Times taken are in ratio:
\[ Anu : Tanu : Manu = 5 : 8 : 10 \]
Let their times be \(5x, 8x, 10x\). Let total work be \(W = 40x\) units.
Individual rates:
\[ \text{Anu's rate} = \frac{W}{5x} = 8, \quad \text{Tanu's rate} = \frac{W}{8x} = 5, \quad \text{Manu's rate} = \frac{W}{10x} = 4. \]
Together, rate = \(8 + 5 + 4 = 17\) units/hr. They finish job in 32 hours:
\[ 40x = 17 \times 32 \implies x = 13.6. \]
Anu and Tanu work for 6 days, 6 hours and 40 minutes per day (\(=\frac{20}{3}\) hours):
\[ \text{Total hours} = 6 \times \frac{20}{3} = 40. \]
Work done by Anu and Tanu:
\[ (8 + 5) \times 40 = 13 \times 40 = 520 \text{ units}. \]
Remaining work:
\[ 40x - 520 = 544 - 520 = 24 \text{ units}. \]
Time Manu takes (\(y\)) to finish remaining work at 4 units/hr:
\[ 4y = 24 \implies y = 6 \text{ hours}. \]
Manu will take 6 hours to complete the remaining job.
Given time ratios:
\[ Anu : Tanu : Manu = 5 : 8 : 10 \]
Efficiencies (units per hour) are inverse of times (assuming total work = \(40x\)):
\[ 8 : 5 : 4 \]
Total efficiency:
\[ 8 + 5 + 4 = 17 \text{ units/hr} \]
Total time when all work together:
\[ 8 \text{ hours/day} \times 4 \text{ days} = 32 \text{ hours} \]
Total work completed:
\[ 40x = 17 \times 32 = 544 \]
Anu and Tanu work for 6 days, 6 hours 40 minutes/day = \( \frac{20}{3} \) hours/day:
\[ 6 \times \frac{20}{3} = 40 \text{ hours} \]
Work done by Anu and Tanu:
\[ (8 + 5) \times 40 = 13 \times 40 = 520 \]
Remaining work:
\[ 544 - 520 = 24 \]
Manu's rate is 4 units/hr, so time taken to finish remaining work:
\[ 4y = 24 \implies y = 6 \text{ hours} \]
When $10^{100}$ is divided by 7, the remainder is ?