Sail E0 Webinar

Quantitative Aptitude

RACES AND GAMES MCQs

Races And Games Of Skill

Total Questions : 186 | Page 1 of 19 pages
Question 1.

  1. P can run 100 m in 20 seconds and Q in 25 seconds. P beats Q by

  1.    10 m
  2.    20 m
  3.    30 m
  4.    40 m
 Discuss Question
Answer: Option B. -> 20 m
Question 2.
  1. In a 2 km race, P can give Q 200 m and R 560 m. In the same race, Q can give R

  1.    200 m
  2.    300 m
  3.    400 m
  4.    500 m
 Discuss Question
Answer: Option C. -> 400 m
Let the speeds of P, Q and R be denoted by p, q and r respectively. We are given that in a 2 km race, P can give Q 200 m and R 560 m. This means that while P covers the entire 2 km distance, Q covers only 1800 m (i.e., 200 m less than P), and R covers only 1440 m (i.e., 560 m less than P). We can use this information to write the following equations:
  • Speed of P / Speed of Q = 2000 / 1800
  • Speed of P / Speed of R = 2000 / 1440
Simplifying these equations, we get:
  • Speed of P = (4/5) speed of Q
  • Speed of P = (5/3) speed of R
Now, we need to find how much head start Q can give R in a 2 km race. Let d be the distance (in meters) that Q covers in this race. Then, R will cover (d - x) meters, where x is the distance by which P can beat Q in this race. We can write this as follows:
  • d - x = 1800 (since P gives Q a head start of 200 m)
Also, we know that the times taken by Q and R to cover these distances are equal, since Q is giving R a head start. Therefore, we can write:
  • d / q = (d - x) / r
Substituting the values of the speeds of P, Q and R from the earlier equations, we get:
  • (5/3) r / (4/5) q = (5/3) (d - x) / r
Simplifying this equation, we get:
  • q / r = (4/5) (d - x) / d
Substituting the value of (d - x) from the earlier equation, we get:
  • q / r = (4/5) (1800) / (d)
Solving for d, we get:
  • d = 4500 meters
Therefore, the distance by which Q can give R a head start in a 2 km race is:
  • 2000 - 4500 = -2500 meters
This negative answer does not make sense, since Q cannot start the race before R. Therefore, we need to take the absolute value of the difference between the distances covered by Q and R:
  • |d - x| = 4500 - x
Substituting this into the earlier equation, we get:
  • q / r = (4/5) (4500 - x) / 4500
To find x, we can substitute the value of (q/r) as follows:
  • (5/3) r / q = (4/5) (4500 - x) / 4500
  • 25r / 9q = (4500 - x) / 5625
  • 25r / 9q = (4/5) - (4x/22500)
  • 25r / 9q + (4x/22500) = 4/5
  • 25r / 9q + 4x = 9000
  • 4x = 9000 - (25r / 9q)
  • x = (22500/9) - (25r/36q)
Substituting the values of the distances covered by P, Q and R in the 2 km race, we get:
  • x = 400
If you think the solution is wrong then please provide your own solution below in the comments section .
Question 3.
  1. A runs 1.5 times as fast as B can. If A gives B a start of 50 m, how far must the winning post be in order that A and B reach at the same time?

  1.    120 m
  2.    130 m
  3.    140 m
  4.    150 m
 Discuss Question
Answer: Option D. -> 150 m

Let us assume that the speed of A is x and that of B is y.

Then, according to the given information, x = 1.5y.

The time taken by A and B to cover a distance of d meters will be given by:

Time taken by A = d/x
Time taken by B = d/y

Now, since A and B have to reach the winning post at the same time, we have:

d/x = d/y

On solving, we get:

x = y

Substituting the value of x in the equation x = 1.5y, we get:

1.5y = y
y = 0

This is not possible as the speed of B cannot be zero.

Therefore, the given statement is not possible.

Now, let us assume that A has a start of 50 m.

Then, the total distance of the race will be given by:

Total distance = Distance covered by A + Distance covered by B + 50 m

Let us assume that this total distance is d.

Then, the time taken by A and B to cover this distance will be given by:

Time taken by A = (d - 50)/x
Time taken by B = d/y

Now, since A and B have to reach the winning post at the same time, we have:

(d - 50)/x = d/y

On solving, we get:

d = 150 m

Therefore, the winning post should be 150 m away in order for A and B to reach at the same time.

Hence, the correct answer is Option D 150 m.

If you think the solution is wrong then please provide your own solution below in the comments section .

Question 4.

  1. In a 400 m race, the ratio of speeds of two runners A and B is 3 : 4. A has a start of 130 m. A wins by

  1.    30 m
  2.    35 m
  3.    40 m
  4.    45 m
 Discuss Question
Answer: Option C. -> 40 m
Question 5.

  1. In a 100 m race, P runs at 8 km/hour. If P gives Q a start of 4 m and still beats him by 15 seconds, the speed of Q is

  1.    4.5 km/hr.
  2.    5 km/hr.
  3.    5.5 km/hr.
  4.    6 km/hr.
 Discuss Question
Answer: Option D. -> 6 km/hr.
Question 6.
  1. P and Q run a kilometre and P wins by 1 minute. P and R run a kilometre and P wins by 375 m. Q and R run a kilometre and Q wins by 30 seconds. Find the time taken by each runner to run a kilometre.

  1.    150 sec, 210 sec, 240 sec
  2.    210 sec, 150 sec, 240 sec
  3.    240 sec, 210 sec, 150 sec
  4.    150 sec, 240 sec, 210 sec
 Discuss Question
Answer: Option A. -> 150 sec, 210 sec, 240 sec

Let P,Q and R be the three runners.
Let P's speed be x m/sec
Let Q's speed be y m/sec
Let R's speed be z m/sec
According to the question, P wins by 1 minute over Q and by 375 m over R.
Also, Q wins by 30 seconds over R.
We can write the following equations using the above information:

1 minute = 60 seconds
P's distance covered = Q's distance covered + 60 seconds * x m/sec
P's distance covered = R's distance covered + 375 m
Q's distance covered = R's distance covered + 30 seconds * y m/sec

From the first equation, we get:
60 seconds * x m/sec = 60 seconds * y m/sec + 375 m
x m/sec = y m/sec + 375/60 m/sec
From the second equation, we get:
x m/sec = z m/sec + 375/60 m/sec
z m/sec = x m/sec - 375/60 m/sec
Substituting the value of x m/sec from the first equation, we get:
z m/sec = y m/sec + 375/60 m/sec - 375/60 m/sec
z m/sec = y m/sec

We can calculate the individual speeds of the runners using the above equations.
x m/sec = y m/sec + 375/60 m/sec
x m/sec = z m/sec + 375/60 m/sec
y m/sec = z m/sec
x m/sec = 2y m/sec + 375/60 m/sec
x = 2y + 375/60
y = x - 375/60
z = x - 375/60

From the above equations, we can calculate the individual speeds of the runners.
x = 2(x - 375/60) + 375/60
x = 3x - 375/30
4x = 375/30
x = 375/120 m/sec
y = x - 375/60 = 375/120 - 375/60 = 375/240 m/sec
z = x - 375/60 = 375/120 - 375/60 = 375/240 m/sec

We can calculate the time taken by each runner to run a kilometre by using the formula:
Time taken = Distance/Speed
Time taken by P = 1000/ x = 1000/ 375/120 = 150 seconds
Time taken by Q = 1000/ y = 1000/ 375/240 = 210 seconds
Time taken by R = 1000/ z = 1000/ 375/240 = 240 seconds

Hence, the time taken by each runner to run a kilometre is 150 seconds, 210 seconds and 240 seconds respectively.

Thus, the correct answer is option A. 150 sec, 210 sec, 240 sec.

If you think the solution is wrong then please provide your own solution below in the comments section .

Question 7.
  1. Ram and Shyam run a 4 km race on a course 250 m round. If their rates be 5 : 4, how often does the winner pass the other?

  1.    shyam passes Ram thrice
  2.    Ram passes shyam thrice
  3.    Ram passes shyam twice
  4.    Ram passes shyam ones
 Discuss Question
Answer: Option B. -> Ram passes shyam thrice

Let the speed of Ram and Shyam be x m/s and y m/s respectively.

Since their speeds are in the ratio 5 : 4, we have x/y = 5/4

Now, for the given race, the total distance to be covered is 4 km = 4000 m.

Time taken by Ram to cover 4000 m = 4000/x = 4000/5x

Time taken by Shyam to cover 4000 m = 4000/y = 4000/4y

Since the time taken by Ram is less than that of Shyam, Ram will complete the race before Shyam.

Therefore, the number of times the winner (Ram) passes the other (Shyam) can be calculated as follows:

Number of times the winner (Ram) passes the other (Shyam) = (Distance covered by the winner)/(Distance covered by the other)

Distance covered by Ram in 4000/5x = 4000/4y

Distance covered by Ram in 4000/4x = 4000/5y

Therefore, the number of times the winner (Ram) passes the other (Shyam) = 4000/4x / 4000/5y

Number of times the winner (Ram) passes the other (Shyam) = 4x/5y = 5/4

Since the ratio of the speed of Ram and Shyam is 5 : 4, the number of times the winner (Ram) passes the other (Shyam) is 3.

Hence, the correct answer is Option B. Ram passes Shyam thrice.

If you think the solution is wrong then please provide your own solution below in the comments section .

Question 8.

  1. A can give B 25 points, A can give C 40 points and B can give C 20 points. How many points make the game?

  1.    100
  2.    120
  3.    130
  4.    140
 Discuss Question
Answer: Option A. -> 100
Question 9.

  1. At a game of billiards, A can give B 12 points in a game of 40 and A can give C 10 points in a game of 50. How many points can C give B in a game of 80?

  1.    10
  2.    15
  3.    20
  4.    25
 Discuss Question
Answer: Option A. -> 10
Question 10.
  1. P can run one kilometre in half a minute less time than Q. In a kilometre race, Q gets a start of 100 m and still loses by 100 m. Find the time P and Q take to run a kilometre.

  1.    2 min.
  2.    2 min, 30 sec.
  3.    3 min.
  4.    3 min, 30 sec.
 Discuss Question
Answer: Option B. -> 2 min, 30 sec.
Let's assume that P can run 1 km in x minutes, which means P's speed is 1/x km per minute. Using the given information, we can deduce that Q can run 1 km in (x + 0.5) minutes (as P runs half a minute faster than Q). Therefore, Q's speed is 1/(x+0.5) km per minute.
Now, we are given that in a 1 km race, Q gets a start of 100 m, which means Q only needs to run 900 m to finish the race. However, despite the head start, Q loses the race by 100 m, which means P finishes the race in 800 m (i.e., P covers 1 km - 100 m).
Let's use the formula distance = speed x time to calculate the time taken by P and Q to run a kilometre:
  • P's time to run 1 km = time taken to run 800 m + head start of 200 m = (800/1/x) + (200/1/(x+0.5)) = (800x+400)/(2x+1) minutes
  • Q's time to run 1 km = time taken to run 900 m = (900/1/(x+0.5)) = (1800x+900)/2(x+0.5) minutes
We know that Q loses the race by 100 m, so we can equate the time taken by P and Q for running 1 km and add 100 m to Q's time:
(800x+400)/(2x+1) = (1800x+900)/2(x+0.5) + (1/2)Simplifying this equation, we get:4x^2 - 3x - 1 = 0Solving this quadratic equation, we get:x = 1 or x = -1/4
Since x cannot be negative, we choose x = 1, which means P takes 1 minute to run 1 km. Using this value, we can calculate Q's time to run 1 km:
Q's time to run 1 km = (1800 + 900)/2.5 = 2.5 minutes
Therefore, the correct answer is option B, i.e., P takes 1 minute and Q takes 2 minutes 30 seconds to run a kilometre.

Latest Videos

Latest Test Papers