Practice Crossing both platform and man pole - quantitative aptitude Online Quiz (set-1) For All Competitive Exams
Q-1) 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)
(b)
(c)
(d)
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
Q-2) A train crosses a pole in 15 seconds and a platform 100 metres long in 25 seconds. Its length (in metres) is
(a)
(b)
(c)
(d)
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
Q-3) A train crosses a platform in 30 seconds travelling with a speed of 60 km/h. If the length of the train be 200 metres, then the length (in metres) of the platform is
(a)
(b)
(c)
(d)
Rule 10 and Rule 1,
Speed of train = 60 kmph
= $(60 × 5/18)$ m/sec.
= $50/3$ m/sec.
If the length of platform be = x metre, then
Speed of train
= $\text"Length of (train + platform)"/ \text"Time taken in crossing"$
$50/3 = {200 + x}/30$
50 × 10 = 200 + x
x = 500 - 200 = 300 metre
Q-4) A train passes by a lamp post on a platform in 7 sec. and passes by the platform completely in 28 sec. If the length of the platform is 390 m, then length of the train (in metres) is
(a)
(b)
(c)
(d)
Rule 10 and Rule 1,
Let the length of train be x metre,
then, Speed of train
= $x/7 = {x + 390}/28$
$x = {x + 390}/4$
4x - x = 390
$x = 390/3$ = 130 metres
Q-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)
(b)
(c)
(d)
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
Q-6) Points 'A' and 'B' are 70 km apart on a highway. A car starts from 'A' and another from 'B' at the same time. If they travel in the same direction, they meet in 7 hours, but if they travel towards each-other, they meet in one hour. Find the speed of the two cars (in km/hr).
(a)
(b)
(c)
(d)
Let speed of car starting from A be x kmph
and speed of car starting from B be y kmph
Case I
When cars meet at P,
7x = AP = AB + BP = 70 + 7y
7x - 7y = 70
x - y = 10 ...(i)
Case II
When cars meet at Q,
x + y = 70 ...(ii)
On adding these equations,
x = 40 kmph
Putting the value of x in equation (i),
y = 40 - 10 = 30 kmph
Q-7) 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)
(b)
(c)
(d)
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.
Q-8) A train leaves a station A at 7 am and reaches another station B at 11 am. Another train leaves B at 8 am and reaches A at 11.30 am. The two trains cross one another at
(a)
(b)
(c)
(d)
Let both trains meet after t hours since 7 a.m.
Distance between stations A and B = x Km.
$x/4 × t + x/{7/2} × (t - 1) = x$
[Speed = $\text"Distance"/ \text"Time"$]
$t/4 + {2(t - 1)}/7$ = 1
${7t + 8t - 8}/28$ = 1
15 t - 8 = 28
15 t = 28 + 8 = 36
t = $36/15 = 12/5$ hours
= 2 hours 24 minutes
∴ Required time = 9 :24 a.m.
Q-9) A train moves past a telegraph post and a bridge 264 m long in 8 seconds and 20 seconds respectively. What is the speed of the train ?
(a)
(b)
(c)
(d)
Rule 10 and Rule 1,
Let the length of train be x metres.
Then, speed of train when it passes a telegraph post
= $x/8$ m/sec. and speed of train, when it passes the bridge = ${x + 264}/20$
Clearly, $x/8 = {x + 264}/20$
$x/2 = {x +264}/5$
5x = 2x + 528
3x = 528 ⇒ $x = 528/3$ = 176m
Speed of train
= $176/8$ = 22 m/sec.
= 22 × $18/5$ Kmph = 79.2 kmph
Q-10) A moving train passes a platform 50 metres long in 14 seconds and a lamp-post in 10 seconds. The speed of the train is
(a)
(b)
(c)
(d)
Rule 10 and Rule 1,
Suppose length of train be x
According to question
${x + 50}/14 = x/10$
14x = 10x + 500
4x = 500
$x = 500/4 = 125 m$
Therefore, speed
=$125/10 × 18/5 = 45$ kmph