SOLUTION: A Chemist needs 170 milliliters of a 36% solution but has only 24% and 58% solutions available. Find how many milliliters of each that should be mixed to get the desired solution.

Question 1116078: A Chemist needs 170 milliliters of a 36% solution but has only 24% and 58% solutions available. Find how many milliliters of each that should be mixed to get the desired solution.
Found 3 solutions by greenestamps, ikleyn, josgarithmetic:
Answer by greenestamps(13200)   (Show Source): You can put this solution on YOUR website!

This kind of problem, where two ingredients are being mixed, is most easily solved using the fact that the ratio in which the ingredients need to be mixed is exactly determined by where the percentage of the mixture lies between the percentages of the two ingredients.

Here are the simple calculations required for your example.

(1) 58-24 = 34; 36-24 = 12. The target percentage of 36% is 12/34 = 6/17 of the way from 24% to 58%.

(2) That means 6/17 of the mixture must be the higher percentage ingredient.

(3) 6/17 of 170 ml = 60 ml, so 60ml of the 58% solution. That leaves 11/17 of 170 ml or 110 ml of the 24% ingredient.

Answer: 60ml of 58%; 110ml of 24%
Answer by ikleyn(52794)   (Show Source): You can put this solution on YOUR website!
.

Let x = an amount of the 24% solution, im milliliters (mL). Then the amount of the 58% solution is (170-x) mL. The balance equation is 0.24x + 0.58*(170-x) = 0.36*170 0.24x + 0.58*170 - 0.58*x = 0.36*170 -0.34x = 0.36*170 - 0.58*170 x = = 110. Answer. 110 mL of the 24% solution and (170-110) = 60 mL of the 58% solution. Check. 0.24*110 + 0.58*60 = 61.2 = 0.36*170. ! Correct !

Solved.

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It is a standard and typical mixture problem.

For introductory lessons covering various types of mixture word problems see
    - Mixture problems
    - More Mixture problems
    - Solving typical word problems on mixtures for solutions
    - Typical word problems on mixtures from the archive
in this site.

You will find there ALL TYPICAL mixture problems with different methods of solutions,
explained at different levels of detalization, from very detailed to very short.

Read them and become an expert in solution mixture word problems.

Also, you have this free of charge online textbook in ALGEBRA-I in this site
    - ALGEBRA-I - YOUR ONLINE TEXTBOOK.

The referred lessons are the part of this textbook in the section "Word problems" under the topic "Mixture problems".

Save the link to this online textbook together with its description

Free of charge online textbook in ALGEBRA-I
https://www.algebra.com/algebra/homework/quadratic/lessons/ALGEBRA-I-YOUR-ONLINE-TEXTBOOK.lesson

to your archive and use it when it is needed.

Answer by josgarithmetic(39618)   (Show Source): You can put this solution on YOUR website!
A solution all in variables and using only ONE unknown variable can be like this:

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A Chemist needs M milliliters of a T% solution but has only L% and H% solutions available. Find how many milliliters of each that should be mixed to get the desired solution.
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x, quantity of L% material
M, quantity of finished mixture
L, low percent materialconcentration
H, high percent material concentration
T, target percent mixture concentration

Equation to account for the amount of dissolved material in the mixture:

Solve for x.

Doing the steps to find x, and the fact that and , the solution will be .
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