Aptitude Discussion

Q. |
A man driving his bike at 24 kmph reaches his office 5 minutes late. Had he driven $25%$ faster on an average he would have reached 4 minutes earlier than the scheduled time. How far is his office? |

✖ A. |
24 km |

✖ B. |
72 km |

✔ C. |
18 km |

✖ D. |
Data Insufficient |

**Solution:**

Option(**C**) is correct

Let $x$ km be the distance between his house and office.

While travelling at 24 kmph, he would take \(\dfrac{x}{24}\) hours,

While travelling at $25\%$ faster speed,

i.e. $24+25\text{% of }24=$\(24\times \left(\dfrac{1}{4}\right)\)

= 30 kmph, he would take \(\dfrac{x}{30}\) hours

Now as per the problem, time difference = 5 min late + 4 min early = 9 min

\(\Rightarrow \dfrac{x}{24}-\dfrac{x}{30}=9\) min

$⇒ x$ = **18 km**

**Sandeep**

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**Bipul Thakur**

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try Distance= Product of speed *(difference of time)/ Difference of speeds

The question is not clear, since it is mentioned that 25 faster , not 25% faster which is missing...please update the question.

If a person travelling between two points reaches P hours late travelling at a speed of U kmph and reaches q hours early travelling at V kmph , the distance between the two points is given by VU/V-U ( p+q).

V = 24 and U = 30 p = 5 min and q = 4 mins late

24*30/6 ( 9/60 ) , we are dividing by 60 because time is in mins and speed is in kmph, so converting to hours.

D = 18