Question
Asghar traveled from his house to his office at 10 km/hr and returned at rate of 9 km/hr. Rohit travelled through the same path both ways at 12 km/hr and took 10 minutes less than Asghar. What is the distance between Asghar’s house and his office?
Answer: Option C
AVERAGE SPEED FOR RAJ IS: V = (2*9*10)/(9+10) = 180/19 KM/HR = 50/19 M/S
AVERAGE SPEED FOR ROHIT = 12 KM/HR = 10/3 M/S
NOW, V = D/T
AS D IS SAME, V*T = CONSTANT.
LET, T BE THE TIME TAKEN BY ROHIT IN SECONDS. HENCE, TIME TAKEN BY RAJ IS (T +600)S
50/19 * (T + 600) = 10/3 * T
150*(T + 600) = 190T
T = 2250S
LET D BE THE DISTANCE BETWEEN THE HOUSE AND THE OFFICE
2D = 2250 * 10/3
2D = 7500 M
D = 3750M
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AVERAGE SPEED FOR RAJ IS: V = (2*9*10)/(9+10) = 180/19 KM/HR = 50/19 M/S
AVERAGE SPEED FOR ROHIT = 12 KM/HR = 10/3 M/S
NOW, V = D/T
AS D IS SAME, V*T = CONSTANT.
LET, T BE THE TIME TAKEN BY ROHIT IN SECONDS. HENCE, TIME TAKEN BY RAJ IS (T +600)S
50/19 * (T + 600) = 10/3 * T
150*(T + 600) = 190T
T = 2250S
LET D BE THE DISTANCE BETWEEN THE HOUSE AND THE OFFICE
2D = 2250 * 10/3
2D = 7500 M
D = 3750M
Was this answer helpful ?
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