Aptitude - Problems on Trains - Discussion
Discussion Forum : Problems on Trains - General Questions (Q.No. 27)
27.
A train overtakes two persons who are walking in the same direction in which the train is going, at the rate of 2 kmph and 4 kmph and passes them completely in 9 and 10 seconds respectively. The length of the train is:
Answer: Option
Explanation:
| 2 kmph = |  |
2 x | 5 |  |
m/sec = | 5 | m/sec. |
| 18 | 9 |
| 4 kmph = | |
4 x | 5 | |
m/sec = | 10 | m/sec. |
| 18 | 9 |
Let the length of the train be x metres and its speed by y m/sec.
| Then, |  |
x |  |
= 9 and | |
x | |
= 10. |
|
|
9y - 5 = x and 10(9y - 10) = 9x
9y - x = 5 and 90y - 9x = 100.
On solving, we get: x = 50.
Length of the train is 50 m.
Discussion:
106 comments Page 2 of 11.
Mihir gandhi said:
5 years ago
Assume speed for s and find length means (distance)
D1=D2
(S-2)*T1=(s-4)*T2.
(S-2)*9/3600=(s-4)*10/3600.
(S-2)*9=(s-4)*10.
9s-18=10s-40.
10s-9s=40-18.
S=22.
So find length
L=D=(s-2)*T1.
=(22-2)*9/3600,
=20*9/3600,
=2*9/360,
=18/360,
=1/20,
=0.05km.
=50 meter.
D1=D2
(S-2)*T1=(s-4)*T2.
(S-2)*9/3600=(s-4)*10/3600.
(S-2)*9=(s-4)*10.
9s-18=10s-40.
10s-9s=40-18.
S=22.
So find length
L=D=(s-2)*T1.
=(22-2)*9/3600,
=20*9/3600,
=2*9/360,
=18/360,
=1/20,
=0.05km.
=50 meter.
Nisma said:
5 years ago
There is a simple formula to do this type of questions.
If a train covers 2 persons with speed S1 and S2 in time T1 and T2, then the length of train is:
(S1-S2 ÷ T1-T2) * T1 * T2.
So applying it in this question;
{(4-2)5/18÷10-9} * 10 * 9 =50m.
Note:(4-2) is multiplied by 5/18 so as to convert km to m.
If a train covers 2 persons with speed S1 and S2 in time T1 and T2, then the length of train is:
(S1-S2 ÷ T1-T2) * T1 * T2.
So applying it in this question;
{(4-2)5/18÷10-9} * 10 * 9 =50m.
Note:(4-2) is multiplied by 5/18 so as to convert km to m.
Sathees said:
9 years ago
Equ 1 is => x = (9y - 5) /9 * 9 = 9 y - 5 => 5 = 9y - 5.
Equ 2 is => x = (9y - 10) /9 *10.
9 x = (9y - 10) * 10.
9x = (90y - 100) => 100 = 90y - 9 x.
90y - 9 x = 100 (person - 2 is walk fast).
90y - 10 x= 50 (equ 2 is multiplied by 10).
------------------
x = 50.
------------------
Equ 2 is => x = (9y - 10) /9 *10.
9 x = (9y - 10) * 10.
9x = (90y - 100) => 100 = 90y - 9 x.
90y - 9 x = 100 (person - 2 is walk fast).
90y - 10 x= 50 (equ 2 is multiplied by 10).
------------------
x = 50.
------------------
Swapnil jagadale said:
1 decade ago
Take Length of Train = X meter.
And Speed of train = Y m/s.
2kmph = 5/9m/s, 4kmph = 10/9 m/s.
For first train X/(Y-5/9) = 9.
That is X-9Y = -5.....eq1.
For Second train X/(Y-10/9) = 10.
That is 9X-90Y = -100.....eq2.
Solving eq1 and eq2.
X = 50 meter.
And speed Y = 22 kmph or 55/9 m/s.
And Speed of train = Y m/s.
2kmph = 5/9m/s, 4kmph = 10/9 m/s.
For first train X/(Y-5/9) = 9.
That is X-9Y = -5.....eq1.
For Second train X/(Y-10/9) = 10.
That is 9X-90Y = -100.....eq2.
Solving eq1 and eq2.
X = 50 meter.
And speed Y = 22 kmph or 55/9 m/s.
Sumaira said:
9 years ago
@Kapil.
Thanks for the trick I tried to simplify more.
(S - 2) * 5/18 = L/9..........(1)
Similarly
(S - 4) * 5/18 = L/10..........(2)
9(s - 2) = 10(s - 4)
9s - 18 = 10s - 40
S = 40-18
S = 22
Put value of s in any one equal , suppose in (1)
(22 - 2) * 5/18 = L/9.
L = 50.
Thanks for the trick I tried to simplify more.
(S - 2) * 5/18 = L/9..........(1)
Similarly
(S - 4) * 5/18 = L/10..........(2)
9(s - 2) = 10(s - 4)
9s - 18 = 10s - 40
S = 40-18
S = 22
Put value of s in any one equal , suppose in (1)
(22 - 2) * 5/18 = L/9.
L = 50.
Ranu said:
1 decade ago
See
9=(x/y-5/9)
when we solve dis..the LCM comes as 9 which get cancelled by 9 on the other side of the equation.
x*9/9y-5=9
9x=9(9y-5)
x=9y-5
But in case of 10=x/y-10/9, the LCM 9 is not cancelled with 10 on the other side & thus
x*9/9y-10=10
9x=10(9y-10)
9x=90y-100
9=(x/y-5/9)
when we solve dis..the LCM comes as 9 which get cancelled by 9 on the other side of the equation.
x*9/9y-5=9
9x=9(9y-5)
x=9y-5
But in case of 10=x/y-10/9, the LCM 9 is not cancelled with 10 on the other side & thus
x*9/9y-10=10
9x=10(9y-10)
9x=90y-100
Jagan Reddy said:
3 years ago
S = D/T
S-2 = D/9
9(S-2) = D ------> (1)
S-4 = D/10
10(S-4) = D ------> (2)
Distances are equal as the same cross both, equating 1,2.
10S - 40 = 9S - 18.
1S = 22(Speed of train).
Taking any one person's reference;
22-2 = D/9.
20 * 9 = D,
D = 180 * 5/18,
D = 50mtrs.
S-2 = D/9
9(S-2) = D ------> (1)
S-4 = D/10
10(S-4) = D ------> (2)
Distances are equal as the same cross both, equating 1,2.
10S - 40 = 9S - 18.
1S = 22(Speed of train).
Taking any one person's reference;
22-2 = D/9.
20 * 9 = D,
D = 180 * 5/18,
D = 50mtrs.
(16)
Karthik said:
5 years ago
Simplify to the easy method :
Distance we don't know:
s = s.
(s-2)*9/60*60=(s-4)*10/60*60---> Equation 1.
9s-18=10 - 40.
s=22--->Equation 2
Substitute equation 2 into 1.
Ans:180
Because we convert into meter & multiply by the 5/18 then we got.
180*5/18 = 50m.
Distance we don't know:
s = s.
(s-2)*9/60*60=(s-4)*10/60*60---> Equation 1.
9s-18=10 - 40.
s=22--->Equation 2
Substitute equation 2 into 1.
Ans:180
Because we convert into meter & multiply by the 5/18 then we got.
180*5/18 = 50m.
(1)
K.Anitha said:
1 decade ago
2 kmph = 2*(5/18) = 5/9 (m/s).
4 kmph = 4*(5/18) = 10/9 (m/s).
Let the speed of the train be x.
(x-5/9)9 = (x-10/9)10 (because the distance is same).
On solving we get.
x = 55/9 (m/s).
Train length = (x-5/9)9.
= (55/9-5/9)9.
= (50/9)*9.
= 50 m.
4 kmph = 4*(5/18) = 10/9 (m/s).
Let the speed of the train be x.
(x-5/9)9 = (x-10/9)10 (because the distance is same).
On solving we get.
x = 55/9 (m/s).
Train length = (x-5/9)9.
= (55/9-5/9)9.
= (50/9)*9.
= 50 m.
Farooqui mehtab said:
8 years ago
let x be the speed of train;
(x-2) *(5/18) *9 = (x-4) *(5/18) *10
9x-18 = 10x-40
x = 22 m/sec.
Now, substitute this in,
Length of train = (x-2) *(5/18) *9,
= (22-2) *(5/18) *9 = 20*5/2,
= 50 mins.
(x-2) *(5/18) *9 = (x-4) *(5/18) *10
9x-18 = 10x-40
x = 22 m/sec.
Now, substitute this in,
Length of train = (x-2) *(5/18) *9,
= (22-2) *(5/18) *9 = 20*5/2,
= 50 mins.
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