Question

Find three largest consecutive natural numbers such that the sum of one-third of first, one-forth of second and one fifth of the third is atmost 25

Answer 1

let the first no be x
second consecutive no be x+1
third consecutive no be x+2

one third of first + one fourth of second + one fifth of third = 25
x/3 +(x+1)/4 + (x+2)/5 =25
x/3 + x/4 + 1/4 + x/5 + 2/5 = 25
x/3 + x/4 + x/5 + 1/4 + 2/5 = 25
(20x + 15x + 12x) / 60 + [(5+8)/20] = 25
47x/60 + 13/20 = 25
47x/60 = 25-(13/20)
47x/60 = 487/20
x = 487/20 × (60/47)
x= 31.0851063829