Q.1 A bus left point X for point Y. two hours later a car left point X for Y and arrived at the same time as the bus. If the car and bus left simultaneously from the opposite ends X and Y towards each other they would meet 1.33 hours after the start . How much time did the bus take to travel from X to Y
Q.2 Sardar khan went out wasseypur at 10:00 am at a speed of 50km/hr. Ramadhir finds out about it at 11:00 am & sends his man to kill sardar khan. Ramadhir’s men travels at a speed of 60km/hr. What will be the distance between them 1hr before they meet & at what time they will meet ?
A) Distance- 10 km , time-4 pm
B) Distance- 11 km , time-5 pm
C) Distance- 11.5 km , time-5.30 pm
D) Distance-12 km , time- 6 pm
11 am sardar khan would be 50km
12pm he would be 100km
1pm he would be 150km
11am ramadhir men (gap=50km)
12pm Ramadhir men would be 60km (gap=40km)
1pm Ramadhir men would be 120km (gap=30km)
Relative speed =60-50 = 10km/hr
Or 50/10 = 5hr
Now calculate 5hr from 11am; i.e. 4pm
Distance between them = 10km
Q.3 A man sets out to cycle from Delhi to Rohtak & at the same time another man starts from Rohtak to delhi on cycle. After passing each other they complete their journey in 3 1/3 & 4 4/5hrs. At what rate does the second man cycle if the first cycles at 8km/hr ?
A) 6 2/3
B) 7 2/3
C) 8 2/3
D) 9 2/3
If two persons start from same time in opposite direction from two points, then
A’s speed : B’s speed = vb : va
8 /3 = v4 4/5 / v3 9/3
Or B’s speed = 8*5/6 = 6 2/3
Q.4 When a person covers a distance between his house & office at 50km/hr he is late by 20 min. but when he travels at 60km/hr he reaches 10min. early. What is the distance between his house & office ?
Product of speed / difference of speed * difference between arrival time
60*50 / 60-50 * 60 / 60
Q.5 Running at 5/4 of his usual speed an athlete improves his timing by 5 min.How much time he usually takes to run the same distance ?
A) 25 min.
B) 26 min.
C) 27 min.
D) 28 min.
Usual time = Improvement in time / 1- old speed/new speed
5 / 1-1/ 5/4