June 23, 2019
Two trains running in opposite directions cross a man standing
on the platform in 27 seconds and 17 seconds respectively and
they cross each other in 23 seconds. The ratio of their speeds is:
1 : 3
3 : 2
3 : 4
None of these
Let the speeds of the two trains be
x m/sec and y m/sec respectively.
Then, length of the first train = 27x metres,
and length of the second train = 17y metres.
(27x+ 17y)/(x+y ) =23
=>27x+ 17 y = 23x+ 23y =4x =6y
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A man performs 2/15 of the total journey by train 9/20 by bus
and the remaining 10 km on foot. His total journey in km is
If the total journey be x km, then
By walking at 3/4 of his usual speed, a man reaches his
office 20 minutes later than usual. His usual time is
New speed = 3/4 × usual speed
∴ New time = 4/3 × usual time.
∴ 1/3 × usual time = 20 minutes
⇒ Usual time = 3 × 20
= 60 minutes
Only a single rail track exists between station A and B on a
railway line. One hour after the northbound super fast
train N leaves station A for Station B, a southbound passenger
train S reaches station A from station B. The speed of the super
fast train is twice that of a normal express train E,
while the speed of a passenger train is half that of E.
On a particular day N leaves for station B from Station A,
20 minutes behind the normal schedule. In order to maintain the
schedule, both N and S increased their speed. If the super fast
train doubles its speed, what should be the ratio (approximately)
of the speed of passenger train to that of the super fast train
so that passenger train S reaches exactly at the scheduled time
at station A on that day?
As per condition , initial ratio of the speed
of passenger train to that of the super fast train is 1:4.
Initially, passenger train takes 4/5 or 48 mins hr
to travel from B to A and superfast train train
takes 1/5 hr or 12 mins to reach stn A from Stn B.
when On a particular day N leaves for station B from
Station A, 20 minutes behind the normal schedule.
In order to maintain the schedule, both N and S
increased their speed. If the super fast train doubles its speed,
it covers distance in 6 mins and remains 14 mins behind schedule.
to maintain schedule, passenger train has to cover the
distance in 48-14 = 34 mins.
so final ratio of speeds will be 6:34 or 1:6 (approx)
An Engine length 1000 m moving at 10 m/s. A bird is flying
from the engine to end with 'x' m/s and coming back at '2x' m/s.
Take total time of bird traveling as 187.5 s. Find x and 2x in km/ hr?
21.4 and 42.8
25.2 and 50.44
31.4208 and 62.8416
33.12 and 66.2
Birds speeds in mtrs/sec is 'x' and '2x' .
While flying from Engine to end, relative speed
= (x+10) m/sec
from end to engine, flying speed = (2x - 10) mtr/sec
1000/(x+10) + 1000/(2x-10) = 187.5 secs
solving it, we get
so x = 8.728 m/sec and 2x= 17.456 m/sec
x = 31.4208 km/hr and 2x = 62.8416 km/hr.
A single pipe of diameter x has to be replaced by six pipes of
diameters 12 cm each. The pipes are used to covey some liquid in a
laboratory. If the speed/flow of the liquid is maintained the
same then the value of x is?
Volume of water flowing through 1 pipe of diameter x =
Volume discharged by 6 pipes of diameter 12 cms.
As speed is same, area of cross sections should be same.
Area of bigger pipe of diameter x =
Total area of 6 smaller pipes of diameter 12
i.e πR2 = 6πR12
Here R = R and R1 = 12/2 = 6
⇒R2 = 6×6×6
R = 14.696
=> D = X = 14.696 * 2 = 29.3938 cm.
a,b,c running on a race a is 15m ahead than b,,and 19m ahead of c.
Then 3700m race wat is d distance b/w b and c
A is 15m away from B and 19m away from C
which means distance between B and C
Walking at the rate of 4 kmph a man cover certain distance in 2 hr
45 min. Running at a speed of 16.5 kmph the man will cover the
same distance in.
Distance = Speed × time
Here time = 2hr 45 min = 11/4 hr
Distance = 4×11/4=11 km
New Speed =16.5 kmph
Therefore time = D/S=11/16.5=40 min
Excluding stoppages, the speed of a bus is 54 kmph and including
stoppages, it is 45 kmph. For how many minutes does the bus stop
Due to stoppages, it covers 9 km less.
Time taken to cover 9 km = 9(/54)×60 = 10 min.
A train covers a distance in 50 min, if it runs at a speed
of 48 kmph on an average. The speed at which the train
must run to reduce the time of journey to 40 min will be
Time = 50/60=56 hr
Speed = 48 mph
Distance = S×T=48×(5/6)= 40 km
Time = 40/60 hr
New speed = 40×32=60 kmph
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