John bought z hamburgers at $3 each and paid with a 20 dollar note. How much change did he get?
2 answers:
You might be interested in
So, they alloy is hmmm some piece of metal.
it has tin, among other metals, regardless of what other metals it may have, we know that for the first one, 15% is tin, the rest something else, and we also know that the alloy itself, weights 20 lbs.
now, how much is just the weight of the tin metal in it alone? well, is 15% of 20 lbs, or (15/100) * 20, or 3 lbs, so of all those 20 lbs, only 3 lbs or 15% is tin.
for the second alloy, let's say we'll add "x" lbs, we know the second alloy is 10% tin, how much of that weight is tin? well 10% of x, or (10/100) * x, or 0.10x lbs.
now, we're looking for a mixture, say the resulting lbs of mix will be "y" lbs, and we know the mixture will be 12% of tin, how much is 12% of y? (12/100) * y, or 0.12y.
now, regardless of what "x" and "y" is, we know that 20 + x = y, and that 3 + 0.10x = 0.12y, thus
it has tin, among other metals, regardless of what other metals it may have, we know that for the first one, 15% is tin, the rest something else, and we also know that the alloy itself, weights 20 lbs.
now, how much is just the weight of the tin metal in it alone? well, is 15% of 20 lbs, or (15/100) * 20, or 3 lbs, so of all those 20 lbs, only 3 lbs or 15% is tin.
for the second alloy, let's say we'll add "x" lbs, we know the second alloy is 10% tin, how much of that weight is tin? well 10% of x, or (10/100) * x, or 0.10x lbs.
now, we're looking for a mixture, say the resulting lbs of mix will be "y" lbs, and we know the mixture will be 12% of tin, how much is 12% of y? (12/100) * y, or 0.12y.
now, regardless of what "x" and "y" is, we know that 20 + x = y, and that 3 + 0.10x = 0.12y, thus