Practice Trains problems - quantitative aptitude Online Quiz (set-1) For All Competitive Exams
Q-1) If a man reduces his speed to 2/ 3, he takes 1 hour more in walking a certain distance. The time (in hours) to cover the distance with his normal speed is :
(a)
(b)
(c)
(d)
Since man walks at $2/3$ of usual speed, time taken wil be $3/2$ of usual time.
$3/2$ of usual time
= usual time + 1 hour.
$(3/2 –1)$ of usual time = 1
usual time = 2 hours.
Q-2) Two trains 180 metres and 120 metres in length are running towards each other on parallel tracks, one at the rate 65 km/ hour and another at 55 km/hour. In how many seconds will they be clear of each other from the moment they meet ?
(a)
(b)
(c)
(d)
Required time
= $\text"Sum of the lengths of trains"/\text"Relative speed"$
Relative speed = 65 + 55 = 120 kmph
= ${120 × 5}/18$ m/sec
Required time = ${180 + 120}/{{120 × 5}/18}$
= ${300 × 18}/{120 × 5}$ = 9 seconds
Q-3) A train is 250m long. If the train takes 50 seconds to cross a tree by the railway line, then the speed of the train in km/hr is :
(a)
(b)
(c)
(d)
Using Rule 1,
Speed of train = $\text"Length of train"/ \text"Time taken in crossing"$
= $250/50$ = 5 m/sec.
= $(5 × 18/5)$ kmph = 18 kmph
Q-4) A student goes to school at the rate of 2$1/2$ km/h and reaches 6 minutes late. If he travels at the speed of 3 km/h. he is 10 minutes early. The distance (in km) between the school and his house is
(a)
(b)
(c)
(d)
Let the required distance be x km.
$x/{5/2} - x/3 = 16/60$
${2x}/5 - x/3 = 4/15$
${6x - 5x}/15 = 4/15 ⇒ x = 4$ km.
Using Rule 10,
Here, $S_1 = 2{1}/2 , t_1 = 6, S_2 = 3, t_2$ = 10
Distance = ${(S_1 × S_2)(t_1 + t_2)}/{S_2 - S_1}$
= ${5/2 × 3(6 + 10)}/{3 - 5/2}$
= $15 × 16/60$ km = 4 km
Q-5) Two trains 125 metres and 115 metres in length, are running towards each other on parallel lines, one at the rate of 33 km/ hr and the other at 39 km/hr. How much time (in seconds) will they take to pass each other from the moment they meet ?
(a)
(b)
(c)
(d)
Relative speed
= (33 + 39) kmph = 72 kmph
= $({72 × 5}/18)$ m/sec. = 20 m/sec.
Time taken in crossing
= $\text"Length of both trains"/ \text"Relative speed"$
= ${125 + 115}/20 = 240/20$
= 12 seconds
Q-6) A train crosses a platform in 30 seconds travelling with a speed of 60 km/h. If the length of the train be 200 metres, then the length (in metres) of the platform is
(a)
(b)
(c)
(d)
Rule 10 and Rule 1,
Speed of train = 60 kmph
= $(60 × 5/18)$ m/sec.
= $50/3$ m/sec.
If the length of platform be = x metre, then
Speed of train
= $\text"Length of (train + platform)"/ \text"Time taken in crossing"$
$50/3 = {200 + x}/30$
50 × 10 = 200 + x
x = 500 - 200 = 300 metre
Q-7) A train passes by a lamp post on a platform in 7 sec. and passes by the platform completely in 28 sec. If the length of the platform is 390 m, then length of the train (in metres) is
(a)
(b)
(c)
(d)
Rule 10 and Rule 1,
Let the length of train be x metre,
then, Speed of train
= $x/7 = {x + 390}/28$
$x = {x + 390}/4$
4x - x = 390
$x = 390/3$ = 130 metres
Q-8) A boy is late by 9 minutes if he walks to school at a speed of 4 km/hour. If he walks at the rate of 5 km/hour, he arrives 9 minutes early. The distance to his school is
(a)
(b)
(c)
(d)
Let the required distance be x km.
According to the question,
$x/4 - x/5 = 18/60$
${5x - 4x}/20 = 3/10$
$x = 3/10 × 20$ = 6 km
Using Rule 10,
Here, $S_1 = 4, t_1 = 9, S_2 = 5, t_2$ = 9
Distance = ${(S_1 × S_2)(t_1 + t_2)}/{S_2 - S_1}$
= ${(4 × 5)(9 + 9)}/{5 - 4}$
= $20 × 18/60 = $6 km
Q-9) Two trains of equal length are running on parallel lines in the same direction at 46 km/h and 36 km/h. The faster train passes, the slower train in 36 seconds. The length of each train is :
(a)
(b)
(c)
(d)
Let the length of each train be x metre.
Relative speed
= 46 – 36 = 10 kmph
= ${10 × 5}/18$ metre/second
= $25/9$ metre/second
${2x}/{25/9} = 36$
2x = ${36 × 25}/9 = 100$
x = 50 metre
Q-10) Two trains of equal length are running on parallel lines in the same direction at the rate of 46 km/hr and 36 km/hr. The faster train passes the slower train in 36 seconds. The length of each train is
(a)
(b)
(c)
(d)
Length of each train = x metre
Relative speed
= 46 – 36= 10 kmph
= $(10 × 5/18)$ m/sec = $25/9$ m/sec
Time taken in crossing
= $\text"Length of both trains"/ \text"Relativespeed"$
36 = ${2x}/{25/9}$
$2x = 36 × 25/9$ = 100
$x = 100/2$ = 50 metre
Q-11) A passenger train 150m long is travelling with a speed of 36 km/ hr. If a man is cycling in the direction of train at 9 km/hr., the time taken by the train to pass the man is
(a)
(b)
(c)
(d)
Using Rule 5,Let 'a' metre long train is running with the speed 'x' m/s. A man is running in same direction and with the speed 'y' m/s, then time taken by the train to cross the man = $a/{(x - y)}$seconds. And a = (x - y)t
Relative speed of train
= (36 - 9) kmph = 27 kmph
= ${27 × 5}/18$ m/sec = $15/2$ m/sec
Required time
= $\text"Length of the train"/ \text"Relative speed"$
= ${150 × 2}/15 = 20$ seconds
Q-12) How many seconds will a 500 metre long train take to cross a man walking with a speed of 3 km/hr. in the direction of the moving train if the speed of the train is 63 km/hr ?
(a)
(b)
(c)
(d)
Using Rule 5,
Relative speed of train
= 63–3 = 60 kmph
= $60 × 5/18$ m / sec
Time = $\text"Length of train"/ \text"Relative Speed"$
= ${500 × 18}/{60 × 5}$ = 30 sec.
Q-13) Points 'A' and 'B' are 70 km apart on a highway. A car starts from 'A' and another from 'B' at the same time. If they travel in the same direction, they meet in 7 hours, but if they travel towards each-other, they meet in one hour. Find the speed of the two cars (in km/hr).
(a)
(b)
(c)
(d)
Let speed of car starting from A be x kmph
and speed of car starting from B be y kmph
Case I
When cars meet at P,
7x = AP = AB + BP = 70 + 7y
7x - 7y = 70
x - y = 10 ...(i)
Case II
When cars meet at Q,
x + y = 70 ...(ii)
On adding these equations,
x = 40 kmph
Putting the value of x in equation (i),
y = 40 - 10 = 30 kmph
Q-14) A train is moving at a speed of 80 km/h and covers a certain distance in 4.5 hours. The speed of the train to cover the same distance in 4 hours is
(a)
(b)
(c)
(d)
Distance = Speed × Time
= 80 × 4.5 = 360 km
∴ Required speed = $360/4$ = 90 kmph.
Q-15) A passenger train running at the speed of 80 kms./hr leaves the railway station 6 hours after a goods train leaves and overtakes it in 4 hours. What is the speed of the goods train?
(a)
(b)
(c)
(d)
Let the speed of goods train be x kmph.
Distance covered by goods train in 10 hour = distance covered by passenger train in 4 hours
10x = 80 × 4
$x = {80 × 4}/10$ = 32 kmph.
Q-16) The time taken by a train 160 m long, running at 72 km/hr, in crossing an electric pole is
(a)
(b)
(c)
(d)
Distance covered in crossing a pole
= Length of train Speed of train = 72 kmph
= $({72 × 5}/18)$ m./sec. = 20 m./sec.
Required time = $160/20$ = 8 seconds
Q-17) A 120 m long train takes 10 seconds to cross a man standing on a platform. What is the speed of the train ?
(a)
(b)
(c)
(d)
Using Rule 1,
In crossing a man standing on platform, train crosses its own length.
Speed of train = $120/10 = 12$m/s
Q-18) In what time will a train, 60 metre long, running at the rate of 36 km/hr pass a telegraph post ?
(a)
(b)
(c)
(d)
Speed of train = 36 kmph
= $({36 × 5}/18)$ m./sec. = 10 m./sec.
Required time = $\text"Length of train"/ \text"Speed of train"$
= $60/10$ = 6 seconds
Q-19) A train 150m long passes a km stone in 30 seconds and another train of the same length travelling in opposite direction in 10 seconds. The speed ot the second train is :
(a)
(b)
(c)
(d)
Speed of train A
= $150/30$ = 5 m/sec.
Speed of train B = x m/sec.
Relative speed = (5+x) m/sec.
Length of both trains
= Relative speed × Time
300 = (5 + x) × 10
5 + x = $300/10$ = 30
x = 30 - 5 = 25 m/sec.
= $({25 × 18}/5)$ kmph. = 90 kmph.
Q-20) A train 100 metre long is running at a speed of 120 km/hr. The time taken to pass a person standing near the line is
(a)
(b)
(c)
(d)
Speed of train = 120 kmph.
= $({120 × 5}/18)$ m./sec. = $100/3$ m./sec.
Required time = $\text"Length of train"/ \text"Speed of train"$
= $(100/{100/3})$ seconds
= $(100/100 × 3)$ seconds = 3 seconds