Dell Speed, Distance & Time Questions

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
=x/y =3/2

If the total journey be x km, then
2x/15+9x/20+10=x
=>x-2x/15-9x/20=10
=> 60x-8x-27x/60=10
=> 25x/60=10
=> X=60*10/25=24km

New speed = 3/4 × usual speed
∴ New time = 4/3 × usual time.
∴ 1/3 × usual time = 20 minutes
⇒ Usual time = 3 × 20
= 60 minutes

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)

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
so
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.

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 is 15m away from B and 19m away from C
which means distance between B and C
= 19m-15m
= 4m

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

Due to stoppages, it covers 9 km less.
Time taken to cover 9 km = 9(/54)×60 = 10 min.

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