Let the speed of the rickshaw and the bus are x and y km/h, respectively
Now, she has taken time to travel 2 km by rickshaw , t1=2xh [∴Speed=distancetime]
and she has taken time to travel remaining distance i.e., (14 – 2) = 12 km by bus t2=12yh
similarly, by Second condition;
t3=4xh
t4=10yh
t3+t4=12+960=12+320
⇒ 4x+10y=1320
Let 1x=u 1y=v, then Eqs (i) and (ii) becomes
2u + 12v = 12
And 4u + 10v =1320
On multiplying in Eq. (iii) by 2 and then subtracting, we get
4u + 24v = 1
4u + 10v = 1320
_ _ _
14v=1−1320=720
2v=120⇒v=140
Now, put the value of v in Eq (iii), we get
2u+12=(140)=12
⇒ 2u=12−310=5−310
∴ 1x=u
⇒ 1x=110⇒10 km/h
and 1y=v⇒1y=140
⇒ y=40 km/h
Hence, the speed of rickshaw and the bus are 10 km/h and 40 km/h, respectively.