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:
233 comments Page 3 of 24.
Gaurav singh said:
1 decade ago
Time x speed = distance.
Let the speeds of two train be x and y.
Then distances will be 27x for 1st train.
17y for 2nd train.
Now when they both cross each other fully. Time is 23 seconds.
Distance will be length of 1st train + length of 2nd train.
Speed will be as we assumed above is x and y (sum of both the speed).
So,
23 = (27x+17y)/x+y.
23(x+y) = 27x+17y.
23y-17y = 27x-23x.
6y = 4x.
3y = 2x.
Hence the answer 3:2.
Let the speeds of two train be x and y.
Then distances will be 27x for 1st train.
17y for 2nd train.
Now when they both cross each other fully. Time is 23 seconds.
Distance will be length of 1st train + length of 2nd train.
Speed will be as we assumed above is x and y (sum of both the speed).
So,
23 = (27x+17y)/x+y.
23(x+y) = 27x+17y.
23y-17y = 27x-23x.
6y = 4x.
3y = 2x.
Hence the answer 3:2.
Harry Joshi said:
1 decade ago
@Hardik Mistry : When trains move in opposite direction their speeds are added, but when they go in the same direction their speeds are subtracted. Imagine you are going on road by bike and another bike is coming from front (i. e opposite direction) , you feel that the other bike is coming faster. But if it comes from back and tries to overtake you (i.e same direction) then it overtakes you slowly as compared to its speed.
Nidhi said:
1 decade ago
Lets speed of 1st train = x and of 2nd = y.
when they cross man then.....
so length of 1st = 27x and second 17y.
total length = 27x + 17y
now when they cross each other in opp direction they take 23 sec and we know when 2 object run in opp direction their rel speed is
speed of 1st + speed of 2nd = x+y so now length = time * speed = 23(x+y)
Compare both :
27x + 17y = 23(x+y)
27x + 17y = 23x + 23y
4x = 6y => x/y = 3/2.
when they cross man then.....
so length of 1st = 27x and second 17y.
total length = 27x + 17y
now when they cross each other in opp direction they take 23 sec and we know when 2 object run in opp direction their rel speed is
speed of 1st + speed of 2nd = x+y so now length = time * speed = 23(x+y)
Compare both :
27x + 17y = 23(x+y)
27x + 17y = 23x + 23y
4x = 6y => x/y = 3/2.
(1)
Vincent said:
1 decade ago
Let the speeds of the two trains be x m/sec and why m/sec respectively.
Reference to man standing on the platform.
Length of the first train = 27X metres,
Length of the second train = 17Y metres.
Total speed of train when crossing opposite direction is (x+y) m/s.
Two trains are crossing each other in 23 seconds.
So length of two trains using this condition is (x+y) 23 m.
So 27X+17Y = 23X+23Y.
4X = 6Y.
X/Y = 3/2.
Reference to man standing on the platform.
Length of the first train = 27X metres,
Length of the second train = 17Y metres.
Total speed of train when crossing opposite direction is (x+y) m/s.
Two trains are crossing each other in 23 seconds.
So length of two trains using this condition is (x+y) 23 m.
So 27X+17Y = 23X+23Y.
4X = 6Y.
X/Y = 3/2.
Sumitra said:
1 decade ago
Here the time taken by two trains itself 27 & 17 sec rt.
So we know the formula that speed=d/t. Let us take the speed of the two trains is x & y.
Then the corresponding distances are 27x and 17y.
Then t=d/s as per formula.
Time taken to cross each other is 23 sec.
Equate that time as 23 sec and to cross that man by both trains.
i.e (27x+17y)/(x+y) = 23.
Then solve the equation we will get the answer.
So we know the formula that speed=d/t. Let us take the speed of the two trains is x & y.
Then the corresponding distances are 27x and 17y.
Then t=d/s as per formula.
Time taken to cross each other is 23 sec.
Equate that time as 23 sec and to cross that man by both trains.
i.e (27x+17y)/(x+y) = 23.
Then solve the equation we will get the answer.
Rajkumar said:
1 decade ago
It is very simple to understand the logic of 27x+17y/x+y=23 first we came across to know logic first of all we know the, speed=x/t in this way we assume that the sum of the two train lengths divided by assuming speeds is equal to relative crossing time is 23 seconds.
One of the question arise in your mind that is why we took length s of trains in sum because they cross each other with respect to man.
Thank you.
One of the question arise in your mind that is why we took length s of trains in sum because they cross each other with respect to man.
Thank you.
Ankit Choudhary said:
1 decade ago
As we know that,when two train of length l1 and l2 travels in opposite direction with speed let s1 km/hr and s2 km/hr respectively then time taken to cross each other is given by formula
t=(l1+l2)/(s1+s2)
So by using this formula we can easily solve the above problem.
here l1=27*s1=27s1.
l2=17*s2=17s2.
and t is given in problem equal to 23.
therefore,
23=(27s1+17s2)/(s1+s2)=>3:2
t=(l1+l2)/(s1+s2)
So by using this formula we can easily solve the above problem.
here l1=27*s1=27s1.
l2=17*s2=17s2.
and t is given in problem equal to 23.
therefore,
23=(27s1+17s2)/(s1+s2)=>3:2
Anubhav Bharti said:
2 years ago
Let the speed of the two trains be x m/sec. and y m/sec.
Then, the length of the first train = 27x metres,
and length of the second train = 17y metres,
Time taken by the train to cross each other is 23 sec.
Time taken by the train to cross each other = Total Distance of both trains/ Total of Both trains.
so, 23 = (27x + 17y) / (x + y).
23x + 23y = 27x + 17y.
6y = 4x.
x/y = 6/4 => 3:2.
Then, the length of the first train = 27x metres,
and length of the second train = 17y metres,
Time taken by the train to cross each other is 23 sec.
Time taken by the train to cross each other = Total Distance of both trains/ Total of Both trains.
so, 23 = (27x + 17y) / (x + y).
23x + 23y = 27x + 17y.
6y = 4x.
x/y = 6/4 => 3:2.
(33)
RAHUL RAJ PRASAD said:
5 years ago
Train A length=x;time A=27;speed A=x/27.
Train B length=Y;time B=17;speed B=Y/17.
Now relative speed between train A and B is=(x+Y)/23.
Now we equating these equation through relative speed;
(x+Y)/23=x/27+Y/17.
=> (x+Y)/23=(17x+27Y)/459,
=>459x+459Y=391x+621Y,
=>68x=162Y,
=>x/Y=162/68,
=>(x/27)/(Y/17)=(162/27)/(68/17),
=>speed A/speed B=3/2,
=>speed A : speed B = 3:2.
Train B length=Y;time B=17;speed B=Y/17.
Now relative speed between train A and B is=(x+Y)/23.
Now we equating these equation through relative speed;
(x+Y)/23=x/27+Y/17.
=> (x+Y)/23=(17x+27Y)/459,
=>459x+459Y=391x+621Y,
=>68x=162Y,
=>x/Y=162/68,
=>(x/27)/(Y/17)=(162/27)/(68/17),
=>speed A/speed B=3/2,
=>speed A : speed B = 3:2.
Samiddho said:
7 months ago
Time taken by train 1 = 27s.
Time taken by train 2 = 17s.
Time taken for them to cross each other = 23s
Let the velocity of train 1 be v1 and that of train 2 is v2, and their lengths be l1 and l2 resp.
Net velocity when they cross each other v = v1 + v2.
Thus;
(l1+l2) = (v1+v2)x23---> i
and
l1 = v1 * 27 and l2 = v2 * 17---> ii
By Replacing ii in i.
27v1 + 17v2 = 23(v1+v2)
4v1 = 6v2
v1:v2 = 3:2.
Time taken by train 2 = 17s.
Time taken for them to cross each other = 23s
Let the velocity of train 1 be v1 and that of train 2 is v2, and their lengths be l1 and l2 resp.
Net velocity when they cross each other v = v1 + v2.
Thus;
(l1+l2) = (v1+v2)x23---> i
and
l1 = v1 * 27 and l2 = v2 * 17---> ii
By Replacing ii in i.
27v1 + 17v2 = 23(v1+v2)
4v1 = 6v2
v1:v2 = 3:2.
(27)
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