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illsy illsy
wrote...
Posts: 42
Rep: 1 0
11 years ago
this is a self taught class so please no crap about doing it myself. I already am.

    Miss Bass manages a herbicide plant. she wants a solution of 61% herbicide to be mixed with a solution of 50% herbicide to form 44 liters of a 53%solution. How many liters of the 61% solution must Miss Bass use?

Please not just the answer; I would like to know how to solve this type of problem on my own.
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wrote...
11 years ago
hahaha my teacher teaches me that and i don't even get it sorry!
wrote...
11 years ago
44*.50 = X *.61 + (44 -X) * .50

solve for X
wrote...
11 years ago
I think what you would do for this is set up a proportion.  

44/.53 = X/.61  This is because 44 L is 53% solution.  SO we need to know 61% solution.

The next step is cross multiplying.  44(.61)=.53X

When doing this you get  X= 50.64 L
wrote...
11 years ago
The total number of liters of solution is 44.

Let x be the number of liters of 61% solution
so (44 - x) is the number of liters of 50% solution

0.61x + 0.50(44 - x) = 0.53(44)
0.61x - 0.50x + 22 = 23.32
0.11x = 1.32
x = 1.32 / 0.11
 = 12

So she needs 12liters of 61% solution (and 32 liters of 50%)
wrote...
11 years ago
You do these types of problems via mass balance. That is, the amount of herbicide in the solutions you mix is equal to the amount of herbicide in the final mixture.

Next we have to assume what your percentages mean: they could be based on volume or they could be based on mass. Let's assume they're based on volume because gthis will make the problem easier.

Let S be the volume of 61% herbicide solution. Then 0.61S must be the amount of herbicide in the solution.

Let D be the volume of 50% herbicide solution. Then 0.5D must be the amount of herbicide in this solution.

The total amount of herbicide is 0.61S + 0.50D

Now look at the final mixture. It contains 53% herbicide. Since there are 44 liters, then the herbicide is 0.53*44 = 23.32 liters of herbicide.

This amount must equal what went in

0.61S + 0.50D = 23.32

So far, so good. But we have 1 equation with 2 unknowns. We need another equation.

That equation mus be that the total mix must be the sum of the total from the two smaller amounts

S+D = 44

You now have two equations and 2 unknowns. You can solve this by multipying the second equation by 0.5 and then subtracting it from the first:

0.61S + 0.50D = 23.32
0.5S  + 0.5D    = 22
___________________

0.11S = 1.32

S = 12 liters

S referred to the 61% solution so we're done.
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