Question

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

Answer 1

Answer: 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

$\frac{11}{15}\times 60=11\times 4=44\ ml$

Quantity of 50% solution in the mixture is given by

$\frac{4}{15}\times 60=4\times 4=16\ ml$

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

Therefore, Option 'B' is correct.