# Question Bank - 100 CAT level questions on Time, Speed and Distance topic

• Q100) Rishi Kapoor can swim a certain course against the river flow in 84 minutes; he can swim the same course with the river flow in 9 minutes less than he can swim in still water. How long would he take to swim the course with the river flow?

• total steps be T
T/N -T/(N+4) =6
3N^2+12N-2T=0 N=-12+rt (144+24T) /6 --(1)
and T/(G-4)-T/G = 6
3G^2 -12G-2T=0
G= 12+rt (144+24T) /6 ---(2)
relative speed = N+G
so T= 4N+4G --(3)
using 1 and 2,
T = 4 * rt (144+24T)/3
9T^2 = 16 (144+24T)
9T^2 - 384T-2304 = 0
T = (384 + 480)/18 = 48 steps

• P's distance be Y, Q's be X
Y/50 = (X-50)/25
Y=2X-100
say Y=100, X=100
distance at any time=d
d^2 = (100-50t)^2 + (100-25t)^2
diff wrt t for minima
(100-50t)*2 +(100-25t) = 0
t=12/5
d^2 = 20^2 + 40^2 = 2000 so d= 20 rt 5

• Train speed= T, men speed: v1, v2. length L.
so L/(T-v1)=20 and L/(T+v2) = 18
so 20T-20v1=18T+18v2
T= 9v2+10v1 --(1)
Ditsane between the men = 600 (T+v2)
Distance covered in these 10 mins = 600(v1+v2)
so reqd time = {600(T+v2)-600(v1+v2)}/(v1+v2)
= (6000-600)= 5400 sec= 90 mins ( using 1)

• A=2x, B=x
length= 2pi*r = 49pi = 154 m.
so given, 154/3x = 14 x=11/3 so 2x=22/3

• @mbatious 9 hour 54 min is it?

• @mbatious 6:24 am?

• Let the distance between Home and School be D and the uniform speed during onward journey be S.
While coming back, 2D/7 was covered with a speed of 2S/7
Remaining distance = 5D/7 and 3/5th of this is 3D/7 which is covered with a speed of 3S/7.
He finally reduces the speed to 3S/7 - 12S/42 = S/7 (as 66.67% is equivalent to 2/3rd) and cover the remaining distance, 2D/7.

Average speed for the total journey = 14 kmph
2D/14 = D/S + (2D/7) * (7/2S) + (3D/7) * (7/3S) + (2D/7) * (7/S)
D/7 = D/S + D/S + D/S + 2D/S
1/7 = 5/S
S = 35 kmph

• @mbatious option d

• @mbatious 180

• @mbatious pls provide answer key. asap. so that it is easier for me to prepare.

• pawan's speed = 4x m/s
escalator speed = y m/s
400x - 100y = D
rishabh's speed = 3x m/s
120x + 40y = D
400x - 100y = 120x + 40y
280x = 140y
ratio = 2:1
it means P has to cover twice of step
P's speed = 4x
E's speed = 2x
R's speed = 3x
so , distance = 154/2 = 77
77/(2x+5x) = 11/x sec
so they will meet after 11/x secs that means (11/x) * 2x = 22 steps from bottom

• D = 40 m
T = 5 m/ sec
R = 8 m/s
1st meet at 25/3 s
distance by R = 8 * 25/3 = 200/3 = 40 + 2 * 40/3
had both started from same point then they would meet at exactly 3 points
2 * 40/3 , 1 * 40/3 and 3 * 40/3
Time difference between 1st meet and 3rd meet = 2 * 40/3 = 80/3
total 25/3 + 80/3 = 35

• @Akshata

A good set which can give you a headache under exam pressure and if treated with calm, can fetch you 4 correct answers in some 5 minutes.

First we know distance PM and MQ are in the ratio 1 : 2
So lets consider PQ = 3d => PM = d and MQ = 2d
Now let say Kedar took t minutes to cover PM.
So, A would take t + 50 and B would take t + 25 (as they all started at different time and reached simultaneously) to cover PM.
So time taken by K to cover PQ = twice the time he took to cover PM = 2t
Similarly B will take 2t + 50

K reached Q and ran back 20 more mins to meet with A and B again at S.
So total time travelled = 2t + 20
So B will take (2t + 50) - (2t + 20) = 30 minutes to reach Q.
=> B will take 30 minutes to cover what K covered in 20 minutes.
Ratio of speed = 2 : 3
(d/t) / (d/t+25) = 2/3
(t + 25)/t = 2/3
t = 50

Question 2:
So K will take t = 50 minutes to reach M.
So All will reach M at 7.50 + 50 minutes = 8.40 am

Question 4:
If speed of K is 12 kmph = 1/5 km/minute
Kedar has took t + 2t = 3t = 150 minutes to reach PQ
So distance PQ = 150/5 = 30 KM

Question 1:
Total distance is 30 KM.
Distance SQ = distance covered by K in 20 minutes = 20/5 = 4 km.
Speed of A (before increasing it) = 10km/(100 minutes) = 1/10 km/minute
When K took 100 minutes to reach Q from M, and in this time A would have covered 100/10 = 10 km from M.
Now as A should cover another 20 - 10 - 4 = 6 km in 20 minutes to reach S with all others.
A would have increased it speed to 6/20 km/min = 18 kmph.

Question 3:
Speed of A = 1/10
Speed of K = 1/5
When they meet, together they would have covered a total distance of 2MQ = 2 * 20 = 40 KM
As ratio of speed is 1 : 2, distance will also be covered in same ratio
So K will cover 80/3 and A will cover 40/3.
Time taken by A to cover 40/3 = (40/3) / (1/10) = 400/3 = 133 minutes + 20 seconds
So K will overtake A again after 8.40 + 133 minutes + 20 seconds = 10:53:20 AM!

Happy Learning!

• @mbatious it will be 35

• @mbatious what is the answer to this team?

• @mbatious 6mins.

22

200

2

127

62

207

45