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
(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
Discuss Form
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 »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 »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 × TimeAnd Time = $\text"Length of train"/\text"Speed"$ andSpeed = $\text"Length of train"/\text"Time"$
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 »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 »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 »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 »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
trains Shortcuts and Techniques with Examples
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Model 1 Train Vs Train in same direction
Defination & Shortcuts … -
Model 2 Train Vs Train in opposite direction
Defination & Shortcuts … -
Model 3 Train Vs Bridge/Platform
Defination & Shortcuts … -
Model 4 Train Vs Pole/Signal Post/Man
Defination & Shortcuts … -
Model 5 Train Vs Both platform and a man/a pole
Defination & Shortcuts … -
Model 6 Change in speed Vs Change with time travel
Defination & Shortcuts …
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