How Much Pure Disinfectant is Needed to Increase Solution Strength by 25%?

[mathdad]

How much pure disinfectant must be added to 30 gallons of an 8% solution to increase its strength by 25%?

Let x = pure disinfectant to be added

The word PURE tells me that 100 will be in the equation somewhere.

30 gallons of 8 percent = 0.08(30)

Must be added to some unknown = plus x

My equation is 0.08(30) + x = 0.25(x + 100)

Correct? If not, can someone break this mixture problem step by step leading to the right equation?

 

[MarkFL]

The LHS of your equation is correct, however for the RHS, we want the concentration to be increase BY 25%, not TO 25%, which means instead of 8%, we want 1.25*8% = 10%. Also, the amount of the final solution will be x + 30, not x + 100...so the correct equation, at least for the way I am interpreting the problem, is:

0.08(30) + x = 0.1(x + 30)

 

[mathdad]

0.08(30) + x = 0.1(x + 30)

2.4 + x = 0.1x + 3

x - 0.1x = 2.4 + 3

0.9x = 5.4

x = 5.4/0.9

x = 6

So, 6 pure disinfectant must be added.

Correct?

 

[MarkFL]

0.08(30) + x = 0.1(x + 30)

2.4 + x = 0.1x + 3

x - 0.1x = 2.4 + 3

You've subtracted 2.4 from the LHS, but added 2.4 to the RHS...:D

 

[mathdad]

0.08(30) + x = 0.1(x + 30)

2.4 + x = 0.1x + 3

x - 0.1x = -2.4 + 3

0.9x = 0.6

x = 0.6/0.9

x = 0.66

Right?

 

[MarkFL]

0.08(30) + x = 0.1(x + 30)

2.4 + x = 0.1x + 3

x - 0.1x = -2.4 + 3

0.9x = 0.6

x = 0.6/0.9

x = 0.66

Right?

If you are going to round to 2 decimal places then x ≈ 0.67 gal., otherwise x = 2/3 gal. :D

x = 0.6/0.9 = 6/9 = 2/3

 

[mathdad]

If you are going to round to 2 decimal places then x ≈ 0.67 gal., otherwise x = 2/3 gal. :D

x = 0.6/0.9 = 6/9 = 2/3

Great as always.