Ineed setting up this problem. me.

715 mL pure water; 455 mL 90% solution

Step-by-step explanation:

For mixture problems, I find it convenient to let a variable (x) represent the quantity of the greatest contributor to the mix. Here, that is the quantity of 90% solution. Then the quantity of 0% solution (pure water) will be (1170-x), where quantities are in mL.

The problem conditions let you write a relation describing the amount of alcohol in the mix.

0.90x + 0.00(1170 -x) = 0.35(1170)

The solution is easily found by dividing by the coefficient of x.

0.35·1170/0.90 = x = 455 . . . . mL of 90% solution

1170 -455 = 715 . . . . . . . . . . . . .mL of pure water

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Comment on the equation

The term 0.00(1170 -x) represents the contribution of alcohol to the mix made by the quantity of pure water, which is a 0% alcohol solution. (Water contributes no alcohol to the mix—it all comes from the 90% solution.)

answer: 10 meter

step-by-step explanation:

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