Assume the consecutive integral multiple of 5 be 5n and 5(n + 1) where n is a positive integer.
ATQ
5n x 5(n + 1) = 300
n2 + n – 12 = 0
(n – 3) (n + 4) = 0
n = 3 and n = – 4.
As n is a positive integer so n =3.
Hence the required numbers are 15 and 20.