Deloitte Time, Speed and Distance Questions and Answers Quiz-1

Since the train and man are moving in opposite directions,

The relative speed = 60-(-6) kph = 60 + 6 = 66 kph

66 km/h in m/s = 66*5/18 = 55/3 m/s

Total distance to be covered = 110 m (length of the train)

So, time = distance\div speed = 110 \div 55 \div 3 = 6 s.

So, the required time is 6 seconds.

Let the length of the train be L meters

Speed of first train = \frac{3600}{18}\times Lm/hour = 200L m/hour

Speed of second train = \frac{3600}{12}\times L m/hour = 300L m/hour

When running in opposite directions, relative speed = 200 L + 300 L m/hour = 500 L m/hour

Distance to be covered = L + L = 2L meter

Time taken = \frac{2L}{500L} \times 3600 sec = 14.4 sec

Let the speed of the cyclist be x km/h and wind be y km/h
\frac{1}{(x+y)}=\frac{3}{60}
x+y=20..........(i)

And
\frac{1}{(x-y)}=\frac{4}{60}
x-y=15.........(ii)
On solving both the equations we get:
x=\frac{35}{2} , y=\frac{35}{2}

Now, time taken by cyclist without wind: \frac{2}{35} \times 60=\frac{24}{7} min

 

\frac{(x+100)}{25} = \frac{x}{15}

=> 15(x + 100) = 25x

=> 15x + 1500 = 25x

=> 1500 = 10x

=> x = 150 m.

Given speed = 63 km/hr = 63×\frac{5}{18}=17.5 m/s

Let the length of the bridge = x mts

Given time taken to cover the distance of (170 + x)mts is 30 sec.

We know speed = distance\times time m/s

⇒17.5=\frac{170+x}{30}
--> x = 355 mts.

Let the speed of the current be x km/hr
Thus upward speed =(4.5+x) km/hr
and downward speed =(4.5−x) km/hr

Let distance travelled be y, then
y \times (4.5−x) = 2y \times (4.5+x)

⇒ x= 1.5 km/hr

Let the total distance covered be S km.
Total time taken= \frac {S}{2×40} +\frac {S}{2×60} = \frac{5S}{240} hr
Average speed = S\times \frac {240}{5S} = 48 km/hr

Let the speed of the stream be x km/hr. Then,

Speed downstream = (15 + x) km/hr

Speed upstream = (15 - x) km/hr

\frac{30}{15 + x} +\frac{30}{15 - x} = \4tfrac{1}{2}
\frac{900}{225 - x^{2}} = \frac{9}{2}
9x^{2} = 225

x^{2} = 25

x = 5 km/hr.

Let the speed of goods train be x km/hr.

Distance = 187.5 m

Time = 9 sec

Then, (50+x)\times \frac{5}{18} = \frac{187.5}{9}

⇒ x= 25 km/hr

Speed of boat downstream = Distance covered/Time taken= 3/2 = 15 km/h

Now, speed of boat upstream = Distance covered/Time taken= 3/6 = 5 km/h

Now, speed of boat = Speed downstream+Speed upstream/2

= (15+5)/2=20/2 = 10 km/h