A chemist wants to have a 32% saline solution. She has a 48 ounce of a 23% saline solution and a 56% saline solution available. How many ounces of the 56% solution should be used in the mixture?

Question

Jacoby Gill

Mathematics

12 August, 01:41

A chemist wants to have a 32% saline solution. She has a 48 ounce of a 23% saline solution and a 56% saline solution available. How many ounces of the 56% solution should be used in the mixture?

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Hector Kirby · Answer 1

Often, a mixture problem can be worked very rapidly using an X diagram. The desired mixture value is put in the middle of the X, with the component values on the left side. Differences from the component values are written on the right side along the diagonals of the X.

Here, we have a mixture value of 32%, and components of 56% and 23%. The differences are 56-32 = 24 and 32-23 = 9.

The numbers 9 : 24 tell us the proportions of the components written on the same horizontal line. That is, the ratio of 56% to 23% solution must be 9 : 24. Since we have 48 ounces of 23% solution, each unit of these ratio values must stand for 2 ounces of solution. That is, we need ...

... 9*2 ounces = 18 ounces of 56% solution