Math Help - Problem of Mixture

1. 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?

2. You add x units of A and y units of B, and you want to find $\frac{x}{y}$.

Try to get this equation somehow and solve it for $\frac{x}{y}$:

100x + 70y = 88(x + y)

3. 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?

4. Perfect.