A man walking with ¾ of his usual speed,reaches office 20 minutes late. His usual time is
We know, $$s_{1}\ t_{1} = s_{2}\ t_{2}$$
$$s_{1}\ t_{1} = (\frac{3}{4}) s_{1}\ (t_{1} + 20)$$
$$t_{1} = (\frac{3}{4}) (t_{1} + 20)$$
$$4 t_{1} = (3 t_{1} + 60)$$
$$t_{1} = 60$$ minutes
Hence, option D is the correct answer.
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