SOLUTION: Two metal alloys, containing 6% zinc and 16% zinc are to be mixed together. How much of each is needed to get to 150 kg of an 14% zinc alloy?

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Question 1184584: Two metal alloys, containing 6% zinc and 16% zinc are to be mixed together. How much of each is needed to get to 150 kg of an 14% zinc alloy?
Found 3 solutions by ikleyn, greenestamps, josgarithmetic:
Answer by ikleyn(52794)   (Show Source): You can put this solution on YOUR website!
.
Two metal alloys, containing 6% zinc and 16% zinc are to be mixed together.
How much of each is needed to get to 150 kg of an 14% zinc alloy
~~~~~~~~~~~~~~~ 

x kilograms of the 16% alloy and

(150-x) kilograms of the 6% alloy.



The total zinc in two alloys is  0.16x + 0.06*(150-x) kilograms.

The total zinc in the mixture is  0.14*150 kilograms.



The equation is

    0.16x + 0.06*(150-x) = 0.14*150.



The solution is

    x =  = 120.



ANSWER.  120 kg of the 16% alloy and the rest, (150-120) = 30 kg of the 6% alloy.


CHECK.  Calculate the concentration of the final alloy  C =  = 0.14 = 14%.      ! Correct !

Solved.

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It is a standard and typical mixture problem.

In this site, there is a bunch of lessons,  covering various types of mixture problems.  See introductory lessons
    - Mixture problems
    - More Mixture problems
    - Solving typical word problems on mixtures for solutions
    - Word problems on mixtures for antifreeze solutions
    - Word problems on mixtures for alloys
    - Typical word problems on mixtures from the archive

Read them and become an expert in solution the 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 online textbook 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 greenestamps(13200)   (Show Source): You can put this solution on YOUR website!


You should know and understand the standard formal algebraic method for solving 2-part mixture problems, as shown by the other tutor.

If a formal algebraic solution is not required, here is a simple non-algebraic way for solving 2-part mixture problems like this. When the numbers are "nice", as they are in this problem, the answer can be obtained in a matter of seconds.

(1) Observe/calculate that 14% is 4/5 of the way from 6% to 16%

(2) That means 4/5 of the mixture is the higher percentage ingredient.

ANSWER: 4/5 of 150kg, or 120kg, of 16% zinc; 1/5 of 150kg, or 30kg, of 6% zinc.

CHECK:
.16(120)+.06(30)=19.2+1.8=21
.14(150)=21
Answer by josgarithmetic(39618)   (Show Source): You can put this solution on YOUR website!
Two metal alloys, containing 6% zinc and 16% zinc are to be mixed together. How much of each is needed to get to 150 kg of an 14% zinc alloy?
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v kilograms of the 16%
150-v kilograms of the 6%
14% of 150, how much zinc in the resulting alloy


Simplify and solve,...
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