SOLUTION: How much of an alloy that is 30% copper should be mixed with 200 ounces of an alloy that is 70 % copper in order to get an alloy that is 50 % copper this is what I came up with.

Algebra ->  Percentage-and-ratio-word-problems -> SOLUTION: How much of an alloy that is 30% copper should be mixed with 200 ounces of an alloy that is 70 % copper in order to get an alloy that is 50 % copper this is what I came up with.      Log On


   

Question 855701: How much of an alloy that is 30% copper should be mixed with 200 ounces of an alloy that is 70 % copper in order to get an alloy that is 50 % copper
this is what I came up with...is this right
x(30%)+(200)(70%)=200+x(50)
.3x+140=100+.5x
40=.2x
x=200
Answer by josgarithmetic(39617) About Me  (Show Source):
You can put this solution on YOUR website!
x how much to add or use

%2830x%2B200%2A70%29%2F%28x%2B200%29=50

Your initial setup seems good. Maybe you let the switch to decimal fractions confuse you?

30x%2B200%2A70=50%28x%2B200%29
3x%2B200%2A7=5%28x%2B200%29
3x%2B1400=5x%2B1000
highlight%28x=200%29, and in fact this works as correct when tried as a check in our original equation.

(Seemed strange to me also, ...until examining the AVERAGE of 30 and 70. Their middle value IS 50.)