11th-grade girls at Stumpville High School have an average GPA of 3.1; 52 girls have GPA = 3.1
overall 11th-grade average GPA is 3.05; 104 students have GPA = 3.05
all of the boys enrolled in Honors Chemistry are in the 11th grade and those boys have an average GPA of 3.8;
6% boys out of 52 total 11th-grade boys enrolled for honors chem; 12 boys have GPA = 3.8
lets say remaining 40 boys have GPA = x

overall 11th-grade average GPA is 3.05;
(52*3.1 + 12*3.8 + 40*x)/104 = 3.05
x= 2.76

Sorry my bad, a little more attention is needed from me!

Where are you getting 12 boys from? 6% of the boys are in honors chem and have a 3.8.

6% of 52 is 3.12

Where are you getting 12 boys from?

as mentioned in the graph; the percentage of boys and girls of the total strength of boys and girls (including all grades)
from 9tht to 12th, there are 200 boys 6% of 200 = 12

For 11th grade, number of boys and girls are same, so 11th grade boy's gpa is 3.0. (3.05 x 2 = 3.1 + 3.0)
Total gpa of 11th boys are 3 x 52 = 156.


The number of all of boys is 200 = 46+52+52+50.

6% of 200 attended "Honors Chemistry" so 12 boys whose grade is 11th attended to "Honors Chemistry".
Their average gpa is 3.8 so 3.8x12 = 45.6

What we want is so , (156 -45.6 ) / (52-12) = 110.4 / 40 = 2.76

Answer: B

y-x > x-y , x cannot be equal to y.

A1) (100 -x) (100+y)
B1) (100+x) (100-y)

If we assume x = y, A)=B) .
A2) (100-x ) * (100+x) = 10000 -x^2
B2) (100+x ) * (100-x) = 10000 -x^2

If we apply y>x,
A3) (100-x)(100+y) > A2) ,
B3) (100+x)(100-y) < A2) as A2=B2.
So A3) > B3) with y>x,

Answer : A

Sorry , seems to be different question. I will look into how I can delete this.

what is right answer is ?.....is it "a" or "b"...

The OA is B

You are right, Sorry , seems to be different question. I will look into how I can delete my post..
Thank you for pointing out.

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