Alligation with 3 strengths

by Sandra
(Sudbury, Ontario, Canada)

This question involves 3 different strengths of product to result in a final needed %strength.

I know how to solve the problem using 2 products but not 3. Please show me the correct solution.

You need to use 3 solutions of drug "H" containing respectively 50%, 20% and 5% to make 600mL of 10% drug "H". How much of the 50% will you use?
A. 95mL
B. 50mL
C. 25mL
D. None of the above

Reply (by Keith)

For this I would use an Alligation.

Here's a video how:

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Alligation with 3 strengths

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the problem contains 2 possible answers.
by: Mike

If the problem is absolutely ignoring to whether how much amount in volume of either 20% or 5% is needed, then the answer could be either 50mL or 25mL in preparing the 10% of 600mL solution.

If we use 450mL for 5%, then 25mL of 50% MUST be used.
If we use 500mL for 5%, then 50mL of 50% MUST be used.

Whether which case we are to choose, they both could prepare the same solution of 10% in 600mL.

As the matter of fact, the range of the amount could be used for 50% to prepare the 10% of 600mL solution is discretely at least 1mL to 66mL maximum in volume.
@Mike
by: Keith

You're exactly right. but...

The purpose is to show how to use alligation to solve this.

You are solving it using Algebra / Trig and are indeed correct with what you are saying. However, since probably less than 20% of people who use the math help here know upper Algebra or Trig, the answer is there to reflect what you'll get when solving with alligation. Hence, the title of the post is "Alligation with 3 strengths."

This kind of thing probably won't be on the PTCB. The reason it's on the site is to demonstrate the versatility of using an Alligation grid to solve solution mixture problems.

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