Aptitude Discussion

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

✖ A. |
2 hr |

✔ B. |
1 hr |

✖ C. |
3 hr |

✖ D. |
1.5 hr |

**Solution:**

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).

Given,

\(\dfrac{3T}{4}-T=\dfrac{20T}{3}\)

$⇒ T=60$ minutes

= **1 hour**

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

**Edit 2:** Based on the comment by **Vaibhav Gupta,** word 'place' in the question has been replaced by 'pace'.

**Edit 3:** For yet another alternative solution, check comment by **Roy.**

**Saikat**

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**Roy**

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We know that the speed is inversely proportional to time:

If he is walking at $\left(\dfrac{3}{4}\right)v$ then he must be taking $\left(\dfrac{4}{3}\right)t$

Extra time $= \left(\dfrac{4}{3}\right)t-t =\left(\dfrac{1}{3}\right)t = 20$

So $\textbf{t = 60 minutes}$

**VENI**

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I cannot understand this problem. Please explain simply the question

Can you specify which part is not clear to you?

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**

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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

s1:s2=4:3. so, t1:t2=3:4

now, t2=t1+20.

t1/t1+20=3/4.

t1=60(1hr)