A chemist needs 500 milliliters of a 20% acid and 80% water mix for a chemistry experiment. The chemist combines X milliliters of a 10% acid and 90% water mix and Y milliliters of a 30% acid and 70% water mix to make the 20% acid and 80% water mix.
The chemists also needs 500 milliliters of a 15% acid and 85% water mix. Does the chemist need more of the 10% acid and 90% water mix than the 30% acid and 70% water mix to make this new mix? Explain.
Ok so first I would create two equations:
Equation one based off of total amount of acid needed. 20% of 500 =100 so to figure out acid take .1x +.3Y = 100. Because the first mix is ten percent acid the other mix is 30 percent acid. The total mix should be 500 ML so X+Y=500
Now its simple substitution. 500-Y=X Plug that into the other equation you get. .1(500-Y)+.3Y = 100
Y=250. Plug 250 into the equation you find out X = 250. SO in your first problem the chemists uses 250 ML of both mixes to get 20%. Which makes sense half of a mix of 10% acid plus half of a mix of 30% acid equals a total mix of 20%
If the chemists wants to dilute the mixture down to 15% than obviously they would need more of the 10% mix. So if the answer is just asking to explain that should explain but if it wants the exact mixture you would just go about figuring it out the same way.
15% of 500 is 75
Multiply both sides of first equation by 10 you get X + 3Y = 750
Which is x = 750-3Y
Plug that into the other equation you get 750-3Y+Y=500
So this mixture would be a result of 375 ML of 10% acid mixture and 125 ML of 30% acid mixture.