how many ml of a 20 percent solution should be added to 50 ml of a 40 percent solution to obtain a 25 pecent s

Answer 1

50 ml of a 40% solution of the chemical in question contains
(50 ml)(.4) = 20 ml of that chemical.

Suppose we add z ml of a 20% solution. The solution we add contains .2 z ml of the chemical. We end up with a total of
(50 + z) ml of the mixture, and the mixture contains (20 + .2 z) ml of the chemical.

We want the mixture to be a 25% solution. Therefore,
(20 + .2 z) / (50 + z) = .25.
Then 20 + .2 z = (50 + z)(.25) = 12.5 + .25 z, or
20 - 12.5 = (.25 - .2) z.
Thus, 7.5 = .05 z, or z = 7.5 / .05 = 150.

You must add 150 ml of the 20% solution.

The solution you add contains (150)(.2) = 30 ml of the chemical, and your final mixture is 50 + 150 = 200 ml total, containing
20 + 30 = 50 ml of chemical. Since 50 / 200 = .25 = 25%, this is the mixture you want.

Answer 2

Let x = # of ml of 20% solution

Notice that 50 ml = # of ml of 40% solution

and

x + 50 = # of ml of 25% solution

The amount of ml of alcohol can be written in 2 different ways: 20% x + 40% (50) and 25% (x+50). So we can set up an equation as follows:

20% x + 40% (50) = 25% (x+50)

0.2 x + 0.4 (50) = 0.25 (x+50)

Multiplying by 100:

20 x + 40 (50) = 25 (x+50)

20 x + 2000 = 25 x + 1250

5 x = 750

or

x = 150 ml

Answer 3

Let x be the number of ml of a 20% solution.

Balance by the solute,
20%x + 40%(50) = 25%(x+50)

Solve for x,
x = 150 ml

Answer 4

150 ml

Answer 5

150ml is correct

See this website for info how to calculate under Lesson 2 - "alligation."

Answer 6

should be added 100ml,


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