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 5 of 24.
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.
Rajkumar said:
1 decade ago
Let the speed of the two trains be a=x m/sec and b=y m/sec,
crossing time=23sec
formula :distance=speed*time
for 1st train,distance,u=x*27=27x
for 2nd train,distance,v=y*17=17x
suppose in opposite direction the formula is time=u+v/a+b
therefore,27x+17x/x+y=23
27x+17y=23(x+y)
27x+17y=23x+23y
27x-23x=23y-17y
4x=6y
x/y=6/4
x/y=3/2
x:y=3:2
crossing time=23sec
formula :distance=speed*time
for 1st train,distance,u=x*27=27x
for 2nd train,distance,v=y*17=17x
suppose in opposite direction the formula is time=u+v/a+b
therefore,27x+17x/x+y=23
27x+17y=23(x+y)
27x+17y=23x+23y
27x-23x=23y-17y
4x=6y
x/y=6/4
x/y=3/2
x:y=3:2
Maha said:
1 decade ago
Distance=speed*time.
1st train time =27sec. 2nd train time =17sec.
two trains can cross the each other in opposite direction is 23 sec.
so total time=(total distance)/totalspeed;
total distance,
1st&2nd train have speed x,y;
total distance=27x+17y; total speed=x+y;
23=(27x+17y)/x+y;
23x+23y=27x+17y;
6y=4x;
6/4=x/y;
3/2=x/y;
3:2
1st train time =27sec. 2nd train time =17sec.
two trains can cross the each other in opposite direction is 23 sec.
so total time=(total distance)/totalspeed;
total distance,
1st&2nd train have speed x,y;
total distance=27x+17y; total speed=x+y;
23=(27x+17y)/x+y;
23x+23y=27x+17y;
6y=4x;
6/4=x/y;
3/2=x/y;
3:2
Mohit sahu said:
1 decade ago
Friends its easy to undrstnd as the qustn says that d trains r moving in oppste drtn that means the total distance travveld by the trains wen they crosses the person will be equal to the total distance travlld by dem in crossng each other.
Mathematly we can also better undrstnd ths by wrtng tht equation as
27x + 17y = 23(x + y).
Mathematly we can also better undrstnd ths by wrtng tht equation as
27x + 17y = 23(x + y).
Rahul said:
1 decade ago
@Vadivel.
There may be possible that the lengths are same but in question given that the crossing time is 23. So we will apply the formula of crossing time (a+b)/(u+v). And for length a & b we will use the man crossing conditions.
Please don't apply anything from your side only use the conditions which are given in question.
There may be possible that the lengths are same but in question given that the crossing time is 23. So we will apply the formula of crossing time (a+b)/(u+v). And for length a & b we will use the man crossing conditions.
Please don't apply anything from your side only use the conditions which are given in question.
Sujay said:
8 years ago
---23------>----4-----|27
23<--6sec---|<---17sec----- Man
Train which takes 27 sec to cross a man takes 4 more seconds to cross man(27-23)
Train which takes 17 sec to cross a man takes 6 more seconds to cross train(23-17)
Let distance between crossover of train and man standing to be X
Speed ratio=x/4:x/6.
i,e 3:2.
23<--6sec---|<---17sec----- Man
Train which takes 27 sec to cross a man takes 4 more seconds to cross man(27-23)
Train which takes 17 sec to cross a man takes 6 more seconds to cross train(23-17)
Let distance between crossover of train and man standing to be X
Speed ratio=x/4:x/6.
i,e 3:2.
Sonu Shah said:
2 years ago
1) Train A speed = x
2) Train B speed = y.
Train A time = 27
Train B time = 17.
Time to cross each other trains T= 23.
So,
Train Length of A= 27x
Train Length of B= 17y.
27x + 17y/x+y = 23.
27x + 17y = 23(x+y).
27x + 17y = 23x+23y.
27x - 23x = 23y-17y.
4x = 6y,
x/y = 6/4,
So, x/y = 3/2.
So, the final answer = 3:2.
2) Train B speed = y.
Train A time = 27
Train B time = 17.
Time to cross each other trains T= 23.
So,
Train Length of A= 27x
Train Length of B= 17y.
27x + 17y/x+y = 23.
27x + 17y = 23(x+y).
27x + 17y = 23x+23y.
27x - 23x = 23y-17y.
4x = 6y,
x/y = 6/4,
So, x/y = 3/2.
So, the final answer = 3:2.
(58)
Shoeb said:
1 decade ago
If 2 Train are moving in opposite direction(train a 27 & train b 17)then Formula applied is
The time taken by the trains to cross each other( 23 )given =(a + b)/(u + v)sec.
* 27u + 17v = 23
-----------
u + v
* 27u + 17v = 23u + 23v
* 27u - 23u = 23v - 17v
* 4u = 6v
* u/v = 6/4
* 3:2 is the answer.
The time taken by the trains to cross each other( 23 )given =(a + b)/(u + v)sec.
* 27u + 17v = 23
-----------
u + v
* 27u + 17v = 23u + 23v
* 27u - 23u = 23v - 17v
* 4u = 6v
* u/v = 6/4
* 3:2 is the answer.
Mainul bangladesh khilkhet said:
8 years ago
A train length= 27x.
B train length= 17y.
X meter goes for 1 second.
1 ---> 1/x second.
27x ---> 27x/x second.
Again same as; 17y/y second for 17y.
So, {(27x)/x + (17y)/y} second = 23 second.
=> (27x+17y) meter = (23x + 23y) meter.
=> 4x meter = 6y meter.
=> So, ratio of meter x/y = 3/2.
B train length= 17y.
X meter goes for 1 second.
1 ---> 1/x second.
27x ---> 27x/x second.
Again same as; 17y/y second for 17y.
So, {(27x)/x + (17y)/y} second = 23 second.
=> (27x+17y) meter = (23x + 23y) meter.
=> 4x meter = 6y meter.
=> So, ratio of meter x/y = 3/2.
Harishankaran said:
7 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.
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