Let the number required is x
According to the question we can write as
(4/5) x-10 = (2/3) x
On rearranging,
(4/5) x-(2/3) x = 10
Now taking the L.C.M of 3 and 5 is 15
(12x-10x)/15=10
On cross multiplication,
2x=150
x=150/2=75
So the required number is 75