First 120 km of the road the car is moving at speed rate of 90 kph. Next one hour and 15 minutes the car is moving at speed rate of 64kph . What needs to be the speed rate for the rest of the road so that the speed rate of the whole trip would be80 kph if \frac{1}{6} of the road is left?
Answers
Given :- First 120 km of the road the car is moving at speed rate of 90 kph. Next one hour and 15 minutes the car is moving at speed rate of 64kph .
To Find :- What needs to be the speed rate for the rest of the road so that the speed rate of the whole trip would be 80 kph if (1/6) of the road is left ?
Answer :-
→ Speed in first case = 90 km/h
→ Distance covered = 120 km.
so,
→ Time taken = D/S = 120/90 = (4/3) hours.
now,
→ Speed in second case = 64 km/h
→ Time = 1(15/60) = 1(1/4) = (5/4) hours.
so,
→ Distance covered = 64 * (5/4) = 80 km.
given that, (1/6) th of the road is still left .
→ (5/6) distance covered = 80 + 120 = 200 km.
→ Total distance = 200 * (6/5) = 240 km .
then,
→ Distance left = 40 km .
now, Let time taken for left distance is x hours.
so,
→ Average speed = 80 km/h
→ Total distance / total time = 80
→ 240/[(4/3) + (5/4) + x) = 80
→ (4/3) + (5/4) + x = 3
→ (16 + 15 + 12x)/12 = 3
→ 31 + 12x = 36
→ 12x = 36 - 31
→ 12x = 5
→ x = (5/12) hours.
therefore,
→ Speed for the left distance = D/T = 40/(5/12) = 40 * (12/5) = 96 km/h .
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