Aptitude - Problems on Trains - Discussion
Discussion Forum : Problems on Trains - General Questions (Q.No. 3)
3.
The length of the bridge, which a train 130 metres long and travelling at 45 km/hr can cross in 30 seconds, is:
Answer: Option
Explanation:
Speed = | ![]() |
45 x | 5 | ![]() |
= | ![]() |
25 | ![]() |
18 | 2 |
Time = 30 sec.
Let the length of bridge be x metres.
Then, | 130 + x | = | 25 |
30 | 2 |
2(130 + x) = 750
x = 245 m.
Video Explanation: https://youtu.be/M_d8WufJWKc
Discussion:
245 comments Page 23 of 25.
Vichithra said:
8 years ago
@Abd.
I have solution to ur question,
Speed = 45 x (5/18) m/sec = (25/2) m/sec.
Time = 30 sec.
distance = speed x time.
=(25/2)*30.
=750/2.
Multiplication 25 with 30, is 750.
I have solution to ur question,
Speed = 45 x (5/18) m/sec = (25/2) m/sec.
Time = 30 sec.
distance = speed x time.
=(25/2)*30.
=750/2.
Multiplication 25 with 30, is 750.
Thameem ansari said:
8 years ago
45*5/18=75/6*30=375-130=245m.
Pobitro Ghosh said:
8 years ago
Hi,
I solved it differently.
Like,
Total speed= 45*5/18=25/2*30=375,
Actual speed of the train= 375-130=245.
Ans: 245.
I solved it differently.
Like,
Total speed= 45*5/18=25/2*30=375,
Actual speed of the train= 375-130=245.
Ans: 245.
DURAI said:
8 years ago
@Rahul.
Your explanation is clear my doubt thank you.
Your explanation is clear my doubt thank you.
Syam said:
7 years ago
Length of the train l1 = 130m;
Speed s = 45kmph;
time t = 30 sec ;
length of the bridge l2 = ?
Here total distance D= (length of the train + length of the bridge)
so d=l1+l2;
we know the formula of distance (d)=speed(s)*time(t);
but the speed in kmph we have to convert into mph so speed (s) = 45*(5/18)=25/2;
then d=s*t;
then substitute all the values in it.
(l1+l2)=(25/2)*30;
we know the l1 value l1=130m;
(130+l2)=(25*15)
(130+l2)=375
then l2 =375-130
length of the bridge is l2=245;
Speed s = 45kmph;
time t = 30 sec ;
length of the bridge l2 = ?
Here total distance D= (length of the train + length of the bridge)
so d=l1+l2;
we know the formula of distance (d)=speed(s)*time(t);
but the speed in kmph we have to convert into mph so speed (s) = 45*(5/18)=25/2;
then d=s*t;
then substitute all the values in it.
(l1+l2)=(25/2)*30;
we know the l1 value l1=130m;
(130+l2)=(25*15)
(130+l2)=375
then l2 =375-130
length of the bridge is l2=245;
Kuma said:
7 years ago
Speed=Distance/Time.
45*5/18=Distance/30,
375= Distance,
Length= 375-130=245.
45*5/18=Distance/30,
375= Distance,
Length= 375-130=245.
Ravi said:
7 years ago
How do you take 5/18?
Harichandan Sabat said:
7 years ago
1 km --- 1000 metre & 1 hour --- 3600sec whenever you divide 1km/1 hour then 1000 metre/3600 sec equals to 5/18 m/s.
Sarang said:
7 years ago
Distance = speed*time,
D= 45* 5/18*30,
D= 5*5/2* 30,
D= 5 * 5 * 15 = 375,
D= 375- 130= 245.
D= 45* 5/18*30,
D= 5*5/2* 30,
D= 5 * 5 * 15 = 375,
D= 375- 130= 245.
Ani said:
7 years ago
@Sarang you are correct this is the easiest way to solve this problem.
As
Distance= time * speed.
D=45*5/18=25/2(speed)*30(time),
Actual distance=25 * 15=375.
Then bridge distance=Actual distance (375)- train Distance (130)=245.
As
Distance= time * speed.
D=45*5/18=25/2(speed)*30(time),
Actual distance=25 * 15=375.
Then bridge distance=Actual distance (375)- train Distance (130)=245.
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