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

B

Explanation:

Answer 2

Answer: Option 'B' is correct.

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

`(11)/(15)* 60=11* 4=44\ ml`

Quantity of 50% solution in the mixture is given by

`(4)/(15)* 60=4* 4=16\ ml`

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

Therefore, Option 'B' is correct.