Question

Two persons are walking beside a railway track at respective speeds of 2 and 4 km per hour in the same direction. A train came from behind them and crossed them in 90 and 100 seconds, respectively. The time, in seconds, taken by the train to cross an electric post is nearest to

• A. 87
• B. 82
• C. 75
• D. 78

Answer 1

The Correct Option is B

The correct option is (B): \(82\) 

Let the length of the train be \(l\) and its speed be \(s\).

Given:

\[ \frac{l}{(s - 2) \times \frac{5}{18}} = 90, \quad \frac{l}{(s - 4) \times \frac{5}{18}} = 100 \]

Equating both expressions for \(l\):

\[ 90(s - 2) \times \frac{5}{18} = 100(s - 4) \times \frac{5}{18} \] \[ \Rightarrow 90(s - 2) = 100(s - 4) \] \[ \Rightarrow 90s - 180 = 100s - 400 \] \[ \Rightarrow 100s - 90s = 400 - 180 = 220 \] \[ \Rightarrow s = 22 \]

∴ Length of the train:

\[ l = 90(s - 2) \times \frac{5}{18} = 90 \times 20 \times \frac{5}{18} = 500 \text{ m} \]

Time to cross a lamp post:

\[ \frac{500}{22 \times \frac{5}{18}} = \frac{500 \times 18}{110} = 81.81 \approx \boxed{82 \text{ sec}} \]

Answer 2

Let the length of the train be \(l\) km and the speed be \(s\) km/h. 

According to the question:

\[\frac{l}{s - 2} = \frac{90}{3600} \quad \text{……… (i)}\]\[\frac{l}{s - 4} = \frac{100}{3600} \quad \text{……… (ii)}\]

Dividing equation (ii) by equation (i):

\[\frac{\frac{l}{s - 4}}{\frac{l}{s - 2}} = \frac{100}{90} \Rightarrow \frac{s - 2}{s - 4} = \frac{100}{90} \Rightarrow \frac{s - 2}{s - 4} = \frac{10}{9}\]

Cross-multiplying:

\[9(s - 2) = 10(s - 4) \Rightarrow 9s - 18 = 10s - 40 \Rightarrow s = 22 \text{ km/h}\]

Substitute \(s = 22\) in equation (i):

\[\frac{l}{22 - 2} = \frac{90}{3600} \Rightarrow \frac{l}{20} = \frac{90}{3600} \Rightarrow l = \frac{90}{3600} \times 20 = \frac{1}{2} \text{ km} = 500 \text{ m}\]

Now, time to cross a lamp post =

\[\frac{500}{22 \times \frac{5}{18}} = \frac{500 \times 18}{110} = 81.81 \approx 82 \text{ seconds}\]

So, the correct option is (B): \(82\) seconds.