A and B together can do a piece of work in 4 days, B and C can do it in 6 days, A and C can do it in 8 days. Then A, B and C together can do the same work in :-
48/13 days
This problem involves calculating the time taken by three individuals, A, B, and C, to complete a piece of work together, given the time taken by them in pairs.
The key concept here is the work rate. If a person or a group can complete a work in 'd' days, their work rate (the amount of work done in one day) is \( \frac{1}{d} \).
Given the information:
Based on the work rate concept:
Now, if we add the work done by all pairs in one day, we get:
\[ (\text{Work by A in 1 day} + \text{Work by B in 1 day}) + (\text{Work by B in 1 day} + \text{Work by C in 1 day}) + (\text{Work by A in 1 day} + \text{Work by C in 1 day}) \] \[ = \frac{1}{4} + \frac{1}{6} + \frac{1}{8} \]
This sum is equal to twice the work done by A, B, and C together in one day, because each person's work rate is included twice (A is in A+B and A+C; B is in A+B and B+C; C is in B+C and A+C).
So, \( 2 \times (\text{Work by A + B + C in 1 day}) = \frac{1}{4} + \frac{1}{6} + \frac{1}{8} \)
To add the fractions \( \frac{1}{4}, \frac{1}{6}, \) and \( \frac{1}{8} \), we find the least common multiple (LCM) of the denominators 4, 6, and 8. The LCM of 4, 6, and 8 is 24.
\[ \frac{1}{4} = \frac{1 \times 6}{4 \times 6} = \frac{6}{24} \] \[ \frac{1}{6} = \frac{1 \times 4}{6 \times 4} = \frac{4}{24} \] \[ \frac{1}{8} = \frac{1 \times 3}{8 \times 3} = \frac{3}{24} \]
Now, adding the fractions:
\[ \frac{6}{24} + \frac{4}{24} + \frac{3}{24} = \frac{6 + 4 + 3}{24} = \frac{13}{24} \]
So, \( 2 \times (\text{Work by A + B + C in 1 day}) = \frac{13}{24} \)
To find the work done by A, B, and C together in 1 day, we divide the sum by 2:
\[ \text{Work by A + B + C in 1 day} = \frac{1}{2} \times \frac{13}{24} = \frac{13}{48} \]
The total time taken by A, B, and C together to complete the work is the reciprocal of their combined work rate per day.
Time taken by A, B, and C together = \( \frac{1}{\text{Work by A + B + C in 1 day}} = \frac{1}{\frac{13}{48}} = \frac{48}{13} \) days.
Therefore, A, B, and C together can do the same work in \( \frac{48}{13} \) days.
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