Let the required number be x.Then,
= (x/5) + 5 = (x/4) – 5
Transposing (x/4) to LHS and it becomes – (x/4) and 5 to RHS it becomes -5
= (x/5) – (x/4) = -5 – 5
= (4x – 5x)/20 = -10
= -x/20 = -10
Multiplying both side by (-20)
= (-x/20) × (-20) = (-10) × (-20)
= x = 200
∴ The required numbers are 200