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. Select the correct option
A) 60  B) 37  C) 26  D) 28  E) 11

Rationale: 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 = 70 v = 50 + 20 = 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 = 70 v = 70 km / hr , the travel time is x − 3060 x − 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 = 7 x − 5 x 72 + 72 = 7 x − 5 x 2 x = 72 x = 7 x = 72 x = 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 . answer : a
Answer: A