text: # Dilution of a Solution Strengths
  
  #### frgeorgeh
  
  ##### New member
  I want to mix 4 strengths of solutions. I thought to save time I would mix 1 a 0mg strength and a second at 18mg strength. How do I use those two solutions to make a 6mg, 12 mg strength solution?
  
  The ratio of (OV)*(OS)=(NV)*(NS) where OV=Old Volume, OS=Old Strength,, NV=New Volume, NS=New Strength, does not work since I know all the results. For example, if I wanted to make the following:
  
  I have: 1 gal (3,785ml) of 0mg and 1gal of 18mg
  
  I want to make: 30ml bottles at 6mg and 12mg.
  
  I hope this makes sense.
  
  Thanks
  
  #### frgeorgeh
  
  ##### New member
  I want to mix 4 strengths of solutions. I thought to save time I would mix 1 a 0mg strength and a second at 18mg strength. How do I use those two solutions to make a 6mg, 12 mg strength solution?
  
  The ratio of (OV)*(OS)=(NV)*(NS) where OV=Old Volume, OS=Old Strength,, NV=New Volume, NS=New Strength, does not work since I know all the results. For example, if I wanted to make the following:
  
  I have: 1 gal (3,785ml) of 0mg and 1gal of 18mg
  
  I want to make: 30ml bottles at 6mg and 12mg.
  
  I hope this makes sense.
  
  Thanks
  __________________________________
  
  OK, I think I am over thinking this. It seems to me, to get a 6mg strength, I add 2/3 of the 0MG and 1/3 of the 18mg. To get 12mg, I add 1/3 0mg and 2/3 of 18mg. What do u think?
  
  #### MarkFL
  
  Staff member
  I am assuming the solution strengths are in milligrams per milliliters. So, if you have a solution with a strength of $18$, to get a solution of volume $V$ with a strength of $18k$, where:
  
  $$\displaystyle 0\le k\le1$$
  
  You would need to use $$\displaystyle kV$$ of the solution of strength $18$, with the rest consisting of the strength $0$ solution.
  
  For example, since 12 is 2/3 of 18, then you would put 2/3 of 30 mL which is 20 mL with 10 mL of strength 0 to get 30 mL of strength 12.
  
  So yes, your reasoning is correct.
  
  #### frgeorgeh
  
  ##### New member
  I am assuming the solution strengths are in milligrams per milliliters. So, if you have a solution with a strength of $18$, to get a solution of volume $V$ with a strength of $18k$, where:
  
  Thanks for the reply! I am an Engineer but always seem to over complicate my math.
explanation: The document contains a math question related to diluting solutions to achieve specific strengths, and it also includes an answer to this question. The user initially asks how to mix two solutions (0 mg and 18 mg) to create 6 mg and 12 mg solutions. They then provide their own solution, which is confirmed as correct by another user.
distilabel_metadata.content: Evaluate the user given text extract and determine if it contains a math question and optionally and answer to this question.
  
  Output rules:
  
  1. Briefly describe the document and mention if it contains a question or a question and answer.
  2. Finalize the output with: Label: no-question,question,question-and-answer,multiple-question,multiple-question-and-answer
distilabel_metadata.role: system
distilabel_metadata.content: # Dilution of a Solution Strengths
  
  #### frgeorgeh
  
  ##### New member
  I want to mix 4 strengths of solutions. I thought to save time I would mix 1 a 0mg strength and a second at 18mg strength. How do I use those two solutions to make a 6mg, 12 mg strength solution?
  
  The ratio of (OV)*(OS)=(NV)*(NS) where OV=Old Volume, OS=Old Strength,, NV=New Volume, NS=New Strength, does not work since I know all the results. For example, if I wanted to make the following:
  
  I have: 1 gal (3,785ml) of 0mg and 1gal of 18mg
  
  I want to make: 30ml bottles at 6mg and 12mg.
  
  I hope this makes sense.
  
  Thanks
  
  #### frgeorgeh
  
  ##### New member
  I want to mix 4 strengths of solutions. I thought to save time I would mix 1 a 0mg strength and a second at 18mg strength. How do I use those two solutions to make a 6mg, 12 mg strength solution?
  
  The ratio of (OV)*(OS)=(NV)*(NS) where OV=Old Volume, OS=Old Strength,, NV=New Volume, NS=New Strength, does not work since I know all the results. For example, if I wanted to make the following:
  
  I have: 1 gal (3,785ml) of 0mg and 1gal of 18mg
  
  I want to make: 30ml bottles at 6mg and 12mg.
  
  I hope this makes sense.
  
  Thanks
  __________________________________
  
  OK, I think I am over thinking this. It seems to me, to get a 6mg strength, I add 2/3 of the 0MG and 1/3 of the 18mg. To get 12mg, I add 1/3 0mg and 2/3 of 18mg. What do u think?
  
  #### MarkFL
  
  Staff member
  I am assuming the solution strengths are in milligrams per milliliters. So, if you have a solution with a strength of $18$, to get a solution of volume $V$ with a strength of $18k$, where:
  
  $$\displaystyle 0\le k\le1$$
  
  You would need to use $$\displaystyle kV$$ of the solution of strength $18$, with the rest consisting of the strength $0$ solution.
  
  For example, since 12 is 2/3 of 18, then you would put 2/3 of 30 mL which is 20 mL with 10 mL of strength 0 to get 30 mL of strength 12.
  
  So yes, your reasoning is correct.
  
  #### frgeorgeh
  
  ##### New member
  I am assuming the solution strengths are in milligrams per milliliters. So, if you have a solution with a strength of $18$, to get a solution of volume $V$ with a strength of $18k$, where:
  
  Thanks for the reply! I am an Engineer but always seem to over complicate my math.
distilabel_metadata.role: user
distilabel_metadata.raw_output_q_a_detector_0: The document contains a math question related to diluting solutions to achieve specific strengths, and it also includes an answer to this question. The user initially asks how to mix two solutions (0 mg and 18 mg) to create 6 mg and 12 mg solutions. They then provide their own solution, which is confirmed as correct by another user.
  
  Label: question-and-answer