Races and games
- In a game of billiards, A can give B 20 points in the game of 120 points and he can give C 30 points in the game of 120 points, How many points can B give C in a game of 90 ?
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If A scores 120 points, then B scores 100 points and C scores 90 points.
When B scores 100 points, then C scores 90 points.
When B scores 90 points than C scores (90/100) x 90 points = 81 pointsCorrect Option: A
If A scores 120 points, then B scores 100 points and C scores 90 points.
When B scores 100 points, then C scores 90 points.
When B scores 90 points than C scores (90/100) x 90 points = 81 points
∴ B can give C, 9 points in a game of 90.
- Arun and Bhaskar start from place P at 6:00 am and 7:30 am, respectively and run in the same direction. Arun and Bhaskar run at 8 km/h and 12 km/h. respectively. Bhaskar overtakes
Arun at ?
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Distance between Arun and Bhaskar at 7:30 am = 8 x 11/2 = 12 km
Time taken by Bhaskar in covering a distance of 12 km = 12/(12 - 8) = 3 hCorrect Option: A
Distance between Arun and Bhaskar at 7:30 am = 8 x 11/2 = 12 km
Time taken by Bhaskar in covering a distance of 12 km = 12/(12 - 8) = 3 h
∴ Required time = 10 : 30 am
- A, B and C walk 1 km in 5 min, 8 min and 10 min, respectively, C starts walking from a points at a certain time, B starts from the same point 1 min later and A starts from the same
point 2 min later than C, Then, A meet B and C at times ?
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A walks 1 km in 5 min.
Then, distance covered by A in 1 min = 1000/5 = 200 m
B walks 1 km in 8 min.
Then, distance covered by B in 1 min = 1000/8 = 125 m
C walks 1 km in 10 min.
Then, distance covered by C in 1 min = 1000/101 = 100 m
Then, according to the question,
Distance covered by C in(x + 2) min = Distance covered by A in x min
Distance covered by B in (y + 1) min = Distance covered by A in y minCorrect Option: C
A walks 1 km in 5 min.
Then, distance covered by A in 1 min = 1000/5 = 200 m
B walks 1 km in 8 min.
Then, distance covered by B in 1 min = 1000/8 = 125 m
C walks 1 km in 10 min.
Then, distance covered by C in 1 min = 1000/101 = 100 m
Let A meets B and C in x and y min, respectively.
Then, according to the question,
Distance covered by C in(x + 2) min = Distance covered by A in x min
⇒ 100(x + 2) = 200 x
⇒100 x + 200 = 200 x
⇒ 200 = 100 x
∴ x = 200/100 = 2 min
Now, for A and B
Distance covered by B in (y + 1) min = Distance covered by A in y min
⇒ 125(y + 1) = 200y
⇒ 125y + 125 = 200y
⇒ 125 = 200y - 125y
⇒ 125 = 75y
∴ y = 125/75 = 5/3 min.
- A 10 km race is organised at 800 m circular race course. P and Q are the contestants of the race. If the ratio of the speed of P and Q is 5 : 4, how many times will the winner overtake the loser ?
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Speed of P : Speed of Q = 5 : 4
Time taken by P to cover 5 rounds = Time taken by Q to cover 4 rounds
Distance covered by P in 5 rounds = 5 x (800/1000) = 4 km
Distance covered by Q in 4 rounds = 4 x (800/1000) = 16/5 kmCorrect Option: C
Speed of P : Speed of Q = 5 : 4
Time taken by P to cover 5 rounds = Time taken by Q to cover 4 rounds
Distance covered by P in 5 rounds = 5 x (800/1000) = 4 km
Distance covered by Q in 4 rounds = 4 x (800/1000) = 16/5 km
In 5 rounds, P will overtake Q every time. It means that after covering 4 km, P will overtake Q one time .
∴ After covering 10 km P will overtake Q, = (1/4) x 10 = 21/2 times
- A runs 12/3 times as fast as B. If A gives B a start of 40 m, how far must the winning post be, so that A and B might reach it at the same time ?
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Ratio of the speeds of A and B = 5 : 3
Thus, in a race of 5 m, A gains 2 m over B.Correct Option: C
Ratio of the speeds of A and B = 5 : 3
Thus, in a race of 5m, A gains 2 m over B.
2 m are gained by A in a race of 5 m.
40 m will be gained by A in a race of (5/2) x 40 m = 100 m
∴ Winning post is 100 m away from the starting point.