A cab driver picks 1st passenger who is 0.8 km away and then he picks another passenger who is 0.5 km away from the first passenger. Then drops the 1st passenger to a place that is 2 km away from the second pick up point and drops the 2nd passenger to a place 1.7 km away from the place where he dropped the 1st passenger. If the average speed throughout the journey is 40 kmph, find out the time taken to complete the whole journey.

7.5 m i n u t e s

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0.15 h

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9 m i n u t e s

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0.125 h

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Solution

The correct options are
A $7.5 m i n u t e s$
D $0.125 h$
Total distance covered by the driver
$= 0.8 + 0.5 + 2 + 1.7 = 5 k m$
Average speed $= 40 k m p h$
So total time taken $= \frac{total distance}{average speed} = \frac{5}{40} h$ $= 0.125 h$
$= \frac{5}{40} \times 60 = 7.5 minutes$ .

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The correct options are A 7.5 minutes D 0.125 hTotal distance covered by the driver =0.8+0.5+2+1.7=5 km Average speed =40 kmph So total time taken =total distanceaverage speed=540 h =0.125 h =540×60=7.5 minutes.

The correct options are
A $7.5 m i n u t e s$
D $0.125 h$
Total distance covered by the driver
$= 0.8 + 0.5 + 2 + 1.7 = 5 k m$
Average speed $= 40 k m p h$
So total time taken $= \frac{total distance}{average speed} = \frac{5}{40} h$ $= 0.125 h$
$= \frac{5}{40} \times 60 = 7.5 minutes$ .