Time and Work

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$A$ can complete a project in 20 days and $B$ can complete the same project in 30 days. If $A$ and $B$ start working on the project together and $A$ quits 10 days before the project is completed, in how many days will the project be completed?









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Option(A) is correct

If $A$ can do complete a project in 20 days, then $A$ will complete \(\left(\dfrac{1}{20}\right)^{th}\)of the project in a day.

Similarly, $B$ will complete \(\left(\dfrac{1}{30}\right)^{th}\) of the project in a day.
Let the total number of days taken to complete the project be $x$ days.

Then, $B$ would have worked for all $x$ days, while $A$ would have worked for $(x–10)$ days.

Therefore, $A$ would have completed  $\dfrac{x-10}{20}$ of the project and B would have completed $\dfrac{x}{30}$ of the project.

$\Rightarrow \dfrac{x-10}{20} + \dfrac{x}{30}=1$

Solving for $x$, we get $x$ = 18.

(3) Comment(s)

Ranjit Mishra

I am getting 27 with a different approach.

A and B can complete work in one day together = 1/20+1/30 = 1/12

So they can complete work in 12 days together.

Since A left before 10 days so work done together in 2 days = (1/12)*2 = 1/6

Work left to be done = 1-(1/6) = 5/6

Days taken by B = (5/6)/(1/30) = 25 days

Total time taken = 25 +2 = 27 days. .

how is this possible?? have I missed something


Is the following equation correct for the problem:



Yes that is correct equation.