### Brain Teasers

# Student Averages ...

Fun: (2.15)
Difficulty: (2.86)
Puzzle ID: #19274

Submitted By: lesternoronha1 Corrected By: MarcM1098

Submitted By: lesternoronha1 Corrected By: MarcM1098

The average marks of girls in a class is equal to the number of boys and the average marks of boys in a class is equal to the number of girls. If the class average is 4 less than the average of the boys' average marks and the girls' average marks, find the minimum double digit number of students in the class.

### Answer

The minimum double digit strength of the class is 18.Let the number of girls in the class be X and the number of boys in the class be Y.

Then the average marks of the girls and boys is Y and X respectively.

Class average

= (Total marks/Total number of students)

= (Total marks scored by girls and boys/Total number of students in the class)

= ((Y*X + X*Y)/(Y+X))

= ((2*X*Y)/(X+Y))

The average of the boys' average and the girls' average

= (X+Y)/2

It is given that the class average is 4 less than the average of the boys' average marks and the girls' average marks.

Hence

(X+Y)/2 - ((2*X*Y)/(X+Y)) = 4

Multiplying the above equation throughout by 2*(X+Y), we get

(X+Y)^2 - 4*X*Y = 8*(X+Y)

X^2 + Y^2 + 2*X*Y - 4*X*Y = 8*(X+Y)

X^2 + Y^2 - 2*X*Y = 8*(X+Y)

(X-Y)^2 = 8*(X+Y)

In order for the Right Hand Side in the above equation to be a perfect square, the value of (X+Y) should be of the type 2k^2 i.e. 2, 8, 18, 32, 50, 72 etc.

It is given that the minimum double digit strength of the class has to be found.

Therefore the minimum double digit strength of the class = X+Y = 18.

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