Model 5 Train Vs Both platform and a man/a pole Section-Wise Topic Notes With Detailed Explanation And Example Questions

MOST IMPORTANT quantitative aptitude - 6 EXERCISES

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The following question based on trains topic of quantitative aptitude

Questions : The time for a train of length 110 metre running at the speed of 72 km/hr to cross a bridge of length 132 metre is

(a) 12.1 seconds

(b) 12.42 seconds

(c) 14.3 seconds

(d) 9.8 seconds

The correct answers to the above question in:

Answer: (a)

Speed of train = 72 kmph.

= $({72 × 5}/18)$ m./sec. = 20 m./sec.

Required time

= $\text"Length of train and bridge"/ \text"Speed of train"$

= $({110 +132}/20)$ seconds

= $(242/20)$ seconds = 12.1 seconds

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Read more crossing both platform and man pole Based Quantitative Aptitude Questions and Answers

Question : 1

A train crosses a pole in 15 seconds and a platform 100 metres long in 25 seconds. Its length (in metres) is

a) 100

b) 150

c) 200

d) 50

Answer: (b)

Rule 10 and Rule 1,

Let the length of train be x metre.

$x/15 = {x + 100}/25$

$x/3 = {x + 100}/5$

5x = 3x + 300

2x = 300

$x = 300/2$ = 150 metres

Question : 2

A train passes a 50 metres long platform in 14 seconds and a man standing on the platform in 10 seconds.The speed of the train is :

a) 36km/hr

b) 40 km/hr

c) 45 km/hr

d) 24 km/hr

Answer: (c)

Using Rule 10,
If a train of length x m crosses a platform/tunnel/bridge of length y m with the speed u m/s in t seconds, then,
t = ${x + y}/u$

Using Rule 1,
If a train crosses an electric pole, a sitting/standing man, km or mile stone etc. then distance = Length of train. Then,
Length of train = Speed × Time
And Time = $\text"Length of train"/\text"Speed"$ and
Speed = $\text"Length of train"/\text"Time"$

Let the length of train be x metres

According to question

Speed of the train = $x/10$ m/ sec.

Also, the speed of the train

= $({x + 50}/14)$ m/sec.

[Since, It passes the platform in 14 seconds]

Both the speeds should be equal,

i.e., $x/10 = {x + 50}/14$ = or 14x = 10x + 500

or 14x - 10x = 500

or 4x = 500

x = 125 metres

Hence, Speed = $125/10 = 12.5$ m/sec.

= ${12.5 × 18}/5$ km/hr. = 45 km/hr.

Question : 3

A moving train crosses a man standing on a platform and a bridge 300 metres long in 10 seconds and 25 seconds respectively. What will be the time taken by the train to cross a platform 200 metres long ?

a) 18 seconds

b) 20 seconds

c) 22 seconds

d) 16$2/3$ seconds

Answer: (b)

Rule 10 and Rule 1,

Let the length of the train be x metre

Speed of train when it crosses man = $x/10$

Speed of train when it crosses platform = ${x + 300}/25$

According to the question,

Speed of train

= $x/10 = {x + 300}/25$

25x = 10x + 3000

15x = 3000

$x = 3000/15$ = 200 metres

Length of train = 200 metre

Speed of train = $x/10 = 200/10$

= 20 m/sec

Time taken in crossing a 200 m long platform

= ${200 + 200}/20$ = 20 seconds

Question : 4

A train passes a man standing on a platform in 8 seconds and also crosses the platform which is 264 metres long in 20 seconds. The length of the train (in metres) is :

a) 176

b) 175

c) 96

d) 188

Answer: (a)

Rule 10 and Rule 1,

Let length of train be x m

Speed of train = ${x + 264}/20$

Also, speed of train = $x/8$

Obviously, $x/8 = {x + 264}/20$

$x/2 = {x + 264}/5$

5x = 2x + 528

5x - 2x = 528

x = 528 ÷ 3 = 176 m

Question : 5

A train passes a platform 110 m long in 40 seconds and a boy standing on the platform in 30 seconds . The length of the train is

a) 110 m

b) 220 m

c) 330 m

d) 100 m

Answer: (c)

Rule 10 and Rule 1,

Let the length of the train be x metres.

Speed of train in crossing boy = $x/30$

Speed of train in crossing platform = ${x +110}/40$

According to the question,

${x +110}/40 =x/30$

${x +110}/4 = x/3$

4x = 3x + 330

x = 330 metres

Question : 6

Two trains 100 metres and 95 metres long respectively pass each other in 27 seconds when they run in the same direction and in 9 seconds when they run in opposite directions. Speed of the two trains are

a) 52 km/hr, 26 km/hr

b) 36 km/hr. 18 km/hr

c) 40 km/hr, 20 km/hr

d) 44 km/hr, 22 km/hr

Answer: (a)

Let the speed of trains be x

and y metre/sec respectively,

${100 + 95}/{x - y} = 27$

$x - y = 195/27 = 65/9$ ...(i)

Again,

$195/{x + y} = 9$

$x + y = 195/9$ ...(ii)

By equation (i) + (ii)

$2x = 65/9 + 195/9 = 260/9$

$x = 260/{2 × 9} = 130/9$ m/sec.

= $(130/9 × 18/5)$ kmph = 52 kmph

From equation (ii),

$y = 195/9 - 130/9 = 65/9$ m/sec.

= $65/9 × 18/5$ = 26 kmph

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