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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.

Soumee
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2 Answers2

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

lab bhattacharjee
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  • 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? – Soumee Dec 16 '17 at 10:22
  • @Soumee, In the first case,as $A$ starts he/she will work on the odd days like $1,3,5,\cdots,15,17$ – lab bhattacharjee Dec 16 '17 at 10:24
  • 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! – Soumee Dec 16 '17 at 10:31
  • Can you please help me with this question: https://math.stackexchange.com/q/2567879/394202 – Soumee Dec 16 '17 at 10:34
1

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.

user
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