In 2000, the internatonal price of gold was $270 per ounce, and the price of silver was $5 per ounce. How much silver could be mixed with 9 oz of gold to obtain a mixture that costs $164 per ounce?
total weight of silver + gold = s+g
total value of s+g = 5s + 270g
So, now for the mixture
5s + 270g = (s+g)*164
Bu, g=9
5s + 270*9 = 164(s+9)
5s + 2430 = 164s + 1476
159s = 954
s = 6
To solve this problem, we need to find out how much silver can be mixed with 9 oz of gold to obtain a mixture that costs $164 per ounce.
Let's assume that x ounces of silver are mixed with 9 oz of gold.
The price of gold is given as $270 per ounce, and the price of silver is given as $5 per ounce.
The cost of the gold in the mixture is 9 oz * $270/oz = $2,430.
The cost of the silver in the mixture is x oz * $5/oz = $5x.
The total cost of the mixture is the sum of the cost of gold and silver, which is $2,430 + $5x.
Since we want the mixture to cost $164 per ounce, we can set up the following equation:
($2,430 + $5x) / (9 + x) = $164
To solve this equation, we can multiply both sides by (9 + x) to get rid of the denominator:
$2,430 + $5x = $164(9 + x)
Now, let's simplify the equation:
$2,430 + $5x = $1,476 + $164x
Subtract $164x from both sides:
$2,430 - $1,476 = $164x - $5x
$954 = $159x
Now, we can solve for x by dividing both sides by $159:
x = $954 / $159
x ≈ 6
Therefore, approximately 6 ounces of silver can be mixed with 9 ounces of gold to obtain a mixture that costs $164 per ounce.