If Adam can do a job in 10 days and Ben's speed is twice of Adam's spe

Question stem, \(\frac{W_{A}}{W_{Total}}\)*100=?

Given, \(r_{A}\)=\(\frac{1}{10}\) & \(r_{B}\)=2*\(\frac{1}{10}\)=\(\frac{1}{5}\)
\(t_A+t_B=7\)

We know, work=Rate*Time & Total work done=1

So, \(r_A*t_A+r_B*t_B=1\)
Or, \(\frac{1}{10}*t_A+\frac{1}{5}*(7-t_A)=1\)
Or, \(14-t_A=10\)
Or, \(t_A=4\)

Now, \(\frac{W_{A}}{W_{Total}}\)*100=\(\frac{r_A*t_A}{1}\)*100=\(\frac{1}{10}\)*4*100=40%

Ans. (D)

Both work one after another??? both work one after another for 1 day each or 2 day each?? no clarity

That doesn't matter at all..
Since we have to ensure that the work has to be completed in 7 days and both have different speed, IT is possible only if they do certain% of work ..
The answer will not change if A works for 10 h and then B works for 1hr or there is NO pattern in their working. Finally the work jaa to be completed in exact 7 daya

A reasonable test-taker might interpret this information as follows:
If Adam works one day, then Ben must work the next day.
Earlier posts suggest that a different interpretation is intended.
I believe that the following clarifies the intent of the problem:


Let the job = 10 units.
Since Adam takes 10 days to complete the 10-unit job, Adam's rate = \(\frac{w}{t} = \frac{10}{10} = 1\) unit per day.
Since Ben is twice as fast as Adam, Ben's rate = 2 units per day.
For the 10-unit job to be completed in exactly 7 days, only one case is possible:
Adam works at his rate of 1 unit per day for 4 days, producing a total of 4 units.
Ben works at his rate of 2 units per day for 3 days, producing a total of 6 units.
Thus:
\(\frac{Work-by-Adam}{Total-work} = \frac{4}{10} = 40\)%.


An algebraic way to determine the number of days worked by Adam:
Let \(a=\) Adam's number of days and \(b=\) Ben's number of days.
Since Adam produces 1 unit per day, Ben produces 2 units per day, and a total of 10 units are produced, we get:
\(a+2b=10\)
Since a total of 7 days are worked, we get:
\(a+b=7\)
\(2a+2b=14\)
Subtracting the red equation from the blue equation, we get:
\((2a+2b)-(a+2b) = 14-10\)
\(a=4\)

Worked on the problem assuming they worked in the sequence of their names respectively.

Adam takes 10 days to do whole work, Adam's rate -\(\frac{1}{10}\) and Ben takes 5 days to do whole work, Ben's rate -\(\frac{1}{5}\)
\(\frac{1}{10}\) + \(\frac{1}{5}\)+\(\frac{1}{10}\)+\(\frac{1}{5}\)+\(\frac{1}{10}\)+\(\frac{1}{5}\)+\(\frac{1}{10}\) - since work is completed in 7 days

Total work by Adam in 7 days \(\frac{1}{10}\) +\(\frac{1}{10}\) +\(\frac{1}{10}\) +\(\frac{1}{10}\) Hence % work \(\frac{4}{10}*100\) =40%

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