Speed, Distance and Time Questions and Answers

Relative speed of cop with respect to robbers = (11 - 10) = 1 Km/hr.

Now the relative distance covered by cop in 6 min

= Speed x Time = 1 x 6/60

=1/10km = 100m

The distance between the Cop and Robbers after 6 min = 200 - 100 = 100 m.

Assume the time in which he travelled on foot = x hour
Time required to travel on cycle = (9 - x) hr

the total distance = 66 km
Hence,
4x + 9(9-x) = 66
=> 5x = 15
=> x = 3

so distance travelled on foot = 4(3) = 12 km. Option 3 is the correct answer.

The Airbus A350 flying north would have travelled for 2 hours at 2 PM.

It travels at 500 kmph.

Hence, the Airbus flying north would have covered 1000 kms from the city at 2 PM.

The Spice jet flying east would have travelled for 1.5 hours at 2 PM.

It travels at 400 kmph.

Hence, the Spice jet flying east would have covered 1.5*400 = 600 kms from the city at 2 PM.

Both the Airbus and Spice jet are flying at right angles to each other.

The shortest distance between the two planes is the length of the hypotenuse of the right triangle whose legs are the distances covered by both of them.

Hence, distance =   = 1166 km

Therefore the nearest 100 km distance between the two planes is equal to 1200 kms.

Assume the distance to be X km.
=>Time for the first rider = X/4.
Similarly, time for the second cyclist = X/5.
Given that: X/4 – X/5 =X/20
=> X/20= 1/2 => X=10 km = 10000 m. So Option 3 is the correct answer.

Let us assume,
Speed of water be x.
Now,
Distance =50 miles, Speed=20 mph (in still water)
Speed of the ship along with speed of water=10+x
Speed of the ship against the speed of water=10-x
Time=5 hr 30 minutes.
T=D/S
5.5= \frac{50}{20+x} + \frac{50}{20-x}
5.5=  \frac {50(20-x)+50(20+x)}{(20+x)(20-x) }
5.5 = \frac {2000}{400- x^{2}}
2200-5.5x ^{2}=2000
5.5x ^{2}=200
X^{2}=36.36
X=6.06 rounding off to 6

Since the distance covered is same,
Average speed = 2*a*b/ (a+b), where a & b are respective speeds of to and fro journey.
Avg speed = 2*3.5*7.5/ (3.5+7.5) = 4.77 km/hr (approx.)
The total time = 6 hrs 37 minutes.
Rest = 45 minutes
Hence, travel time = 6 hrs 37 minutes - 45 minutes = 5 hrs 52 minutes =5\frac{52}{60} = 5\frac{13}{15}
D = S * T = 4.77 *5\frac{13}{15}  = 4.77 *\frac{88}{15}  = 28 Km
This is the total distance, that is, to and fro.
Hence distance to town = 28/2 = 14 Km.
Option 1 is the correct answer.

In one hour without stopping, the bus would have travelled 54km. Once stopping is factored in, the bus actually travels 45km. So it travels 9km less far as a consequence.
Distance/Speed = Time
9km/54kmph = 0.16 hours
0.16 * 60 = 10 minutes
So the bus stops for ten minutes in every hour on average.
Option 3 is the correct answer.

Let ‘s’ km/h be the speed of Jack’s yacht.
Then, 1(s+4) = 2(s-4) → s=12 km/h
So option 2 is the correct answer.

m miles in h - 1/2 = (2h - 1)/2 hours
=> 2m/ (2h - 1) miles per hours
So option 3 is the correct answer.

Romita travelled totally a distance of = 80 -20 = 60
Supriya covered a distance = 80 + 20 = 100
Let‘s’ be the speed of Romita
Time taken by Romita => t1 = 60/s
Time taken by Supriya => t2 = 100/(s+4)
t1=t2
Thus, 60/s = 100/(s+4) => 4s = 24 => s = 6Km/hr
So option 2 is the correct answer.

Length of the platform = speed of train * extra time taken to cross the platform.
So length of platform = 72 kmph * 12 seconds
Convert 72 kmph into m/sec
1 kmph = 5/18 m/s
Therefore 72 kmph = 5/18∗72 = 20 m/sec
Hence the length of the platform = 20 m/s * 12 sec = 240 meters.

Option 1 is the correct answer.