TCS Ninja Speed Time Distance Question Page-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

Video solution:

 

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.

Video solution:

sol -
(1) & (2) Cross with Torch 2
(1) Returns with Torch 1
(5) & (10) Cross with Torch 10
(2) Returns with Torch 2
(1) & (2) Cross with Torch 2
TOTAL 17

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

Video solution:

 

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
So option 1) is correct

 

By the data Distance from starting point to T is : 20*2 = 40 kms
Therefore time taken by B to reach T : 40km/40hmph= 1hr
So from 11:00AM to 2:00 PM B travels the distance : 40*3 = 120kms
from 12:00noon to 2:00PM A travels the distance 20*2 = 40kms
So total distance is 160 kms

 

total perimeter= 2(3x+2x)
=10x,
speed = 24*5/18= 20/3 m/sec
distance = (20/3)* (4*6),
=1600 m
10x=1600
x=160
length = 3x= 480m
breadth = 2x= 320m

Video solution:

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 theyll meet 10 minutes after the young man starts his journey, i.e in
10:15am.

Video solution:

 

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

Video solution: