18. a science teacher has a supply of 20% saline solution and a supply of 50% saline solution. how much of each solution should the teacher mix together to get 60 ml of 28% saline solution for an experiment? (1 point)
a) 16 ml of the 20% solution and 44 ml of the 50% solution
b) 44 ml of the 20% solution and 16 ml of the 50% solution
c) 16 ml of the 20% solution and 16 ml of the 50% solution
d) 44 ml of the 20% solution and 44 ml of the 50% solution

Option 'B' is correct.

Step-by-step explanation:

Since we have given that

There is a supply of 20% saline solution and a supply of 50% saline solution.

If a teacher wants to mix together to get 60 ml of 28% saline solution for an experiment.

We will use "Mixture Allegation" to find the quantity of each solution:

Ist type of solution                           II nd type of solution

      20%                                                       50%

                                    28%

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

50%-28%                        :                         28%-20%

 22%                               :                            8%

  11                                  :                             4

Since there is total quantity of 60 ml of solution.

So, Quantity of 20% solution in the mixture is given by

Quantity of 50% solution in the mixture is given by

Hence, there is 44 ml of the 20% solution and 16 ml of the 50% solution.

Therefore, Option 'B' is correct.

the answer will be

(0.2)+(-8)=0.3+-3+0.3m is the equation i guess? ?

step-by-step explanation: