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:
239 comments Page 6 of 24.
Harishankaran said:
8 years ago
@Priyesh Verma.
How can be a train, which taking time than the another, is consider as faster?
In the above example, the time taken by 1 train is 23sec and the other one is 17 sec but the ratio shows that 1train is faster than second.just imagine that length of the train is 54m.
Then,
54/27::54/17~=2:3.
How can be a train, which taking time than the another, is consider as faster?
In the above example, the time taken by 1 train is 23sec and the other one is 17 sec but the ratio shows that 1train is faster than second.just imagine that length of the train is 54m.
Then,
54/27::54/17~=2:3.
GUFRAN said:
1 decade ago
Let, the ist train speed v1=X m/s and T1=27s
2nd trin speed V2=Y m/s and T2=17s
Then S1=V1*T1=27x
S2=V2*T2=17Y
Both are in opposite dir. Then speed is S=S1-S2=27X-17Y
They cross each other means time difference 23(X-Y)=23X-23Y
now...
27X-17Y=23X-23Y
4X=-6Y
X/Y=-3/2=3/2(BECZ speed is not -ve)
X:Y=3:2
2nd trin speed V2=Y m/s and T2=17s
Then S1=V1*T1=27x
S2=V2*T2=17Y
Both are in opposite dir. Then speed is S=S1-S2=27X-17Y
They cross each other means time difference 23(X-Y)=23X-23Y
now...
27X-17Y=23X-23Y
4X=-6Y
X/Y=-3/2=3/2(BECZ speed is not -ve)
X:Y=3:2
Manish said:
8 years ago
27 and 17 are the length of the train and the speed of these train respectively x mtr/sec and y mtr/sec.
So 27x+17x are there relative speed of the train.
(27x+17y)/(x+y)=23.
Here 23 is given time in which the Cross each other's train after solving the above eq...
We get 4x=6y.
x/y=2/3 that's solved.
So 27x+17x are there relative speed of the train.
(27x+17y)/(x+y)=23.
Here 23 is given time in which the Cross each other's train after solving the above eq...
We get 4x=6y.
x/y=2/3 that's solved.
Dhinesh said:
2 years ago
Well, I have something different, 1st train is 27 sec so take this time as a total as 100% minus the 2nd train which is 17 sec so 62. 96 % is the percent of 1st train slower than the second train or vice versa.
So, 62.
96 is a 3 ratio to 100 so the remaining 38.04 is 2.
So, it's 3:2 am I right?
So, 62.
96 is a 3 ratio to 100 so the remaining 38.04 is 2.
So, it's 3:2 am I right?
(31)
Mobarak said:
1 decade ago
If two trains running at the speed x & y respectively in opposit direction then their relative velocity is (x+y) m/s.
Now if two trains cross each other then they have to cross total length of two trains which is (27x+17y).
Now time taken for crssng the total length = (27x+17y)/(x+y) i.e 23.
Now if two trains cross each other then they have to cross total length of two trains which is (27x+17y).
Now time taken for crssng the total length = (27x+17y)/(x+y) i.e 23.
Harry said:
10 years ago
I don't know its correct or not but I got the answer easily.
My step-- it's not telling about the speed so what we have here is 27 sec and 17 sec so its the ratio 27:17.
(27/9) = 3 : (17/9) =1.8 = 2.
So I got 3 : 2.
I don't know whether its right step or not, Please correct me, if it is wrong.
My step-- it's not telling about the speed so what we have here is 27 sec and 17 sec so its the ratio 27:17.
(27/9) = 3 : (17/9) =1.8 = 2.
So I got 3 : 2.
I don't know whether its right step or not, Please correct me, if it is wrong.
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.
Arunkumar said:
1 decade ago
Hi,
Nice explanation for why we need to.
1. Add the both trains speed which were running in opposite Directions with a reference point.
2. Subtract the both trains speed which were running in Same direction with a reference point.
http://en. Wikipedia. Org/wiki/Relative_velocity.
Nice explanation for why we need to.
1. Add the both trains speed which were running in opposite Directions with a reference point.
2. Subtract the both trains speed which were running in Same direction with a reference point.
http://en. Wikipedia. Org/wiki/Relative_velocity.
Priyesh verma said:
8 years ago
Two trains are moving in opposite direction with the speed of 36 k.m. /h and 54 k.m. /h cross each other in 12 seconds. The length of the 2nd train is half of the 1st train. The 1st train crosses a platform in 1. 30 minutes. Then what is the length of the platform? Please solve this.
Vidhyadhar said:
8 years ago
Let x=speed of train 1.
And y=speed of train 2.
Our task=x/y.
So, length of 1st train (d1)=x*27(dist=speed*time)
So, 2nd (d2) = y * 17.
Time taken by trains to cross each other =23 sec.
Time =dist/speed.
23=(d1+d2)/x+y.
23=(27*x+17*y)/x+y.
By solving this we get.
Result = x/y = 3/2
And y=speed of train 2.
Our task=x/y.
So, length of 1st train (d1)=x*27(dist=speed*time)
So, 2nd (d2) = y * 17.
Time taken by trains to cross each other =23 sec.
Time =dist/speed.
23=(d1+d2)/x+y.
23=(27*x+17*y)/x+y.
By solving this we get.
Result = x/y = 3/2
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