A train overtakes two persons who are walking in the same direction in

Train speed = v m/s
Train length = L m

Person #1 (2km/h)
After 9 secs, train travels 9v meter and person #1 travels (2*5/18*9)=5 meter. --> 9v= 5+ L ... Eq.(1)

Person #2 (4km/h)
After 10 secs, train travels 10v meter and person #2 travels (4*5/18*10)=100/9 meter. --> 10v= 100/9 + L ... Eq.(2)

(2)-(1) --> v=55/9 m/s

(1) --> 9v - 5 = L = 9(55/9) - 5 = 50m

FINAL ANSWER IS (B)

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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 seconds and 10 seconds respectively. The length of the train (in metres) is

1st: \(2kmph=(\frac{5}{9})\) meter per second
2nd:\(4kmph=(\frac{10}{9})\) meter per second
--> Let \(x\) and \(y\) be the speed and length of the train: \(y =\)????

if the train passes \(1st\) person in \(9\) seconds:
--> \(x*9= y+ \frac{5}{9}*9\)
\(9x= y+ 5\)

if the train passes \(2nd\) person in \(10\) seconds:
--> \(x*10= y+ (\frac{10}{9})*10\)
\(10x= y+ \frac{100}{9}\)
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--> \(10x -9x = y+ \frac{100}{9} -y -5\)
\(x= \frac{55}{9}\)
--> \(y = 55-5 =50\)

Answer (B)

let t be speed of train and x be the length of the train
We know that speed = distance/time, hence distance = speed * time
Then the length of the train is (t-2)*9/3600 with respect to the person traveling at 2kmph and (t-4)*10/3600 with respect to the person traveling at 4kmph.
But the length is the same,
hence (t-2)*9/3600 = (t-4)*10/3600
9t-18=10t-40
t=22kmph

x = (22-2)*9/3600 = 0.050km = 50m

B is the answer.

let the length of train be x
speed of train be s
two persons who are walking in the same direction in which the train is going, at the rate of 2 kmph and 4 kmph which is 2*5/18 = 5/9m/sec and 4*5/18 = 10/9m/sec
x = 9(s-5/9)
x= 10(s-10/9)
from above we get
s=55/9
then x becomes 50m

OA:B

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 seconds and 10 seconds respectively. The length of the train (in metres) is

A. 45
B. 50
C. 54
D. 62
E. 72
Length of train = L
Speed of train = S
Speed of person 1 = 2 kmph = \(\frac{5}{9}\) m/s
Speed of person 2 = 4 kmph = \(\frac{10}{9}\) m/s

Now,
\(\frac{L}{S-5/9}\) = 9 and
\(\frac{L}{S-10/9}\) = 10

Two Equation and two variables. Solving we have
S = \(\frac{55}{9}\) and L = 50

Answer B.
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Let the length of the train = \(L\) kms & speed = \(v\) kmph

train overtakes them in 9 & 10 seconds respectively
--> \(\frac{9}{3600} = \frac{L}{v - 2}\) --> \(L = \frac{9(v - 2)}{3600}\) ....... (1)
& \(\frac{10}{3600} = \frac{L}{v - 4}\) --> \(L = \frac{10(v - 4)}{3600}\) ....... (2)

From (1) & (2),
\(\frac{9(v - 2)}{3600} = \frac{10(v - 4)}{3600}\)
--> \(9v - 18 = 10v - 40\)
--> \(10v - 9v = 40 - 18\)
--> \(v = 22 kmph\)

From (2),
\(L = \frac{10(v - 4)}{3600} = \frac{10*18}{3600} = \frac{1}{20} kms = \frac{1}{20}*1000 meters = 50 meters\)

Option B

same direction: subtract rates

d/(t-2)=9, d=9t-18
d/(t-4)=10, d=10t-40
9t-18=10t-40, t=22kmh
d/(t-2)=9 secs
d/(t-2)=9/3600 hrs
d=1/400*(22-2)=20/400=1/20km
d=1/400*(22-2)=20/400=50m

Ans (B)

First, let’s convert the rates to meters per hour (m/h): 2 kmph = 2000 m/h and 4 kmph = 4000 m/h. Now, let L = the length of the train, in meters, and r = the speed of the train, in m/h. We can create the equations:

2000 x 9/3600 + L = r x 9/3600

5 + L = r x 9/3600 → Eq. 1

and

4000 x 10/3600 + L = r x 10/3600

100/9 + L = r x 10/3600 → Eq. 2

Subtracting Eq. 1 from Eq. 2, we have:

55/9 = r x 1/3600

55/9 x 3600 = r

22,000 = r

Substituting 22,000 for r in Eq. 1, we have:

5 + L = 22,000 x 9/3600

5 + L = 55

L = 50

Answer: B
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Hi,

9v= 5+ L . Could you please tell me how did you equate this?

Regards
Kunal

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