Let's assume W be the total amount of work.
And g and s be the efficiencies of Gautam and Suhani respectively.
According to the question:
$ \Rightarrow\ g + s = \frac{W}{20} $ (1 day work) ……. (i)
And given that Gautam is doing only 60%: $ \frac{3g}{5} $
Suhani is doing 150%: $ \frac{3s}{2} $
Now, using this, we get:
$ \Rightarrow\ \frac{3g}{5} + \frac{3s}{2} = \frac{W}{20} $ (1 day work)
$ \Rightarrow\ g + s = \frac{3g}{5} + \frac{3s}{2} $
$ \Rightarrow\ \frac{s}{g} = \frac{4}{5} $
This implies that Gautam is more efficient.
By using equation (i), we get:
$ \Rightarrow\ g + \frac{4g}{5} = \frac{W}{20} $
$ \Rightarrow\ \frac{9}{5}g = \frac{W}{20} $
$ \Rightarrow\ g = \frac{W}{36} $
Therefore, Gautam takes 36 days to finish the given work.
So, the correct option is (B) : 36.
For any natural number $k$, let $a_k = 3^k$. The smallest natural number $m$ for which \[ (a_1)^1 \times (a_2)^2 \times \dots \times (a_{20})^{20} \;<\; a_{21} \times a_{22} \times \dots \times a_{20+m} \] is: