Ankita travels 14km to her home partly by rickshaw and partly by bus. She takes half an hour if she travels 2 km by rickshaw, and the remaining distance by bus. On the other hand, if she travel 4 km by rickshaw and the remaining distance by bus, she takes 9 minute longer. Find the speed of the rickshaw and of the bus.

Explanation

Let the speed of the rickshaw and the bus are x and y km/h, respectively.
Now, the has taken time to travel 2 km by rickshaw, `t_(1)=(2)/(x)h . " " `[`:'` speed=`("distance")/("time")`]
and she has taken time to travel remaining distance i.e., (14-2)=12km by bus `=t_(2)=(12)/(y)`h.
By first condition, `t_(1)+t_(2)=(1)/(2) rArr (2)/(x)+(12)/(y)=(1)/(2) " " ...(i)`
Now, she has taken time to travel 4 km by rickshaw, `t_(3)=(4)/(x)h`
and she has taken time to travel remaining distance i.e., (14-4)=10km by bus `=t_(4)=(10)/(y)h`.
By second condition, `t_(3)+t_(4)=(1)/(2)+(9)/(60)=(1)/(2)+(3)/(20)`
`rArr" " (4)/(x)+(10)/(y)=(13)/(20) " " ...(ii)`
Let `(1)/(x)=u` and `(1)/(y)=v`, then Eq. (i) and (ii) becomes
`2u+12v=(1)/(2) " " ...(iii)`
and `" " 4u+10v=(13)/(20) " " ...(iv)`
On multiplying in Eq. (iii) by 2 and then subtracting, we get
`{:(4u+24u=1),(ul(4underset(-)u+1underset(-)0v=(13)/(underset(-)20))),(7x=14):}`
`14v=1-(13)/(20)=(7)/(20)`
`rArr " " 2v=(1)/(20)rArrv=(1)/(40)`
Now, put the value of v in Eq. (iii), we get
`2u+12((1)/(40))=(1)/(2)`
`rArr " " 2u=(1)/(2)-(3)/(10)=(5-3)/(10)`
`rArr " " 2u=(2)/(10)rArru=(1)/(10)`
`:. " " (1)/(x)=u`
`rArr " " (1)/(x)=(1)/(10)rArr x=10`km/h
and `(1)/(y)v rArr (1)/(y)=(1)/(40)`
`rArr " " y=40`km/h
Hence, the speed of rickshaw and the bus are 10km/h and 40 km/h, respectively.