Question

Find the three consecutive whole numbers whose sum is more than 45 but less than 54

Answer 1

Here is your answer-

Let,
the three consecutive number be x, x+1, x+2.
A/C
45< x + x+1 +x+2 < 54
45< 3x+3 < 54
42<3x<51 ( subtracted by 3)
14<x<17 ( divided by 3)
..
So the value of x must be between 14 & 17..
possibilities are 15 and 16.
If x= 15 ..then
x+1= 15+1=16
x+2=15+2=17

Total sum = 15+16+17=48..
it verify 1st condition. .45<48

If x=16. then
x+1=16+1=17
x+2=16+2=18

Total sum = 16+17+18=51
it verify 2nd condition..54>51

Therefore ..the three consecutive no. are 15,16,17 or 16,17,18

Hope it helps..!