Joe has a ring weighing 10 grams made of an alloy of 13% silver and the rest gold. He decides to melt down the rings and add enough silver to reduce the gold content to 69%. How many grams of silver should he add?
My set up is:
10 (.13 + .87) = 10 (.13 + X + .69)
I'm not sure if that's correct.
Silver = 0.13 * 10 = 1.3 grams.
Gold = 10 1.3 = 8.7 grams.
New Alloy: Gold=69% of total solution.
G / (G+S) = 0.69,
8.7 / (8.7 + S) = 0.69,
Cross multiply:
6 + 0.69S = 8.7,
0.69S = 8.7 - 6 = 2.7,
S = 3.91 Grams of silver, total.
3.91 - 1.3 = 2.6 grams of silver added.
Your setup is close! However, there seems to be a small mistake in your equation. Let me help you correct it and explain the proper approach.
Let's break down the problem step by step:
1. Joe's ring initially weighs 10 grams and is made of an alloy of 13% silver and the rest gold.
This means that the amount of gold in the ring is 100% - 13% = 87%.
2. Joe wants to add more silver to reduce the gold content to 69%.
This means that the amount of gold after the addition of silver will be 100% - 69% = 31%.
Now, let's find the amount of silver Joe needs to add:
1. Determine the weight of gold in the ring initially:
Weight of gold = 10 grams * 87% = 8.7 grams
2. Determine the weight of gold after the addition of silver:
Weight of gold after = 10 grams * 31% = 3.1 grams
3. Calculate the weight of silver that needs to be added:
Weight of silver needed = Weight of gold initially - Weight of gold after
Weight of silver needed = 8.7 grams - 3.1 grams = 5.6 grams
Therefore, Joe needs to add 5.6 grams of silver to his ring in order to reduce the gold content to 69%.
I hope this explanation helps you understand the problem better!