A man travels first $50 k m$ at $25 k m p h$ , next $40 k m$ at $20 k m p h$ and then $90 k m$ at $15 k m p h$ . His average speed for the whole journey $(i n k m p h)$ is:

25

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18

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40

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Solution

The correct option is C $18$
To find : average speed
$s p e e d = \frac{D i s t a n c e}{T i m e}$
$ϑ_{1} = \frac{s_{1}}{t_{1}} \Rightarrow t_{1} = \frac{s_{1}}{ϑ_{1}}$
$t_{1} = \frac{50}{25} = 2$ hrs
$t_{2} = \frac{s_{2}}{ϑ_{2}} = \frac{40}{20} = 2$ hrs
$t_{3} = \frac{s_{3}}{ϑ_{3}} = \frac{90}{15} = 6$ hrs
Average Speed $\frac{T o t a l D i s t a n c e}{T o t a l t i m e t a k e n}$
$= \frac{s_{1} + s_{2} + s_{3}}{t_{1} + t_{2} + t_{3}} = \frac{50 + 40 + 90}{2 + 2 + 6}$ km/h
$= \frac{180}{10}$ km/h
= 18 km/h

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A man travels first 50 km at 25 kmph, next 40 km at 20 kmph and then 90 km at 15 kmph. His average speed for the whole journey (in kmph) is:

The correct option is C 18To find : average speedspeed=DistanceTimeϑ1=s1t1⇒t1=s1ϑ1t1=5025=2 hrst2=s2ϑ2=4020=2 hrst3=s3ϑ3=9015=6 hrsAverage Speed TotalDistanceTotaltimetaken=s1+s2+s3t1+t2+t3=50+40+902+2+6 km/h=18010 km/h = 18 km/h

A man travels first $50 k m$ at $25 k m p h$ , next $40 k m$ at $20 k m p h$ and then $90 k m$ at $15 k m p h$ . His average speed for the whole journey $(i n k m p h)$ is: