Q. In a tournament of Chess having 150 entrants, a player is eliminated whenever he loses a match. It is given that no match results in a tie/draw. How many matches are played in the entire tournament?
a. 151 b. 150 c. 149 d. 148 Correct Answer: c. 149
Question from UPSC Prelims 2022 CSAT Paper
Explanation :
In a tournament of Chess
In a knockout chess tournament like this, each player is eliminated after losing a match. Since there are 150 entrants, only one player remains undefeated at the end of the tournament, and this player is the winner.
To find out how many matches are played, we can think about it in terms of elimination: for every player except the winner, they have to lose exactly one match to be eliminated. Since there are 150 entrants and only one winner, 150 – 1 = 149 players need to be eliminated.
Each elimination comes from one match, so there are 149 matches played in the entire tournament.
Q. X and Y run a 3 km race along a circular course of length 300 m. Their speeds are in the ratio 3:2. If they start together in the same direction, how many times would the first one pass the other (the start-off is not counted as passing)?
a. 2 b. 3 c. 4 d. 5 Correct Answer: b. 3
Question from UPSC Prelims 2022 CSAT Paper
Explanation :
X and Y run a 3 km race
The faster runner will cross the slower one when he covers an extra 300 m. Let their speeds be 3 m/sec and 2 m/sec. So, their relative speed = 3 – 2 = 1 m/sec. So, the time taken by the faster runner to cross the slower one = Distance/Relative Speed = 300/1 = 300 seconds.
It basically means that the faster runner will cross the slower one every 300 seconds or 5 minutes. Now, the time taken for the faster racer to complete the entire race = Total Distance/Speed = 3000/3 =1000 seconds.
So, during the entire race, which lasts for 1000 seconds, the faster racer will cross the slower one 3 times – after 300 seconds, 600 seconds, and 900 seconds.
Q. A has some coins. He gives half of the coins and 2 more to B. B gives half of the coins and 2 more to C. C gives half of the coins and 2 more to D. The number of coins D has now, is the smallest two- digit number. How many coins does A have in the beginning?
a. 76 b. 68 c. 60 d. 52 Correct Answer: d. 52
Question from UPSC Prelims 2022 CSAT Paper
Explanation :
A has some coins
Let’s say A had x coins in the beginning. After giving half of the coins and 2 more to B, A has (x/2) – 2 coins left and B has (x/2) + 2 coins. After B gives half of the coins and 2 more to C, B has ((x/2) + 2)/2 – 2 = (x/4) – 1 coins left and C has ((x/2) + 2)/2 + 2 = (x/4) + 3 coins. After C gives half of the coins and 2 more to D, C has ((x/4) +3)/2 – 2 = (x/8) -1/2 coins left and D has ((x/4)+3)/2+2 = (x/8)+7/2 coins.
Since D has the smallest two-digit number which is 10, we can set up an equation: (x/8)+7/2 =10. Solving for x we get: x=52.
Q. A, B and C are three places such that there are three different roads from A to B, four different roads from B to C and three different roads from A to C. In how many different ways can one travel from A to C using these roads?
a. 10 b. 13 c. 15 d. 36 Correct Answer: c. 15
Question from UPSC Prelims 2022 CSAT Paper
Explanation :
A B C are three places
One can travel from A to C in 15 different ways. There are two possible routes: directly from A to C or from A to B and then from B to C.
The number of ways to travel directly from A to C is 3 (since there are 3 different roads). The number of ways to travel from A to B and then from B to C is the product of the number of roads between each pair of places: 3 x 4 = 12.
So, in total, there are 3 + 12 = 15 different ways one can travel from A to C using these roads.
Q. On one side of a 1.01 km long road, 101 plants are planted at equal distance from each other. What is the total distance between 5 consecutive plants?
a. 40 m b. 40.4 m c. 50 m d. 50.5 m Correct Answer: b. 40.4 m
Question from UPSC Prelims 2022 CSAT Paper
Explanation :
101 plants on a 1.01 km road
Let’s say we have a road that is 1.01 km long. On one side of the road, there are 101 plants planted at equal distances from each other. We can convert the length of the road to meters: 1.01 km * 1000 m/km = 1010 m.
Now, imagine that the plants are represented by dots on a number line. The first plant is at position 0 and the last plant is at position 1010. Since there are 101 plants in total, there must be 100 gaps between them.
We can find the distance between two consecutive plants by dividing the total distance (1010 m) by the number of gaps (100): 1010 m / 100 = 10.1 m.
So, if we want to find the total distance between five consecutive plants, we need to multiply this distance (10.1 m) by one less than the number of plants because for 5 Consecutive Plants, gaps between them is 4 (5-1=4): 10.1 m * 4 = 40.4 m.
Q. The letters A, B, C, D and E are arranged in such a way that there are exactly two letters between A and E. How many such arrangements are possible?
a. 12 b. 18 c. 24 d. 36 Correct Answer: c. 24
Question from UPSC Prelims 2022 CSAT Paper
Explanation :
Letters A B C D and E are arranged
The problem asks for the number of arrangements of the letters A, B, C, D and E such that there are exactly two letters between A and E. There are four possible arrangements for A and E: A _ _ E _, _ A _ _ E, E _ _ A _, and _ E _ _ A. For each of these arrangements, there are 3! = 6 ways to arrange the remaining letters B, C and D. So the total number of arrangements is 4 * 6 = 24.