SOLUTION: 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

Question 1024233: 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.

Found 2 solutions by Alan3354, Theo:
Answer by Alan3354(69443) (Show Source): You can put this solution on YOUR website!
No need to mention the water part.
A chemist needs 500 milliliters of a 20% acid for a chemistry experiment. The chemist combines X milliliters of a 10% acid and Y milliliters of 30% acid to make the 20% acid.
--------
You didn't ask a question.
20% is the average of 10% & 30% --> equal parts.
x = y = 250 ml
===========
The chemists also needs 500 milliliters of 15% acid. Does the chemist need more of the 10% acid than the 30% acid to make this new mix? Explain.
---
Yes, he needs more 10%.
Equal parts gives the average, 20%.
More 10% --> lower than average concentration of the mix.
More 30% --> higher than average concentration of the mix.
--
In math terms:
x + y = 500 --> y = 500-x
10x + 30*(500-x) = 500*C where C is the %age of the mix.
15000 - 20x = 500*C
C = 30 - x/25 (0 <= x <= 500)
--> Increasing x decreases C

Answer by Theo(13342) (Show Source): You can put this solution on YOUR website!
x = the number of milliliters of the first solution.
y = the number of milliliters of the second solution.

you want a mixture that's 20% acid and 80% water and you want the total milliliters to be equal to 500.

that means you want 100 milliliters of acid and 400 milliliters of water because 100 = .2 * 500 and 400 = .8 * 500.

you have two mixtures available.

they are a mixture of 10% acid and 90% water, and a mixture of 30% acid and 70% water.

your equations are:

.10 * x + .30 * y = 100
.90 * x + .70 * y = 400

solve these two equations simultaneously and you get:

x = 250 and y = 250.

you can confirm these values are correct by replacing x and y in those equations and you will find that .10 * 250 + .30 * 250 = 100.

you will also find that .90 * 250 + .70 * 250 = 400.

the chemist also needs a mixture of 15% acid and 85% water.

.15 * 500 = 75 milliliters of acid.

.85 * 500 = 325 milliliters of water.

the same mixtures are used as before.

the difference is in what the total milliliters of acid and what the total milliliters of water are required.

the equations are:

.10 * x + .30 * y = 75
.90 * x + .70 * y = 425

solve these two equations simultaneously and you get:

x = 375 and y = 125.

you can confirm this solution is correct by going back to the original equations and finding that .10 * 375 + .30 * 125 = 75.

you can also confirm that .90 * 375 + .70 * 125 = 425.

you had two scenarios.

you wanted to mix 10/90 solution with 30/70 solution to get 20/80 solution of 500 milliliters.

you also wanted to mix 10/90 solution with 30/70 solution to get 15/85 solution of 500 milliliters.

when you solve the first scenario, you had x = 250 and y = 250.

when you solved the second scenario, you had x = 375 and y = 125.

x was the number of milliliters of the first solution and y was the number of milliliters of the second solution.

since the first solution was the 10/30 mix, then you needed more of that to solve for the second scenario.

why?

the first solution has 10% acid which was less than 30% acid that the second solution had.

since the new mixture required less acid, it makes sense that you would need more of the first solution and less of the second solution since the first solution contained less acid than the second solution.

the numbers confirm the logic.

this problem may have been able to solved with just the last bit of logic i gave you, but i went through the calculations anyway, even though i don't think i had to.

regardless, the calculations confirmed the logic, so it looks pretty good.

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