Question

The middle school baskeball team spent 192$ on a new pair of sneakers for the coach so that he would match the rest of the squad. Four boys quit the team so that they didn't have to pay for the sneakers. If they hadn't quit, the remaining boys would have each paid 8$ less. How many boys stayed on the team and payed for the sneakers?

Answer 1

8 boys

Explanation:

Given

Let

`n \to` The initial number of boys

`A \to` Amount paid by each

Required

Number of boys that paid

After four boys left, the amount paid by each is:

`A = (192)/(n - 4)`

Before the four boys left, the amount each would have paid is:

`A -8= (192)/(n)` i.e. $8 less

Make A the subject

`A= (192)/(n) + 8`

Equate both expressions

`A = A`

This gives:

`(192)/(n - 4) =(192)/(n) + 8`

Take LCM

`(192)/(n - 4) =(192+ 8n)/(n)`

Cross multiply

`192n = (192 + 8n)(n -4)`

Open brackets

`192n = 192n - 768 + 8n^2 - 32n`

Subtract 192n from both sides

`0 = - 768 + 8n^2 - 32n`

Rearrange

`8n^2 - 32n - 768 = 0`

Divide through by 8

`n^2 - 4n - 96 = 0`

Expand

`n^2 + 8n - 12n - 96 = 0`

Factorize

`n(n + 8) - 12(n + 8) = 0`

Factor out n + 8

`(n - 12) (n + 8) = 0`

Split

`n - 12 = 0` or
`n + 8= 0`

Solve for n

`n =12` or
`n = -8`

n cannot be negative

So:

`n = 12`

The number of boys that stayed and paid are:

`Boys = n -4`

`Boys = 12-4`

`Boys = 8`