Question

. A bus left from the depot 1 hr late than the usual time. If it covers the delay for a distance of 120 km by moving at 20 kmph extra speed, find the usual speed of the bus. (1) 30 km/h (2) 8 km/h (3) 16 km/h (4) 40 km/h 1 оо 03​

Answer 1

Answer:

there is your answer

Step-by-step explanation:

Let time for first trip be x hrs

For first trip distance travelled = 60x

For round trip distance travelled 40(x+4) = 40x+160

60x = 40x + 160

x = 8 hrs.

Total Round trip distance = 60*8*2 = 960 kilometers

Answer 2

Answer:

(4) 40 km/h

Step-by-step explanation:

Let the usual speed be 'v'.

Let the total distance is travels be 'x' = 120 km.

Let the total time it takes to cover 'x' distance be 't'.

It's moving with speed: v+20. Now, if it leaves 1 hr late, so the time the bus has left to cover 'x' distance will be: t-1

=> x=(v+20)(t-1)

and x=vt

=> (v+20)(t-1)=vt

=> -v+20t-20=0

Now v=x/t=120/t

=> -120/t+20t-20=0

=> 20t²-20t-120=0

=> t²-t-6=0

=> (t-3)(t+2)=0

t=-2 hrs (or) t=3 hrs

time can't be negative

Therefore, t=3 hrs

Now v=120/t = 120/3 = 40 km/h

Please comment if further clarification is required in any of the above given steps.

Thanks