percentage maths problem

Question

A container of production line solvent is mixed at 22% strength and another is mixed at strength 53% strength to obtain 12 liter of a mixture of 25% strength. How many liter should be used from the container that is mixed at 53% strength?

options:

a:4 liter

b:1 5/3  liter

c: 2 16/25 liter

d: 5 35/53 liter

e:4/59

Answer 1

Let X be the amount of 22% solvent used and Y be the amount of 53% solvent used. Then we know the total amount used (X + Y) is equal to 12 liters:

   X + Y = 12

 

We also know that the solvent strengths (0.22*X + 0.53*Y) must add up to 25%:

   0.22*X + 0.53*Y = 0.25

 

From here you would solve this like any other linear system, just with different numbers. I believe you have seen them before. If you need help finishing this problem, let me know.

 

 

 

 

P.S. Keep in mind that your answer will be the value of Y, NOT X.

Answer 2

Hi Avi, the best way to solve mixture problems is to set up a table. Let x represent the amount of 53% solution needed. Then the amount of 22% solution is 12 - x.

 

Solution      Amount       Decimal               Total

------------------------------------------------------------

22%            12 - x           .22              .22 * (12 - x)

 

53%               x               .53              .53 * x

 

25%              12              .25              .25 * 12

 

Equation to solve:

 

The total of the 22% solution plus the total of the 53% solution must equal the total of the 25% solution.

 

.22 * (12 - x) + .53 * x = .25 * 12

 

Questions?