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# A class has 4 sections P, Q, R and S and the average weights of the st

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Re: A class has 4 sections P, Q, R and S and the average weights of the st [#permalink]
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In order to maximize the mean we need to equalize the difference of numbers between the two poles of the data set.

45 has a difference of 10 from 55 and
50 has difference of 5 from 55 therefore, the left side of the mean has a total difference of 15 at the least.

65 has a difference of 10 from the mean, the least difference possible. Therefore, the least average of combined weight to the left should be equal to the least combined weight of the right.

$$(1)10 + 2 (5) = 2 (10) ==> 10 + 10= 20$$

Therefore, a total of 5 students in other sections which bring the number of students in Section R to 25.
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Re: A class has 4 sections P, Q, R and S and the average weights of the st [#permalink]
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Bunuel wrote:
A class has 4 sections P, Q, R and S and the average weights of the students in the sections are 45lb, 50lb, 55lb and 65lb respectively. What is the maximum possible number of students in section R if there are 40 students in all sections combined and the average weight of the all students across all the sections is 55lb? It is known that each section has at least one student.

(A) 18
(B) 20
(C) 25
(D) 35
(E) 37

Can someone explain me this question in the most easiest way to understand?
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Re: A class has 4 sections P, Q, R and S and the average weights of the st [#permalink]
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Call p,q,r,s are the number of students in each section respectively. We have p+q+r+s= 40
Total W=45p+50q+55r+65s=55*40
Rewrite the W with respect to average weight of R (55) W=(55-10)p+(55-5)q+55r+(55+10)s=55(p+q+r+s)+10s-5q-10p=W+10s-5q-10p
---> 10s-5q-10p=0 hence s:q:p=2:1:2
Each section must have at least 1 student and max r --> min q =1 --> min s = min p=2 --> max r = 40 -5=35 -->D
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Re: A class has 4 sections P, Q, R and S and the average weights of the st [#permalink]
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Bunuel wrote:
A class has 4 sections P, Q, R and S and the average weights of the students in the sections are 45lb, 50lb, 55lb and 65lb respectively. What is the maximum possible number of students in section R if there are 40 students in all sections combined and the average weight of the all students across all the sections is 55lb? It is known that each section has at least one student.

(A) 18
(B) 20
(C) 25
(D) 35
(E) 37

What to look in an average problem is distance of a object from its average... so here average is 55lb which is section R

distance of P from section R= -10lb (negative as I consider R as centre since average lies at R)
distance of Q from R = - 5lb (negative as I consider R as centre since average lies at R)
distance of S from R = 10lb

Given: Each section has at least one student ..
So 37 is ruled out since if we take 1 from P, 1 from Q, 1 from S and 37 from R....average will not be 55lb....Lets see how?
1 from P = -10 distance from R.....1 from S = +10 distance from R...so both negate each other but we have additional -5 from P...so average will be slightly below 55lb now

Lets take another case 1 from P = -10lb
2 from Q = 2*-5lb = -10lb
and 2 from S= 2*+10lb= 20lb

So overall 1 from P + 2from Q = 2 from S...and average still remains at 55lb in section R

So total students taken= 1from P + 2 from Q + 2 from S = 5

Remaning students in R = 40-5= 35, which is the answer

Please give kudos if it is useful for you to understand..
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Re: A class has 4 sections P, Q, R and S and the average weights of the st [#permalink]
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Re: A class has 4 sections P, Q, R and S and the average weights of the st [#permalink]
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