Question

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Solve the above problem.
#Explanation Required​

Answer 1

Say there are x boys and y girls. It might be helpful later, to remember that x + y = 8

Since the expense of each boy is 2 less than number of girls, i.e. y-2, the expense of all boys must be the number of boys times the expense of one boy, i.e. x(y-2), and similarly, the expense of all girls combined is y(x-1)

If x(y-2):y(x-1) = 4:3, this would mean

3xy - 6x = 4xy - 4y (cross multiplication you call it, I guess)

4y - 6x = xy

Remember from the previous result that x and y add up to 8.

4y - 6(8-y) = (8-y)y

4y + 6y - 48 = 8y - y²

y² + 2y - 48 = 0

y² + 8y - 6y - 48 = 0

y(y+8) -6(y+8) = 0

(y+8)(y-6) = 0

y = -8 or 6

Number of boys can't be negative (unless those are some special superboys) so y = 6

The required answer is y-2, i.e. the expense of each boy, in essence, 4.

Answer 2

Answer:

Above answer is correct mark it as brainliest

Step-by-step explanation:

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