Practice Trains in opposite direction - quantitative aptitude Online Quiz (set-1) For All Competitive Exams
Q-1) Two trains are running in opposite direction with the same speed. If the length of each train is 120 metres and they cross each other in 12 seconds, the speed of each train (in km/hour) is
(a)
(b)
(c)
(d)
Using Rule 3,
Let the speed of each train be x kmph.
Their relative speed
= x + x = 2x kmph.
Time taken
= $\text"Total length of trains"/\text"Relative Speed"$
= $12/{60 × 60} = {240 × {1/1000}}/{2x}$
= $1/300 = 120/{1000x}$
$x = {300 × 120}/1000$ = 36
The required speed = 36 kmph.
Q-2) Two trains of equal length take 10 seconds and 15 seconds respectively to cross a telegraph post. If the length of each train be 120 metres, in what time (in seconds) will they cross each other travelling in opposite direction ?
(a)
(b)
(c)
(d)
Using Rule 3,
When a train crosses a telegraph post, it covers its own length.
Speed of first train
= $120/10$ = 12 m/sec.
Speed of second train
= $120/15$ = 8 m/sec.
Relative speed
= 12 + 8 = 20 m/sec.
Required time
= $\text"Total length of trains"/ \text"Relative speed"$
= ${2 × 120}/20$ = 12 seconds.
Q-3) P and Q starting simultaneously from two different places proceed towards each other at a speed of 20 km/hour and 30 km/hour respectively. By the time they meet each other. Q has covered 36 km more than that of P. The distance (in km.) between the two places is
(a)
(b)
(c)
(d)
Let P and Q meet after t hours.
Distance = speed × time
According to the question,
30t - 20t = 36
10t = 36 ⇒ t = $36/10$ = 3.6 hours
Distance between P and Q
= 30t + 20t
= 50t = (50 × 3.6) km. = 180 km.
Using Rule 13,From stations A and B, two trains start travelling towards each other at speeds a and b, respectively. When they meet each other, it was found that one train covers distance d more than that of another train. The distance between stations A and B is given as$({a + b}/{a - b}) × d$
Here, a = 30, b = 20, d = 36
Required distance
= $({a + b}/{a - b}) × d$
= $({30 + 20}/{30 - 20}) ×$ 36
= $50/10 × 36$ = 180 km
Q-4) Two trains start from station A and B and travel towards each other at speed of 16 miles/ hour and 21 miles/ hour respectively. At the time of their meeting, the second train has travelled 60 miles more than the first. The distance between A and B (in miles) is :
(a)
(b)
(c)
(d)
Let the trains meet after t hours
Then, 21t - 16t = 60
5t = 60 ⇒ t = 12 hours
Distance between A and B
= (16 + 21) × 12
= 37 × 12 = 444 miles
Using Rule 13,From stations A and B, two trains start travelling towards each other at speeds a and b, respectively. When they meet each other, it was found that one train covers distance d more than that of another train. The distance between stations A and B is given as$({a + b}/{a - b}) × d$
Here, a = 21, b = 16, d = 60
Distance between A and B
= $({a + b}/{a - b}) × d$
= $({21 + 16}/{21 - 16}) × 60$
= $37/5 × 60$ = 37 × 12 = 444 miles
Q-5) A train running at the speed of 84 km/hr passes a man walking in opposite direction at the speed of 6 km/hr in 4 seconds. What is the length of train (in metre) ?
(a)
(b)
(c)
(d)
Using Rule 6,Let 'a' metre long train is running with the speed 'x' m/s. A man is running in the opposite direction of train with the speed of 'y' m/s. Then, time taken by the train to cross the man = $(a/{(x + y)})$seconds.
Relative speed
= (84 + 6) = 90 kmph
= $(90 × 5/18)$ m/sec. = 25 m/sec.
Length of train
= Relative speed × Time
= 25 × 4 = 100 metre
Q-6) A train travelling at 48 km/hr crosses another train, having half its length and travelling in opposite direction at 42 km/hr, in 12 seconds. It also passes a railway platform in 45 seconds. The length of the railway platform is
(a)
(b)
(c)
(d)
Let the length of the train travelling at 48 kmph be x metres.
Let the length of the platform be y metres.
Relative speed of train
= (48 + 42) kmph
= ${90 × 5}/18$ m./sec. = 25 m./sec.
and 48 kmph = ${48 × 5}/18 = 40/3$ m./sec.
According to the question,
${x + {x/2}}/25$ = 12
${3x}/{2 × 25}$ =12
3x = 2 × 12 × 25 = 600
x = 200 m.
Also, ${200 + y}/{40/3}$ = 45
600 + 3y = 40 × 45
3y = 1800 - 600 = 1200
y = $1200/3$ = 400 m.
Q-7) Two trains of length 70 m and 80 m are running at speed of 68 km/hr and 40 km/hr respectively on parallel tracks in opposite directions. In how many seconds will they pass each other ?
(a)
(b)
(c)
(d)
Using Rule 3,
Relative speed
= (68 + 40) kmph = 108 kmph
= $({108 × 5}/18)$ m/s or 30 m/s
Required time
= $\text"Sum of the lengths of both trains"/ \text"Relative speed"$
= $({70 + 80}/30)$ second = 5 seconds
Q-8) Two trains 108 m and 112 m in length are running towards each other on the parallel lines at a speed of 45 km/hr and 54 km/ hr respectively. To cross each other after they meet, it will take
(a)
(b)
(c)
(d)
Using Rule 3,
Relative speed
= 45 + 54 = 99 kmph
= $(99 × 5/18)$ m/sec. or $55/2$ m/sec.
Required time = ${108 + 112}/{55/2}$
= ${220 × 2}/55 = 8$ seconds
Q-9) Two trains start at the same time from Aligarh and Delhi and proceed towards each other at the rate of 14 km and 21 km per hour respectively. When they meet, it is found that one train has travelled 70 km more than the other. The distance between two stations is
(a)
(b)
(c)
(d)
Let the trains meet each other after t hours.
Distance = Speed × Time
According to the question,
21t - 14t = 70
7t = 70 ⇒ t = $70/7$ = 10 hours
Required distance
= 21t + 14t = 35t = 35 × 10 = 350 km.
Using Rule 13,
Here, a = 21, b = 14, d = 70
Required distance
= $({a + b}/{a - b}) × d$
= $({21 + 14}/{21 - 14}) × 70$
= $35/7 × 70$ = 350 km.
Q-10) Two trains of length 137 metre and 163 metre are running with speed of 42 km/hr and 48 km/hr respectively towards each other on parallel tracks. In how many seconds will they cross each other?
(a)
(b)
(c)
(d)
Using Rule 3,
Relative speed
= 42 + 48 = 90 kmph
= $({90 × 5}/18)$ m/s = 25 m/s
Sum of the length of both trains
= 137 + 163 = 300 metres
Required time = $300/25$ = 12 seconds