1. 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?  Reply With Quote

2.

3. 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)  Reply With Quote

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?  Reply With Quote

5. Originally Posted by RTCNTC 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...   Reply With Quote

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?  Reply With Quote

7. Originally Posted by RTCNTC 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. x = 0.6/0.9 = 6/9 = 2/3  Reply With Quote Originally Posted by MarkFL If you are going to round to 2 decimal places then x ≈ 0.67 gal., otherwise x = 2/3 gal. x = 0.6/0.9 = 6/9 = 2/3
Great as always.  Reply With Quote

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