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
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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