Aptitude - Problems on Trains - Discussion
Discussion Forum : Problems on Trains - General Questions (Q.No. 4)
4.
Two trains running in opposite directions cross a man standing on the platform in 27 seconds and 17 seconds respectively and they cross each other in 23 seconds. The ratio of their speeds is:
Answer: Option
Explanation:
Let the speeds of the two trains be x m/sec and y m/sec respectively.
Then, length of the first train = 27x metres,
and length of the second train = 17y metres.
![]() |
27x + 17y | = 23 |
x+ y |
27x + 17y = 23x + 23y
4x = 6y
![]() |
x | = | 3 | . |
y | 2 |
Discussion:
234 comments Page 13 of 24.
Bijay said:
1 decade ago
Length = speed*time.
Means 27x+17y = (x+y)*23.
=>(27x+17x)/(x+y) = 23.
Means 27x+17y = (x+y)*23.
=>(27x+17x)/(x+y) = 23.
SURESH KASTURI said:
1 decade ago
There are 2 trains .
Here is some experiment ,one man is standing on platform.
Trian 1:
=========
we assume X is train 1 speed.
this train cross that man in 27X meters.
Trian 2:
===========
We assume Y is train 2 speed.
This train cross that man in 17Y meters.
We concentrate trains are moving in opposite direction.
So speed is X+Y.
Distance is (train_1_length+trian_2_length) = 27X+17Y.
Time = 23 seconds.
Formule : distance = speed*time.
27X+17Y = (X+Y)*23.
X/Y = 3/2.
Here is some experiment ,one man is standing on platform.
Trian 1:
=========
we assume X is train 1 speed.
this train cross that man in 27X meters.
Trian 2:
===========
We assume Y is train 2 speed.
This train cross that man in 17Y meters.
We concentrate trains are moving in opposite direction.
So speed is X+Y.
Distance is (train_1_length+trian_2_length) = 27X+17Y.
Time = 23 seconds.
Formule : distance = speed*time.
27X+17Y = (X+Y)*23.
X/Y = 3/2.
KD Phatak said:
1 decade ago
Let the speed of the two trains bee x and y meters.
Then a/c to the formula,
Speed = distance/time.
Distance = speed*time.
Distance of one train becomes 17x.
Second become 27y.
a/c to formula distance + distance/speed = time.
17x+27y/x+y = 23.
Now by solving you can get the answer.
Then a/c to the formula,
Speed = distance/time.
Distance = speed*time.
Distance of one train becomes 17x.
Second become 27y.
a/c to formula distance + distance/speed = time.
17x+27y/x+y = 23.
Now by solving you can get the answer.
Karthik said:
1 decade ago
Velocity = distance/time.
Let velocities are x, y m/s.
x = d1/27; y = d2/17;
d1 = 27x; d2 = 17y;
Step 2: According to relative speed, since they are running in opposite direction their speeds are added.
x+y = d1+d2/23.
Sub d1 and d2; find answer.
Let velocities are x, y m/s.
x = d1/27; y = d2/17;
d1 = 27x; d2 = 17y;
Step 2: According to relative speed, since they are running in opposite direction their speeds are added.
x+y = d1+d2/23.
Sub d1 and d2; find answer.
Sidhes said:
1 decade ago
Let speed of train1: x m/s.
Speed of train2: y m/s.
Then relative speed of train1 = (x+y) m/s............(1).
Train are crosses a man means they cross their length.
Length of train1: 27x m.
Length of train2: 17y m.
According to Qst:
Train crosses Each other in 23sec.
Means train1 crosses his length & also the length of train2.
=> Total length crosses by train1 = Length of train1+Length of train2.
=>Total length covered by train1(D)=27x+17y...........(2).
We knows,
Speed(s) = Distance(D)/Time(t)............Eqn(3).
Put Values OF eqn(1) &(2) in eqn(3).
x+y = (27x+27y)/23.
=> 23x+23y = 27x+17y.
=> 27x-23x = 23y-17y.
=> 4x = 6y.
=> x/y = 6/4 = 3/2..........Answer.
Speed of train2: y m/s.
Then relative speed of train1 = (x+y) m/s............(1).
Train are crosses a man means they cross their length.
Length of train1: 27x m.
Length of train2: 17y m.
According to Qst:
Train crosses Each other in 23sec.
Means train1 crosses his length & also the length of train2.
=> Total length crosses by train1 = Length of train1+Length of train2.
=>Total length covered by train1(D)=27x+17y...........(2).
We knows,
Speed(s) = Distance(D)/Time(t)............Eqn(3).
Put Values OF eqn(1) &(2) in eqn(3).
x+y = (27x+27y)/23.
=> 23x+23y = 27x+17y.
=> 27x-23x = 23y-17y.
=> 4x = 6y.
=> x/y = 6/4 = 3/2..........Answer.
GrandMaster said:
1 decade ago
Very short form is:
(27-23):(23-17),
4:6,
2:3.
So the ratio is 2:3 or 3:2.
(27-23):(23-17),
4:6,
2:3.
So the ratio is 2:3 or 3:2.
Harish said:
1 decade ago
I have solved it in yet simple(crude) way.
Don't worry about their crossing each other in 23 seconds or whatever.
On the face of it, one can make out that he has asked the simplified ration for 27:17.
Simply take them to next higher level so that both of them are divisible by common highest divisor.
i.e. 30/20 = 3/2.
(but recommended not to follow it, as it doesn't apply to every problem?).
Don't worry about their crossing each other in 23 seconds or whatever.
On the face of it, one can make out that he has asked the simplified ration for 27:17.
Simply take them to next higher level so that both of them are divisible by common highest divisor.
i.e. 30/20 = 3/2.
(but recommended not to follow it, as it doesn't apply to every problem?).
SWETHA.T said:
1 decade ago
SPEED = DISTANCE/TIME.
DISTANCE/SPEED = TIME.
i.e 27X+17Y/X+Y = 23.
DISTANCE/SPEED = TIME.
i.e 27X+17Y/X+Y = 23.
Het said:
1 decade ago
@Arashan has describe very well with formula of speed.
Sarath said:
1 decade ago
Lets assume,
The train 1 speed= x m/sec ==> distance traveled by the train 1= 27*x m.
Similarly assume the speed of the train 2 is y m/sec then the train 2 has traveled a distance of 17*y meter.
The total distance (x+y)*23 m.
Therefore,
23(x+y) = 27x+17y.
23x+23y=27x+17y.
==> 4x=6y.
==> x/y=3/2.
Therefore x:y= 3:2. Answer B.
The train 1 speed= x m/sec ==> distance traveled by the train 1= 27*x m.
Similarly assume the speed of the train 2 is y m/sec then the train 2 has traveled a distance of 17*y meter.
The total distance (x+y)*23 m.
Therefore,
23(x+y) = 27x+17y.
23x+23y=27x+17y.
==> 4x=6y.
==> x/y=3/2.
Therefore x:y= 3:2. Answer B.
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