To solve this problem, we first need to understand how much work each person completes per day. This is known as their individual work rate.
Initially, A, B, and C started working together. We can find their combined work rate by adding their individual rates.
Combined rate of A, B, and C = Rate of A + Rate of B + Rate of C
Combined rate = $\frac{1}{12} + \frac{1}{15} + \frac{1}{20}$
To add these fractions, we find the least common multiple (LCM) of the denominators (12, 15, and 20). The LCM is 60.
Combined rate = $\frac{5}{60} + \frac{4}{60} + \frac{3}{60} = \frac{5+4+3}{60} = \frac{12}{60}$
Simplifying the fraction $\frac{12}{60}$, we get $\frac{1}{5}$. So, A, B, and C together complete $\frac{1}{5}$ of the work per day.
A, B, and C worked together for the first 2 days before C left.
Work done in 2 days = Combined rate $\times$ Number of days worked
Work done = $\frac{1}{5} \times 2 = \frac{2}{5}$
This means $\frac{2}{5}$ of the total work was completed in the first 2 days.
After C left, there was a portion of the work that still needed to be completed.
Remaining work = Total work - Work done in the first 2 days
Remaining work = $1 - \frac{2}{5} = \frac{5}{5} - \frac{2}{5} = \frac{3}{5}$
So, $\frac{3}{5}$ of the work is remaining.
Now, A and B are responsible for finishing the remaining $\frac{3}{5}$ of the work. Let's calculate their combined work rate.
Combined rate of A and B = Rate of A + Rate of B
Combined rate (A+B) = $\frac{1}{12} + \frac{1}{15}$
The LCM of 12 and 15 is 60.
Combined rate (A+B) = $\frac{5}{60} + \frac{4}{60} = \frac{5+4}{60} = \frac{9}{60}$
Simplifying this fraction, the combined rate of A and B is $\frac{3}{20}$ of the work per day.
To find out how many days A and B will take to complete the remaining work, we divide the remaining work by their combined rate.
Time = $\frac{\text{Remaining Work}}{\text{Combined Rate of A and B}}$
Time = $\frac{\frac{3}{5}}{\frac{3}{20}}$
Dividing fractions involves multiplying the first fraction by the reciprocal of the second fraction:
Time = $\frac{3}{5} \times \frac{20}{3}$
Time = $\frac{3 \times 20}{5 \times 3} = \frac{60}{15}$
Time = 4 days.
Thus, A and B will finish the remaining work in 4 days.
Mohan can do a piece of work in 10 days and Sohan in 15 days. They started working together, but after 3 days Mohan left the work . What time will Sohan take to finish the work?
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Had been one menless, then the number of days required to do a piece of work would have been one more. If the number of Man. Days required to complete the work is 56, how many workers were there?
A can do a piece of work in 16 hours, B and C can do it in 8 hours while A and C can do it 12 hours. How long will B alone take to do it?