Discussion

Explanation:

Let the efficiency of man be x unit’s per day.

Day 1: Only 1 person is working
Work done = x

Day 2: 2 persons are working

Work done = x + $\frac{x}{2}$ = $\frac{3x}{2}$

Day 3: 3 persons are working
Work done = x + $\frac{x}{2}$ + $\frac{x}{4}$ = $\frac{7x}{4}$

...

Day n: n persons are working
Work done = $\left(x+\frac{x}{2}+\frac{x}{4}+\dots +\frac{x}{2^{n-1}}\right)$ = $\frac{\left(2^n-1\right)x}{2^{n-1}}$ = $\left(2-\frac{1}{2^{n-1}}\right)x$

∴ Total work done in n days = $\left(2-\frac{1}{1}\right)x$ + $\left(2-\frac{1}{2}\right)x$ + $\left(2-\frac{1}{4}\right)x$ + .... + $\left(2-\frac{1}{2^{n-1}}\right)x$

= $\left(2n-\left(1+\frac{1}{2}+\frac{1}{4}+\dots +\frac{1}{2^{n-1}}\right)\right)$

One man can complete the work in 20 days, i.e., total work done = x × 20 = 20x

⇒ $\left(2n-\left(1+\frac{1}{2}+\frac{1}{4}+\dots +\frac{1}{2^{n-1}}\right)\right)$x ≥ 20x

⇒ 2n - $\frac{\left(2^n-1\right)}{2^{n-1}}$ ≥ 20

⇒ 2n - $\left(2-\frac{1}{2^{n-1}}\right)$ ≥ 20

⇒ 2n + $\frac{1}{2^{n-1}}$ ≥ 22

⇒ n + $\frac{1}{2^n}$ ≥ 11

⇒ n ≥ 11

11th day from Monday will be Thursday.

Hence, option (b).

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