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Bunuel
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Let x be the population of class

90x is the total marks
x/3 students have marks = 96 * x/3

Rest of the class is 2/3x
The marks of x/3 + marks of 2x/3 = marks of x

i.e. 96X/3 + 2/3*avg = 90X
2/3 marks = (90 - 96/3)X
marks = 58 *3 /2 X
avg = 87
Ans D
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Bunuel
The grade point average of the entire class is 90. If the average of one third of the class is 96, what is the average of the rest?

A. 92.
B. 89.
C. 88.
D. 87.
E. 86.

Let the average of 2/3 rd class = a

And, number of students in class= x

Total = number *average

Total points of whole class= total points of 1/3 class +total points of 2/3rd class
90x= 96 *1/3x + a* 2/3x
3/2x(90x-32x)= a
a= 87

D is the answer
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equation1: t/n=90
equation2: (t-x)/(n/3)=96➡t/n-x/n=32
subtracting 2 from 1,
x/n=58
because ratio seems high and we know n is a multiple of 3,
assume n=3
then x=58*3=174
174/2=87=average of other 2 students

(checking: 90*3-174=96)
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Bunuel
The grade point average of the entire class is 90. If the average of one third of the class is 96, what is the average of the rest?

A. 92.
B. 89.
C. 88.
D. 87.
E. 86.


We can solve this question using mixtures type problem

90 is average of the entire class.

1/3 students avg 96 and the rest 2/3 avg is X ( let's assume this).

96 X

90

The ratio between the 2/1

=> 2/1 = 6 / 90 - X

=> 3 = 90 - X

X = 87.

IMO D is correct.
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Bunuel,
Can you please explain how come the ans is 88?
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nipunjain14
Bunuel,
Can you please explain how come the ans is 88?
______________
Edited. Thank you.
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can be solved within 5 sec

96...............x
........90

1...............3
3..............6

90-3 =87
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Let the number of students =3
Hence one of them will have score of 96
and Sum(3)= 270
So Sum(2)=174
Hence Mean(2)=> 174/2 => 87

Hence D
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