
Problem of Mixture
Coffee A normally costs $ 100 per lb. It is mixed with coffee B, which normally costs $ 70 per lb, to form a mixture that costs $ 88 per lb. If there are 10 lbs of the mix, how many pounds of coffee A is used in the mixture?
Actually I don't get this problem properly.
What I got so far,
Coffee A costs $ 100 per lb, coffee B costs $ 70 per lb. Their mixture in certain ratio costs $ 88 per lb. Then how I count the ratio from this?

You add x units of A and y units of B, and you want to find $\displaystyle \frac{x}{y}$.
Try to get this equation somehow and solve it for $\displaystyle \frac{x}{y}$:
100x + 70y = 88(x + y)

Then I got, x:y = 3:2
Therefore, the amount of coffee A in the mixture is 10 x 3/5 = 6 lbs.
Is that the right way to do?
