Question 189945: How many ounces of pure gold must be added to five ounces of something that is 1 % gold in order that the mixture is 6% gold? Answer by cutepiscean5(19) (Show Source):
You can put this solution on YOUR website! Let the amount of pure gold to be added by x ounces,
and we know that the amount of something that is 1% gold is 5 ounces.
The total amount of the mixture would then be ounces
We know that pure gold is 100 % gold, and this x amount will be added to 5 ounces of 1 % gold to make the total amount of 6 % gold. So the resultant equation will be:
100 % (x) + 1 % (5) = 6 % (x+5)
using this equation we solve for x.
The above equation becomes:
Canceling the percent from the whole equation or multiplying the whole equation by 100 we get:
100x + 5 = 6(x+5)
=> 100x + 5 = 6x + 30 (using distributive law to open the brackets on the right hand side)
=> 100x - 6x = 30 - 5 (shifting the x-terms to the LHS and constant terms to the RHS.
=> 94x = 25
=> x = 25/94
or, x = 0.266 ounces
Thus, 0.266 ounces of pure gold must be added to five ounces of something that is 1 % gold in order that the mixture is 6% gold.
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