Find three consecutive integers that have a sum of 96.

Let the smallest number = n

Short 31 32 33
Let the second number = n + 1
Let the 3rd number = n + 2

Let the sum = s
s = 96

s = n + n + 1 + n + 2 Collect like terms.
s = 3n + 3
96 = 3n + 3  Subtract 3 from both sides
96 - 3 = 3n  Collect like terms on the left
93 = 3n  Divide by 3
93/3 = n
n = 31 

So the 3 consecutive numbers are
31 , 32, 33 

For a 0.25 pfor b o.3pp (0.25+0.3)=0.28