A dishonest milkman professes to sell his milk at cost price but he mixes it with water and thereby gains $25$%. The percentage of water in the mixture is
Let's assume milkman has $100$ liter of milk, If he added $x$ liter of water, the percentage of water in the mixture$=\dfrac {x}{(100+x)}\times 100$
The milk gain $25$%, he must have added $25$ liter water in $100$ liter of milk. Then the percentage of water in the mixture$=\dfrac {25}{(100+25)}\times 100=20\%$
In the new mixture, If the milk is $80$% then $80$% of total mixture should be $100$ liter
$(100+x)\times \dfrac{80}{100}=100$
$8x=200$
$x=25$
Then, the percentage of water in the mixture $=\dfrac {25}{(100+25)}\times 100=20\%$