Question

If a man walks from his house at 5 km/hr, he
reaches office 5 minutes late. If he walks at 6
km/hr, he reaches 5 minutes early. Then the s
distance from his house to the office is​

Answer 1

The distance from his house to the office is 5 km.

Step-by-step explanation:

Required Formula:

• Distance = Speed * Time

Let the distance from the man’s house to the office be denoted as “d” km and the usual time taken to cover this distance be “t” hours.

When the man is walking with a speed of 5 km/hr, he is reaching office 5 minutes late i.e., he is taking 5 minutes more than the usual time, so

Distance, d = [t + (5/60)]*5 …… (i)

When the man is walking with a speed of 6 km/hr, he is reaching office 5 minutes early i.e., he is taking 5 minutes less than the usual time, so

Distance, d = [t - (5/60)]*6 …… (ii)

Since the distance is constant, therefore, equating (i) & (ii), we get

[t + (5/60)]*5 = [t - (5/60)]*6

⇒ 5t + (5/12) = 6t – (½ )  

⇒ t = (5/12) + (½)  

⇒ t = (11/12) hr …… (iii)

Substituting (iii) in (i), we get

d = [(11/12) + (5/60)]*5

⇒ d = [(55+5)/60] * 5

⇒ d = [60/60] * 5

⇒ d = 5 km ← the distance from the man’s house to his office  

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Answer 2

The  distance from his house to the office is​ 5 km.

Step-by-step explanation:

Given:

Let distance = x km.

Time taken at 5 km/hr = $\frac{distance}{speed}=\frac{x}{5} = 5 minutes$

Time taken at 6 km/hr = $\frac{distance}{speed}=\frac{x}{6} = 5 minutes$

Now, difference between time taken = 5-(-5) = 5+5

                                                                           = 10 mins

                                                                           = $\frac{10}{60} hours$

$\frac{x}{5} -\frac{x}{6}=\frac{10}{60}$

$\frac{6x-5x}{30}=\frac{1}{6}$

$\frac{x}{30} =\frac{1}{6}$

$x=\frac{30}{6}$

x = 5 km.

Therefore the  distance from his house to the office is​ 5 km.