Coast To Coast Tickets Interview Questions, Process, and Tips

13 Kigs & 6 Libs –> 510tors in 10hrs
” –> 51 tors/hr
14 Libs –> 484 tors in 12hrs
” –> 121/3 tors/hr
1 Lib –> 121/42 tors/hr
==> 6 Libs –> 121/7 tors/hr
Now,
13 kigs and 6 Libs –> 51 tors/hr
Only 13 Kigs –> 51 – (121/7) tors/hr
= 236/91 tors/hr

Day = Night
White = Black

Let’s assume the length of each train is ‘L’ and the speeds of the two trains are ‘V₁’ and ‘V₂’ respectively.

When the trains are moving in the opposite direction, their relative speed is the sum of their individual speeds. The total distance they need to cover is the sum of their lengths. Since they cross each other completely in 5 seconds, we can set up the following equation:

(V₁ + V₂) × 5 = 2L

When the trains are moving in the same direction, their relative speed is the difference between their individual speeds. The total distance they need to cover is the difference between their lengths. Since they cross each other completely in 15 seconds, we can set up the following equation:

(V₁ – V₂) × 15 = 2L

Now, let’s solve these equations to find the ratio of their speeds.

From the first equation, we have:
(V₁ + V₂) × 5 = 2L
V₁ + V₂ = (2L) / 5

From the second equation, we have:
(V₁ – V₂) × 15 = 2L
V₁ – V₂ = (2L) / 15

Let’s add these two equations together:
V₁ + V₂ + V₁ – V₂ = (2L) / 5 + (2L) / 15
2V₁ = (6L + 2L) / 15
2V₁ = (8L) / 15
V₁ = (4L) / 15

So, the speed of the first train is (4L) / 15.

Now, let’s substitute this value back into the first equation to find V₂:
(4L) / 15 + V₂ = (2L) / 5
V₂ = (2L) / 5 – (4L) / 15
V₂ = (6L – 4L) / 15
V₂ = (2L) / 15

Therefore, the speed of the second train is (2L) / 15.

The ratio of their speeds is given by:
(V₁ / V₂) = ((4L) / 15) / ((2L) / 15)
(V₁ / V₂) = 4L / 2L
(V₁ / V₂) = 2

So, the ratio of their speeds is 2:1.

none of these
there will be 20 men

100/8=12-4
12/8=1-4
ans:144

16 poles per min
48kmph=48000 meters per hr
=800 meters per min
800/50=16 poles per min

6.40 am

Relative speed = 60 + 90 = 150 km/hr.

= 150 x 5/18 = 125/3 m/sec.

Distance covered = 1.10 + 0.9 = 2 km = 2000 m.

Required time = 2000 x 3/125 = 48 sec.

47.5

(x+2)^2 -x^2 = 84
X=20
So (20,22)
Sum= 42

127.179(app)
given distance of the train along the wind is 695
and againt the wind is 498
and time = distance/speed
as we know that time is equal in both the cases hence equate
695/s1=498/s2———–(1);
where s1=speed of the plane + speed of the wind
and s2=speed of the plane -speed of the wind
given that speed of the wind is 21k/h
s1=sp+21
s2=sp-21
substitu in eq 1
we get the answer as 27.17(app)

At 130 years.