Time and Distance- Aptitude Questions and Answers

Time taken to cover 60 km = 1 hour
that is 60000 m = 60 minutes (1 km=1000 m,1 hour=60 minutes)
Time taken to cover 1 m = 60/60000 minutes = 1/1000 minutes
Time taken to cover 200 m = 200 * 1/1000 minutes
= 1/5 minutes.

karan cover a km distance in 1/40 hour and Arjun take 1/60 hour
Hence ratio of time taken by karan and arjun would be 60/40 =3/2

Let the distance is x.then
Time to go to school + Time taken in return = 5
x/3 + x/2 = 5
2x+3x / 6 = 5
5x = 30
So ditance of school is , x = 6 km

he walk = 24 - 9 = 15 hours per day
so the total time he walks would be = 15 * 50 = 750 hours

let the speed is s,distance is d and time is t.
t = d/s = 750 hours
he doubles his speed(2*s) and will cover the double distance(2*d), then new time would still be same i.e.=2*d/2*s=d/s=750 hours

he rests twice more than he did before, so he would walk = 24 - (2*9) = 6 hours per day

So,Required number of days = 750/6 = 125 days

Distance(d)=Speed(S)*Time(t)
Let the Speed is s and Time is t

distance coverd by Malleshwari in both case is equal.then
s * t = (3/4) * s *(t + 2)
t = (3/4) * (t + 2)
t - 3*t/4 = 3/2
t/4 = 3/2
t = 6

Speed(S) = Distance(d)/Time(t)
= 1500/ 5 m/mins
= 300 m/mins
= 300 * (60/1000 ) km/hr
= 18 km/hr

Speed(S) = Distance(d)/Time(t)
Rohit's speed = 10 Km/hr = (10 * 5/18) m/sec = 50/18 m/sec

Time taken to cover 800 m = 800 / (50/18) sec = 800 *18 / 50 sec = 288 sec = 4 min 48 sec

Speed(S) = Distance(d)/Time(t)
Let total distance be d km.
Then , (2d / 3) / 4 + ( d / 3) / 5 = 42/60
d/6 + d/ 15 = 7/10
5d + 2d = 21
hence, Required distance d = 3 km

When d1 = d2 , Average speed = 2*S1*S2/(S1+S2), where S1 and S2 are the speeds for covering d1 and d2 respectively

Average speed = ( 2*35*45) / (35 + 45) km/hr. = 3150 / 80 = 315 / 8 km/hr

Let distance between his house and office be x km
Time taken to go to office =x/2 hr
Time taken to return =x/3 hr

x/2 + x/3=5
2x+3x=30
5x=30
x=6

i.e., distance between house and office = 6 km

Let Abhay's speed be X km/hr and Sameer's speed be Y km/hr
As is given, Abhay takes 2 hours more than Sameer
30/X - 30/Y = 2 ........ (i)

Abhay doubles his speed, i.e his speed becomes 2x, then he would take 1 hour less than Sameer
30/Y - 30/2X = 1 ......... (ii)

After adding above equations
30/X - 30/2X = 3
30/2X=3
X = 5 km/h

Speed(s) = Distance(d)/Time(t)
T1 =D/5
T2=D/4
T1/T2 = 4 / 5

Total distance = 315 km = 315000 m
Total time = 2.8 * 60 * 60 sec
Average speed = Total distance/Total time
= 315000/2.8 * 60 * 60
= 31.25 m/s

Let us assume that the time to taken by the policeman to catch the thief is t.
Now, distance run by police in time, t = t*10

Time is same for the thief and policeman.

According to question, 10t = 8t + 100/1000 km
2t = 0.1
t = 0.05

Distance run by thief in 0.05 Hrs =8*0.05 km = 0.4 km = 400 m

Let the distance is X km , Speed= Distance/Time

Time taken to travel by Ruhan = X/45

Yuvan speed = 5 km/hr slower than Ruhan= 45 - 5 km/hr = 40 km/hr

Time taken to travel by Yuvan = X/40

As is given, X/40 - X/45 = 1 (an hour)

(9*X - 8*X) / 360 = 1

X = 360

When d1 = d2 , Average speed = 2S1S2/(S1+S2), where S1 and S2 are the speeds for covering d1 and d2 respectively

Average speed = 2S1S2/(S1+S2) = 2*40*24/(40+24)= 30 km/hr