Time, Speed & Distance

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Walking at 3/4 of his usual place, a man reaches his office 20 minute late. Find his usual time?


2 hr


1 hr


3 hr


1.5 hr

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Option(B) is correct

Let the original speed be $S$ and time be $T$

If new speed = \(S\times \dfrac{3}{4}\), then new time would be \(T\times \dfrac{4}{3}\) ($D = ST$ = Constant).



$⇒ T=60$ minutes
1 hour

Edit: For an alternate solution, check comment by Vaibhav Gupta.

(4) Comment(s)


I cannot understand this problem. Please explain simply the question


Can you specify which part is not clear to you?

Vaibhav Gupta

In question read the word 'place' as 'speed' as word 'place' is wrongly written over there and the question will be crystal clear to you

Vaibhav Gupta

Alternate solution:

$s=\dfrac{d}{t}$ So $d=st$ ..........(1)

New speed $=\dfrac{3s}{4}$

New time $= t+20$

So, $d=\dfrac{3s}{4}×(t+20)$ .......(2)

Comparing (1) and (2),

We get, $t = \textbf{60 minutes}$ i.e. 1 hour