Time and work. Arithmetic 
3 cooks have to make 80 burgers.They are known to make 20 pieces every
  minute working together.The 1st cook began working alone and made 20
  pieces having worked for sometime more than 3 mins.The remaining part
  of the work was done by second and 3rd cook working together.It took a
  total of 8 minutes to complete the 80 burgers.How many minutes would
  it take the 1st cook alone to cook 160 burgers?

Let the rate at which the three cooks A,B,C prepare burgers be a burgers/min, b  burgers/min, c  burgers/min.
Therefore ATQ:
$$a+b+c= 20 \tag{1}$$
Let the amount of time in which the first cook A worked alone be t.
This implies:
$$at=20 \tag{2}$$
also:
$$at+(b+c)(8-t)=80$$
This implies:
$$20+(b+c)(8-\frac{20}{a})=80 \tag{3}$$
So I have two equations (1) and (3) involving the variables a,b and c.
I can't form the third equation.
And also, how to solve the equations wherein variable 'a' in eq 3 looks so scary!
 A: Let the time required to cook $160$ burgers by the first cook be $T.$ This gives $a = \frac{160}{T}$. Then, the time required to cook twenty burgers equals: $$\frac{20}{\frac{160}{T}}=\frac{T}{8}$$
This gives us: $$b+c = \frac{80-20}{8-\frac{T}{8}}=\frac{480}{64-T}$$
Now, as $a+b+c=20$, we have the quadratic equation: $$512-48T+T^2=0 \implies T= 16, 32$$
As, $\frac{T}{8}>3$, we choose the solution $T=32$.
A: Let $x_{1}$ be rate of work of the first cook (burgers per minute)
Let $x_{2}$ be rate of work of the first cook (burgers per minute)
Let $x_{3}$ be rate of work of the first cook (burgers per minute)
And let $t$ be the time (measured in minutes) it takes the first cook to make $20$ burgers
These are our conditions expressed in mathematical form:
$$
x_{1}+x_{2}+x_{3}=20 \text{ (when they work together)}\\
t\cdot x_{1}=20 \text{ (first cook makes 20 burgers)}\\
(8-t)(x_{2}+x_{3})=60 \text{ (rest of work done by 2nd and 3rd cooks)}\\
t\cdot x_{1}+(8-t)(x_{2}+x_{3})=80 \text{ (8 minutes to complete the job)}
$$
Solution:
$$
x_{1}+x_{2}+x_{3}=20\implies x_{2}+x_{3}=20-x_{1}\\
\text{therefore}\\
t\cdot x_{1}+(8-t)(20-x_{1})=80
$$
Rate of work of the first cook:
$$
t\cdot x_{1}+(8-t)(20-x_{1})=80\implies x_{1}=\frac{t10-40}{t-4}
$$
Now, solve for $t$:
$$
t\cdot \left(\frac{t10-40}{t-4}\right)=20\implies t=2 \text{ and } t=4\\
$$
$t=4$ is the only solution of this equation that makes sense here because according to one of the problem's statements, the time it takes the first cook to make $20$ burgers is more than $3$ minutes (and $t=4 \text{ minutes}$ is that three and a little bit minutes it takes him to make $20$ burgers).
The rate of work of the first cook therefore:
$$
t\cdot x_{1}=20\implies\\\
x_{1}=\frac{20 \text{ burgers}}{4 \text{ minutes}}\implies\\
x_{1}=5 \text{ burgers / minute}
$$
For the first cook to make $160$ burgers, we have to solve this simple equation where $M$ is how many minutes it will take him to do this job:
$$
M\cdot x_{1}=160\\
M=\frac{160}{x_{1}}\\
M=\frac{160 \text{ burgers}}{5 \text{ burgers / minute}}\\
M=32 \text{ minutes}
$$
Answer: it will take the first cook $32$ minutes to make $160$ burgers.
A: Cooks 1, 2 and 3.
Respective speeds: $a,b,c$.
First part (1st cook) : $(3+x) \times a=20$.
Second part (2nd and 3rd cooks working together) : $b \times (8-(3+x)) + c \times (8-(3+x))=60$. I.e. $b+c= \dfrac {60} { 5-x}$.
We know that : $a+b+c=20$ and thus: $b+c= 20-a$.
So:

$a= 20- \dfrac {60} { 5-x}$.

Substituting $a$ in the first equation and simplifying, we get:


$x^2-1=0$.


We need the solution $x=1$, that means that the time the 1st cook worked alone is $4$ minutes. This implies: $a=\dfrac {20} {4} =5$.
Conclusion: How many minutes would it take the 1st cook alone to cook $160$ burgers? 
