Previous Year Topic Wise Questions With Solutions: Speed, Time & Distance | CSAT Preparation - UPSC PDF Download

Ans: (d)
Team X scored a total of N runs in 20 overs. Team Y tied the score in 10% less overs, i.e. in 18 overs.
Had team Y’s average run rate (runs per over) been 50% higher, it would have scored 1.5N runs in 18 overs.
Or, in 12 overs it would have scored 1.5N × (12/18) = 1.5N × (2/3) = N runs
This information is not enough to find the value of N. 
So, the number of runs scored by team X can’t be determined.

Ans: (c)
The city map:

If a person wants to travel from D to F, he can opt one of the following two routes: 
Route I: DG (5 km + 14 km) + GF (12 km) = 19 km + 12 km = 31 km
Route II: DC (6 km) + CA (10 km) + AB (6 km) + BF (9 km) = 6 km + 10 km + 6 km + 9 km = 31 km
In both cases, he will cover the same distance of 31 km.

Ans: (a)
The path taken by them has been represented below:
It’s pretty clear that they are 40 + 60 = 100 m apart at the end of their run.

Ans: (d)
Total time taken by the man to come back home = 16 – 14.5 = 1.5 hours = 90 minutes
Out of which he stayed in the village for 25 minutes.
So, his total travelling time = 90 – 25 = 65 minutes
The return route was 1.25 times the initial route. So, time taken must have increased by 25% too. 
So, if initial time was 100 units, now it must be 125 units.
But it is also given that while returning he drove twice as fast. So, time taken must have been halved.
So, time taken while returning back = 125/2 = 62.5 units
So, 100 + 62.5 = 65 minutes
Or 162.5 units = 65 minutes
So, 100 units = (65/162.5) × 100 = 40 minutes
So, the man took 40 minutes to reach to the village.
So, the actual time at that moment = 14:30 + 40 minutes = 15:10 hours
It’s pretty evident that the village clock is 15:15 – 15:10 = 5 minutes fast