question_answer

A bus travelling the first one-third distance at a speed of 10 km/h, the next one fourth at 20km/h and the remaining at 40 km/h. Find the average speed of the bus.

Answer:

\[{{V}_{ave}}=\frac{total\,\,dis\tan ce\,\,travelled}{total\,\,time\,\,taken}\] \[{{V}_{ave}}=\frac{\frac{s}{3}+\frac{s}{4}+\frac{5s}{12}}{{{t}_{1}}+{{t}_{2}}+{{t}_{3}}}\]                               [s = distance] \[\left[ \because s-\left( \frac{s}{3}+\frac{s}{4} \right)=s-\frac{7s}{12}=s\left[ \frac{12-7}{12} \right]=\frac{5}{12}s \right]\] but here \[{{t}_{1}}=\frac{{{s}_{1}}}{{{V}_{1}}}=\frac{s/3}{20km/h}=\frac{s}{80}hr\]        ……(1) time \[{{t}_{2}}=\frac{{{s}_{2}}}{{{V}_{2}}}=\frac{s/3}{20km/h}=\frac{s}{80}hr\]                 ……(2) \[{{t}_{3}}=\frac{{{s}_{3}}}{{{V}_{3}}}=\frac{\frac{5s}{12}}{40km/h}=\frac{5s}{480}=\frac{s}{96}hr\]          …..(3) \[{{V}_{ave}}=\frac{s}{\frac{s}{30}+\frac{s}{80}+\frac{s}{96}}=\left[ \frac{1}{\frac{1}{30}+\frac{1}{80}+\frac{1}{96}} \right]\] \[=\frac{480}{27}km/hr=17.7\approx 18km/hr\]