In a group of 5 boys the $$2^{nd}$$ boy is twice as efficient as the $$1^{st}$$ boy. The $$3^{rd}$$ boy is twice as efficient as the $$2^{nd}$$boy and so on. All of them working together will take 5 days to complete a job. How much extra time will the $$2^{nd}$$ and $$4^{th}$$ boys take working together as compared to the $$5^{th}$$ boy working alone to complete the same job approximately

Ratio of efficiency of $$1^{st}$$ boy to $$2^{nd}$$ boy = 1: 2

Ratio of efficiency of $$2^{nd}$$ boy to $$3^{rd}$$ boy = 1: 2

And so on...

So ratio of efficiency of $$1^{st}$$, $$2^{nd}$$, $$3^{rd}$$, $$4^{th}$$ and $$5^{th}$$tboy = 1: 2: 4: 8: 16

Let unit of work done by $$1^{st}$$, $$2^{nd}$$, $$3^{rd}$$, $$4^{th}$$ and $$5^{th}$$boy is 1, 2, 4, 8 and 16 respectively. 

Since, job is completed by all the five boys together in 5 days. Hence,total work = 5 * (1 + 2 + 4 + 8 + 16) = 155

Time taken by $$2^{nd}$$ and $$4^{th}$$ boys together to do the work = $$\frac{155}{2 + 8} = 15.5$$ days

Time taken by $$5^{th}$$ boy alone to do the work = $$\frac{155}{16}$$ days

Required diffrence = 15.5 - $$\frac{155}{16} = \frac{93}{16} = 5.8125$$ days = 6 days (Approx)