Answer:
^@ 13 ^@ days
- 10 = rem
m = 10
n = 5 - Barbara takes ^@10^@ days to finish the work. So in ^@1^@ day, she finishes ^@\dfrac { 1 }{ 10 }^@ of the work.
- David takes ^@20^@ days to finish the work. So in ^@1^@ day, he finishes ^@\dfrac { 1 }{ 20 }^@ of the work.
- Since the work is done on alternate days, so the work done in two days ^@= \dfrac{ 1 }{ 10 } + \dfrac{ 1 } { 20 } = \dfrac{ 2 + 1 }{ 20 } = \dfrac{ 3 }{ 20 } ^@
- Multiply ^@ \dfrac{ 3 }{ 20 } ^@ by number such that it is less than or equal to ^@1.^@
The work done in ^@ 12 ^@ days ^@ = 6 \times \dfrac{ 3 }{ 20 } = \dfrac { 9 }{ 10 } ^@ - Remaining work ^@ = 1 - \dfrac { 9 }{ 10 } = \dfrac { 1 }{ 10 } ^@
- Since, they work in alternate days, on ^@ 13^{th} ^@ day remaining work ^@ \dfrac { 1 }{ 10 } ^@ is done by Barbara.
- Thus, they will take ^@ 13 ^@ days to finish the work.