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:
237 comments Page 10 of 24.
Sarat said:
9 years ago
TIME = DISTANCE /SPEED.
We don't know how much distance apart
27x is train1 distance at which it crosses man
17y is 2nd train.
We know when they meet 23 sec.
23 = 27x + 17y/ x + y.
We don't know how much distance apart
27x is train1 distance at which it crosses man
17y is 2nd train.
We know when they meet 23 sec.
23 = 27x + 17y/ x + y.
Vadivel said:
1 decade ago
Why should both trains have different length. Even sometimes they can have same length and can run in different speeds right. Then how can we do like that. Can anyone explain please?
DURAI said:
5 years ago
Short method is;
x person 27.
y person 17 add both distances and divide the total distance and multiple the same for both persons.
27+17/23 *2.
37/23 = 1.6 *2.
1.6*2.
Ans : 3.2.
x person 27.
y person 17 add both distances and divide the total distance and multiple the same for both persons.
27+17/23 *2.
37/23 = 1.6 *2.
1.6*2.
Ans : 3.2.
(2)
Sourabh kumar said:
2 years ago
Let 1st train be x = 27.
2nd train be y = 17.
Two trains crossed each other=23(x+y).
27x + 17y = 23x + 23y.
How do you change the sign of x and y?
Anyone please explain to me.
2nd train be y = 17.
Two trains crossed each other=23(x+y).
27x + 17y = 23x + 23y.
How do you change the sign of x and y?
Anyone please explain to me.
(30)
Vipul said:
1 decade ago
First we are doing KM Hr to MS.
Then Formula is a*5/18.
Then,
45*5/18=75/6 MS.
Distance = Speed*Time.
= 75/6*30.
= 375.
Distance = 375-130.
= 245.
Then Formula is a*5/18.
Then,
45*5/18=75/6 MS.
Distance = Speed*Time.
= 75/6*30.
= 375.
Distance = 375-130.
= 245.
Gokulakisan S said:
1 decade ago
You can use this simple method:
Time to cross men = tm.
Time to cross each other = te.
So 27(tm)+23(te) = 60.
17(tm)+23(te) = 40.
Then 60/40 = 3/2.
Therefore ratio is 3:2.
Time to cross men = tm.
Time to cross each other = te.
So 27(tm)+23(te) = 60.
17(tm)+23(te) = 40.
Then 60/40 = 3/2.
Therefore ratio is 3:2.
NICK said:
1 decade ago
Can any one tell me how to calculate mod(23-17) : mod(23-27)
I forgot how to calculate mod and if we are calculating same mod(23-17) : mod(23-27) then how it comes to 6:4????
I forgot how to calculate mod and if we are calculating same mod(23-17) : mod(23-27) then how it comes to 6:4????
Adesh Katiya said:
2 years ago
u=27s
v=17s
Cross = 23s.
l1 = 27x.
l2 = 17y.
23 = l1 + l2/x + y
23 = 27x + 17y/x + y.
23x + 23y = 27x + 17y.
4x - 6y = 0.
4x = 6y.
x/y = 6/4.
x/y = 3/2.
Hence, option (b).
v=17s
Cross = 23s.
l1 = 27x.
l2 = 17y.
23 = l1 + l2/x + y
23 = 27x + 17y/x + y.
23x + 23y = 27x + 17y.
4x - 6y = 0.
4x = 6y.
x/y = 6/4.
x/y = 3/2.
Hence, option (b).
(14)
Carlo said:
1 decade ago
Let Train A distance = 27x Train A speed = x.
Train B distance = 17y Train B speed = y.
Time = distance/speed.
23 = 27x + 17y / x + y.
x/y = 3/2 ----> Answer.
Train B distance = 17y Train B speed = y.
Time = distance/speed.
23 = 27x + 17y / x + y.
x/y = 3/2 ----> Answer.
Prabhakaran said:
2 decades ago
Im not clear with above step can anyone give more explanation about this problem. I have doubt in (27x+17y) / (x+y) =23 what this statement mean by. Anyone help me please.
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