SOLUTION: A jeweler has 3 rings, each weighing 15 g, made of an alloy of 25 percent silver and 75 percent gold. He decides to melt down the rings and add enough silver to reduce the gold con

Algebra.Com
Question 812764: A jeweler has 3 rings, each weighing 15 g, made of an alloy of 25 percent silver and 75 percent gold. He decides to melt down the rings and add enough silver to reduce the gold content to 65 percent. How much silver should he add?
a) Write an equation for the total amount of gold in the mixture. Use s to represent the amount of silver added.
b) Solve this equation for s. Include the appropriate units in your answer.

Answer by josgarithmetic(39617)   (Show Source): You can put this solution on YOUR website!
Focus on concentration of Gold (but this is arbitrary).
Let g = grams of Silver to add.



If you could read and analyze the description and obtain that equation, or an equivalent one, then you have solved most of the problem. The rest is only a few arithmetic steps.






grams of silver

RELATED QUESTIONS

A jeweler has five rings, each weighing 18g, made of an alloy of 10 percent silver and 90 (answered by richwmiller)
A jeweler has five rings, each weighing 18 g, made of an alloy of 10% silver adn 90%... (answered by Paul)
A jeweler has five rings, each weighing 18 grams made of an alloy of 10% silver and 90%... (answered by ankor@dixie-net.com)
Hello. I'm having some difficulty figuring out the following word problem, Can you... (answered by scott8148)
A jeweler has 5 rings, each weighing 16 g, made of an alloy of 15% silver and 85% gold.... (answered by stanbon)
A jeweler has four rings, each weighing 24 g, made of an alloy of 15% silver and 85%... (answered by stanbon)
A jeweler has 3 rings, each weighing 18 g, made of an alloy of 7% silver and 93% gold.... (answered by josmiceli)
A jeweler has four rings, each weighing 20 grams, made of an alloy of 5% silver and 95%... (answered by ankor@dixie-net.com)
Mixture Problem: A jeweler has five rings, each weighing 18g, made of an alloy of 10%... (answered by ankor@dixie-net.com)