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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?

Blue
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user160370
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    If you know how to solve the problem with a quadratic equation, you should include that in your question. That'll save readers some time and effort in seeking an alternative approach – Blue Jul 27 '18 at 18:42
  • Not really. Dealing with rates of time and sharing work loads makes for an equation in which one must solve for variables in the reciprical. An the only way to solve those is to invert by multiplying by a common multiple of the denominaters. As the denominaters has a linear combination of variable, the common multiple will be a quadratic. – fleablood Jul 27 '18 at 19:10
  • @user160370 Please recall that if the OP is solved you can evaluate to accept an answer among the given, more details here https://meta.stackexchange.com/questions/5234/how-does-accepting-an-answer-work – user Sep 06 '18 at 20:13

2 Answers2

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

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user160370
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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.

user
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