A chemist has 300 grams of 20% hydrochloric acid solution. He wishes to drain some and replace it with an 80% solution so as to obtain a 25% solution. How many grams must he drain and replace with the 80% solution?PLZ HELP I NEED IT FAST PLZ!!!!!!!!!!!!

The chemist has to drain 25 grams of the 20% hydrochloric acid solution and replace with the 80% solutionSolution:Let us first set up a table and fill in the known values givenThe table is attached belowLet "x" be the amount in grams for 20% acid solutionLet "y" be the amount in grams for 80% acid solutionFrom the given table, we can set up two equationsSum of values of two acids = Value of mixture  0.2x + 0.8y=75  For convenience, we'll multiply the entire equation by 10, 2 x + 8 y = 750x + 4y = 375 ------ eqn (1)  Now, Sum of amounts of each acid = Amount of mixture  x + y = 300 --------- eqn (2) Subtracting equation (2) from (1),  ( + ) x + 4 y = 375 ( − )  x +  y = 300 − − − − − − − − ( = ) 0 + 3y = 75  Thus, 3y = 75y = 25  Substituting y = 25 in eqn (2),x + 25  = 300x = 300 – 25 = 275   So, we have x = 275 and y = 25 Here "y = 25" represents amount in grams for 80% acid solution.We can conclude that he has to drain 25 grams of the 20% acid solution.