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 509509 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.
Choose the correct choice from the following answers
A) 60  B) 37  C) 26  D) 28  E) 11
First we will determine the speed of the bus following its increase. The speed is increased by 509509 m/sec =50⋅60⋅6091000=50⋅60⋅6091000 km/hr =20=20 km/hr. Therefore, the new speed is V=50+20= 
<gadget id="calculator">50+20</gadget>
<output>70</output>
70V=50+20= 
<gadget id="calculator">50+20</gadget>
<output>70</output>
70 km/hr. If xx 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+4260)(x+4260) hr. When the speed of the bus is V=70V=70 km/hr, the travel time is x−3060x−3060 hr. Then   50(x+4260)=70(x−3060)50(x+4260)=70(x−3060)   5(x+710)=7(x−12)5(x+710)=7(x−12)   72+72=7x−5x72+72=7x−5x   2x=72x=7   x=72x=72 hr.   So, the bus is scheduled to make the trip in 33 hr 3030 min.   The distance between the two towns is 70(72−12)=70⋅3=21070(72−12)=70⋅3=210 km and the scheduled speed is 210/7/2=60 km/hr.   60nswer:60
<result>A</result>
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 509509 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.