1. A can do a piece of work in 120 days and B can do it in 150 days. They work together for 20 days. Then A leaves and B continues the work. 12 days after that, C joins the work and the work is completed in 48 more days. In how many days C can do it alone?
A. 225
B. 220
C. 230
D. 160
Answer - Option D
Explanation -
A and B per day work = [latex]\frac {1}{120} + \frac {1}{150} = \frac {27}{1800}[/latex]
A and B work in 20 days = [latex]\frac {20 \times 27}{1800}[/latex]
B work in 12 days = [latex]\frac {12}{150} = \frac {144}{1800}[/latex]
Remaining work = 1 - [latex](\frac {540}{1800} + \frac {144}{1800})[/latex]
= [latex] \frac {1116}{1800} [/latex]
B and C, 1 day work = [latex] \frac {1116}{\frac {1800}{48}} [/latex]
C per day work = [latex] \frac {1116}{1800 \times 48} - \frac {1}{150}[/latex]
[latex]\frac {540}{1800 \times 48} = \frac {1}{160}[/latex]
So no. of days by C to complete the work = 160
2. A and B can do a piece of work in 10 days, B and C in 15 days and C and A in 20 days. They all work at it for 6 days, and then A leaves, and B and C go on together for 4 days more. If B then leaves, how long will C take to complete the work?
Answer - Option C
Explanation -
Amount of work done by A, B and C together in aday is [latex] \frac {6 + 4 + 3}{2 \times 60} = \frac {13}{120} [/latex]
Work done by all in 6 days = [latex] \frac {13}{120} [/latex]
Work done by B and C in 4 days = [latex]\frac {4}{15}[/latex]
Remaining work = 1 - [latex](\frac {13}{20} + \frac {4}{15})[/latex]
= [latex]\frac {1}{2}[/latex], which is to be done C
Now, from the question,
C alone can do the whole work in [latex]\frac {\frac {120}{18}\times {10}}{10 - \frac {120}{18}}[/latex] = 120 days
therefore, [latex]\frac {1}{12}[/latex] of the work is done by C [latex]\frac {120}{12}[/latex] = 10 days