$A, B \text{ and } C$ started working together on a project, but $A$ left the work $9$ days before the completion and $B$ left the work $7$ days before completion. Had $A$ worked for one more day, the work would have been completed two days before the actual time. Had $B$ worked for two days more, the work would have been completed three days before the actual time. In how many days can $A, B \text{ and}$ $C$ together do a different work which can be done by $C$ alone in $18$ days?
Let in the first case the total number of days taken be $t$ days.Let the number of days taken by $A, B \text{ and } C$ to complete the work individually be $a, b \text{ and } c$ respectively.
As per the question ;
$\frac{t-9}{a}+\frac{t-7}{b}+\frac{t}{c}=1$
I am getting stuck after this as to how to form the equation for the other cases. In the second case (where $A$ worked for one more day), the number of days will be $t-2$ days. Now the equation should be like ;
$\frac{p}{a}+\frac{q}{b}+\frac{t-2}{c}=1$ and I am not able to figure out what should be the values for $p$ and $q$ in the equation.
Also how to create the equation for the $3rd$ case where $B$ worked for two more days.
Please help me resolve this problem!!!
Thanks in advance !!!
Hint: when $A$ works one day more, he works $t-8$ days. How many days do $B$ and $C$ work in the second case?