Question : A man walks from his house at an average speed of 5 km/h and reaches his office 6 minutes late. If he walks at an average speed of 6 km/h, he reaches 2 minutes early. The distance of the office from his house is:
Option 1: 6 km
Option 2: 9 km
Option 3: 12 km
Option 4: 4 km
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Correct Answer: 4 km
Solution : $\text{Distance}=\frac{(S_1 \times S_2)(t_1+ t_2)}{S_2– S_1}$ where $S_1$ and $S_2$ are two speeds for the same distance and $t_1$ is the time by which man reaches late than the actual time and $t_2$ is the time by which man reaches early than actual time. $S_1$ = 5 km/h and $t_1$ = 6 minutes. $S_2$ = 6 km/h and $t_2$ = 2 minutes. The formula used here is, $\text{Distance}=\frac{(S_1\times S_2)\frac{(t_1+t_2)}{60}}{S_2–S_1}$. $=\frac{(5 \times 6)(6+2)}{6–5}\times \frac{1}{60}$ $=30\times \frac{8}{60}=4$ km Hence, the correct answer is 4 km.
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