A car travelled first third of a distance at 10 km/h, second third of the distance at 20 km/h and last third of the distance at 60 km/h. Mean speed of the car over entire distance is

30 km/h

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20 km/h

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18 km/h

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54 km/h

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Solution

The correct option is C 18 km/h
$v = \frac{d}{t} = \frac{d}{t_{1} + t_{2} + t_{3}}$
for first part of travelling;
$10 = \frac{d}{3 t_{1}} t_{1} = \frac{d}{3 \times 10}$
for second part of travelling;
$t_{2} = \frac{d}{3 \times 20}$
for third part of travelling;
$t_{3} = \frac{d}{3 \times 60}$
mean speed $(v)$ = $\frac{d}{\frac{d}{3} (\frac{1}{10} + \frac{1}{20} + \frac{1}{60})} v = 18 km/h$
Hence, $v$ = 18 km/h

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A car travelled first third of a distance at 10 km/h, second third of the distance at 20 km/h and last third of the distance at 60 km/h. Mean speed of the car over entire distance is

The correct option is C 18 km/hv=dt=dt1+t2+t3 for first part of travelling; 10=d3t1t1=d3×10 for second part of travelling; t2=d3×20 for third part of travelling; t3=d3×60 mean speed (v)=dd3(110+120+160)v=18 km/h Hence, v= 18 km/h