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What I have understood is that:

In the first case:

A starts the work and A finishes the work.

In the second case:

B starts the work and B finishes the work.

Otherwise, the time taken to complete the work would have been same in both the cases.

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2 Answers 2

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Hint:

Let the parts of the work completed in a day by $A,B,C$ be $a,b,c$

$100=9a+8b=9b+\dfrac{26a}3\iff a=3b$

$\implies9(3b)+8b=100\implies b=?,a=3b=?$

Again, $\dfrac{100}5=a+b+c\implies c=?$

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  • $\begingroup$ What is the logic for deciding how many days A would work and how many days B would work in the first and second cases respectively? $\endgroup$
    – Soumee
    Dec 16, 2017 at 10:22
  • $\begingroup$ @Soumee, In the first case,as $A$ starts he/she will work on the odd days like $1,3,5,\cdots,15,17$ $\endgroup$ Dec 16, 2017 at 10:24
  • $\begingroup$ Then in the second case, $B$ works in the odd number of days, and ends his work on the 17th day. Therefore, the last $\frac{2}{3}rd$ days work should have been completed by $A$. Understood! Thanks! $\endgroup$
    – Soumee
    Dec 16, 2017 at 10:31
  • $\begingroup$ Can you please help me with this question: math.stackexchange.com/q/2567879/394202 $\endgroup$
    – Soumee
    Dec 16, 2017 at 10:34
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Let's indicate with 100 the work to be done and with x,y,z the productivity for A,B,C.

We know that

$$9x+8y=100$$

$$\left(8+\frac23\right)x+9y=100$$

$$5(x+y+z)=10$$

From the above equations you can find x,y,z: $x=\frac{60}{7}$, $y=\frac{20}{7}$, $z=\frac{60}{7}$

Thus

$17\frac23$ days need for B and C to complete the work.

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