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 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

$\frac{512}{640+x}*100=67.5\phantom{\rule{0ex}{0ex}}51200=67.5*(640+x)\phantom{\rule{0ex}{0ex}}67.5x+67.5*640=51200\phantom{\rule{0ex}{0ex}}67.5x+43200=51200\phantom{\rule{0ex}{0ex}}67.5x=51200-43200\phantom{\rule{0ex}{0ex}}67.5x=8000\phantom{\rule{0ex}{0ex}}x=\frac{8000}{67.5}=118.52lt$

hope this helps you

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