Question

Amal and Vimal together can complete a task in 150 days, while Vimal and Sunil together can complete the same task in 100 days. Amal starts working on the task and works for 75 days, then Vimal takes over and works for 135 days. Finally, Sunil takes over and completes the remaining task in 45 days. If Amal had started the task alone and worked on all days, Vimal had worked on every second day, and Sunil had worked on every third day, then the number of days required to complete the task would have been how many?

• A. 138 days
• B. 101 days
• C. 143 days
• D. 166 days
• E. 139 days

Hint:

When dealing with cyclic patterns, find the Least Common Multiple (LCM) of the cycles to define a repeat period. Calculate work per period and then find the number of full periods and the leftover work.

Answer 1

Answer: (E)

Step 1: Define Variables:
Let the total work be $W$. Let the efficiencies (work per day) of Amal, Vimal, and Sunil be $a$, $v$, and $s$ respectively.
Step 2: Form Equations:
From the given information: a + v &= \fracW150 (1)
v + s &= \fracW100 (2) Also, from the second work pattern: $75a + 135v + 45s = W$.
Step 3: Solve for Individual Efficiencies:
We can treat $W$ as 1 (unit work) for convenience. Let $W = 1$ unit. Then $a+v = \frac{1}{150}$ and $v+s = \frac{1}{100}$. Substitute into $75a + 135v + 45s = 1$. $75a + 90v + 45(v+s) = 1$ $75a + 90v + 45 \times \frac{1}{100} = 1$ $75a + 90v + 0.45 = 1$ $75a + 90v = 0.55 = \frac{55}{100} = \frac{11}{20}$. Multiply the equation $a+v = \frac{1}{150}$ by 75: $75a + 75v = \frac{75}{150} = \frac{1}{2} = 0.5$. Subtract this from $75a + 90v = 0.55$: $(75a+90v) - (75a+75v) = 0.55 - 0.5 \implies 15v = 0.05 \implies v = \frac{0.05}{15} = \frac{1}{300}$. Then $a = \frac{1}{150} - \frac{1}{300} = \frac{1}{300}$. And $s = \frac{1}{100} - \frac{1}{300} = \frac{2}{300} = \frac{1}{150}$. So, $a = \frac{1}{300}$, $v = \frac{1}{300}$, $s = \frac{1}{150}$.
Step 4: New Work Pattern:
Amal works on all days. Vimal works on every second day (days 2, 4, 6, ...). Sunil works on every third day (days 3, 6, 9, ...). We need to find the number of days $N$ such that the total work done is at least 1.
Step 5: Calculate Work in a Cycle:
Consider a cycle of 6 days (LCM of 1, 2, 3). Day 1: Amal only. Work = $a = \frac{1}{300}$. Day 2: Amal + Vimal. Work = $a+v = \frac{1}{300}+\frac{1}{300} = \frac{2}{300}$. Day 3: Amal + Sunil. Work = $a+s = \frac{1}{300}+\frac{1}{150} = \frac{1+2}{300} = \frac{3}{300}$. Day 4: Amal + Vimal. Work = $\frac{2}{300}$. Day 5: Amal only. Work = $\frac{1}{300}$. Day 6: Amal + Vimal + Sunil. Work = $a+v+s = \frac{1}{300}+\frac{1}{300}+\frac{1}{150} = \frac{1+1+2}{300} = \frac{4}{300}$. Total work in 6 days = $\frac{1+2+3+2+1+4}{300} = \frac{13}{300}$.
Step 6: Determine Number of Days:
Number of complete cycles = $\left\lfloor \frac{1}{\frac{13}{300}} \right\rfloor = \left\lfloor \frac{300}{13} \right\rfloor = 23$ cycles. Work done in 23 cycles ($23 \times 6 = 138$ days) = $23 \times \frac{13}{300} = \frac{299}{300}$. Remaining work = $1 - \frac{299}{300} = \frac{1}{300}$. Day 139 is the first day of the next cycle. On day 139, Amal works (as it's day 1 of new cycl(e). Work done on day 139 = $\frac{1}{300}$, which exactly completes the work. Total days required = $138 + 1 = 139$.