| 1. P starts chasing Q, who is 200 m away from P. The speeds of P and Q are of 30 km/hr and 24 km/hr respectively. After how much time will P catch Q? | Easy |
| A. 2 mins B. 5 mins C. 6 mins D. None of these |
View Answer
Answer: Option A
Explanation:
Relative Speed = 30-24 = 6km/hr
Distance = 200 m = 1/5 km
Time = (1/5)/6 = 1/30 hr = 2mins
| 2. Train P overtakes train Q 350 m long and a cyclist in 50 and 20 seconds, respectively. If the speeds of train Q and the cyclist are 54 km/hr and 36 km/hr, respectively, find the length and speed of train P.
A. 400 m, 20 m/s |
Easy |
View Answer
Answer: Option B
Explanation:
Let the length of train P be x metres and its speed be y m/s.
Speed of train Q = 54 km/hr = 54*(5/18) = 15 m/s
Relative speed of the trains = (y-15) m/s
-> 50 = (x+350)/(y-15)—– ❶
Speed of cyclist = 36 km/hr = 36*(5/18) = 10m/s
Also, train P overtakes the cyclist in 20 seconds.
-> x= (y-10)*20 or x = 20y-200 —– ❷
From ❶ & ❷, y = 30m/s
x = 20(30-10) = 400 metres
| 3. Train I leaves city P for city Q at 4:30 am. It travels at 50 km/hr towards Q which is 100 kms away. At 5:00 am Train II leaves Q for P and travels at 40 km/hr. At 5:30 am the traffic policeman at Q discovers that they are on the same track. How much time does he have to avoid a collision? A. 20 mins B. 10 mins C. 30 mins D. None of these |
Medium |
View Answer
Answer: Option A
Explanation:
Train I covers 50 kms in an hour. At 5:30 am it is 50 km away from Q.
Train II starts at 5:00 am -> By 5:30 am it covers 20 km, as its speed is 40 km/hr.
At 5:30 am distance between the two trains = 50-20 = 30 km
Relative Speed = 50+40 = 90 km/hr
Time = 30/90 = 1/3rd of an hour = 20 mins
| 4. A man is riding his bicycle at a speed of x km/hr to complete a 1 km distance. Cycling against the wind, which lowers his speed by y km/hr, he covers this 1 km stretch in s hours. When he travels back the same distance, it takes him t hours. Which of the following statements is correct?
I. 1/s = x-y II. x = 1/2(1/s+1/t) A. I only |
Medium |
View Answer
Answer: Option C
Explanation:
I. Speed of the man against the wind = x-y
Time taken to cover 1 km = 1/(x-y) = s -> 1/s = x-y —– ❶
Hence, I is true.
II. Along the direction of the wind, speed of the man = x+y
-> 1/(x+y) = t -> 1/t = x+y —– ❷
From ❶ & ❷, 1/s + 1/t = x+y+x-y = 2x
-> x = 1/2 (1/s + 1/t)
Hence, II is true.
| 5. A and B start running in clockwise direction at speeds of 2 m/s and 5 m/s respectively, along a circular track of circumference 240 m. They start at the same time. C runs at 6 m/s from the same point and in the same direction. When will they: i. meet for the first time ii. meet for the first time at the starting point A. 240 seconds, 240 seconds B. 240 seconds, 60 seconds C. 120 seconds, 120 seconds D. None of these |
Medium |
View Answer
Answer: Option A
Explanation:
i. They will meet for the first time after the LCM of time taken by fastest person (C) to gain one complete round over the other two.
C meets B for the first time in 240/(6-5) = 240 s
C meets A for the first time in 240/(6-2) = 60 s
LCM of 240 and 60 is 240. Hence, they meet for the first time after 240 seconds.
ii. They will meet for the first time at the starting point after the LCM of the time taken by each to complete one round.
Time taken by A to complete one round = 240/2 =120 s
Time taken by B to complete one round = 240/5 =48 s
Time taken by B to complete one round = 240/6 =40 s
-> Required time = LCM of 120, 48 and 40 = 240 seconds
| 6. Alex and Bandy run on a circular track at constant speeds. Alex takes 5 minutes less than Bandy to complete a lap. If they start from the same point and run in the same direction, they will meet after 1/6 of an hour. How frequently will they meet in an hour if they run in opposite directions? | Difficult |
| A. 10 times B. 15 times C. 18 times D. None of these |
View Answer
Answer: Option C
Explanation:
Let the length of the track be L meters, and the speeds of Alex and Bandy be Sa and Sb meters/minute, respectively. They meet after 1/6 of an hour or 10 minutes if they run in the same direction.
Then, L/(Sa-Sb) = 10 or (Sa/L) – (Sb/L) = 1/10 —– ❶
Also, (L/Sb) – (L/Sa)= 5 —– ❷
Let Sa/L = x and Sb/L = y —– ❸
From ❶ & ❸, x-y = 1/10 —– ❹
Also, from ❷ & ❸, 1/y – 1/x = 5 —– ❺
From ❹& ❺, x-y = 5xy
-> xy = (x-y)/5 = (1/10)/5 = 1/50
Now, (x+y)2 = (x-y)2+4xy = (1/10)2+4(1/50) = 9/100
-> x+y = 3/10
-> Sa/L + Sb/L = 3/10
-> (Sa + Sb)/ L = 3/10
-> L/(Sa + Sb ) = 10/3
-> They will meet every 10/3 minutes if they run in opposite directions. Hence, they will meet 60/(10/3) = 18 times in an hour.
| 7. A overtakes B in the middle of the 4th round in a race around a circular track. What is the ratio of A’s speed to B’s speed? | Easy |
| A. 5:4 B. 7:5 C. 5:7 D. None of these |
View Answer
Answer: Option B
Explanation:
A overtakes B in the middle of the 4th round implies, when A has completed three and a half rounds, B has completed two and a half rounds.
-> A’s Speed/B’s Speed = (7/2)/(5/2) = 7:5
Note:
A overtakes B at (1/n)th of xth round means, when A has completed (x-(1/n)) rounds, B has completed ((x-1)-(1/n)) rounds. Also, (A’s speed)/(B’s speed) = (Rounds completed by A in a given time)/ (Rounds completed by B in the same time)
| 8. A can run 50 metres while B can run 45 metres per minute. If B has a 5-minute start in a race, how much time will A take to get level with him? | Easy |
| A. 25 mins B. 30 mins C. 50 mins D. None of these |
View Answer
Answer: Option D
Explanation:
B has a 5-minute head-start over A, in which he would cover (5*45) = 225 m. He is thus ahead of A by 225 m at the start of the race.
Now, in every minute when A runs 50 m, B runs 45 m
-> A gains 5 m over B in 1 minute. Thus, A gains 225 m over B in (225/5) = 45 minutes
| 9. Three men A, B and C, are running around a circular field 3780 metres in circumference at speeds of 108 m/min, 126 m/min and 135 m/min respectively. If they start from the same point at the same time in the same direction, then A, B and C are together after every x hours, and A meets C after every y minutes. Find x and y. | Medium |
| A. 7, 140 B. 7. 120 C. 10, 140 D. None of these |
View Answer
Answer: Option A
Explanation:
A meets C after C has gained one complete round over A i.e. after 3780/(135-108) = 140 minutes. Thus, y = 140.
Now, A,B and C are together after the LCM of the time taken by C to gain one complete round over both A and B.
C gains one complete round over A in 140 minutes (from above); C gains one complete round over B in 3780/(135-126) = 420 minutes.
-> C gains one round over both A and B after the LCM of 140 and 420 i.e., 420 minutes or 7 hours. Thus, x = 7
| 10. A runs 100 m in 15 seconds whereas B can run the same distance in 14 seconds. What start can B give A in a half kilometre race? | Easy |
| A. 22.2 m B. 33.3 m C. 66.6 m D. None of these |
View Answer
Answer: Option B
Explanation:
LCM of 14 and 15 = 210
Now, in 210 seconds, A runs 1400 m and B runs 1500 m
-> In a 1500 m race, B gives A a start of 100 m
-> In a 500 m race, B gives A a start of (100*500)/1500 = 100/3 = 33 1/3 m