Getting a solution that contains 40% iodine

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To create a 50-liter solution with 40% iodine using a 30% and an 80% iodine solution, a calculation is necessary. The equation 50 * 0.4 = 0.3x + 0.8y is established, where x is the volume of the 30% solution and y is the volume of the 80% solution. The limiting factor in this scenario is the amount of the 80% solution available. After working through the problem, the solution indicates that 40 liters of the 30% iodine solution is needed. The discussion concludes with confirmation of this solution.
Richay
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For solving this problem?

Two containers are on the shelf. The first one contains a 30% iodine solution and the other contains an 80% iodine solution. How much of the 30% solution should be used to get 50 liters of a solution that is 40% iodine?

Thanks for any help.
 
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you should really attepmt the problem before posting here.. but-


okay... how many litres of the 80% stuff is there.. that's the limiting, otherwise you might as well use 25 litres of it diluted with water to 50 litres


or if you can't dilute with distilled water, like any reasonable person would... then i suppose i would set the thing to equal 50*.4 then the "thing" as i ahve described it would be .3*x and .8*y

...so 50*.4=.3*x plus .8*y

then you get some limiting somehting or other prob in the equation...
 
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I attempted the problem but my freaken fomula kept being weird.. uh weird results

Ohkay I get it thanks.

The answer is 40 ^__^
 
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