Problems on Trains Aptitude Questions

Answer

A

Solution

\(Speed=\left ( 144*\frac{5}{18} \right )m/sec=40m/sec\)

time taken=(100/40)sec=2.5 sec

Answer

A

Solution

\(speed=\left ( 45*\frac{5}{18} \right )m/sec=\frac{25}{2}m/sec\)

Total distance covered = (360+140)m = 500 m

\(Required\, time=\left ( 500*\frac{2}{25} \right )sec=40\, sec\)

Answer

E

Solution

\(Speed=\left ( 60*\frac{5}{18} \right )m/sec=\left ( \frac{50}{3} \right )m/sec\)

Length of the train = (Speed * Time) \(=\left ( \frac{50}{3}*9 \right )m=150\, m\)

Answer

C

Solution

\(Speed=\left ( 45*\frac{5}{18} \right )m/sec=\left ( \frac{25}{2} \right )m/sec;\) Time = 30 sec

Let the length of bridge be x metres.

Then, \(\frac{130+x}{30}=\frac{25}{2}\)

2(130+x) = 750

x= 245 m

Answer

B

Solution

\(speed=\left ( \frac{300}{18} \right )m/sec=\frac{50}{3}m/sec\)

Let the length of the platform be x metres.

Then, \(\frac{x+300}{39}=\frac{50}{3}\)

3(x+300)=1950

x = 350

Answer

C

Solution

\(Speed=\left ( 78*\frac{5}{18} \right )m/sec=\left ( \frac{65}{3} \right )m/sec\)

Time = 1 minute = 60 sec

Let the length of the tunnel be x meters.

Then, \(\frac{800+x}{60}=\frac{65}{3}\)

3(800+x) = 3900

x = 500

Answer

C

Solution

Speed of train relative to jogger = (45-9) kmph = 36 kmph

\(=\left ( 36*\frac{5}{18} \right )m/sec=10\, m/sec\)

Distance to be covered = (240+120) m = 360 m

Time taken = (\(\frac{360}{10}\)) = 36 sec

Answer

C

Solution

Let the speed of the slower train be x m/sec

Then, speed of the fastre train = 2x m/sec

Relative speed = (x+2x) m/sec = 3x m/sec

\(\frac{(100+100)}{8}=3x\)

24x = 200

x = \(\frac{25}{3}\)

So, speed of the fastre train = \(\frac{50}{3}m/sec=\left ( \frac{50}{3}*\frac{18}{5} \right )kmph=60kmph\)

Answer

B

Solution

Let us name the train as A and B. Then

(A's speed):(B's speed) = \(\sqrt{b}:\sqrt{a}=\sqrt{16}:\sqrt{9}=4:3\)

Answer

B

Solution

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

Length of the train = 300 m

Clearly, II is sufficient to get the answer.

Also, I is not sufficient to get the answer.

Answer

D

Solution

Time taken to cross a signal pole = \(=\frac{length \, of\, the\, train}{speed \, of\, the\, train}\)

Time taken to cross a platform = \(=\frac{(length \, of\, the\, train+length \, of\, the\, platform)}{speed\, of\, the\, train}\)

Length of the train = 330 m

I and III give, \(18=\frac{330}{x}\Rightarrow x=\frac{330}{18}m/s=\frac{55}{3}m/s\)

II and III give, \(36=\frac{2*330}{x}\Rightarrow x=\frac{660}{36}m/s=\frac{55}{3}m/s\)

Answer

D

Solution

III. Gives, speed = \(\frac{200}{10}m/s=20m/s=\left ( 20*\frac{18}{5} \right )kmph=72\, kmph\)

II gives, time taken = \(\left ( \frac{558}{72} \right )hrs=\frac{31}{4}hrs=7\frac{3}{4}hrs=7\, hrs\, 45\, min\)

So, the train will reach city X at 3 p.m.

Hence, I is redundant.

Answer

A

Solution

Distance covered by the train in 5 hours = 42*5 = 210 km = 210000 m

Number of pillars counted by the man = \(\frac{210000}{60}+1\) = 3500+1 = 3501

Answer

C

Solution

Let the length of the train be x metres.

Time taken to cover x metres = 5 min = 5*60 = 300 sec

Speed of the train = \(\frac{x}{300}\) m/sec

Length of the second train = 2x metres

Length of the platform = 2x metres

Required time = \(\frac{2x+2x}{\frac{x}{300}}\) = \(\frac{4x*300}{x}\) = 1200 sec = \(\frac{1200}{60}\) min = 20 min

Answer

D

Solution

Let the lengths of the train and the bride be x and y m respectively

Speed of the first train = 90 km/h = 90*\(\frac{5}{18}\)  = 25 m/s

Speed of the second train = 45 km/h = 45*\(\frac{5}{18}\)  = \(\frac{25}{2}\) m/s

Then, \(\frac{x+y}{36}\) =25

x+y=900

Required time = \(\frac{x-100+y}{25/2}\) = \(\frac{x+y-100}{25/2}\)  

800*\(\frac{2}{25}\) = 64 sec 

Answer

B

Solution

At the time of meeting,let the distance travelled by the first train be x km

Then, distance traveled by the second train is (x+60) km

\(\frac{x}{16}\) = \(\frac{x+60}{21}\)

21x=16x+960

5x=960

x=192

Hence, distance between two stations = 192+192+60 = 444 km

Answer

A

Solution

Let the train meet after t hours

Speed of train A = 75 km/h

Speed of train B = 50 km/h

Distance covered by train A = 75*t = 75t

Distance covered by train B = 50*t = 50t

Distance = Speed*Time

According to question,

75t-50t = 175

25t=175

t=\(\frac{175}{25}\) = 7 hours

Distance between A and B = 75t+50t=125t

125*7=875 km

Answer

B

Solution

Ratio of speeds = \(\sqrt{4}\) : \(\sqrt{6\frac{1}{4}}\) : \(\sqrt{4}\) : \(\sqrt{\frac{25}{4}}\) = 2 : \(\frac{5}{2}\) = 4:5

Let the speed of the two trains be 4 x and 5 x km/h respectively.

Then, time taken by trains to meet each other = \(\frac{270}{4x+5x}\) = \(\frac{270}{9x}\) = \(\frac{30}{x}\)

Time taken by slower train to travel 270 km = \(\frac{270}{4x}\) hr

\(\frac{270}{4x}\)=\(\frac{30}{x}\)+6\(\frac{1}{4}\)

\(\frac{270}{4x}\)-\(\frac{30}{x}\)=\(\frac{25}{4}\)

\(\frac{150}{4x}\) = \(\frac{25}{4}\)

100x=600

x=6

Hence, speed of slower train = 4x=24 km/h

Answer

A

Solution

Suppose the two trawlers start from a point O and move inn the directions OA and OB respectively.

Let the speeds of the two sea trawlers be x km/hr and y km/hr respectively.

Then, [x*\(\frac{1}{2}]^{2}\)+ [y*\(\frac{1}{2}]^{2}\) = \(17^{2}\)

 \(\frac{x^{2}}{4}\)+ \(\frac{y^{2}}{4}\)=289

\(x^{2}+y^{2}\)=1156           ..i

And, x* \(\frac{3}{4}\)-y*\(\frac{3}{4}\)=10.5

x-y=10.5*\(\frac{4}{3}\)=14       ...ii

Now, [\(x+y]^{2}\)+[\(x-y]^{2}\)=2[\(x^{2}+y^{2}\)]

[\(x+y]^{2}\)=2*1156-\(14^{2}\) = 2312-196 = 2166

x+y=\(\sqrt{2116}\) = 46     ..iii

Adding ii and iii, we get: 2x=60 or x=30

Putting x=30 in ii we get, y= 16

Hence, the speeds of the two sea-trawlers are 30 km/hr and 16 km/hr

Answer

B

Solution

Ratio of speeds = \(\sqrt{4}\) : \(\sqrt{6\frac{1}{4}}\) : \(\sqrt{4}\) : \(\sqrt{\frac{25}{4}}\) = 2 : \(\frac{5}{2}\) = 4:5

Let the speed of the two trains be 4 x and 5 x km/h respectively.

Then, time taken by trains to meet each other = \(\frac{270}{4x+5x}\) = \(\frac{270}{9x}\) = \(\frac{30}{x}\)

Time taken by slower train to travel 270 km = \(\frac{270}{4x}\) hr

\(\frac{270}{4x}\)=\(\frac{30}{x}\)+6\(\frac{1}{4}\)

\(\frac{270}{4x}\)-\(\frac{30}{x}\)=\(\frac{25}{4}\)

\(\frac{150}{4x}\) = \(\frac{25}{4}\)

100x=600

x=6

Hence, speed of slower train = 4x=24 km/h