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Question 98
If a person goes to his office at $$\frac{3}{4}$$ of his usual speed, he reaches the office late by 15 minutes. If he goes at usual rate. then the time taken to reachhis office, in hours, is
Solution
$$\Rightarrow t+15=\dfrac{4d}{3v}-----------(ii)$$
From the equation (i) and (ii)
$$\Rightarrow t+15=\dfrac{4t}{3}$$
$$\Rightarrow 3t+45=4t$$
$$\Rightarrow t=45min$$
$$\Rightarrow t=\dfrac{45}{60} hour = \dfrac{3}{4}hour$$
Let the usual speed of the man is v, the distance between the home and office is d and time taken by the man is t min.
$$t=\dfrac{d}{v}-----(i)$$
As per the condition given in the question,
$$\Rightarrow t+15=\dfrac{d}{\dfrac{3v}{4}}$$$$\Rightarrow t+15=\dfrac{4d}{3v}-----------(ii)$$
From the equation (i) and (ii)
$$\Rightarrow t+15=\dfrac{4t}{3}$$
$$\Rightarrow 3t+45=4t$$
$$\Rightarrow t=45min$$
$$\Rightarrow t=\dfrac{45}{60} hour = \dfrac{3}{4}hour$$
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