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One alloy contains 3 times as much copper as silver. Another contains 5 times as much silver as copper.
How much of each must be used to make 14kg in which there is twice as much copper as silver?
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It is a standard mixture/alloy problem in a non-standard formulation.
First what I will do is to re-phrase it to the STANDARD formulation.
It is done in three lines:
First alloy is 25% silver.
The second alloy is , or % silver.
The resulting alloy is , or % silver.
Now let "x" be an amount of the first alloy to be mixed, in kilograms,
and let "y" be an amount of the first alloy to be mixed, in kilograms.
Then your equations are
x + y = 14, (1) (for total amount)
= . (2) (for the silver content)
Simplify the system (1), (2). For it, multiply the both sides of (2) by 12. You will get
x + y = 14, (1')
3x + 10y = 56. (2')
To solve this system, express x = 14-y from (1') and substitute it into (2'). You will get a single equation for "y"
3(14-y) + 10y = 56.
From this point please complete the assignment on your own.
There is entire bunch of 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
in this site.
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 textbook in the section "Word problems" under the topic "Mixture problems".
The textbook contains many other solved word problems, as well as many other interesting and useful things.