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PLS HELP ASAP WILL GIVE BRAINlEST AND 100 POINTS
A bus travels from town A to town B. If the bus's speed is 50 km/hr, it will arrive in town B 42 min later than scheduled. If the bus increases its speed by \displaystyle \frac{50}{9}
9
50
​
m/sec, it will arrive in town B 30 min earlier than scheduled. Find:
A) The distance between the two towns;
B) The bus's scheduled time of arrival in B;
C) The speed of the bus when it's on schedule.

Respuesta :

saraalvi00

Answer:

60 km/hr

Step-by-step explanation:

First we will determine the speed of the bus following it's increase. The speed is increased by:

50m/9sec=50.60.60/9/1000 .km/hr=20 km/hr

Therefore, the new speed is V = 50 + 20 + = 70

km/hr. If x is the number of hours according to the schedule, then at the speed of 50 km/hr the bus travels from A to B within ( x + 42/60 ) hr. When the speed of the bus is V = 70 km/hr, the travel time is x -30/60 hr. Then:

50(x+42/60)=70(x-30/60)

5(x+7/10)=7(x-1/2)

7/2+7/2=7x-5x

2x = 7

x=7/2 hr

So, the bus is scheduled to make the trip in 3 he 30 minutes.

The distance between the two towns is 70 ( 7/2 – 1/2 ) = 70 • 3 = 210 km and the scheduled speed is:

270/7/2=60km/hr

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