Problem: susan drives from city a to city b . after two hours of driving she noticed that she covered 80 km and calculated that , if she continued driving at the same speed , she would end up been 15 minutes late . so she increased her speed by 10 km / hr and she arrived at city b 36 minutes earlier than she planned . find the distance between cities a and b .
Rationale: let xx be the distance between a and b . since susan covered 80 km in 2 hours , her speed was v = 802 = 40 v = 802 = 40 km / hr . if she continued at the same speed she would be 1515 minutes late , i . e . the planned time on the road is x 40 − 1560 x 40 − 1560 hr . the rest of the distance is ( x − 80 ) ( x − 80 ) km . v = 40 + 10 = 50 v = 40 + 10 = 50 km / hr . so , she covered the distance between a and b in 2 + x − 80502 + x − 8050 hr , and it was 36 min less than planned . therefore , the planned time was 2 + x − 8050 + 36602 + x − 8050 + 3660 . when we equalize the expressions for the scheduled time , we get the equation : x 40 − 1560 = 2 + x − 8050 + 3660 x 40 − 1560 = 2 + x − 8050 + 3660 x − 1040 = 100 + x − 80 + 3050 x − 1040 = 100 + x − 80 + 3050 x − 104 = x + 505 x − 104 = x + 505 5 x − 50 = 4 x + 2005 x − 50 = 4 x + 200 x = 250 x = 250 so , the distance between cities a and b is 250 km . answer : c
annotated_formula: add(divide(subtract(add(subtract(divide(36, const_60), divide(80, add(divide(80, const_2), 10))), const_2), divide(15, const_60)), subtract(divide(const_1, divide(80, const_2)), divide(const_1, add(divide(80, const_2), 10)))), const_100)