Question : C1, C2 and C3 can do work alone in 10, 12, and 15 days respectively. All three of them began the work together but C2 left 2 days before the completion of the work. In how many days was the work completed?
Option 1: 3
Option 2: 7
Option 3: 5
Option 4: 4
Correct Answer: 5
Solution :
C1, C2 and C3 can do a work alone in 10,12 and 15 days, respectively.
C1's 1 day's work = $\frac{1}{10}$
C2's 1 day's work = $\frac{1}{12}$
C3's 1 day's work = $\frac{1}{15}$
(C1 + C2 + C3)'s 1 day's work = $\frac{1}{10}+\frac{1}{12}+\frac{1}{15}=\frac{1}{4}$
So, they will complete the work in 4 days.
C2 left 2 days before the completion of the work.
So, remaining work = $1-\frac{1}{2}=\frac{1}{2}$
(C1 + C3)'s 1 day's work = $\frac{1}{10}+\frac{1}{15}=\frac{1}{6}$
C1 and C3 together will complete the remaining work in $\frac{\frac{1}{2}}{\frac{1}{6}}=3$ days
Total days = 2 + 3 = 5 days
Hence, the correct answer is 5.
Related Questions
Know More about
Staff Selection Commission Multi Tasking ...
Application | Cutoff | Selection Process | Preparation Tips | Eligibility | Exam Pattern | Admit Card
Get Updates Brochure
Your Staff Selection Commission Multi Tasking Staff Exam brochure has been successfully mailed to your registered email id “”.
