To do a certain work, A and B work on alternate days with B beginning the work on the first day. A alone can complete the same work in 24 days. If the work gets completed in $11\frac{1}{3}$ days. then B alone can complete $\frac{7}{9}$th part of the original work in:

6 days

Given:

Time taken by A to finish a task alone = 24 days

Calculation:

Let the total work be = 1

A alone can finish the task in 24 days

⇒ A's one-day work = \(\frac{1}{24}\)

A and B complete the whole task in = \( {11}_{3 }^{ 1} \)​ days

A and B work on alternate days, with B beginning so, we can say B will work only 6 days

⇒ A will work only \( {11}_{3 }^{ 1} \) - 6 = ​\( {5}_{3 }^{ 1} \) days

If A's one day work =  \(\frac{1}{24}\)of work A completes in 1 day

⇒ A's \( {5}_{3 }^{ 1} \) days work =  \(\frac{1}{24}\) × \( {5}_{3 }^{ 1} \)​ =  \(\frac{1}{24}\) ×  \(\frac{16}{3}\)

⇒  \(\frac{2}{9}\)

Remaining work = 1 - \(\frac{2}{9}\) = \(\frac{7}{9}\)

Therefore, B does the \(\frac{7}{9}\)th part of the work in 6 days.