Let number of $250 bonus = x
Let number of $750 bonus = 7
Given that total bonus amount awarded is $3000. So we can write equation:
250x+750y=3000
750y=3000-250x
[tex]y=\frac{3000-250x}{750}[/tex]
[tex]y=\frac{300-25x}{75}[/tex]
[tex]y=\frac{12-x}{3}[/tex]
Given that both type of bonus awarded are at least one.
So that means y>1 and x>1
Now we have to find any one possible solution.
Number of awards can only be whole numbers so we can plug x=2,3,4,... to see which one satisfies above equation and gives x as whole number.
I found that x=3 is working good.
at x=3, we get:
[tex]y=\frac{12-x}{3}[/tex]
[tex]y=\frac{12-3}{3}[/tex]
[tex]y=\frac{9}{3}[/tex]
y=3
Hence final answer is given by:
Number of $250 bonus = 3
and number of $750 bonus = 3
