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Difficulty:
85%
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Question Stats:
62% (02:57) correct
38%
(03:12)
wrong
based on 105
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A, B and C can do a work in 20 days, 15 days and 12 days respectively. They all started the work together, but C left the job three days before its completion, and B left the job two days before C. In how many days, did the work get completed?
A. \(6 \frac{11}{12}\)
B. \(7 \frac{11}{12}\)
C. \(8 \frac{11}{12}\)
D. \(9 \frac{11}{12}\)
E. \(10 \frac{11}{12}\)
A. \(6 \frac{11}{12}\)
B. \(7 \frac{11}{12}\)
C. \(8 \frac{11}{12}\)
D. \(9 \frac{11}{12}\)
E. \(10 \frac{11}{12}\)
Method 1: Using efficiency
First, find the LCM (Least Common Multiple) of \((A, B, C) = lcm(20, 15, 12) = 60\).
Consider the LCM as "the total work" to be completed by \((A+B +C)\) when all of them worked together.
Next, find the efficiency of A, B, C, i.e., \(\dfrac{\text{Total work}}{\text{Time to complete the work individually}}\)
\(A = \dfrac{60}{20} = 3, B = \dfrac{60}{15} = 4, C = \dfrac{60}{12} = 5\).
Now to find the work completed by \(A\) in the absence of \(C\) and \(B\):
\(C\) left the work \(3\) days before, which was completed by A. So, the amount of A's work in the absence of C is increased and becomes \((\text{A's efficiency} \times days) = 5\times 3 = 15\).
Similarly, B left the work 2 days before C (i.e. \(2 + 3 = 5 days\)), and the amount of A's work was increased with B's work (in the absence of B), i.e. \(4 \times 5 = 20\).
Remaining total work which was completed by \((A+B+C)\)
Now, add the total work is \(60 + 15 + 20 = 95\) which was completed by \((A+B+C)\) .
Total efficiency of \((A+B+C)\): \((A+B+C) = 3 + 4 + 5 = 12\)
Time to complete the total work \( = \dfrac{95}{12} = 7\frac{11}{12}\)
B
First, find the LCM (Least Common Multiple) of \((A, B, C) = lcm(20, 15, 12) = 60\).
Consider the LCM as "the total work" to be completed by \((A+B +C)\) when all of them worked together.
Next, find the efficiency of A, B, C, i.e., \(\dfrac{\text{Total work}}{\text{Time to complete the work individually}}\)
\(A = \dfrac{60}{20} = 3, B = \dfrac{60}{15} = 4, C = \dfrac{60}{12} = 5\).
Now to find the work completed by \(A\) in the absence of \(C\) and \(B\):
\(C\) left the work \(3\) days before, which was completed by A. So, the amount of A's work in the absence of C is increased and becomes \((\text{A's efficiency} \times days) = 5\times 3 = 15\).
Similarly, B left the work 2 days before C (i.e. \(2 + 3 = 5 days\)), and the amount of A's work was increased with B's work (in the absence of B), i.e. \(4 \times 5 = 20\).
Remaining total work which was completed by \((A+B+C)\)
Now, add the total work is \(60 + 15 + 20 = 95\) which was completed by \((A+B+C)\) .
Total efficiency of \((A+B+C)\): \((A+B+C) = 3 + 4 + 5 = 12\)
Time to complete the total work \( = \dfrac{95}{12} = 7\frac{11}{12}\)
B
26 Jan 2026, 21:01
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