Mathematics paper in our class......

Let the number of boys be $a$ . Let the number of girls be $b$ .

$60b=$ Marks obtained by girls

$80a=$ Marks obtained by boys

Marks of all students $= 40 \times 65$ $=2600$

Hence,

$80a+60b=2600$

$20\left ( 4a+3b \right )=2600$

$4a+3b=130$

$a+3a+3b=130$

Here we know that $a+b=40$

Then,

$a+3\left ( a+b \right )=130$

$a+120=130$

$a=10$

So,

$b=30$

$\therefore$ Different between the number of boys and girls is $\boxed{20}$ .

This is a great problem! Let's let $g$ be the number of girls, and $b$ be the number of boys. We can set up:

$g+b=40$

The hard part about this is about the averaging of both male and female. Because the sum of all girls scores is $60g$ , and the sum of all boys scores is $80b$ , then the average of all of their scores is

$(60g+80b)/40=65$

Solving for $g$ and $b$ , we get $g=30$ and $b=10$ . Thus the difference between them is $\boxed{20}$ .

Hello,

let b = total number of boys , g = total number of girls, and we know that b + g = 40,

given, let x = total marks for boys, y = total marks for girls,

x / b = 80 ----> x = 80b

y / g = 60 ----> x = 60g

as x + y / 40 = 65

80b + 60g = 40 x 65 ----> 4b + 3g = 130 compare with b + g = 40,

4b + 3g = 130 ----> by elimination method, 4b+ 4g = 160

-g = - 30

g= 30

b = 40 - 30 =10

Therefore, g - b = 30 -10 = 20,

thanks....

let the number of girls be x and the number of boys be 40-x

total marks of girls=60x

total marks of boys=80(40-x)

total marks of all students=65x40=2600

so, 60x + 80(40-x) = 2600

on solving we get, x=30

so, number of girls=30 and number of boys=10

difference=20

G = (80 - 65)/(80 - 60) x 40 =30 B = (65 - 60)/(80 - 60) x 40 = 10. Therefore diff. is G-B=30-10 =20

number of girls = (80 - 65)/(80 - 60) x 40 = 30,
number of boys = (65 - 60)/(80 - 60) x 40 = 10. The difference of the no. of girls & the no. of boys is 20