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Question 1162518: A foundry has been commissioned to make souvenir coins. The coins are to be made from an alloy that is 40% silver. The foundry has on hand two alloys, one with 50% silver content and one with a 25% silver content. How many kilograms of each alloy should be used to make 5 kilograms of the 40% silver alloy?
Found 2 solutions by ikleyn, greenestamps:
Answer by ikleyn(52810)   (Show Source):
You can put this solution on YOUR website! .
Let x be the mass of the 50% silver alloy (in kilograms).
Then the mass of the 25% silver alloy is (5-x) kilograms.
The amount of the pure silver in ingredients is 0.5x + 0.25*(5-x) kilograms.
It should be equal to the amount of the pure silver in the resulting 40% alloy
0.5x + 0.25*(5-x) = 0.4*5 kilograms.
From this equation,
x =  = 3 kilograms.
ANSWER. 3 kilograms of the 50% silver alloy and the rest, (5-3) = 2 kilograms of the 25% silver alloy.
CHECK. 0.5*3 + 0.25*2 = 2 kilograms of the pure silver in ingredients, and
0.4*5 = 2 kilograms of the pure silver in the 40% alloy. ! Correct !
Solved.
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It is a standard and typical mixture word problem.
There is entire bunch of introductory lessons covering various types of mixture problems
- 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
- Advanced mixture problems
- Advanced mixture problem for three alloys
- Unusual word problem on mixtures
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.
A convenient place to quickly observe these lessons from a "bird flight height" (a top view) is the last lesson in the list.
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 greenestamps(13203)  (Show Source):
You can put this solution on YOUR website!
Here a solution by a quick and easy method for solving "mixture" problems like this if a formal algebraic solution is not required.
You are mixing 25% and 50% to obtain 40%:
(1) 40% is 3/5 of the way from 25% to 50%. (Picture the three percentages on a number line; 25 to 50 is a difference of 25; 25 to 40 is a difference of 15; 15/25 = 3/5.)
(2) That means 3/5 of the mixture has to be the higher percentage ingredient.
ANSWER: 3/5 of 5kg, or 3kg, of 50% silver; the other 2kg of 25% silver.
CHECK:
.50(3)+.25(2) = 1.5+.5 = 2
.40(5) = 2
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