A legal clerk walks from his residence to the courthouse at 5 km/h and arrives 15 minutes late. If he sets a pace of 7.5 km/h, he arrives 5 minutes early. Find the distance. MCQ with Answer and Explanation
A legal clerk walks from his residence to the courthouse at 5 km/h and arrives 15 minutes late. If he sets a pace of 7.5 km/h, he arrives 5 minutes early. Find the distance.
A. 4.5 km
B. 5.0 km
C. 5.5 km
D. 6.0 km
Answer: Option B
Solution (By JKSSB Mock Tests)
Time difference = 15 - (-5) = 20 minutes = 20/60 = 1/3 hours. Let distance be d. d/5 - d/7.5 = 1/3 => (1.5d - d) / 7.5 = 1/3 => 0.5d / 7.5 = 1/3 => d / 15 = 1/3 => d = 5 km.
Explanation:
New speed = 5/6 of normal pace => New time = 6/5 of normal time. Difference = 1/5 of normal time = 10 minutes. Normal time = 10 * 5 = 50 minutes.
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