A mixture of 17 parts of A, 3 parts of B and 4 parts of C weight 72 gm. How many gm of substance B are in the mixture?
A mixture of 17 parts of A, 3 parts of B and 4 parts of C weight 72 gm. How many gm of substance B are in the mixture? Correct Answer 9
Total parts = 17+3+4 = 2424 parts = 72 gm1 parts = 72/24 = 3gmnow, total substances of B = 3B = 3*3 = 9 (Ans.)