Discussion

Explanation:

Let the efficiency of each man and woman be ‘m’ and ‘w’ units/day.

∴ Work done by 2 men and 3 women in 70 days  = (2m + 3w) × 70   …(1)
Work done by a men and 6 women in d days  = (2m + 6w) × d   …(2)

(1) = (2)
⇒ (2m + 3w) × 70 = (2m + 6w) × d

⇒ d = $\frac{2\mathrm{m}+3\mathrm{w}}{2\mathrm{m}+6\mathrm{w}}$ × 70 = $\frac{2\mathrm{m}+3\mathrm{w}}{\mathrm{m}+3\mathrm{w}}$ × 35 = $\left(\frac{\mathrm{m}}{\mathrm{m}+3\mathrm{w}}+\frac{\mathrm{m}+3\mathrm{w}}{\mathrm{m}+3\mathrm{w}}\right)$ × 35

⇒ d = $\left(\frac{1}{1+3{\displaystyle \frac{\mathrm{w}}{\mathrm{m}}}}+1\right)$ × 35

Now, d will be maximum when w/m is minimum i.e., when a woman is only twice as efficient as a man.
⇒ d_max = $\left(\frac{1}{1+3\times 2}+1\right)$ × 35 = 40 days

Now, d will be minimum when w/m is maximum i.e., when a woman is thrice as efficient as a man.
⇒ d_min = $\left(\frac{1}{1+3\times 3}+1\right)$ × 35 =  days

∴ 2 men and 6 women can complete the task in 38.5 - 40 days.

Hence, option (a).

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