Question

Question #0f4ca

Answer 1

a) `44 4/9` km/hr.

Explanation:

Let the total distance of the journey be `x`km .

`50%` of Distance is `x/2`. ------Let this be `d_1`
This is covered at a speed of 40 km/hr.
So, time `t`, required for this part of journey is :

`t =( distance) / (speed)`

`t_1 = (x/2)/40 = x/80` hrs.

`30%` of `x` =>`30/100 xx x` = `0.3x`----- Let this be `d_2`

This distance is covered with speed of 60 km/hr, so,

`t_2 = (0.3x)/60`

And remaining journey= `x- 0.5x -0.3 = 0.2x`------let this be `d_3` is travelled by40 km/hr. So,

`t_3 = (0.2x) /40`

Total distance is `x` and total time taken is :

`t= t_1 +t_2+t_3`

`t = x/80 +(0.3x)/ 60 +(0.2x)/40`

`t = x/80 xx(3/3) +(0.3x)/ 60 xx(4/4) +(0.2x)/40 xx (6/6)`

`t = (3x)/240 +(1.2x)/240 + (1.2x)/240`

`t= (3x+1.2x+1.2x)/240 = (5.4x)/240`

Average speed = (total distance) / (total time taken)

`S_(avg) = x/( (5.4x)/240)`

`S_(avg) = cancelx xx 240/(5.4cancelx)`

`S_(avg) = 2400/54 = 1200/27 =400 /9= 44 4/9`

Answer = a) `44 4/9` km/hr.

Answer 2

Average speed is 44 4/9# km/hour [Option -A]

Explanation:

Let `x` km is the whole journey then the man covers `50/10x=0.5x`

km at `40`km/hour and he covers `30/10x=0.3 x` km at

`60` km/hour and rest `x-(0.5+0.3x) =0.2x` km covers at

`40` km/hour. Therefore he covers `(0.5+0.2)x=0.7x` km covers

at `40`km/hour and covers `0.3 x`km at `60`km/hour.

Time taken to cover `0.7x` km journey is `(0.7x)/40` hour and

time taken to cover `0.3x` km journey is `(0.3x)/60` hour.

Hence total time taken to cover distance `(D=x)` km journey is

`T=(0.7x)/40+(0.3x)/60 = (2.1x+0.6x)/120= (2.7x)/120` hour.

Average speed is `S_a=D/T=x/((2.7x)/120)=(120cancelx)/(2.7cancelx)`

`:. S_a= (120*10)/27 = 1200/27=44 12/27 = 44 4/9` km/hour

Average speed is 44 4/9# km/hour [Option -A] [Ans]