If a boy walks from his house to town at a speed of 5 km/hr. he is late by 42 minutes. However if he walks at 6 km/hr. he is late by 5 minutes. What is the distance of town from his house.
A. 12.5 km
B. 18.5 km
C. 15.5 km
D. 21.5 km
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Let usual speed was S and usual time was T and distance is D.
ST = D ------------------------ (1)
With speed of 5 kmph, he late by 42 min
Time taken = T + 42/60 = T + 0.7
So, 5 * (T + 0.7) = D ------------------------- (2)
Comparing (1) and (2)
ST = 5 * (T + 0.7)
ST = 5T + 3.5 ------------------------ (3)
For 6 kmph, late by 5 min.
Time taken = T + 5/60 = T + 0.083
6 * (T + 0.083) = D ----------- (4)
Equating equation (1) and (4), we get
ST = 6T + 0.5 -------------- (5)
Now, (5) - (3)
ST - ST = 6T + 0.5 - 5T - 3.5
T = 3
Required distance,
5 * (T + 0.7) = D
D = 15 + 3.5 = 18.5 km