1. A train is running with the speed of 45 km per hour? what is its speed in meter per second ?
(a) 12.5 m/s (b) 13 m/s
(c) 16 m/s (d) 81 m/s (e)None of the
Answer: (a) 12.5 m/s
Solution:
Speed of train = 45 km per hour
Convert km to meters= 45 x 1000 = 45000 m ( because 1 km = 1000 m)
Convert Hour to seconds=
60 x 60 seconds = 3600 seconds (because 1 hour = 60 min, 1 min = 60 seconds)
Therefore 45 km/hr = (45000/3600) meter per second = 12.5 meter per second
Short Cut:
1 Km/hr = (5/18) m/s
Simply Multiply the speed in km/hr by (5/18) to get the answer in m/s.
here, 45 km/hr = (45)(5/18) = 12.5 m/s
2. A train 100 m long is running at the speed of 21 km /hr and another train 150 m long is running at the speed of 36 km /hr in the same direction.How long will the faster train take to pass the first train ?
(a) 2 min (b)9 min
(c) 1 min (d) 7 min (e) None of the
Answer: (c) 1 min
Solution:
The faster train is the one that is running at 36 km/hr.
The difference in speed of the two trains is (36 - 21 ) km/hr = 15 km/hr
And the combined length of the train is (100 + 150 ) m = 250 m.
Therefore, the time taken is = (250m )/(15 km/hr) = (250 m)/ (15 x 5/18 m/s)
= 60 seconds
= 1 minutes.
3. A plat form is 131 meter long if the speed of the train, 67 meter long, is 45 km per hour, how long will it take to pass the platform ?
(a) 15.84 sec (b) 59 sec
(c) 439 sec (d) 36.7 sec (e) None of the
Solution:
Here, the platform is stationary. There its speed = 0
The difference in speed between the two objects = 45 - 0=45 km/hr=45 (5/18) m/s= 12.5 m/s
And the combined length of the platform and the train = (131 + 67) = 198 m
Thus the time taken by the train to pass the platform = 198m/12.5m/s = 15.84 seconds
4. Find the distance covered by a man walking for 10 minutes at a speed of 6 km/hr.
(a) 3 km (b) 4 km
(c) 8 km (d) 1 km
Answer: (d) 1 km
Solution:
Total distance covered = speed x time = (6 km/hr)x(10minutes) = (6km/hr)(10/60)hr
= 1 Km
5. A motor car does a journey in 10 hrs for the first half at 21 km/hr and the second half at 24 km/ hr. Find the distance?
(a) 244 km (b) 238 km
(c) 268 km (d)224 km
Answer: (d)224 km
Solution:
Total distance covered = speed x time
Let the Total distance be X.
Then, for the first half (X/2) speed was 21 Km/hr.
Therefore the time taken for the first half = (X/2)/(21 Km/hr) = X/42 hrs --------- (i)
And the time taken for the second half is = (X/2)/(24) = X/48 hrs --------- (ii)
But given that total time taken is 10 hrs. Equating (i) and (ii) to get
X/42 + X/48 = 10
=> X(8 + 7)/336 = 10
=> X(15) = 3360
=> X = (3360)/15
=> X = 224 km/hr
6. Walking at (3/4)th of his usual speed a person is 10 min late to office. Find his usual time to cover the distance.
(a) 35 min (b) 30 min
(c) 38 min (d) 25 min
Answer: (b) 30 min
Solution:
Let the distance between his house and office is X km and his usual speed be S.
Then, his usual time to reach office = X/S
And if he is walking at (3/4)th of his usual speed then the time taken is= (X)/ (3S/4) = 4X/3S
Again, It is given that he was 10 min or 1/6 hours late.
Therefore the equation thus forms is
4X/3S - X/S = 1/6
=> X/S( 4/3 - 1) = 1/6
=> X/S (1/3)= 1/6
=> X/S = (1/6)(3/1) = 3/6 = (1/2) km/hr
Therefore, the time taken walking at 1/2 km/hr = 2X hours
And time taken walking at (3/4)(1/2) or 3/8 km/hr = 8X/3 hours.
Again it is given walking at (3/4) speed he is late by 10 min or 1/6 hours
=> (8X/3) - 2X = 1/6
=> X(2/3) =1/6
=> X = (1/6)(3/2) = 1/4 km
And usual time = (1/4)/(1/2) = 1/2 hour or 30 minutes
And Since time taken is inversely proportional to the speed then his time taken would be (4/3) of his usual time.
Then, Let the time be T,
then (4T/3) - T = 1/6
(1/3)T = 1/6
T = 3/6 = 1/2 hours or 30 minutes
7. Two men A and B start from a place P walking at 3 km and 3.5 km an hour respectively. How many km will they be apart at the end of 3 hours if they walk in same direction ?
(a) 6.5 km (b) 1.5 km
(c) 7.7 km (d) 3.2 km
Answer: (b) 1.5 km
Solution:
Distance Covered by A in 3 Hours = (3)(3.5) = 10.5 km
Distance Covered by B in 3 Hours = (3)(3) = 9 km
Thus they will be 10.5 - 9 = 1.5 km apart walking in the same direction.
8. Two men A and B walk from P to Q , a distance of 21 km, at 3 and 4 km an hour respectively B reaches Q, returns immediately and meets A at R. Find the distance from P to R .
(a) 18 km (b) 34 km
(c) 20 km (d) 22 km
Answer: (a) 18 km
Solution:
Time Taken by B to reach Q is 21/4 hours.
In this time, A covered distance of = (21/4)(3) = 63/4 km.
He is still short of 21 - 63/4 = (84-63)/4 = 21/4 km.
Now, after reaching Point Q, B turns back to go to point P.
The relative speed of A and B is = 3 + 4 = 7 km
And distance Between them = 21/4 km.
Therefore, time taken to cover this distance = (21/4)/7 = 3/4 hours.
And distance traveled by A in this time to reach point R is = (3/4)3 = 9/4 km.
Thus, the distance between Point P and R = (63/4) + (9/4) = 72/4 = 18 km.
9. A man covers a certain distance by car driving at 70 km/hr and he returns back to the starting point riding on a scooter at 55 km/hr. Find his average speed for the whole journey.
(a) 31.5 km/hr (b) 71.5 km/hr
(c) 11.5 km/hr (d) 61.5 km/hr
Answer: (d) 61.5 km/hr
Solution:
For such questions, where the distance covered is same but at different speeds the formula for the average speed is
average speed = (2xy)/(x + y), where x and y are the different speeds.
=> average speed = (2 x 70 x 55)/(70 + 55)
= 7700/125
= 61.5 km/hr
10. Peter can cover a certain distance in 1 hour 24 minutes by covering two third of the distance at 4 km/hr and rest at 5 km/hr. Find the total distance.
(a) 6 km (b) 19 km
(c) 22 km (d) 32 km
Answer: (a) 6 km
Solution:
Let the total distance be X km.
Then Peter covered (2/3)X of the distance at 4 km/hr in time = (2X/3)/4 = X/6 hours
And Peter covered (1/3)X of the distance at 3 km/hr in time = (X/3)/5 = ( X/15) hours
Given that total time taken was 1 hour 24 minutes = 1 + 24/60 = 1 + 2/5 = 7/5 hours
There the equation thus formed is X/6 + (X/15) = 7/5
X(5 + 2 )/30=7/5
X(7/30) = 7/5
X = (7/5)(30/7)
X = 6 km
11. An hero plane flies along the four sides of a square at the speed of 200, 400, 600 and 800 km/hr . Find the average speed of the plane around the field .
(a) 384 km (b) 333 km
(c) 324 km (d) 312 km
Answer:(a) 384 km
Solution:
For such questions , where the distance covered is the same for all the speeds, the formula for 4 different speeds is
4abcd/(abc + bcd + cda + dab)
where, a,b,c,d are the different speeds to cover the same distance
Therefore, average speed =
{4(200)(400)(600)(800)}/{(200)(400)(600) + (400)(600)(800) + (600)(800)(200) + (800)(200)(400)}
= (400)(2)(4)(6)(8)/{(2)(4)(6) + (4)(6)(8) + (6)(8)(2)+(8)(2)(4)}
= (153600)/( 48 + 192 + 96 + 64)
= (153600)/(240 + 160) = 384 km
12. A man is walking at a speed of 12 km per hour. After every km he takes rest for 12 minutes. How much time will he take to cover a distance of 36 km ?
(a) 30 hours (b) 10 hours
(c) 25 hours (d) 40 hours
Answer: (b) 10 hours
Solution:
Total time taken to cover the distance of 36 km (without the rest time) = 36/12 = 3 hours
Now, number of rest he will take to complete the journey = 36 - 1 = 35. 35 because he would have covered the entire distance by the time he takes 36th rest break.
Therefore total rest taken = 35 x 12 = 420 min = 7 hours.
Thus Total time taken = 3 + 7 = 10 hours.
13. I walk a certain distance, ride back taking a total time of 37 minutes . I could walk both ways in 55 minutes. How long would it take me to ride both ways .
(a) 22 min (b) 28 min
(c) 26 min (d) 19 min
Answer: (d) 19 min
Solution:
Given that
riding time + walking time = 37 minutes ----------(i)
and walking time + walking time = 55 minutes
=> 2 x walking time = 55
=> walking time = 55/2 = 27.5 minutes
Put this in the equation (i)
Therefore, 27.5 minutes + riding time = 37 minutes
riding time = 37 - 27.5 = 9.5 minutes
Therefore, total time taken to ride back = 2 x 9.5 = 19 minutes
14. A man travels a distance of 18 km from his house to an exhibition by tonga at 15 km/hr and returns back on cycle at 10 km/hr . Then the average speed for the whole journey is
(a) 12 km/hr (b) 10 km/hr
(c) 15 km/hr (d) 18 km/hr
Answer:
Solution:
Since the distance traveled is the same for both the journey, therefore using the formula mentioned in the Short Cut formula for time speed and distance . we have
Average Speed = 2(15)(10)/(15+10) = 2(15)(10)/(25) = 12 km/hr
15. A gun is fired at a distance of 1.34 km from Geeta. She hears the sound after 4 secs. The speed at which the sound travels is
(a) 330 m/s (b) 300 m/s
(c) 325 m/s (d) 335 m/s
Answer: (d) 335 m/s
Solution:
Speed of sound = (1.34 x1000 m)/ (4 sec)
= (1340 m)/ 4 sec
= 335 m/s
16. If I walk at 5 km/hr , I miss a train by 7 minutes. However, if I walk at 6 km/hr I reach
the station 5 minutes before the departure of the train. The distance between my house and the station is
(a) 7 km (b) 6 km
(c) 5 km (d) 6.5 km
Answer:
Solution:
Using the formula mentioned in the Short Cut formula for time speed and distance, the required distance is
= {(5x6)/(6-5)} x {(7+5)/60}
= (30) x {(12)/60}
= 6 km
17. Two trains starts from the stations A and B and travel towards each other at speeds of 50 kmph and 60 kmph respectively. At the time of their meeting, the second train has traveled 120 km more than the first. The distance between A and B is
(a) 1200 km (b) 1440 km
(c) 1320 km (f) 990 km
Answer:(c) 1320 km
Solution:
The trains were traveling towards each other then the relative speed is (50+60)=110 kmph
Again it is given that the second train traveled 120 km more than first.
The difference in their speed = 60 - 50 = 10 kmph
And time taken to cover 120 kmph = 120/10 = 12 hours.
Thus, it took 12 hours for the trains to cover as distance of 110x12 = 1320 km
5. Tom is chasing Jerry. In the same interval of time Tom jumps 8 times while Jerry jumps 6 times. But the distance covered by Tom in 7 jumps is equal to the distance covered by Jerry in 5 jumps. The ratio of speed of Tom and Jerry is
(a) 48:35 (b) 28:15
(c) 24:20 (d) 20:21
Answer: (b) 28:15
Solution:
Distance covered by Tom in 7 jump = Distance covered by Jerry in 5 jumps.
Distance covered by Tom in 1 jump = (5/7) jumps of Jerry.
Therefore, ratio of speed of Tom to Jerry = (8x1)/{6x(5/7)}
= 56/30 = 28/15
(a) 12.5 m/s (b) 13 m/s
(c) 16 m/s (d) 81 m/s (e)None of the
Answer: (a) 12.5 m/s
Solution:
Speed of train = 45 km per hour
Convert km to meters= 45 x 1000 = 45000 m ( because 1 km = 1000 m)
Convert Hour to seconds=
60 x 60 seconds = 3600 seconds (because 1 hour = 60 min, 1 min = 60 seconds)
Therefore 45 km/hr = (45000/3600) meter per second = 12.5 meter per second
Short Cut:
1 Km/hr = (5/18) m/s
Simply Multiply the speed in km/hr by (5/18) to get the answer in m/s.
here, 45 km/hr = (45)(5/18) = 12.5 m/s
2. A train 100 m long is running at the speed of 21 km /hr and another train 150 m long is running at the speed of 36 km /hr in the same direction.How long will the faster train take to pass the first train ?
(a) 2 min (b)9 min
(c) 1 min (d) 7 min (e) None of the
Answer: (c) 1 min
Solution:
The faster train is the one that is running at 36 km/hr.
The difference in speed of the two trains is (36 - 21 ) km/hr = 15 km/hr
And the combined length of the train is (100 + 150 ) m = 250 m.
Therefore, the time taken is = (250m )/(15 km/hr) = (250 m)/ (15 x 5/18 m/s)
= 60 seconds
= 1 minutes.
3. A plat form is 131 meter long if the speed of the train, 67 meter long, is 45 km per hour, how long will it take to pass the platform ?
(a) 15.84 sec (b) 59 sec
(c) 439 sec (d) 36.7 sec (e) None of the
Solution:
Here, the platform is stationary. There its speed = 0
The difference in speed between the two objects = 45 - 0=45 km/hr=45 (5/18) m/s= 12.5 m/s
And the combined length of the platform and the train = (131 + 67) = 198 m
Thus the time taken by the train to pass the platform = 198m/12.5m/s = 15.84 seconds
4. Find the distance covered by a man walking for 10 minutes at a speed of 6 km/hr.
(a) 3 km (b) 4 km
(c) 8 km (d) 1 km
Answer: (d) 1 km
Solution:
Total distance covered = speed x time = (6 km/hr)x(10minutes) = (6km/hr)(10/60)hr
= 1 Km
5. A motor car does a journey in 10 hrs for the first half at 21 km/hr and the second half at 24 km/ hr. Find the distance?
(a) 244 km (b) 238 km
(c) 268 km (d)224 km
Answer: (d)224 km
Solution:
Total distance covered = speed x time
Let the Total distance be X.
Then, for the first half (X/2) speed was 21 Km/hr.
Therefore the time taken for the first half = (X/2)/(21 Km/hr) = X/42 hrs --------- (i)
And the time taken for the second half is = (X/2)/(24) = X/48 hrs --------- (ii)
But given that total time taken is 10 hrs. Equating (i) and (ii) to get
X/42 + X/48 = 10
=> X(8 + 7)/336 = 10
=> X(15) = 3360
=> X = (3360)/15
=> X = 224 km/hr
6. Walking at (3/4)th of his usual speed a person is 10 min late to office. Find his usual time to cover the distance.
(a) 35 min (b) 30 min
(c) 38 min (d) 25 min
Answer: (b) 30 min
Solution:
Let the distance between his house and office is X km and his usual speed be S.
Then, his usual time to reach office = X/S
And if he is walking at (3/4)th of his usual speed then the time taken is= (X)/ (3S/4) = 4X/3S
Again, It is given that he was 10 min or 1/6 hours late.
Therefore the equation thus forms is
4X/3S - X/S = 1/6
=> X/S( 4/3 - 1) = 1/6
=> X/S (1/3)= 1/6
=> X/S = (1/6)(3/1) = 3/6 = (1/2) km/hr
Therefore, the time taken walking at 1/2 km/hr = 2X hours
And time taken walking at (3/4)(1/2) or 3/8 km/hr = 8X/3 hours.
Again it is given walking at (3/4) speed he is late by 10 min or 1/6 hours
=> (8X/3) - 2X = 1/6
=> X(2/3) =1/6
=> X = (1/6)(3/2) = 1/4 km
And usual time = (1/4)/(1/2) = 1/2 hour or 30 minutes
Short cut:
Given that he is walking at 3/4 of his usual speed.And Since time taken is inversely proportional to the speed then his time taken would be (4/3) of his usual time.
Then, Let the time be T,
then (4T/3) - T = 1/6
(1/3)T = 1/6
T = 3/6 = 1/2 hours or 30 minutes
7. Two men A and B start from a place P walking at 3 km and 3.5 km an hour respectively. How many km will they be apart at the end of 3 hours if they walk in same direction ?
(a) 6.5 km (b) 1.5 km
(c) 7.7 km (d) 3.2 km
Answer: (b) 1.5 km
Solution:
Distance Covered by A in 3 Hours = (3)(3.5) = 10.5 km
Distance Covered by B in 3 Hours = (3)(3) = 9 km
Thus they will be 10.5 - 9 = 1.5 km apart walking in the same direction.
8. Two men A and B walk from P to Q , a distance of 21 km, at 3 and 4 km an hour respectively B reaches Q, returns immediately and meets A at R. Find the distance from P to R .
(a) 18 km (b) 34 km
(c) 20 km (d) 22 km
Answer: (a) 18 km
Solution:
Time Taken by B to reach Q is 21/4 hours.
In this time, A covered distance of = (21/4)(3) = 63/4 km.
He is still short of 21 - 63/4 = (84-63)/4 = 21/4 km.
Now, after reaching Point Q, B turns back to go to point P.
The relative speed of A and B is = 3 + 4 = 7 km
And distance Between them = 21/4 km.
Therefore, time taken to cover this distance = (21/4)/7 = 3/4 hours.
And distance traveled by A in this time to reach point R is = (3/4)3 = 9/4 km.
Thus, the distance between Point P and R = (63/4) + (9/4) = 72/4 = 18 km.
9. A man covers a certain distance by car driving at 70 km/hr and he returns back to the starting point riding on a scooter at 55 km/hr. Find his average speed for the whole journey.
(a) 31.5 km/hr (b) 71.5 km/hr
(c) 11.5 km/hr (d) 61.5 km/hr
Answer: (d) 61.5 km/hr
Solution:
For such questions, where the distance covered is same but at different speeds the formula for the average speed is
average speed = (2xy)/(x + y), where x and y are the different speeds.
=> average speed = (2 x 70 x 55)/(70 + 55)
= 7700/125
= 61.5 km/hr
10. Peter can cover a certain distance in 1 hour 24 minutes by covering two third of the distance at 4 km/hr and rest at 5 km/hr. Find the total distance.
(a) 6 km (b) 19 km
(c) 22 km (d) 32 km
Answer: (a) 6 km
Solution:
Let the total distance be X km.
Then Peter covered (2/3)X of the distance at 4 km/hr in time = (2X/3)/4 = X/6 hours
And Peter covered (1/3)X of the distance at 3 km/hr in time = (X/3)/5 = ( X/15) hours
Given that total time taken was 1 hour 24 minutes = 1 + 24/60 = 1 + 2/5 = 7/5 hours
There the equation thus formed is X/6 + (X/15) = 7/5
X(5 + 2 )/30=7/5
X(7/30) = 7/5
X = (7/5)(30/7)
X = 6 km
11. An hero plane flies along the four sides of a square at the speed of 200, 400, 600 and 800 km/hr . Find the average speed of the plane around the field .
(a) 384 km (b) 333 km
(c) 324 km (d) 312 km
Answer:(a) 384 km
Solution:
For such questions , where the distance covered is the same for all the speeds, the formula for 4 different speeds is
4abcd/(abc + bcd + cda + dab)
where, a,b,c,d are the different speeds to cover the same distance
Therefore, average speed =
{4(200)(400)(600)(800)}/{(200)(400)(600) + (400)(600)(800) + (600)(800)(200) + (800)(200)(400)}
= (400)(2)(4)(6)(8)/{(2)(4)(6) + (4)(6)(8) + (6)(8)(2)+(8)(2)(4)}
= (153600)/( 48 + 192 + 96 + 64)
= (153600)/(240 + 160) = 384 km
12. A man is walking at a speed of 12 km per hour. After every km he takes rest for 12 minutes. How much time will he take to cover a distance of 36 km ?
(a) 30 hours (b) 10 hours
(c) 25 hours (d) 40 hours
Answer: (b) 10 hours
Solution:
Total time taken to cover the distance of 36 km (without the rest time) = 36/12 = 3 hours
Now, number of rest he will take to complete the journey = 36 - 1 = 35. 35 because he would have covered the entire distance by the time he takes 36th rest break.
Therefore total rest taken = 35 x 12 = 420 min = 7 hours.
Thus Total time taken = 3 + 7 = 10 hours.
13. I walk a certain distance, ride back taking a total time of 37 minutes . I could walk both ways in 55 minutes. How long would it take me to ride both ways .
(a) 22 min (b) 28 min
(c) 26 min (d) 19 min
Answer: (d) 19 min
Solution:
Given that
riding time + walking time = 37 minutes ----------(i)
and walking time + walking time = 55 minutes
=> 2 x walking time = 55
=> walking time = 55/2 = 27.5 minutes
Put this in the equation (i)
Therefore, 27.5 minutes + riding time = 37 minutes
riding time = 37 - 27.5 = 9.5 minutes
Therefore, total time taken to ride back = 2 x 9.5 = 19 minutes
14. A man travels a distance of 18 km from his house to an exhibition by tonga at 15 km/hr and returns back on cycle at 10 km/hr . Then the average speed for the whole journey is
(a) 12 km/hr (b) 10 km/hr
(c) 15 km/hr (d) 18 km/hr
Answer:
Solution:
Since the distance traveled is the same for both the journey, therefore using the formula mentioned in the Short Cut formula for time speed and distance . we have
Average Speed = 2(15)(10)/(15+10) = 2(15)(10)/(25) = 12 km/hr
15. A gun is fired at a distance of 1.34 km from Geeta. She hears the sound after 4 secs. The speed at which the sound travels is
(a) 330 m/s (b) 300 m/s
(c) 325 m/s (d) 335 m/s
Answer: (d) 335 m/s
Solution:
Speed of sound = (1.34 x1000 m)/ (4 sec)
= (1340 m)/ 4 sec
= 335 m/s
16. If I walk at 5 km/hr , I miss a train by 7 minutes. However, if I walk at 6 km/hr I reach
the station 5 minutes before the departure of the train. The distance between my house and the station is
(a) 7 km (b) 6 km
(c) 5 km (d) 6.5 km
Answer:
Solution:
Using the formula mentioned in the Short Cut formula for time speed and distance, the required distance is
= {(5x6)/(6-5)} x {(7+5)/60}
= (30) x {(12)/60}
= 6 km
17. Two trains starts from the stations A and B and travel towards each other at speeds of 50 kmph and 60 kmph respectively. At the time of their meeting, the second train has traveled 120 km more than the first. The distance between A and B is
(a) 1200 km (b) 1440 km
(c) 1320 km (f) 990 km
Answer:(c) 1320 km
Solution:
The trains were traveling towards each other then the relative speed is (50+60)=110 kmph
Again it is given that the second train traveled 120 km more than first.
The difference in their speed = 60 - 50 = 10 kmph
And time taken to cover 120 kmph = 120/10 = 12 hours.
Thus, it took 12 hours for the trains to cover as distance of 110x12 = 1320 km
5. Tom is chasing Jerry. In the same interval of time Tom jumps 8 times while Jerry jumps 6 times. But the distance covered by Tom in 7 jumps is equal to the distance covered by Jerry in 5 jumps. The ratio of speed of Tom and Jerry is
(a) 48:35 (b) 28:15
(c) 24:20 (d) 20:21
Answer: (b) 28:15
Solution:
Distance covered by Tom in 7 jump = Distance covered by Jerry in 5 jumps.
Distance covered by Tom in 1 jump = (5/7) jumps of Jerry.
Therefore, ratio of speed of Tom to Jerry = (8x1)/{6x(5/7)}
= 56/30 = 28/15
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