$A$, $B$, $C$ do a job $6$ hours faster than $A$ alone, $1$ hour faster than $B$ alone, and in half the time needed by $C$ alone. ...

Question

Three men $A$, $B$, $C$ working together can do a job in $6$ hours less time than $A$ alone, in $1$ hour less time than $B$ alone, and in one-half the time needed by $C$ when working alone. Then $A$ and $B$ together can do the job in how many hours?

Could we solve this without a quadratic equation?

Answer 1

I think it is not possible without a qudratic equation, here is my solution.

Let's indicate with 100 the work to be done and with $x,y,z$ the productivity for A,B,C.

We know that

$$\frac{100}{x+y+z}=\frac{100}x-6 \implies100x=100(x+y+z)-6x(x+y+z)$$

$$\frac{100}{x+y+z}=\frac{100}y-1\implies100y=100(x+y+z)-y(x+y+z)$$

$$\frac{100}{x+y+z}=\frac12\frac{100}z\implies z=x+y$$

then summing up the two first equations we obtain

$$100z=400z-(5x+z)(2z)\implies5x+z=150 \implies z=150-5x$$

and from the first one

$$100x=200z-12xz \implies100x=30000-1000x-1800x+60x^2 $$

$$\implies 3x^2-145x+1500=0$$

$$\implies x=15 \implies z=75 \implies y=60$$

Thus $A$ and $B$ working togheter need $\frac{100}{75}=\frac{4}{3}$h to complete the work.

Answer 2

I did some search over google and found this. They have used minimal steps. But, I couldn't understand most of it.