$A, B$ and $C$ can complete a work in $10, 12\ and\ 15$ days respectively. They started the work together, but $A$ left the work before 5 days of its completion. $B$ also left the work $2$ days after $A$ left. In how many days was the work completed.

Let total work be 60 units. Rates: A=6, B=5, C=4. Let total days be T. A worked for T-5 days, B for T-2 days, C for T days. 6(T-5) + 5(T-2) + 4T = 60. 6T-30 + 5T-10 + 4T = 60. 15T = 100. T = 6.66. Checking the logic: If T=7, A works 2 days, B works 5 days, C works 7 days. 12+25+28 = 65. Close. Re-reading: A left 5 days before completion, B left 2 days after A (so 3 days before completion). 6(T-5) + 5(T-3) + 4T = 60. 15T = 60+30+15 = 105. T = 7.