How many quarts of pure antifreeze must be added to 3 quarts of a 30​% antifreeze solution to obtain a 50% antifreeze​ solution?

Question

homework homework

Answer 1

You currently have 2 ways of doing this problem. Here is a 3rd way.

This method is referred to as "alligation" in pharmacy and other preparations.  You arrange what you HAVE (100% and 30%) and what you W ANT (50%) in a grid and subtract cross-wise to obtain what you NEED, as shown below:

 

HAVE.......WANT.....NEED

100%.....................20...

...............50%..............

30%......................50....

 

So, you need to use 20 PARTs of the 100% and 50 PARTS of the 30%.  Since parts is actually quarts in this problem, and the 50 parts is actually told to you to be 3 quarts, then the 20 parts is 1.2 quarts.  This later calculation would be 

50parts/20parts = 3quarts/x quarts and x = 1.2 quarts.

Answer 2

The 3 quarts of 30% antifreeze solution contains 0.3 x 3 = 0.9 quarts of pure antifreeze solute.

 

So there must be 2.1 quarts of solvent (water? maybe), in order to obtain 50% concentration, we need to have the as much solute as the solvent.

 

So, adding 1.2 quarts (2.1 - 0.9 = 1.2) of pure antifreeze will do the job.  

 

A formal way of doing it will be:

 

Set x is the number of quarts of pure antifreeze needed to make a 50% solution,

 

so 【(3*0.3)+x】/(3+x) = 50%, solve for x, we get the x = 1.2, same answer. 

 

Answer 3

Hi Jordan,

 

The question is how many quarts of pure antifreeze must be added to 3 quarts of a 30% antifreeze solution to obtain a 50% antifreeze solution?

 

Note that there is 30% times 3 quarts or 0.9 quarts of antifreeze in the current mix and we want a 50% solution.

 

The equation to solve becomes, where x is the amount of pure antifreeze to be added

 

(.9 + x )/(3 + x) = .5

(.9 + x ) = .5 * (3 + x)

(.9 + x ) = 1.5 +.5

.5x = .6

x = 1.2

 

so we need to add 1.2 quarts of pure antifreeze to the current solution to make a 50% solution