Model 3 Train Vs Bridge/Platform Practice Questions Answers Test With Solutions & More Shortcuts
TRAINS PRACTICE TEST [6 - EXERCISES]
Model 1 Train Vs Train In Same Direction
Model 2 Train Vs Train In Opposite Direction
Model 3 Train Vs Bridge/Platform
Model 4 Train Vs Pole/Signal Post/Man
Model 5 Train Vs Both Platform And A Man/a Pole
Model 6 Change In Speed Vs Change With Time Travel
Question : 1 [SSC CGL Tier-I 2016]
A train, 500 metre long, running at a uniform speed, passes a station in 35 seconds. If the length of the platform is 221 metre, the speed of the train in km/hr is
a) 74.16
b) 24.76
c) 78.54
d) 72$1/35$
Answer »Answer: (a)
Speed of train
= $\text"Length of train and platform"/ \text"Time taken in crossing"$
= $({221 + 500}/35)$ metre/second
= $(721/35)$ metre/second
= $({721 × 18}/{35 × 5})$ kmph
= 74.16 kmph
Question : 2 [SSC CGL Tier-I 2016]
A moving train passes a platform 50 metre long in 14 seconds and a lamp post in 10 seconds. The speed of the train (in km/h) is :
a) 36
b) 40
c) 45
d) 24
Answer »Answer: (c)
Let the length of train be x metre.
When a train crosses a platform, distance covered by it
= length of train and platform.
Speed of train
= ${x + 50}/14 = x/10$
${x + 50}/7 = x/5$
7x = 5x + 250
7x - 5x = 250
2x = 250 ⇒ x = $250/2$ = 125 metre
Speed of train = $x/10$
= $(125/10)$ m./sec.
= $(125/10 × 18/5)$ kmph = 45 kmph.
Question : 3 [SSC CGL Tier-I 2014]
A train 50 metres long passes a platform of length 100 metres in 10 seconds. The speed of the train in metre/second is
a) 10
b) 15
c) 20
d) 50
Answer »Answer: (b)
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 = ${x + y}/u$
Speed of train
= $\text"Length of (train + platform)"/ \text"Time taken in crossing"$
= ${(50 + 100)}/10 = 150/10$ = 15 m/sec
Question : 4 [SSC CISF Constable 2011]
A train of length 500 feet crosses a platform of length 700 feet in 10 seconds. The speed of the train is
a) 85 ft/second
b) 100 ft/second
c) 120 ft/second
d) 70 ft/second
Answer »Answer: (c)
Speed of train
= $\text"Lengthof(train + platform)"/ \text"Timetaken tocross"$
= $({500 + 700}/10)$ feet/second
= 120 feet/second
Using Rule 10,
Here, x = 500 feet, y = 700 feet
t = 10 seconds, u = ?
using t = ${x + y}/u$
u = ${500 + 700}/10$
= 120ft/second
Question : 5 [SSC CGL Tier-I 2015]
If a man running at 15 kmph crosses a bridge in 5 minutes, the length of the bridge is
a) 500 metres
b) 750 metres
c) 1250 metres
d) 1000 metres
Answer »Answer: (c)
Using Rule 10,
Time = 5 minutes = $1/12$ hour
Length of bridge = Speed × Time
= 15 × $1/12 = 5/4$ km.
= $(5/4 × 1000)$ metre
= 1250 metre
IMPORTANT QUANTITATIVE APTITUDE EXERCISES
-
Top 399+ Trains Based Time and Distance MCQs For BANK SSC »
-
New 500+ Aptitude Problems on Train For All BANK SSC Exam »
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Top 499+ Aptitude MCQs on Trains Crossing Bridge/Platform »
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New 500+ Aptitude MCQs on Train Crossing Pole/Post or Man »
-
Top 499+ Trains Problems Using Time and Distance For SSC »
-
Top 489+ Time And Distance MCQ Problems on Train For BANK »
Model 3 Train Vs Bridge/Platform Shortcuts »
Click to Read...Model 3 Train Vs Bridge/Platform Online Quiz
Click to Start..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 …
QUANTITATIVE APTITUDE CATEGORIES
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» Number System
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» Average
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» LCM & HCF
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» Percentage
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» Profit & Loss
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» Time & Work
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» Pipes & Cisterns
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» Time & Distance
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» Trains
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» Compound Interest
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» Simplification
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» Discount
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» Power
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» Permutations & Combination
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» Ratio & Proportion
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» Advance Math
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» Algebraic Expressions
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» Simple Interest
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» Alligation & Mixtures
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» Boats & Streams
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» Set Theory
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» Linear Equations
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» Quadratic Equations
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