a milk of 80% conc. is diluted at home by the seller by adding some water to it so that milk conc. is reduced btw 65% to 70% . if 640 l of milk of 80% conc. is available how much water has been added??

the amount of milk in 640 lt of milk of con 80% is $640*\frac{80}{100}=512 lt$
the amount of water in 640 lt of milk of 80% con is 640 - 512 = 128 lt.

let the x lt of water has been added,
therefore the concentration of milk in new mixture is ​$\frac{512}{640+x}*100$

let the milk concentration in new mixture is $\frac{65+70}{2}=67.5$ %
therefore
$$\begin{gathered} \frac{512}{640+x}*100=67.5 \\ 51200=67.5*(640+x) \\ 67.5x+67.5*640=51200 \\ 67.5 x + 43200=51200 \\ 67.5 x =51200-43200 \\ 67.5 x=8000 \\ x=\frac{8000}{67.5}=118.52 lt \end{gathered}$$

hope this helps you