance that E is the answer (especially about 90% of 2 by 2 questions where there are more than 3 variables), while C has 15% chance. These two are the majority. In case of common mistake type 3,4, the answer may be from A, B or D but there is only 5% chance. Since E is most likely to be the answer using 1) and 2) separately according to DS definition (It saves us time). Obviously there may be cases where the answer is A, B, C or D.
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Re: 23 students appeared for the Geography paper. What is the su  [#permalink]

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03 May 2016, 19:43
desaichinmay22 wrote:
23 students appeared for the Geography paper. What is the sum of marks obtained by them?

(1) The average mark obtained by the students equals the median mark obtained by the students

(2) The top 12 students have an average of 84 marks and the bottom 12 students have an average of 62 marks

Can anyone please explain me this problem.

1 alone clearly no info at all, so not good.

2 alone -> if 13 students got 84 and 12 got 62 - one sum. if 13 students got 62 and 12 students 84 - then another amount.
2 alone not sufficient.

1+2
(62*12 + 84*12 -x)/23 = x
why -x? because in first 12 we have the middle one (suppose x), then we count again when multiplying by 84, so by adding 62*12 with 84*12, we count x twice
62*12 + 84*12 -x = 23x
(62+84)*12 = 24x
146 = 2x
x=73
since we know x, we can find the sum of the first 11 numbers, then the sum of the last 11 numbers.

1+2 is sufficient.
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Re: 23 students appeared for the Geography paper. What is the su  [#permalink]

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09 Nov 2017, 10:22
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