Arithmetic aptitude :: Problems on Trains :: Data Sufficiency 1

• Problems on Trains - General Questions
• Problems on Trains - Data Sufficiency 1
• Problems on Trains - Data Sufficiency 2
• Problems on Trains - Data Sufficiency 3
• Problems on Trains - Important Formulas and Facts

1. $a$ km/hr $=\left(a\times \frac{5}{18}\right)$ m/s.

2. $a$ m/s $=\left(a\times \frac{18}{5}\right)$ km/hr.

3. Time taken by a train $A$ of length $l$ metres to pass a telegraph post or standing man is equal to the time taken by the train to cover $l$ metres.

4. Time taken by a train $P$ of length $l$ metres to pass a signal post of length $b$ metres is the time taken by the train to cover $(l + b)$ metres.

5. Suppose two trains, $A$ and $B$ are moving in the same direction at $u$ m/s and $v$ m/s, where $u>v$, then their relative speed is $=(u-v)$ m/s.

6. Suppose two trains $P$ and $Q$ are moving in opposite directions at $u$ m/s and $v$ m/s, then their relative speed is $=(u+v)$ m/s.

7. If two trains of length $x$ metres and $y$ metres are moving in opposite directions at $u$ m/s and $v$ m/s, then:

The time taken by the trains to cross each other $=\frac{(x+y)}{(u+v)}$ sec.

8. If two trains of length $x$ metres and $y$ metres are moving in the same direction at $u$ m/s and $v$ m/s, then:

The time taken by the faster train to cross the slower train $=\frac{(x+y)}{u-v}$ sec.

9. If two bodies start at the same time from points $P$ and $Q$ towards each other and after crossing they take $x$ and $y$ sec in

reaching $Q$ and $P$ respectively, then:

($P$'s speed):($B$'s speed) $=\left(\sqrt{y}:\sqrt{x}\right)$

Find the speed of the train whose length is $210$ metres ?

I. The train crosses another train of $300$ meters length running in opposite direction in $10$ seconds.

II. The train crosses another train running in the same direction at the speed of $60$ km/hr in $30$ seconds.

Answer:

Explanation:

Time taken to cross the train, running in opposite directions $=\frac{\left(l_{1}+l_2\right)}{(u+v)}$ sec.

$\Rightarrow 10$ $=\frac{(210+300)}{(u+v)}$ $\Rightarrow u+v$ $=51$.

Time taken to cross the train, running in same direction $=\frac{\left(l_{1}+l_{2}\right)}{(u-v)}$ sec.

$\Rightarrow 30$ $=\frac{(210+300)}{\left(u-60\times\frac{5}{18}\right)}$ $\Rightarrow u$ $=\left(17+\frac{50}{3}\right)$ m/sec.

Thus, $u$ and $v$ can be obtained.

$\therefore$ Correct answer is (E).

What is the speed of the train?

I. $280$ metres long train crosses a signal pole in $18$ seconds

II. $280$ metres long train crosses a platform in $45$ seconds.

Answer:

Explanation:

Speed = Length of the train / Time taken to cross the pole $=\frac{280}{18}$ m/s $=\frac{140}{9}$ m/s.

$\therefore$ I alone gives the answer.

Time taken to cross the platform = (Length of train + Length of platform) / Speed of the train

$\Rightarrow$ Speed $=\frac{(280+p)}{45}$ m/s.

But, p = length of platform, is not given.

$\therefore$ II is not sufficient to give the answer.

$\therefore$ The correct answer is (A).

A trian crosses a pole in $10$ seconds. What is the length of the train ?

I. The train crosses another train running in opposite direction with a speed of $80$ km/hr in $22$ seconds.

II. The speed of the train is $108$ km/hr

Answer:

Explanation:

Time taken to cross a pole $=\frac{Length\quad of\quad train}{Speed\quad of\quad train}$ $\Rightarrow 10$ $=\frac{Length\quad of\quad train}{\left(108\times\frac{5}{18}\right)}$

$\Rightarrow$ Length of the train $=300$ m.

Clearly, II is sufficient to get the answer.

Also, I is not sufficient to get the answer.

$\therefore$ The correct answer is (B).

A train crosses a signal post in $x$ seconds. What is the length of the train ?

I. The train crosses a platform of 100 metres in $y$ seconds.

II. The train is running at the speed of $80$ km/hr.

Answer:

Explanation:

Let the length of the train be $a$ metres.

Time taken to cross a signal post = Length of the train $\div$ Speed of the train $\Rightarrow x=\frac{l}{Speed}$. ----(1)

Time taken to cross the platform $=\frac{(l+100)}{Speed}$ $\Rightarrow y=\frac{l+100}{Speed}$ ----(2)

Thus, from (1) and (2), we can find $l$.

Also II gives, speed $=\left(80\times\frac{5}{18}\right)$ m/s $=\frac{200}{9}$ m/s.

Thus, the data in I or II are sufficient to answer the question.

$\therefore$ The correct answer is (C).

• Problems on Trains - General Questions
• Problems on Trains - Data Sufficiency 1
• Problems on Trains - Data Sufficiency 2
• Problems on Trains - Data Sufficiency 3