Quantitative Aptitude
TRAINS MCQs
Problems On Trains
Total Questions : 842
| Page 84 of 85 pages
Answer: Option B. -> 12
$$\eqalign{
& {\text{Speed}}\,{\text{of}}\,{\text{the}}\,{\text{first}}\,{\text{train}} \cr
& = {\frac{{120}}{{10}}} \,{\text{m/sec}} \cr
& = 12\,{\text{m/sec}} \cr
& {\text{Speed}}\,{\text{of}}\,{\text{the}}\,{\text{second}}\,{\text{train}} \cr
& {\frac{{120}}{{15}}} \,{\text{m/sec}} \cr
& = 8\,{\text{m/sec}} \cr
& {\text{Relative}}\,{\text{speed}} = {12 + 8} = 20\,{\text{m/sec}} \cr
& \therefore {\text{Required}}\,{\text{time}} \cr
& = {\frac{{ {120 + 120} }}{{20}}} \,{\text{ sec}} \cr
& = 12\,{\text{sec}} \cr} $$
$$\eqalign{
& {\text{Speed}}\,{\text{of}}\,{\text{the}}\,{\text{first}}\,{\text{train}} \cr
& = {\frac{{120}}{{10}}} \,{\text{m/sec}} \cr
& = 12\,{\text{m/sec}} \cr
& {\text{Speed}}\,{\text{of}}\,{\text{the}}\,{\text{second}}\,{\text{train}} \cr
& {\frac{{120}}{{15}}} \,{\text{m/sec}} \cr
& = 8\,{\text{m/sec}} \cr
& {\text{Relative}}\,{\text{speed}} = {12 + 8} = 20\,{\text{m/sec}} \cr
& \therefore {\text{Required}}\,{\text{time}} \cr
& = {\frac{{ {120 + 120} }}{{20}}} \,{\text{ sec}} \cr
& = 12\,{\text{sec}} \cr} $$
Answer: Option D. -> 82 km/hr
$$\eqalign{
& {\text{Let}}\,{\text{the}}\,{\text{speed}}\,{\text{of}}\,{\text{the}}\,{\text{second}}\,{\text{train}}\,{\text{be}}\,x\,{\text{km/hr}}. \cr
& {\text{Relative}}\,{\text{speed}}\, \cr
& = \,\left( {x + 50} \right)\,{\text{km/hr}} \cr
& = \left[ {\left( {x + 50} \right) \times \frac{5}{{18}}} \right]\,{\text{m/sec}} \cr
& = {\frac{{250 + 5x}}{{18}}} \,{\text{m/sec}} \cr
& {\text{Distance}}\,{\text{covered}} \cr
& = \left( {108 + 112} \right) = 220\,m \cr
& \therefore \frac{{220}}{{ {\frac{{250 + 5x}}{{18}}} }} = 6 \cr
& \Rightarrow 250 + 5x = 660 \cr
& \Rightarrow x = 82\,{\text{km/hr}} \cr} $$
$$\eqalign{
& {\text{Let}}\,{\text{the}}\,{\text{speed}}\,{\text{of}}\,{\text{the}}\,{\text{second}}\,{\text{train}}\,{\text{be}}\,x\,{\text{km/hr}}. \cr
& {\text{Relative}}\,{\text{speed}}\, \cr
& = \,\left( {x + 50} \right)\,{\text{km/hr}} \cr
& = \left[ {\left( {x + 50} \right) \times \frac{5}{{18}}} \right]\,{\text{m/sec}} \cr
& = {\frac{{250 + 5x}}{{18}}} \,{\text{m/sec}} \cr
& {\text{Distance}}\,{\text{covered}} \cr
& = \left( {108 + 112} \right) = 220\,m \cr
& \therefore \frac{{220}}{{ {\frac{{250 + 5x}}{{18}}} }} = 6 \cr
& \Rightarrow 250 + 5x = 660 \cr
& \Rightarrow x = 82\,{\text{km/hr}} \cr} $$
Answer: Option C. -> $$27\frac{7}{9}$$ m
$$\eqalign{
& {\text{Relative}}\,{\text{speed}} = \left( {40 - 20} \right)\,{\text{km/hr}} \cr
& = \left( {20 \times \frac{5}{{18}}} \right)\,{\text{m/sec}} \cr
& = {\frac{{50}}{9}} \,{\text{m/sec}} \cr
& \therefore {\text{Length}}\,{\text{of}}\,{\text{faster}}\,{\text{train}} \cr
& = \left( {\frac{{50}}{9} \times 5} \right)\,m \cr
& = \frac{{250}}{9}\,m \cr
& = 27\frac{7}{9}\,m \cr} $$
$$\eqalign{
& {\text{Relative}}\,{\text{speed}} = \left( {40 - 20} \right)\,{\text{km/hr}} \cr
& = \left( {20 \times \frac{5}{{18}}} \right)\,{\text{m/sec}} \cr
& = {\frac{{50}}{9}} \,{\text{m/sec}} \cr
& \therefore {\text{Length}}\,{\text{of}}\,{\text{faster}}\,{\text{train}} \cr
& = \left( {\frac{{50}}{9} \times 5} \right)\,m \cr
& = \frac{{250}}{9}\,m \cr
& = 27\frac{7}{9}\,m \cr} $$
Answer: Option B. -> 50 m
$$\eqalign{
& 2\,kmph = \left( {2 \times \frac{5}{{18}}} \right)\,{\text{m/sec}} \cr
& \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = \frac{5}{9}\,{\text{m/sec}} \cr
& 4\,kmph = \left( {4 \times \frac{5}{{18}}} \right)\,{\text{m/sec}} \cr
& \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = \frac{{10}}{9}\,{\text{m/sec}} \cr
& {\text{Let}}\,{\text{the}}\,{\text{length}}\,{\text{of}}\,{\text{the}}\,{\text{train}}\,{\text{be}}\,x\,{\text{metres}}\, \cr
& {\text{and}}\,{\text{its}}\,{\text{speed}}\,{\text{be}}\,y\,{\text{m/sec}} \cr
& {\text{Then}},\, {\frac{x}{{y - \frac{5}{9}}}} = 9\,{\text{and}}\, {\frac{x}{{y - \frac{{10}}{9}}}} = 10 \cr
& \therefore 9y - 5 = x\,{\text{and}}\,10\left( {9y - 10} \right) = 9x \cr
& \Rightarrow 9y - x = 5\,{\text{and}}\,90y - 9x = 100 \cr
& {\text{On}}\,{\text{solving,}}\,{\text{we}}\,{\text{get}}:\,x = 50 \cr
& \therefore {\text{Length}}\,{\text{of}}\,{\text{the}}\,{\text{train}}\,{\text{is}}\,50\,m \cr} $$
$$\eqalign{
& 2\,kmph = \left( {2 \times \frac{5}{{18}}} \right)\,{\text{m/sec}} \cr
& \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = \frac{5}{9}\,{\text{m/sec}} \cr
& 4\,kmph = \left( {4 \times \frac{5}{{18}}} \right)\,{\text{m/sec}} \cr
& \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = \frac{{10}}{9}\,{\text{m/sec}} \cr
& {\text{Let}}\,{\text{the}}\,{\text{length}}\,{\text{of}}\,{\text{the}}\,{\text{train}}\,{\text{be}}\,x\,{\text{metres}}\, \cr
& {\text{and}}\,{\text{its}}\,{\text{speed}}\,{\text{be}}\,y\,{\text{m/sec}} \cr
& {\text{Then}},\, {\frac{x}{{y - \frac{5}{9}}}} = 9\,{\text{and}}\, {\frac{x}{{y - \frac{{10}}{9}}}} = 10 \cr
& \therefore 9y - 5 = x\,{\text{and}}\,10\left( {9y - 10} \right) = 9x \cr
& \Rightarrow 9y - x = 5\,{\text{and}}\,90y - 9x = 100 \cr
& {\text{On}}\,{\text{solving,}}\,{\text{we}}\,{\text{get}}:\,x = 50 \cr
& \therefore {\text{Length}}\,{\text{of}}\,{\text{the}}\,{\text{train}}\,{\text{is}}\,50\,m \cr} $$
Answer: Option B. -> 10 a.m.
$$\eqalign{
& {\text{Suppose}}\,{\text{they}}\,{\text{meet}}\,x\,{\text{hours}}\,{\text{after}}\,{\text{7}}\,{\text{a}}{\text{.m}}. \cr
& {\text{Distance}}\,{\text{covered}}\,{\text{by}}\,{\text{A}}\, \cr
& {\text{in}}\,x\,{\text{hours = 20x}}\,{\text{km}}{\text{.}} \cr
& {\text{Distance}}\,{\text{covered}}\,{\text{by}}\,{\text{B}} \cr
& \,{\text{in}}\,\left( {x - 1} \right)\,{\text{hours}} = 25\left( {x - 1} \right)\,km \cr
& \therefore 20x + 25\left( {x - 1} \right) = 110 \cr
& \Rightarrow 45x = 135 \cr
& \Rightarrow x = 3 \cr
& {\text{So,}}\,{\text{they}}\,{\text{meet}}\,{\text{at}}\,{\text{10}}\,{\text{a}}{\text{.m}}{\text{.}}\, \cr} $$
$$\eqalign{
& {\text{Suppose}}\,{\text{they}}\,{\text{meet}}\,x\,{\text{hours}}\,{\text{after}}\,{\text{7}}\,{\text{a}}{\text{.m}}. \cr
& {\text{Distance}}\,{\text{covered}}\,{\text{by}}\,{\text{A}}\, \cr
& {\text{in}}\,x\,{\text{hours = 20x}}\,{\text{km}}{\text{.}} \cr
& {\text{Distance}}\,{\text{covered}}\,{\text{by}}\,{\text{B}} \cr
& \,{\text{in}}\,\left( {x - 1} \right)\,{\text{hours}} = 25\left( {x - 1} \right)\,km \cr
& \therefore 20x + 25\left( {x - 1} \right) = 110 \cr
& \Rightarrow 45x = 135 \cr
& \Rightarrow x = 3 \cr
& {\text{So,}}\,{\text{they}}\,{\text{meet}}\,{\text{at}}\,{\text{10}}\,{\text{a}}{\text{.m}}{\text{.}}\, \cr} $$
Question 836. A train overtakes two persons walking along a railway track. The first one walks at 4.5 km/hr. The other one walks at 5.4 km/hr. The train needs 8.4 and 8.5 seconds respectively to overtake them. What is the speed of the train if both the persons are walking in the same direction as the train?
Answer: Option D. -> 81 km/hr
$$\eqalign{
& 4.5\,{\text{km/hr}} = \left( {4.5 \times \frac{5}{{18}}} \right)\,{\text{m/sec}} \cr
& \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = \frac{5}{4}\,{\text{m/sec}} \cr
& \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = 1.25\,{\text{m/sec,}}\,{\text{and}} \cr
& 5.4\,km/hr = \left( {5.4 \times \frac{5}{{18}}} \right)\,{\text{m/sec}} \cr
& \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = \frac{3}{2}\,{\text{m/sec}} \cr
& \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = 1.5\,{\text{m/sec}} \cr
& {\text{Let}}\,{\text{the}}\,{\text{speed}}\,{\text{of}}\,{\text{the}}\,{\text{train}}\,{\text{be}}\,x\,{\text{m/sec}} \cr
& {\text{Then}},\,\left( {x - 1.25} \right) \times 8.4 = \left( {x - 1.5} \right) \times 8.5 \cr
& \Rightarrow 8.4x - 10.5 = 8.5x - 12.75 \cr
& \Rightarrow 0.1x = 2.25 \cr
& \Rightarrow x = 22.5 \cr
& \therefore {\text{Speed}}\,{\text{of}}\,{\text{the}}\,{\text{train}} \cr
& = \left( {22.5 \times \frac{{18}}{5}} \right)\,{\text{km/hr}} \cr
& = 81\,{\text{km/hr}} \cr} $$
$$\eqalign{
& 4.5\,{\text{km/hr}} = \left( {4.5 \times \frac{5}{{18}}} \right)\,{\text{m/sec}} \cr
& \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = \frac{5}{4}\,{\text{m/sec}} \cr
& \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = 1.25\,{\text{m/sec,}}\,{\text{and}} \cr
& 5.4\,km/hr = \left( {5.4 \times \frac{5}{{18}}} \right)\,{\text{m/sec}} \cr
& \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = \frac{3}{2}\,{\text{m/sec}} \cr
& \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = 1.5\,{\text{m/sec}} \cr
& {\text{Let}}\,{\text{the}}\,{\text{speed}}\,{\text{of}}\,{\text{the}}\,{\text{train}}\,{\text{be}}\,x\,{\text{m/sec}} \cr
& {\text{Then}},\,\left( {x - 1.25} \right) \times 8.4 = \left( {x - 1.5} \right) \times 8.5 \cr
& \Rightarrow 8.4x - 10.5 = 8.5x - 12.75 \cr
& \Rightarrow 0.1x = 2.25 \cr
& \Rightarrow x = 22.5 \cr
& \therefore {\text{Speed}}\,{\text{of}}\,{\text{the}}\,{\text{train}} \cr
& = \left( {22.5 \times \frac{{18}}{5}} \right)\,{\text{km/hr}} \cr
& = 81\,{\text{km/hr}} \cr} $$
Answer: Option A. -> 400 m
$$\eqalign{
& {\text{Let}}\,{\text{the}}\,{\text{length}}\,{\text{of}}\,{\text{the}}\,{\text{first}}\,{\text{train}}\,{\text{be}}\,x\,{\text{metres}} \cr
& {\text{Then,}}\,{\text{the}}\,{\text{length}}\,{\text{of}}\,{\text{the}}\,{\text{second}}\,{\text{train}}\,{\text{is}}\, {\frac{x}{2}} \,{\text{metres}} \cr
& {\text{Relative}}\,{\text{speed}} = \left( {48 + 42} \right)\,{\text{kmph}} \cr
& \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = \left( {90 \times \frac{5}{{18}}} \right)\,{\text{m/sec}} \cr
& \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = 25\,{\text{m/sec}} \cr
& \therefore \frac{{ {x + \left( {x/2} \right)} }}{{25}} = 12 \cr
& or\,\frac{{3x}}{2} = 300 \cr
& or\,x = 200 \cr
& \therefore {\text{Length}}\,{\text{of}}\,{\text{first}}\,{\text{train}} = 200\,{\text{m}} \cr
& {\text{Let}}\,{\text{the}}\,{\text{length}}\,{\text{of}}\,{\text{platform}}\,{\text{be}}\,y\,{\text{metres}} \cr
& {\text{Speed}}\,{\text{of}}\,{\text{the}}\,{\text{first}}\,{\text{train}} \cr
& = \left( {48 \times \frac{5}{{18}}} \right)\,{\text{m/sec}} \cr
& = \frac{{40}}{3}\,{\text{m/sec}} \cr
& \therefore \left( {200 + y} \right) \times \frac{3}{{40}} = 45 \cr
& \Rightarrow 600 + 3y = 1800 \cr
& \Rightarrow y = 400\,{\text{m}} \cr} $$
$$\eqalign{
& {\text{Let}}\,{\text{the}}\,{\text{length}}\,{\text{of}}\,{\text{the}}\,{\text{first}}\,{\text{train}}\,{\text{be}}\,x\,{\text{metres}} \cr
& {\text{Then,}}\,{\text{the}}\,{\text{length}}\,{\text{of}}\,{\text{the}}\,{\text{second}}\,{\text{train}}\,{\text{is}}\, {\frac{x}{2}} \,{\text{metres}} \cr
& {\text{Relative}}\,{\text{speed}} = \left( {48 + 42} \right)\,{\text{kmph}} \cr
& \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = \left( {90 \times \frac{5}{{18}}} \right)\,{\text{m/sec}} \cr
& \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = 25\,{\text{m/sec}} \cr
& \therefore \frac{{ {x + \left( {x/2} \right)} }}{{25}} = 12 \cr
& or\,\frac{{3x}}{2} = 300 \cr
& or\,x = 200 \cr
& \therefore {\text{Length}}\,{\text{of}}\,{\text{first}}\,{\text{train}} = 200\,{\text{m}} \cr
& {\text{Let}}\,{\text{the}}\,{\text{length}}\,{\text{of}}\,{\text{platform}}\,{\text{be}}\,y\,{\text{metres}} \cr
& {\text{Speed}}\,{\text{of}}\,{\text{the}}\,{\text{first}}\,{\text{train}} \cr
& = \left( {48 \times \frac{5}{{18}}} \right)\,{\text{m/sec}} \cr
& = \frac{{40}}{3}\,{\text{m/sec}} \cr
& \therefore \left( {200 + y} \right) \times \frac{3}{{40}} = 45 \cr
& \Rightarrow 600 + 3y = 1800 \cr
& \Rightarrow y = 400\,{\text{m}} \cr} $$
Answer: Option D. -> 200 meters
Let the length of the train be x meters and its speed be y m/sec.
Then, $$\frac{{\text{x}}}{{\text{y}}}$$ = 20
⇒ y = $$\frac{{\text{x}}}{{20}}$$
∴ $$\frac{{{\text{x}} + 100}}{{30}}$$ = $$\frac{{\text{x}}}{{20}}$$
⇒ 30x = 20x + 2000
⇒ 10x = 2000
⇒ x = 200 meters
Let the length of the train be x meters and its speed be y m/sec.
Then, $$\frac{{\text{x}}}{{\text{y}}}$$ = 20
⇒ y = $$\frac{{\text{x}}}{{20}}$$
∴ $$\frac{{{\text{x}} + 100}}{{30}}$$ = $$\frac{{\text{x}}}{{20}}$$
⇒ 30x = 20x + 2000
⇒ 10x = 2000
⇒ x = 200 meters
Answer: Option A. -> 18 sec
Speed of train relative to man
= (42 - 6) kmph = 36 kmph
= $$\left( {36 \times \frac{5}{{18}}} \right)$$ m/sec
= 10 m/sec
∴ Time taken to pass the man
= $$\frac{{180}}{{10}}$$ sec
= 18 sec
Speed of train relative to man
= (42 - 6) kmph = 36 kmph
= $$\left( {36 \times \frac{5}{{18}}} \right)$$ m/sec
= 10 m/sec
∴ Time taken to pass the man
= $$\frac{{180}}{{10}}$$ sec
= 18 sec
Answer: Option D. -> 252 seconds
Relative speed = (45 - 40) km/hr = 5 km/hr
= $$\left( {5 \times \frac{5}{{18}}} \right)$$ m/sec
= $$\frac{{25}}{{18}}$$ m/sec
Total distance covered = Sum of lengths of trains = (200 + 150) m = 350 m
∴ Time taken
= $$\left( {350 \times \frac{{18}}{{25}}} \right)$$ sec
= 252 seconds
Relative speed = (45 - 40) km/hr = 5 km/hr
= $$\left( {5 \times \frac{5}{{18}}} \right)$$ m/sec
= $$\frac{{25}}{{18}}$$ m/sec
Total distance covered = Sum of lengths of trains = (200 + 150) m = 350 m
∴ Time taken
= $$\left( {350 \times \frac{{18}}{{25}}} \right)$$ sec
= 252 seconds
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