Answer:
8 boys
Step-by-step explanation:
Given
Let
[tex]n \to[/tex] The initial number of boys
[tex]A \to[/tex] Amount paid by each
Required
Number of boys that paid
After four boys left, the amount paid by each is:
[tex]A = \frac{192}{n - 4}[/tex]
Before the four boys left, the amount each would have paid is:
[tex]A -8= \frac{192}{n}[/tex] i.e. $8 less
Make A the subject
[tex]A= \frac{192}{n} + 8[/tex]
Equate both expressions
[tex]A = A[/tex]
This gives:
[tex]\frac{192}{n - 4} =\frac{192}{n} + 8[/tex]
Take LCM
[tex]\frac{192}{n - 4} =\frac{192+ 8n}{n}[/tex]
Cross multiply
[tex]192n = (192 + 8n)(n -4)[/tex]
Open brackets
[tex]192n = 192n - 768 + 8n^2 - 32n[/tex]
Subtract 192n from both sides
[tex]0 = - 768 + 8n^2 - 32n[/tex]
Rearrange
[tex]8n^2 - 32n - 768 = 0[/tex]
Divide through by 8
[tex]n^2 - 4n - 96 = 0[/tex]
Expand
[tex]n^2 + 8n - 12n - 96 = 0[/tex]
Factorize
[tex]n(n + 8) - 12(n + 8) = 0[/tex]
Factor out n + 8
[tex](n - 12) (n + 8) = 0[/tex]
Split
[tex]n - 12 = 0[/tex] or [tex]n + 8= 0[/tex]
Solve for n
[tex]n =12[/tex] or [tex]n = -8[/tex]
n cannot be negative
So:
[tex]n = 12[/tex]
The number of boys that stayed and paid are:
[tex]Boys = n -4[/tex]
[tex]Boys = 12-4[/tex]
[tex]Boys = 8[/tex]
