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Speed Time and Distance-Test 4
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Subject :- Quantitative Aptitude
Chapter :- Speed Time and Distance-Test 4
Questions :- 25
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A person has to reach a place in a certain time and he find that he will be 15minutes
late, if he walks at 4km/h and 10 minutes earlier if he walks at 6km/h. find the
distance he has to cover?
A man can reach a certain place in 30hours. If he reduces his speed by 1/15th, he covers
10km less in that time. Find his speed ?
Correct
Speed = A : B
. 15 : 14
Time = 14 15 (15-1)
but we have to keep time same in B also, so
distance covered in both cases =
A = 15*14 = 210
B = 14*14 =196
210-196 =14
14 =10( 10 km less in question)
210 = 150km
Speed = 150/30 =5km/h
Incorrect
Speed = A : B
. 15 : 14
Time = 14 15 (15-1)
but we have to keep time same in B also, so
distance covered in both cases =
A = 15*14 = 210
B = 14*14 =196
210-196 =14
14 =10( 10 km less in question)
210 = 150km
Speed = 150/30 =5km/h
Question 3 of 25
3. Question
1 points
Ravi and Ajay start simultaneously from the same place. A far B 50km apart. Ravi’s
speed is 5km/h less than that of Ajay. Ajay after reaching B, returns and meet Ravi at a
place 10km apart from B. find Ravi’s speed?
Correct
In the whole journey Ajay covers 20km
more than Ravi .
Then time taken by Ajay = 20/5= 4hrs
(Because in every hour Ajay covers 5km
more than Ravi for 20 km.)
So
Speed of Ajay = 60/4 =15
Ravi’s speed = 15-5= 10km/hr
Incorrect
In the whole journey Ajay covers 20km
more than Ravi .
Then time taken by Ajay = 20/5= 4hrs
(Because in every hour Ajay covers 5km
more than Ravi for 20 km.)
So
Speed of Ajay = 60/4 =15
Ravi’s speed = 15-5= 10km/hr
Question 4 of 25
4. Question
1 points
Walking at 4/5th of his usual speed, a man is 10 minutes late. The usual time taken by
him to cover that distance is ?
Correct
In this case → numerator *
time/(Numerator – denominator)
= 4*10/1 = 40minutes
Incorrect
In this case → numerator *
time/(Numerator – denominator)
= 4*10/1 = 40minutes
Question 5 of 25
5. Question
1 points
A man cover a certain distance in t hours. If he met with an accident at 50km and he
cover remaining distance at 2/3 of his speed. He covered distance in 30 minutes
late. If he met with this accident at 60km he would late by 24 minutes , then find the
distance?
Correct
He saves 6minutes by covering 10km more distance with his normal speed. For
30minutes he cover 50km and 50km are initial
So distance = 50+50 = 100km.
Incorrect
He saves 6minutes by covering 10km more distance with his normal speed. For
30minutes he cover 50km and 50km are initial
So distance = 50+50 = 100km.
Question 6 of 25
6. Question
1 points
A man covers a distance in four equal parts. He covers first part with speed of 60 kmph,
second part with 80 kmph and third part and fourth part with 120 kmph and 80
kmph respectively. Find the average speed of his journey.
Correct
Let x= 240 km (LCM of speed)
Time= 240/60 +240/80 +240/120
+240/80=4+3+2+3=12 hours
Avd speed=total distance/ total
time=240*4/12= 80 kmph
Incorrect
Let x= 240 km (LCM of speed)
Time= 240/60 +240/80 +240/120
+240/80=4+3+2+3=12 hours
Avd speed=total distance/ total
time=240*4/12= 80 kmph
Question 7 of 25
7. Question
1 points
A thief steals a car at 8 PM and starts driving at a speed of 80 kmph. The theft
came into light at 9 PM and police started to chase him at 9 PM at a speed of 100
kmph. At what time will he be caught?
Correct
Thief has moved 80 Km in 1 hour, So
distance=80 km
Time= Distance/ relative speed=80/(100-
80)= 4 hour
9 PM+4= 1AM
Incorrect
Thief has moved 80 Km in 1 hour, So
distance=80 km
Time= Distance/ relative speed=80/(100-
80)= 4 hour
9 PM+4= 1AM
Question 8 of 25
8. Question
1 points
A train starts from P at 8 PM and reaches Q at 11 PM. Another train starts from Q at 6
PM and reaches P at 11 PM. Find at what time they will meet each other?
Correct
Time (A:B)=3:5 => Speed(A:B)=5:3
Let distance=150 km
Speed A= 50 kmph ; Speed B=30 kmph
B starts 6 PM, in 2 hours i.e (till 8 PM
when A starts) it will move2*30=60 km
remaining = 150-60=90 km
Time= distance/relative speed=90/80= 1
hour 7 mins 30 sec
hence time=8 PM + 1 hour 7 mins 30 sec=9:7.5 PM
Incorrect
Time (A:B)=3:5 => Speed(A:B)=5:3
Let distance=150 km
Speed A= 50 kmph ; Speed B=30 kmph
B starts 6 PM, in 2 hours i.e (till 8 PM
when A starts) it will move2*30=60 km
remaining = 150-60=90 km
Time= distance/relative speed=90/80= 1
hour 7 mins 30 sec
hence time=8 PM + 1 hour 7 mins 30 sec=9:7.5 PM
Question 9 of 25
9. Question
1 points
If a person goes to school fro his home at a speed of 4 kmph, he reaches 10 minute late.
If he goes at a speed of 6 kmph he reaches 10 mins early. Find the distance between
school and home
Correct
Direct Formula Distance= [Speed 1*Speed
2/(S1-S2)]* [(Late + Early)//60]
=4*6/(6-4)*[(10+10)/60]=4 km
Incorrect
Direct Formula Distance= [Speed 1*Speed
2/(S1-S2)]* [(Late + Early)//60]
=4*6/(6-4)*[(10+10)/60]=4 km
Question 10 of 25
10. Question
1 points
A man takes 7 hours 30 mins in walking to a certain distance and riding back. He
would have gained 3 hours 10 mins by riding both ways. How long he would take
to walk both ways?
Correct
7 hour 30 mins – 3 hours 10 mins =4 hours
20 mins(When riding both ways)
=> 2 hour 10 mins riding in one way
ride+walk= 7 hour 30 min —(i)
ride= 2 hour 10 min –(ii)
Diff (i)-(ii)=walk one way=5 hour 20 min
walk 2 way= 10 hour 40 mins=640 mins
Incorrect
7 hour 30 mins – 3 hours 10 mins =4 hours
20 mins(When riding both ways)
=> 2 hour 10 mins riding in one way
ride+walk= 7 hour 30 min —(i)
ride= 2 hour 10 min –(ii)
Diff (i)-(ii)=walk one way=5 hour 20 min
walk 2 way= 10 hour 40 mins=640 mins
Question 11 of 25
11. Question
1 points
In covering a distance the speed of A and B are in the ratio 4:5. A takes 40 mins more
than B to reach the destination. The time taken by A to reach the destination is?
Correct
Speed (A:B)=4:5
Time (A:B)=5:4
Time diff=5-4=1
1=40 min
5=200 mins= 3(1/3) hours
Incorrect
Speed (A:B)=4:5
Time (A:B)=5:4
Time diff=5-4=1
1=40 min
5=200 mins= 3(1/3) hours
Question 12 of 25
12. Question
1 points
Two person cover some distance at a speed of 35 kmph and 40 kmph respectively. Find
the distance if one person takes 15 minute more than the other.
Two busses start at same time from two stations and move towards each other at the
rate of 30 kmph and 35 kmph respectively. When they meet one bus has traveled 60
km more than the other. Find the distance between the two bus stations.
Correct
The bus with higher speed moves 35-30=5 km more than the other in 1 hour
means it will move 60 more in 60/5=12 hours
Hence distance=30*12+35*12=780 km
Incorrect
The bus with higher speed moves 35-30=5 km more than the other in 1 hour
means it will move 60 more in 60/5=12 hours
Hence distance=30*12+35*12=780 km
Question 14 of 25
14. Question
1 points
A train crosses a pole in 20 seconds and a platform in 45 seconds. If the length of
platform is 500 meters find the sum of length of train and platform
Correct
Train : Platform= 20 : 25
25=500
20=400
total length=900 m
Incorrect
Train : Platform= 20 : 25
25=500
20=400
total length=900 m
Question 15 of 25
15. Question
1 points
A train overtakes two persons who are walking in the same direction in which the
train is moving, at the rate of 2 kmph and 4 kmph respectively and passes them
completely in 9 seconds and 10 seconds respectively. Find the length of the train.
Correct
x/(y-2)*18/5=10 ——(i)
x/(y-4)*18/5=9 –(ii)
solve and get x=50 m
Incorrect
x/(y-2)*18/5=10 ——(i)
x/(y-4)*18/5=9 –(ii)
solve and get x=50 m
Question 16 of 25
16. Question
1 points
The distance between two towns A and B is 545 km. A train starts from town A at 8 A.M. and travels towards town B at 80 km/hr. Another train starts from town B at 9 : 30 A.M. and travels towards town A at 90 km/hr. At what time will they meet each other?
Correct
With 80 km/hr, distance travelled in 1 n half hours (9:30AM – 8AM) is 3/2 * 80 = 120 Km
Now second train also starts, and at this time distance between both trains is (545-120) = 425 km
Relative speed = 80+90 = 170 km/hr (when travelling in opposite direction, add speed)
So time when they meet = 425/170 = 2.5 hrs
So after 9:30 AM they meet after 2.5 hrs, so 12 AM
Incorrect
With 80 km/hr, distance travelled in 1 n half hours (9:30AM – 8AM) is 3/2 * 80 = 120 Km
Now second train also starts, and at this time distance between both trains is (545-120) = 425 km
Relative speed = 80+90 = 170 km/hr (when travelling in opposite direction, add speed)
So time when they meet = 425/170 = 2.5 hrs
So after 9:30 AM they meet after 2.5 hrs, so 12 AM
Question 17 of 25
17. Question
1 points
A bus can travel 560 km in 8 hours. The ratio of speed to train to that of car is 13 : 8. If the speed of bus is 7/8 of the speed of car, find in how much time train can cover 520 km distance.
Correct
Speed of bus = 560/8 = 70 km/hr
So speed of car = 8/7 * 70 = 80 km/hr
So speed of train = 130 km/hr
So time taken by train to cover 520 km =
520/130 = 4 hours
Incorrect
Speed of bus = 560/8 = 70 km/hr
So speed of car = 8/7 * 70 = 80 km/hr
So speed of train = 130 km/hr
So time taken by train to cover 520 km =
520/130 = 4 hours
Question 18 of 25
18. Question
1 points
A person has to travel from point A to point B in car in a scheduled time at uniform speed.
Due to some problem in car engine, the speed of car has to be decreased by 1/5th of the original speed after covering 30 km. With this speed he reaches point B 45 minutes late than the scheduled time. Had the engine be malfunctioned after 48 km, he would have
reached late by only 36 minutes. Find the distance between points A and B.
Correct
Let total distance be d km, speed = u, and time =
t hours
So case 1:
30 km with speed u, (d-30) with speed 1 – 1/5 =
4/5 of u
If he would have travelled (d-30) by speed u,
then time = (d-30)/u
But now time is = (d-30)/(4u/5) = 5(d-30)/4u
And difference in timings is 45 minutes = 3/4
hour
So 5(d-30)/4u – (d-30)/u = 3/4
Solve (d-30)/u = 3
case 2:
48 km with speed u, (d-48) with speed 1 – 1/5 =
4/5 of u
If he would have travelled (d-48) by speed u,
then time = (d-48)/u
But now time is = (d-48)/(4u/5) = 5(d-48)/4u
And difference in timings is 36 minutes = 3/5
hour
So 5(d-48)/4u – (d-48)/u = 3/5
Solve (d-48)/4u = 3/5
Divide both equations, d = 120 km
Incorrect
Let total distance be d km, speed = u, and time =
t hours
So case 1:
30 km with speed u, (d-30) with speed 1 – 1/5 =
4/5 of u
If he would have travelled (d-30) by speed u,
then time = (d-30)/u
But now time is = (d-30)/(4u/5) = 5(d-30)/4u
And difference in timings is 45 minutes = 3/4
hour
So 5(d-30)/4u – (d-30)/u = 3/4
Solve (d-30)/u = 3
case 2:
48 km with speed u, (d-48) with speed 1 – 1/5 =
4/5 of u
If he would have travelled (d-48) by speed u,
then time = (d-48)/u
But now time is = (d-48)/(4u/5) = 5(d-48)/4u
And difference in timings is 36 minutes = 3/5
hour
So 5(d-48)/4u – (d-48)/u = 3/5
Solve (d-48)/4u = 3/5
Divide both equations, d = 120 km
Question 19 of 25
19. Question
1 points
Towns A and B are 225 km apart. Two cars P and Q travel from towards each other from
towns A and B respectively and meet after 3 hours. If the speed of P be 1/2 of its original
speed and Q be 2/3 of its original speed, they would have met after 5 hours. Find the speed of the faster car.
Correct
Let speeds be x km/hr and y km/hr
So 225/(x+y) = 3
And 225/(x/2 + 2y/3) = 5
Solve, x = 30, y = 45
Incorrect
Let speeds be x km/hr and y km/hr
So 225/(x+y) = 3
And 225/(x/2 + 2y/3) = 5
Solve, x = 30, y = 45
Question 20 of 25
20. Question
1 points
From point A, Priya and Bhavna start cycling towards point B which is 60 km away from A. The speed of Priya is 10 km/hr more than the speed of Bhavna. After reaching point B, Priya returns towards point A and meets Bhavna 12 km away from point B. Find the speed of Bhavna.
Correct
Speed of Bhavna = x km/hr, of priya = (x+10)
km/hr
Distance covered by Priya = 60+12 = 72 km
And by Bhavna = 60-12 = 48 km
So
72/(x+10) = 48/x
Solve, x = 20
Incorrect
Speed of Bhavna = x km/hr, of priya = (x+10)
km/hr
Distance covered by Priya = 60+12 = 72 km
And by Bhavna = 60-12 = 48 km
So
72/(x+10) = 48/x
Solve, x = 20
Question 21 of 25
21. Question
1 points
A train crosses 2 men running in the same direction at speeds 5 km/hr and 8 km/hr in 12
seconds and 15 seconds respectively. Find the speed of the train.
Correct
Let the speed of the train is s km/hr and its
length is a m.
So
a/[(s-5)*(5/18)] = 12; [In same direction relative
speed is obtained by subtracting. Also changing
km/hr to m/s]
Solve 3a = 10s – 50 . . . . . . . . (i)
And also
a/[(s-8)*(5/18)] = 15;
6a = 25s-200 . . . . . . . . . .. . . (ii)
Solve (i) and (ii)
s = 20 km/hr
Incorrect
Let the speed of the train is s km/hr and its
length is a m.
So
a/[(s-5)*(5/18)] = 12; [In same direction relative
speed is obtained by subtracting. Also changing
km/hr to m/s]
Solve 3a = 10s – 50 . . . . . . . . (i)
And also
a/[(s-8)*(5/18)] = 15;
6a = 25s-200 . . . . . . . . . .. . . (ii)
Solve (i) and (ii)
s = 20 km/hr
Question 22 of 25
22. Question
1 points
A train which is travelling at 80 km/hr meets another train travelling in same direction and then leaves it 150 m behind in next 20 seconds. Find the speed of the second train.
Correct
Let speed of the 2nd train is s m/sec.
80 km/hr = (80*5)/18 = 200/9 m/sec.
Trains are travelling in same direction. So
(200/9) – s = 150/20
Solve, s = 265/18 m/sec = 265/18 * 18/5 = 53 km/hr
Incorrect
Let speed of the 2nd train is s m/sec.
80 km/hr = (80*5)/18 = 200/9 m/sec.
Trains are travelling in same direction. So
(200/9) – s = 150/20
Solve, s = 265/18 m/sec = 265/18 * 18/5 = 53 km/hr
Question 23 of 25
23. Question
1 points
In a 500 m race C can beat B by 30 m, and in a 400 m race B can beat C by 20 m. Then in 200 m race A will beat C by how much distance (in m)?
Correct
When A runs 500 m, B runs 470 m
So when A runs 200 m, B runs 470/500 * 200 =
188 m
When B runs 400 m, C runs 280 m
So when B runs 188 m, C runs, 280/400 * 188 =
131.6 m
So A will beat C by (200-131.6) = 68.4 m
Incorrect
When A runs 500 m, B runs 470 m
So when A runs 200 m, B runs 470/500 * 200 =
188 m
When B runs 400 m, C runs 280 m
So when B runs 188 m, C runs, 280/400 * 188 =
131.6 m
So A will beat C by (200-131.6) = 68.4 m
Question 24 of 25
24. Question
1 points
2 towns A and B are 300 km apart. 2 trains start travelling from town A towards town B
such that the second train leaves 8 hours late than the first one. They both arrive at town B simultaneously. If the speed of the faster train is 10 km/hr more than the speed of the slower train, find the time taken by the slower train to complete the journey.
Correct
Let speed of the slower train is x km/hr, then
speed of faster is (x+10) kmph.
Let faster train takes t hours to cover the
distance 300 km, then slower one takes (t+8)
hours.
Distance is same. So
x/(x+10) = t/(t+8)
Solve, 4x = 5t
Incorrect
Let speed of the slower train is x km/hr, then
speed of faster is (x+10) kmph.
Let faster train takes t hours to cover the
distance 300 km, then slower one takes (t+8)
hours.
Distance is same. So
x/(x+10) = t/(t+8)
Solve, 4x = 5t
Question 25 of 25
25. Question
1 points
A man leaves from point A at 4 AM and reaches point B at 6 AM. Another man leaves
from point B at 5 AM and reaches point A at 8 AM. Find the time when they meet.
Correct
Use formula:
4 AM + (6-4)*(8-4)/[(6-4)+(8-5)]
gives 4 AM + 8/5
8/5 hours = 1 3/5 hours = 1 3/5*60 = 1 hour 36
minutes
So 4 AM + 1 hour 36 minutes = 5:36 AM
Incorrect
Use formula:
4 AM + (6-4)*(8-4)/[(6-4)+(8-5)]
gives 4 AM + 8/5
8/5 hours = 1 3/5 hours = 1 3/5*60 = 1 hour 36
minutes
So 4 AM + 1 hour 36 minutes = 5:36 AM
