CRE 5 - Linear Race | Time, Speed & Distance
In a 600 meters race A beats B by 30 meters or 10 seconds. A’s time over the course is (in secs):
- (a)
200
- (b)
190
- (c)
210
- (d)
180
Answer: Option B
Explanation :
30 meters are covered by B in 10 seconds, hence 600 meters are covered by B in 10/30 × 600 = 200 seconds.
∴ Time taken by A = (200 – 10) = 190 seconds
Hence, option (b)
Workspace:
A can run 40 meters while B runs 50 meters. In a km race B beats A by (in meters)?
- (a)
250
- (b)
225
- (c)
200
- (d)
125
Answer: Option C
Explanation :
In 50 meters race, B beats A by 10 meters.
In a km. race B beats A by (10/50 × 1000) = 200 meters
Hence, option (c)
Workspace:
A is times as fast as B. if A gives B a start of 150 meters, how far must be the winning post so that the race ends in a dead heat?
- (a)
100 meters
- (b)
440 meters
- (c)
200 meters
- (d)
550 meters
Answer: Option D
Explanation :
The rates of A and B are 11 : 8 i.e., In the race of 11 meters A gains 3 meters
⇒ 3 meters are gained by J in a race of 11 meters
∴ 150 are gained in 11/3 × 150 = 550 meters.
Hence, option (d).
Workspace:
In a 200 meters race. A can beat B by 50 meters and B can beat C by 8 meters. In a 100 m race, A can beat C by:
- (a)
29 meters
- (b)
21 meters
- (c)
28 meters
- (d)
26 meters
Answer: Option C
Explanation :
Ratio of speeds of A : B = 200 : 150 and B : C = 200 : 192
∴ A : C = A/B × B/C = 200/150 × 200/192 = 100 ∶ 72.
So, A beats C by (100 – 72) = 28 meters
Hence, option (c).
Workspace:
A runs times as fast as B. If A gives B a start of 120 meters, how far must the winning post be in order that A and B reach it at the same time?
- (a)
210 meters
- (b)
160 meters
- (c)
280 meters
- (d)
90 meters
Answer: Option C
Explanation :
Ratio of rates of A and B = 7 : 4
i.e., 3 meters are gained by A in a race of 7 meters
∴ 120 m. are gained by A in a race of (7/3 × 120) = 280 meters.
Hence, option (c).
Workspace:
A can beat B by 62 meters and C by 36 meters in a race of 400 meters. In a race of 700 meters C will beat B by:
- (a)
45.5 meters
- (b)
50 meters
- (c)
15.5 meters
- (d)
39 meters
Answer: Option B
Explanation :
Ratio of rates of A, B and C = 400 : (400 – 62) : (400-36) = 400 : 338 : 364 = 200 : 169 : 182.
i.e., C gains 13 meters in a race of 182 meters.
∴ In a race of 700 meters C will gain (13/182 × 700) = 50 meters.
Hence, option (b).
Workspace:
A and B take part in a 200 meters race. A runs at 5 km. per hour, A gives B a start of 16 meters and still beats him by 16 seconds. Speed of B is?
- (a)
5.15 kmph
- (b)
4.14 kmph
- (c)
4.25 kmph
- (d)
4.4 kmph
Answer: Option B
Explanation :
A’s speed = (5 × 5/18) meters/second = 25/18 meters/second.
Time taken by A to cover 200 meters = 200 × 18/25 = 144 seconds.
∴ B covers 184 meters in (144 + 16) = 160 seconds.
B’s speed = 184/160 × 18/5 = 4.14 km/h.
Hence, option (b).
Workspace:
In a game of 100 points, A can give B 20 points and C 28 points. Then, B can give C how many points in the same game?
- (a)
8 points
- (b)
10 points
- (c)
14 points
- (d)
40 points
Answer: Option B
Explanation :
A can give 20 points to B is same as A giving a headstart of 20 points to B, i.e., when the game begins A starts with 0 points whereas B starts with 20 points.
Hence, when A scores 100 points, B scores (100 - 20 = ) 80 points and C scores (100 - 28 = ) 72 points.
A : B : C = 100 : 80 : 72
∴ B : C = 80/72 = 10/9 = 100/90.
Thus, if B scores 100, ‘C’ scores 90
∴ B can give C 10 points.
Hence, option (b)
Workspace:
At a game of billiards, A can give B 30 points in 120 and A can give C 10 in 30. How many points can B give C in a game of 180?
- (a)
60 points
- (b)
40 points
- (c)
20 points
- (d)
24 points
Answer: Option C
Explanation :
A : B = 120 : 90 and A : C = 30 : 20
A : B : C = 120 : 90 : 80
∴ B : C = 90/80 = 180/160
So, if B scores 180, then C scores 160.
∴ B, can give C 20 points in a game of 180.
Hence, option (c).
Workspace:
In a km race, P gives Q and R starts of 10 meters and 16 meters respectively. When P crosses the finishing line, Q is 12 meters behind him and R is 20 meters behind him. How much start can Q give R in a km race so that they finish together?
- (a)
- (b)
- (c)
- (d)
Answer: Option A
Explanation :
When P completes 1000m, Q completes 1000 - 10 – 12 = 978m
When P completes 1000m, R completes 1000 - 16 – 20 = 964m
i.e., when Q completes 978 m, R completes 964m
Therefore, when Q completes 1000 meters R completes m
Therefore, the start that Q can give R is 1000 -
Hence, option (a).
Workspace: