How much `1.0M HCl` should be mixded with what volume of `0.250 M HCl` in order to prepare `2.0 "litre"` of `0.50M HCl`?

Explanation

Let `a` litre of `1.0M` is mixed, then
`1xxa+(2-a)xx0.250=2xx0.50`
`a= 0.666 "litre"`
Thus volume of `1.0MHCl= 666mL`
Volume of `0.250 MHCl= 1334mL`