TCS NQT Speed Time Distance Question Quiz 1

Let speed of jake = x kmph.
So, speed of paul = ( x - 1.5 ) kmph.
Time difference =1.5hours
Total Distance = 10km
10/(x-1.5) -10/x = 1.5
By solving the above eq.
x = 4 kmph

bus covered a distance in (1.5 hour and 30 min. halt) 2 hour is =
1.5*60 = 90 km
remaing time to reach pune is only 1 hour then bus travels with the speed of
50km/h
now reamin. distance = 1*50 = 50 km
now total 50+90 = 140 km = ans.

A takes (100m) / (10m/s) = 10s to finish one complete round.
B takes (100m) / (8m/s) = 12.5s to finish one complete round.
C takes (100m) / (15m/s) = (20/3)s to finish one complete round.
For them to meet, the time will be LCM of above three times i.e,
LCM(10,12.5,20/3) = LCM(10,25,20)/GCD(1,2,3)=100/1=100sec

Assume that car with 50 km/h covers distance of d km.
in same time another car covers (d+120) km.

so d/50 = (d+120)/60

so, d=600 km.

So. distance between them is d+d+120= 1320 km

First, Shakeel covers a distance of 1800 and Rani covers a distance of 2000. but Shakeel takes 1minuter more. so,
Speed of Rani = 2000/x
Speed of Shakeel = 1800/(x+1)

Second, Shakeel covers certain distance in 6 mins and then Rani starts but cover only 1000m. So total time that Shakel runs is (6+x/2) mins.
therefore,
Distance covered by Shakeel = Speed x time
2000= [1800/(x+1)](6+x/2)
2000(x+1) = 10800 + 900x
2000x - 900x = 10800- 2000
x = 8800/ 1100
x= 8 mins
Rani took 8 mins

Speed in m/sec = 180(5)/18

= 50 m/sec

Time taken to cross the girl = 10 seconds

distance traveled = (50)(10)

= 500 m

= length of the train

Therefore length of the train = 500  m

total perimeter= 2(3x+2x)
=10x,
speed = 24*5/18= 20/3 m/sec
distance = (20/3)* (4*60),
=1600 m
10x=1600
x=160
length = 3x= 480m
breadth = 2x= 320m

From the condition we know that the old man passes 1/30 of the way in
1 minute. So in 5 minutes, he\'ll pass 1/6

So we could think of the problem like this. They both start at the same time,
but the old man need to pass 5/6 of the way, while the young man
should walk whole way. We aleady know that the old man pass
1/30 of the way in a minute, while the young man pass 1/20of the
way in a minute. So we need to find solution to the following equation:

1−(1/20)t=5/6−(1/30)t
1−t/20=5/6−t/30
1−5/6=(1/20−1/30)t

1/6=1/60t
⟹t=10

So they\'ll meet 10 minutes after the young man starts his journey, i.e in
10:15am.

Let distance travelled by taxi be x and by train by y.
x+y=450............................(i)
15x+21y=8130.....................(ii)
solving (i)&(ii)
y=230

Distance covered =Speed\times Time

= 3\times 30 = 90 m

Speed required to cover the same distance in 2 hours = \frac{90}{2} = 45 mph