Aricent Time, Speed & Distance Quiz-1

Distance = Speed × time

Here time = 2hr 45 min = 114 hr

Distance = 4×114=11 km

New Speed =16.5 kmph

Therefore time = DS=1116.5=40 min

Due to stoppages, it covers 9 km less.

Time taken to cover 9 km = 954×60 = 10 min.

In same time, they cover 110 km and 90 km respectively.

For the same time speed and distance is inversely proportional.

So ratio of their speed = 110:90 = 11: 9

Time = 5060=56 hr

Speed = 48 mph

Distance = S×T=48×56=40 km

Time = 4060 hr

New speed = 40×32=60 kmph

Let total distance be S
total time = 1 hr 24 min = 84 min

= 8460=2115 hr

Let Vikas travels from A−>T−>S
A ------------------------ T ------------ S

<---------- (2S3)---------><----(S3)---->

A to T :: speed = 4 kmph

Distance = (23)×S

T to S :: speed = 5 km =

Distance = (1−23)S=13S

Total time:

2115=2/3S4+1/3S4

84=10S+4S {Multiply both sides by 15×4}

S=8414=6 km

Let the normal speed be 's' km/hr

Then new speed = (s+5) km/hr

300s−2=300s+5

On solving this equation we get:

s = 25 km/hr

Time = 10060=53hr

Speed =48 mph

Distance =S×T
=48×53
=80 km
Now, as per the question, journey is to be reduced to 40 min.

So, new time,

=40 min=4060 hr
=23 hr
New speed,

=DistanceNew Time
=802/3
=80×32
=120 kmph

Let the speeds of the two trains be s and 2s m/s respectively.

Also, suppose that the lengths of the two trains are P and Q metres respectively.

Then,

P+Q2s−s=60------(1)

and

P2s−s=40------(2)

On dividing these two equation we get:

P+QP=6040

P:Q = 2 : 1

Let both of them meet after T min

4500 m are covered by Suresh in 60 m.

In T min he will cover 4500T60

Likewise, In T min Suresh's wife will cover 3750T60

Given,

4500T60+3750T60=726

T=5.28 minutes

Average speed of train leaving Delhi = 2004=50 km/hr

Average speed of train leaving Ambala cantt. = 200×27=4007

By the time the other train starts from Ambala cantt, the first train had travelled 100 km

Therefore, the trains meet after:

=200−100(50+400/7)=1415hr

=1415×60=56minutes

Hence they meet at 8:56 am