Question 189945
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  {{{(x+5)}}} 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.