1. Anna left for city A from city B at 5.20 a.m. she traveled at the speed of 80 km/hr for 2 hours 15 minutes.After that speed was reduced to 60 km/hr.If the distance between two cities is 350 kms,at what time did Anna reach city A?

A. 9.20.a.m B. 9.25 a.m C. 9.35 a.m D. 10.25 a.m

Answer Answer: Option D Explanation: Distance covered in 2 hours 15 minutes i.e.2 1/4 hrs = (80*9/4) hrs = 180 hrs
Time taken to cover remaining distance = (350-180/60) hrs = 17/6 hrs = 2 5/6 hrs = 2hrs 50 minutes
Total time taken = 2 hrs 15 minutes + 2 hrs 50 minutes) = 5hrs 5 minutes
So Anna reached city A at 10.25 a.m

2. Mary jogs 9 km at a speed of 6 km/hr. At what speed would see need to jog during the next 1.5 hrs to have an average of 9 km/hr for the entire jogging session?

A. 9 km/hr B. 10 km/hr C. 12 km/hr D. 14 km/hr

Answer Answer: Option C Explanation: Let speed of jogging be X km/hr
Total time take = (9/6 hrs + 1.5 hrs) = 3 hrs
Total distance covered = ( 9+1.5 + x) km
Hence = 9/1.5 x / 3 = 9 = 9 + 1.5 x = 27 = 3/2 x = 18 = x= (18*2/3) = 12 km/hr.

4. The ratio between the speeds of two trains is 7:8. If the second train runs 400 kilometers in 4 hours, then the speed of the first train is?

A. 70 km/hr B. 75 km/hr C. 84 km/hr D. 87.5 km/hr

Answer Answer: Option D Explanation: Let the speeds of two trains be 7x and 8x km/hr.
then 8x = 400/4 = x = 100/8 = 12.5
speed of first train = (7*12.5 = 87.5 km/hr

5. A train running at 7/11 of its own speed reached a place in 22 hrs.How much time could be save if the train would have run at its own speed?

A. 7 hrs B. 8 hrs C. 14 hrs D. 16 hrs

Answer Answer: Option B Explanation: New speed 7/11 of usual speed
New time =11/7 of usual time
So 11/7 of usual time = 22 hrs = usual time = (22*7/11) = 14 hrs
Hence time saved = (22-14) = 8hrs