Time and Work. How much work does C do per hour? 
A, B and C need a certain unique time to do a certain work. C needs 1
  hours less than A to complete the work. Working together, they require
  30 minutes to complete 50% of the job. The work also gets completed if
  A and B start working together and A leaves after 1 hour and B works
  for a further 3 hours. How much work does C do per hour? 
(A)16.66% 
(B)33.33% 
(C)50% 
(D)66.66%

My attempt:
Let the total work be 100 units.
Let the work done by A,B and C be a units/hour,b units/hour,c units/hour respectively.
Let the time taken by A alone to complete the work be t hours.
ATQ:
\begin{align*}
(a+b+c) \cdot \frac{1}{2} & =50 \tag{1}\\
(a+b) \cdot 1+b \cdot 3 & =100 \tag{2}\\
c \cdot (t-1)& =100 \tag{3}\\
at & =100 \tag{4}
\end{align*}
Please help me solve these equations. When I am solving it is getting cumbersome. 
Also if someone tells us some other way of solving, that would be helpful as well. Thanks.

Let us assume, they take $a, b,$ and $c$ hours to finish the job individually, respectively. Note the following points:


*

*They complete half the job in half an hour $\implies$ full job is completed in a hour. Hence, $\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=1$

*Also, A takes 1 hour more time than C. So, $a= c+1\implies \frac{1}{a}=\frac{1}{c+1}$

*If A works for a hour, and B works for four hours, then $\frac{1}{a}+\frac{4}{b}=1 \implies \frac{1}{b}=\frac{1-\frac{1}{a}}{4}=\frac{1-\frac{1}{c+1}}{4}=\frac{c}{4+4c}$.
Now, solving for $c$ gives us the quadratic: $3c^2-4c-4=0 \implies c=2, c= -\frac{2}{3}$. As, $c >0$, we have, the rate of work C does $=0.5$.

Solving by your method: $a =\frac{100}{t}, b = \frac{100-a}{4}=\frac{25t-25}{t}$. Then, we get, $$a +b + c =100 \implies 100+25t-25 + ct = 100t \implies t = \frac{75}{75-c}$$
Putting this in Eq.(3), we get, $$\frac{100}{c}=\frac{c}{75-c}\equiv 7500-100c = c^2 \equiv c^2+100c-7500=0 \implies c = \frac{1}{2}, -\frac{3}{2}$$
Rejecting the negative solution, we have $c =\frac12$, the same answer as obtained earlier. 
No mistake in your approach!
