A smart phone manufacturer uses 10 ounces of gold circuitry to make 5,200 smart phones. If they made 7,800 smart phones last mon
2 answers:
Answer:
Option A. $19,500
Step-by-step explanation:
A smart phone manufacturer uses 10 ounces of gold circuitry to make 5,200 smartphones.
We will solve this problem by unitary method.
In 5200 smartphones use of gold = 10 ounces
in 1 smartphone use of gold = = 0.001923 ounces
In 7,800 smartphones, use of gold = 0.001923 × 7800 = 14.999 ounces rounded to 15 ounces of gold
Last month the value of 1 ounce gold is $1,300
The value of 15 ounces = 1,300 × 15 = $19,500
The value of gold used last month is $19,500.
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<h3>Answer:</h3>
- <u>20</u> kg of 20%
- <u>80</u> kg of 60%
I like to use a little X diagram to work mixture problems like this. The constituent concentrations are on the left; the desired mix is in the middle, and the right legs of the X show the differences along the diagonal. These are the ratio numbers for the constituents. Reducing the ratio 32:8 gives 4:1, which totals 5 "ratio units". We need a total of 100 kg of alloy, so each "ratio unit" stands for 100 kg/5 = 20 kg of constituent.
That is, we need 80 kg of 60% alloy and 20 kg of 20% alloy for the product.
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<em>Using an equation</em>
If you want to write an equation for the amount of contributing alloy, it works best to let a variable represent the quantity of the highest-concentration contributor, the 60% alloy. Using x for the quantity of that (in kg), the amount of copper in the final alloy is ...
... 0.60x + 0.20(100 -x) = 0.52·100
... 0.40x = 32 . . . . . . . . . . .collect terms, subtract 20
... x = 32/0.40 = 80 . . . . . kg of 60% alloy
... (100 -80) = 20 . . . . . . . .kg of 20% alloy