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Speed Time and Distance-Test 1
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Subject :- Quantitative Aptitude
Chapter :- Speed Time and Distance-Test 1
Questions :- 25
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Kiran covers a certain distance with his own speed , but when he reduces his speed
by 10 km/hr his time duration for the journey increases by 40 hrs, while if he
increases his speed by 5 km/hr from his original speed he takes 10 hrs less than the
original time taken . Find the distance covered by him.
Correct
Incorrect
Question 2 of 25
2. Question
1 points
A train met with an accident 60 km away from station A. It completed the remaining
journey at 5/6th of the original speed and reached station B 1 hr 12 mins late. Had the
accident taken place 60 km further, it would have been only 1 hr late. what was the
original speed of the train?
Correct
Let the original speed be 6x.
Travelling 60km at 5/6th of original speed
cost 12 mins etc.
60/5x = 60/6x +12/60
==>x=10
Original speed 6x=60km/hr.
Incorrect
Let the original speed be 6x.
Travelling 60km at 5/6th of original speed
cost 12 mins etc.
60/5x = 60/6x +12/60
==>x=10
Original speed 6x=60km/hr.
Question 3 of 25
3. Question
1 points
Two man start together to walk a certain distance, one at 4 km/hr and another at 5
km/hr. The former arrives half an hour before the latter. Find the distance.
Correct
If the distance be x km, then
x/4-x/5=1/2
(5x-4x)/20=1/2
x/20=1/2==>x=10km
Incorrect
If the distance be x km, then
x/4-x/5=1/2
(5x-4x)/20=1/2
x/20=1/2==>x=10km
Question 4 of 25
4. Question
1 points
In a flight of 600 km, an aircraft was slowed down due to bad weather. Its
average speed for the trip was reduced by 200 km/hr and the time of flight increased
by 30 minutes. The duration of the flight is:
Correct
Let the duration of the flight be x hrs.
Then, 600/x-600/(x+1/2)=200
600/x-1200/2x+1=200
X(2x+1)=3
2×2+x-3=0
X=1hr.
Incorrect
Let the duration of the flight be x hrs.
Then, 600/x-600/(x+1/2)=200
600/x-1200/2x+1=200
X(2x+1)=3
2×2+x-3=0
X=1hr.
Question 5 of 25
5. Question
1 points
Two racers start running towards each other, one from A to B and another from B
to A. They cross each other after one hour and the first racer reaches B, 5/6 hour
before the second racer reaches A. If the distance between A and B is 50 km. what is
the speed of the slower racer?
Correct
Let second racer takes x hr with speed s2
First racer takes x-5/6 hr with speed s1
Total distance = 50km
S1 = 50/(x-(5/6))
S2= 50/x
As they cross each other in 1hr…
Total speed = s1 + s2
Now, T = D / S
50/(s1+s2) = 1
x = 5/2, 1/3
Put x= 5/2 in s2 –> 20km/hr
Incorrect
Let second racer takes x hr with speed s2
First racer takes x-5/6 hr with speed s1
Total distance = 50km
S1 = 50/(x-(5/6))
S2= 50/x
As they cross each other in 1hr…
Total speed = s1 + s2
Now, T = D / S
50/(s1+s2) = 1
x = 5/2, 1/3
Put x= 5/2 in s2 –> 20km/hr
Question 6 of 25
6. Question
1 points
P and Q run at the speed 40m/s and 20m/s resp on the circular track of 800m, as its
circumference , when would the P and Q meet for the first time at the starting point if
they start simultaneously from the same point.
Correct
Time taken by P to complete one round
800/40=20
Time taken by Q to complete one round
800/20=40
LCM of 20 40=40
Every 40sec they would be together at the
starting point.
Incorrect
Time taken by P to complete one round
800/40=20
Time taken by Q to complete one round
800/20=40
LCM of 20 40=40
Every 40sec they would be together at the
starting point.
Question 7 of 25
7. Question
1 points
The speeds of Ram and Raj are 30 km/h and 40 km/h. Initially Raj is at a place L
and Ram is at a place M. The distance between L and M is 650 km. Ram started
his journey 3 hours earlier than Raj to meet each other. If they meet each other at a
place P somewhere between L and M, then the distance between P and M is?
Correct
If the 1st 3hr Ram covers 90km
So the rest 650-90=560km
Now they both travel together towards each
other
So, the time is 560/70=8hr
Then ram travel total 3+8=11hrs
Thus the distance travelled by Ram
11*30=330km
Incorrect
If the 1st 3hr Ram covers 90km
So the rest 650-90=560km
Now they both travel together towards each
other
So, the time is 560/70=8hr
Then ram travel total 3+8=11hrs
Thus the distance travelled by Ram
11*30=330km
Question 8 of 25
8. Question
1 points
The ratio between the speed of a car and a bike is 16 : 15 respectively. Also, a bus
covered a distance of 480 km in 8 hrs. The speed of the bus is three-fourth the speed of
the car. How much distance will the bike cover in 6 h?
Correct
Speed of bus = 480/8 = 60km/ h
Speed of car = (60*4)/3=80 km / h
Speed of car : Speed of bike = 16 : 15
Speed of bike =80/16 * 15 = 75 km/ h
Distance covered by bike in 6 hr = 75 × 6 =450 km
Incorrect
Speed of bus = 480/8 = 60km/ h
Speed of car = (60*4)/3=80 km / h
Speed of car : Speed of bike = 16 : 15
Speed of bike =80/16 * 15 = 75 km/ h
Distance covered by bike in 6 hr = 75 × 6 =450 km
Question 9 of 25
9. Question
1 points
How many seconds will a train 50 m in length, travelling at the rate of 42 km an
hour, rate to pass another train 80 m long, proceeding in the same direction at the rate
of 30 km an hour?
A man rides his bike 20 km at an average speed of 8 km/hr and again travels 45 km at
an average speed of 10 km/hr. What is his average speed for the ride approximately?
Correct
Average speed=total distance/total time
Total time=20/8 + 45/10
Avg speed=(20+45)/(20/8 + 45/10)
= 65/((200+360)/80)
=65*80/560 = 65/7
=9.3km/hr
Incorrect
Average speed=total distance/total time
Total time=20/8 + 45/10
Avg speed=(20+45)/(20/8 + 45/10)
= 65/((200+360)/80)
=65*80/560 = 65/7
=9.3km/hr
Question 11 of 25
11. Question
1 points
Two buses start at same time from Chennai and Bangalore, which are 250 km apart. If
the two buses travel towards each other, they meet after 1 hr and if they travel in
same direction they meet after 5 hrs. What is the speed of the bus starts from Chennai if it is know that the one which started from Chennai has more speed than the other one?
Correct
S=D/T
Here we have two speeds. We get 2
equations as.
250/1hr = C+B—-1 (Travelling in opposite
direction, speed must be added ie C+B)
250/5hr = C-B—-2 (Travelling in same
direction, speed to be subtracted. ie C-B)
solving 2 eqn C=150km/hr.
Incorrect
S=D/T
Here we have two speeds. We get 2
equations as.
250/1hr = C+B—-1 (Travelling in opposite
direction, speed must be added ie C+B)
250/5hr = C-B—-2 (Travelling in same
direction, speed to be subtracted. ie C-B)
solving 2 eqn C=150km/hr.
Question 12 of 25
12. Question
1 points
Car A leaves the city at 5 pmm and is driven at a speed of 30 km/hr. 3 hrs later another
car B leaves the city in the same direction as car A. In how much time will car B be
12 kms ahead of car A if the speed of car B is 50 km/hr?
Correct
Car A travels 3hrs. 3*30=90km
Difference between speeds 50-30=20km/hr
Distance ahead 12km . 90+12=102km
T=D/S ===>102/20=5.1hrs.
Incorrect
Car A travels 3hrs. 3*30=90km
Difference between speeds 50-30=20km/hr
Distance ahead 12km . 90+12=102km
T=D/S ===>102/20=5.1hrs.
Question 13 of 25
13. Question
1 points
Two train starts at the same time from Delhi and Agra and proceed towards each
other at the rate of 40 km/hr and 37 1/2 km/hr. When they meet it is found that
one train has traveled 200 km more than the other train. What is the distance between
Delhi and Agra?
Correct
Speed ratio 40:37 1/2==>40: 75/2==>80:75
ie 16:15
ratio diff betwwen speed is 1[16-15]
1 ===> 200 (more distance)
[16+15]31 ===>?
31*200=6200km.
Incorrect
Speed ratio 40:37 1/2==>40: 75/2==>80:75
ie 16:15
ratio diff betwwen speed is 1[16-15]
1 ===> 200 (more distance)
[16+15]31 ===>?
31*200=6200km.
Question 14 of 25
14. Question
1 points
If a car runs at 45km/hr, it reaches its destination late by 10 min but if runs at
60km/hr it is late by 4min. What is the correct time for the journey?
Correct
Distance = diff in time *(S1*S2)/S1-S2
D=[10-4]/60hr * (45*60)/[60-45] = 6/60 *
45*60/15 ==>18km
time T=D/S(take any one of the speed)
18/45 =2/5hrs = 2/5*60=24min
then correct time is 24-10=14mins.
Incorrect
Distance = diff in time *(S1*S2)/S1-S2
D=[10-4]/60hr * (45*60)/[60-45] = 6/60 *
45*60/15 ==>18km
time T=D/S(take any one of the speed)
18/45 =2/5hrs = 2/5*60=24min
then correct time is 24-10=14mins.
Question 15 of 25
15. Question
1 points
A bike rider starts at 40km/hr and he increases his speed in every 1 hour by
2km/hr. Then the maximum distance covered by him in 24 hrs is:
Correct
Speed of the rider: 40km/hr
He increases his speed in every 1 hr by
2km/hr.
Distance covered by every 1 hr will be
40,42,44….upto 12terms. i.e. (for 24hrs)
Sum of 1st n terms=n/2 ( (2a+(n-1)d)
12/2 * (2*40+11*2)==>12/2 *
(80+22)==>612km
Incorrect
Speed of the rider: 40km/hr
He increases his speed in every 1 hr by
2km/hr.
Distance covered by every 1 hr will be
40,42,44….upto 12terms. i.e. (for 24hrs)
Sum of 1st n terms=n/2 ( (2a+(n-1)d)
12/2 * (2*40+11*2)==>12/2 *
(80+22)==>612km
Question 16 of 25
16. Question
1 points
Two friends Ram and Ravi are travelling from point A to B, which are 600km apart.
Travelling at a certain speed Ram takes 1hr more than Ravi to reach point B. If Ram
doubles his speed he will take 1hr 30mins less than ravi to reach point B. At what
speed was Ram driving from point A to B?
Correct
T = D/S. Let x be the speed
600/x = T+1 , 600/2x = T – 3/2
Equate T
(600-x)/x = (600+3x)/2x
1200-12x = 600+3x ==>x=120km.
Incorrect
T = D/S. Let x be the speed
600/x = T+1 , 600/2x = T – 3/2
Equate T
(600-x)/x = (600+3x)/2x
1200-12x = 600+3x ==>x=120km.
Question 17 of 25
17. Question
1 points
A man takes 4hrs 30min in walking to a certain place and riding back. He would
have gained 2hrs by riding both ways. The time he would take to walk both ways, is:
Correct
W+R = 4hrs 30min ie 9/2hrs .
R+R=2hrs==>R=1hr
then 2W=9/2-1=7/2 , W=7/2*2=7hrs
Incorrect
W+R = 4hrs 30min ie 9/2hrs .
R+R=2hrs==>R=1hr
then 2W=9/2-1=7/2 , W=7/2*2=7hrs
Question 18 of 25
18. Question
1 points
Two trains of equal lengths take 10 seconds and 15 seconds respectively to cross a
telegraph post. If the length of each train be 300 metres, in what time will they cross
each other travelling in opposite direction?
Correct
Speed of the first train = [300/10] m/sec =30 m/sec.
Speed of the second train = [300/15] m/sec=20 m/sec.
speed = (30 + 20) m/sec = 50m/sec.
Required time = (300+300)/50secc = 12sec.
Incorrect
Speed of the first train = [300/10] m/sec =30 m/sec.
Speed of the second train = [300/15] m/sec=20 m/sec.
speed = (30 + 20) m/sec = 50m/sec.
Required time = (300+300)/50secc = 12sec.
Question 19 of 25
19. Question
1 points
The driver of a car sees a school van 60m ahead of him. After 30seconds the school
van is 60m behind. If the speed of the car is 45kmph, what is the speed of the School Van?
Correct
Relative speed = (60+60)/30 = 4m/s = 4 *18/5 = 14.4kmph
Speed of the school van = 45 – 14.4=30.6kmph
Incorrect
Relative speed = (60+60)/30 = 4m/s = 4 *18/5 = 14.4kmph
Speed of the school van = 45 – 14.4=30.6kmph
Question 20 of 25
20. Question
1 points
The distance between two cities A and B is 330km. A train starts from A at 8 AM. and
travels towards B at 60 km/hr. Another train starts from B at 9 AM. and travels towards A at 75 km/hr. At what time do they meet?
Correct
Distance travelled by first train in one hour
= 60 x 1 = 60 km
Therefore, distance between two train at 9AM.
= 330 – 60 = 270 km
Now, Relative speed of two trains = 60 +
75 = 135 km/hr
Time of meeting of two trains =270/135=2hrs.
Therefore, both the trains will meet at 9 + 2= 11 AM.
Incorrect
Distance travelled by first train in one hour
= 60 x 1 = 60 km
Therefore, distance between two train at 9AM.
= 330 – 60 = 270 km
Now, Relative speed of two trains = 60 +
75 = 135 km/hr
Time of meeting of two trains =270/135=2hrs.
Therefore, both the trains will meet at 9 + 2= 11 AM.
Question 21 of 25
21. Question
1 points
A man in a train he can count 31 telephone posts in one minute. If they are known to
be 45 m apart. Find the speed of the train.
Correct
speed of the train = 30*45 = (1350
*60)/1000 = 81 km/hr
Incorrect
speed of the train = 30*45 = (1350
*60)/1000 = 81 km/hr
Question 22 of 25
22. Question
1 points
If a man walks from his house to office at 6 km/hr , he is late by half an hour. However,
if he walks at 8 km/hr , he is late by 10 minutes only. What is the distance of his
office from his house.
Correct
(x/6)- (x/8) = (30-10)/60
=> x = 8 km
Incorrect
(x/6)- (x/8) = (30-10)/60
=> x = 8 km
Question 23 of 25
23. Question
1 points
A candle of 8 cm long burns at the rate of 7 cm in 7 hour and another candle of 10cm
long burns at the rate of 6 cm in 3 hour. What is the time required by each candle to
remain of equal lengths after burning for some hours, when they start to burn
simultaneously with uniform rate of burning?
Correct
(8 – x)=(10 – 2x)
=> x = 2 cm
Incorrect
(8 – x)=(10 – 2x)
=> x = 2 cm
Question 24 of 25
24. Question
1 points
Two trains, 190 m and 170 m long are going in the same direction. The faster train
takes one minute to pass the other completely. If they are moving in opposite
directions, they pass each other completely in 3 seconds. Find the speed of each train.
Correct
let the speed of the faster train be x and the
speed of the slower train be y . Then, When
they move in the same direction , the
relative speed = (x – y)
Total distance = 190 + 170 = 360 m
Now, dist. = speed * time = 360 = (x-y)*60
=> (x – y) = 6———-(1)
When the trains move in opposite direction
= (x+y)
360= (x + y)*3
=> 120 =(x + y) ———–(2)
On solving (1) and (2), we get
x = 63 m/sec. and y = 57 m/sec.
Incorrect
let the speed of the faster train be x and the
speed of the slower train be y . Then, When
they move in the same direction , the
relative speed = (x – y)
Total distance = 190 + 170 = 360 m
Now, dist. = speed * time = 360 = (x-y)*60
=> (x – y) = 6———-(1)
When the trains move in opposite direction
= (x+y)
360= (x + y)*3
=> 120 =(x + y) ———–(2)
On solving (1) and (2), we get
x = 63 m/sec. and y = 57 m/sec.
Question 25 of 25
25. Question
1 points
A motorcyclist covered (2/3)rd of a total journey at his usual speed. He covered the
remaining distance at (3/4)th of his usual speed. As a result, he arrived 30 minutes
later than the time he would have taken at usual speed. If the total journey was 180
km. What was his usual speed?
Correct
Let the usual speed be x km/hr.
(1/3)rd of the journey = 180/3 = 60km
Therefore, 60 / (3x/4) – 60/x = (1/2)
=> x = 40 km/hr.
Incorrect
Let the usual speed be x km/hr.
(1/3)rd of the journey = 180/3 = 60km
Therefore, 60 / (3x/4) – 60/x = (1/2)
=> x = 40 km/hr.