Time and work. Arithmetic. 
$A$, $B$ 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?
 A: Let $A$, $B$ and $C$ be the amounts of work that can be done in a day by each worker. What this means is that $C$, for example, represents the numerical value of the amount of work worker $C$ can accomplish in a day. In one day, they together can do this much work: $A + B + C$. In two days, $2(A+B+C)$. In three days, $3(A + B + C)=\frac{37}{100}$. We also know that during the last $7$ days only workers $A$ and $B$ were working. And we also know the amount of work that they had done: $1-\frac{37}{100} = \frac{100}{100}-\frac{37}{100}=\frac{67}{100}$. This can be expressed like this: $7(A + B)=\frac{63}{100}$. And we also know that the amount of work done in $5$ days by worker $A$ is the same as the amount of work done by worker $B$ in $4$ days (by the way, we already know that worker $B$ works faster than worker $A$). So, here's what we have so far:
$$
3(C + A + B)=\frac{37}{100}\\
7(A + B)=\frac{63}{100}\\
5A=4B
$$
Now let's solve all that. The amount of work $A$ can do per day:
$$
5A=4B\implies B=\frac{5}{4}A\\
7(A + B)=\frac{63}{100}\implies 7\left(A + \frac{5}{4}A\right)=\frac{63}{100}\implies
A=0.04\\
$$
Now we can find the amount of work $B$ can do per day:
$$
B=\frac{5}{4}A\implies B=0.05\\
$$
The amount of work C can do per day:
$$
3(C + A + B)=\frac{37}{100}\implies C=\frac{1}{30}
$$
$0.05 > 0.04 > \frac{1}{30}≈0.03$: the fastest worker is worker $B$ because he can accomplish a larger amount of work per day than the other two workers. The whole job when finished is represented by the quantity $\frac{100}{100}$ which is equivalent to $1$. Now, remember that $0.05$ means that it takes worker $B$ one day to do accomplish this much work: $0.05$. How many days will it take him to complete the entire job ($\frac{100}{100}=1$) with that rate? If we let $x$ be the number of days, then it will take worker $B$ to complete the given work alone:
$$
x \cdot 0.05 = \frac{100}{100}\\
x \cdot 0.05 = 1\\
x = \frac{1}{0.05}\\
x = 20 \text{ days}
$$
Answer: 20 days.
A: I think I interpreted the problem a bit differently, thinking of $a,b$, and $c$ as the proportion of the total job $A$, $B$, and $C$ complete per day, respectively. 
In three days, with $A, B$, and $C$ working, $0.37$ of the work gets done: $$3(a+b+c)=\dfrac{37}{100}$$
The job takes $10$ days to complete, where $A, B$ work all ten days, but $C$ only works $3$: $$10(a+b) + 3c = 1$$
Finally, $A$ does $5$ times the work of $B$: $$5a=4b$$
Solving this system yields $a = \dfrac{4}{100}$, $b=\dfrac{5}{100}$, and $c = \dfrac{1}{30}$, so $B$ is the fastest worker. So, it follows that the number of days it takes $B$ to complete the job solo is given by $$\frac{5}{100}t = 1 \iff t = 20 \, \text{days}.$$
