-> SOLUTION to get the desired solution

Question

-> SOLUTION: A chemist needs 80 milliliters of a 20% solution but has only 16% and 32% solutions available. Find how many milliliters of each that should be mixed to get the desired solution

Answer 1

Hi Jammie, the way to solve this type of problem is to create a table with the information given.

 

Percent solutions, 16%, 32%, and 20%.

 

Amount of 20% solution needed: 80 ml.

 

Let x be the amount of 16% solution needed.

 

Percent solution     Amount      Percent/Decimal             Total

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

16%                         x                   .16                  .16x

 

32%                     80 - x                .32                  .32(80 - x)

 

20%                       80                   .20                  .20 * 80

 

The sum of the 16% solution plus the 32% solution must equal the 20% solution.

 

.16x + .32(80 - x) = .20 * 80

 

Multiply by 100 to clear the decimals

 

16x + 32(80 - x) = 20 * 80

 

Clear parenthesis and multiply:

 

16x + 2560 - 32x = 1600

 

Combine terms:

 

2560 - 16x = 1600

 

Subtract 1600 from both sides:

 

960 - 16x = 0

 

Add 16x to both sides:

 

16x = 960

 

Divide both sides by 16:

 

x = 60

 

Answer:

 

You need 60 ml of the 16% solution.

 

You need 80 - 60, or 20 ml of the 32% solution

 

Check:

 

.16 * 60 + .32(80 - 60) = .20 * 80

 

9.60 + 6.40 = 16.00

 

16.00 = 16.00, values check.

 

Questions? 

Answer 2

Set up eqns  

 

x + y = 80  

 

x*0.16 + y*.32  = 80 *0.20

 

solve 2 eqns in 2 variables.  

 

16X + 32Y =  20*80 =   1600   

 

minus  (16*(x + y) =80))  = 

 

16Y =  1600  - 16*80 =  1600 -1280  =  320       20 Ml of Y  (32% solution)

 

60 ml of X (16%solution)   

 

Check answer      60ml * 0.16  + 20 ML * 0.32  =   80 ml at 20%