$A$, $B$ and $C$ working together completed a job in $10$ days. However, $C$ only worked for the first three days when $\dfrac{37}{100}$ of the job was done. Also, the work done by $A$ in $5$ days equals the work done by $B$ in $4$ days. How many days will the fastest worker take to complete the given work alone?
Is my approach right?
My attempt:
Let the rate of work done by $A,B,C$ be $a \,\text{units/day}, b \,\text{units/day}, c \,\text{units/day}$ respectively.
Since, $C$ works only on the first $3$ days (along with $A$ and $B$), and $B$ and $A$ works together for the rest of the time,
$$\text{Total work}=(a+b+c)\cdot3+(a+b)\cdot7=10(a+b)+3c$$
Also, $$3c=\frac{37}{100}[10(a+b)+3c]$$ so $$189c=370(a+b) \tag{1}$$
and $$5a=4b \tag{2}$$
Where is the third equation that would solve the problem?