A certain ore contains 21 percent gold. How much ore (in tons) is needed to obtain 882 tons of gold?
Well, to find out how much ore is needed, we can do some mathematical clowning!
If 21% of the ore is gold, that's like saying for every 100 tons of ore, we have 21 tons of gold.
Now, if we want 882 tons of gold, we can set up a little equation:
100 tons of ore contains 21 tons of gold,
so x tons of ore contains 882 tons of gold.
Using some magical math wizardry, we can cross-multiply and find that x = (882 * 100) / 21.
So, if you don't mind the clowny calculations, you'll need approximately 4200 tons of ore to obtain 882 tons of gold.
To determine the amount of ore needed to obtain 882 tons of gold, we need to calculate the overall percentage of gold in the ore.
Let's assume we need "x" tons of ore to obtain 882 tons of gold.
We know that the ore contains 21% gold, which means the remaining 79% is other elements.
The amount of gold in x tons of ore can be calculated as:
Gold in ore = 21% of x = 0.21x
According to the problem, this should equal to 882 tons of gold:
0.21x = 882
To solve for x, we divide both sides of the equation by 0.21:
x = 882 / 0.21
x = 4200
Therefore, we need 4200 tons of ore to obtain 882 tons of gold.
To find the amount of ore needed to obtain 882 tons of gold, we can use the concept of proportions.
Let x represent the amount of ore needed in tons.
We are given that the ore contains 21 percent gold, which means 21 percent of x is equal to 882 tons.
We can set up a proportion:
21/100 = 882/x
Now we can cross-multiply to solve for x:
21x = 100 * 882
21x = 88200
Dividing both sides by 21:
x = 88200 / 21
Simplifying:
x ≈ 4200
Therefore, approximately 4200 tons of ore would be needed to obtain 882 tons of gold.