SOLUTION: How many ounces of pure water must be added to 60 oz. of a 8% salt solution to make a 3% salt solution?

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Question 662796: How many ounces of pure water must be added to 60 oz. of a 8% salt solution to make a 3% salt solution?
Found 2 solutions by ewatrrr, DrBeeee:
Answer by ewatrrr(24785)   (Show Source): You can put this solution on YOUR website!
 
Hi,
How many ounces of pure water must be added to 60 oz. of a 8% salt solution to make a 3% salt solution?
water = water

.03x = .05·60
x = 100oz
and...

Answer by DrBeeee(684)   (Show Source): You can put this solution on YOUR website!
I wish I could draw the picture, but can't, so will try to describe the new circle.
Draw the first quadrant and label one unit on the x-axis and one on the y-axis. Now draw a line at 45 degrees from the origin to a measure of one unit. The three one-unit point are on the original one unit radius circle. OK so far?
Now the new circle of radius r, has it's center on the 45 degree line, r units from the original circle. Then the distance from the origin to the center of the new circle is (1 - r). Draw a point at the center of the new circle.
Now the vertical distance from the center point and the x-axis is r, because the new circle "touches the diameter". Likewise, the horizontal distance from the center point to the y-axis is r, for the same reason. Note that these two radii and the x-y axes form a little square with side r in lenght.
Now to get the value of r we need to look at the right triangle formed by the x-axis, the verticl leg of measure r and the hypothneus (1-r) formed by the 45 degree line between the origin and the center of the circle. The pythagorean theorem gives us
(1) (1-r)^2 = r^2 + r^2 or
(2) 1 - 2*r + r^2 = 2*r^2 or
(3) r^2 + 2*r - 1 = 0
Use the quadratic equation to find
(4) r = (-2 +/- 2*sqrt(2))/2
Since r cannot be negative, we choice the positive value of r
(5) r = -1 + sqrt(2)
Sorry for my terrible grammar.

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