680.In track meets both 100 yards and 100 meters are used as distances; by how many meters is 100 meters longer than 100 yards?
0.856 m
8.56 m
0.0856 m
1. 0 m
Explanation:
1 yard = 36 inches
1 inch = 2.54 cm
∴ 1 yard = 2.54 × 36 cm
100 Yards = $\dfrac{2.54 \times 36}{100}$ =91.44m
Difference (100m – 100 yards) = (100 –91.44)m = 8.56 m.
1 inch = 2.54 cm
∴ 1 yard = 2.54 × 36 cm
100 Yards = $\dfrac{2.54 \times 36}{100}$ =91.44m
Difference (100m – 100 yards) = (100 –91.44)m = 8.56 m.
1224.X walked 35 meters towards South, then turned left and walked 25 metres, and again turned left and walked for 35 meters. He then turned to his right and walked 20 metres. At what distance and direction from the starting point is X now standing?
20m West
45m East
20m East
45m North
Explanation:
From the starting to finishing point,
the distance D= ( 25 + 20 ) meters
=45 meters
Direction towards east
the distance D= ( 25 + 20 ) meters
=45 meters
Direction towards east
1245.A boy runs 200 metres in 24 seconds. What is his speed ?
28 hours
30 hours
20 km/hr
24 km/hr
Explanation:
According to the given question,
d = distance = 200 m
t = time = 24 sec
s = speed
And we are asked to find speed in km/h.
We know,
speed = distance/time
s = d/t
Putting the values we get,
s = 200/24
s = 8.33 m/s
Now, to convert it into km/h,
We know,
1 km = 1000 m
1 m = 1/1000 km
Similarly,
1 sec = 1/3600 hr
s = 8.33 m/s
s = 8.33 ×3600 /1000
s = 29.98 km/h
Thus speed of the boy is 29.98 km/h or 30 hours
d = distance = 200 m
t = time = 24 sec
s = speed
And we are asked to find speed in km/h.
We know,
speed = distance/time
s = d/t
Putting the values we get,
s = 200/24
s = 8.33 m/s
Now, to convert it into km/h,
We know,
1 km = 1000 m
1 m = 1/1000 km
Similarly,
1 sec = 1/3600 hr
s = 8.33 m/s
s = 8.33 ×3600 /1000
s = 29.98 km/h
Thus speed of the boy is 29.98 km/h or 30 hours
1246.If the speed of a man is 45 km per hour, then what is the distance traveled by him in 30 seconds ?
28.5 km/hr
30 km/hr
275m
375m
Explanation:
speed= 45km/hour=5/18×45= 12.5m/second.
then in 30 second, he travels= 30×12.5= 375m.
then in 30 second, he travels= 30×12.5= 375m.
1247.An escalator moves towards the top level at the rate of 11 ft.sec and its length is 140 feet. If a person walks on the moving escalator at the rate of 3 feet per second towards the top level, how much time does he take to cover the entire length.
375 m
420 m
14 sec
10 sec
Explanation:
Now we need to use the concept of relative velocity.
According to the question the person moves at the rate of 3 ft/s and 11ft/s which will be 14 ft/s
Length of the level is 140 ft
So speed = distance / time
Or time taken = distance / speed
= 140 / 14
= 10 secs
So the time taken to cover the entire length will be 10 secs.
According to the question the person moves at the rate of 3 ft/s and 11ft/s which will be 14 ft/s
Length of the level is 140 ft
So speed = distance / time
Or time taken = distance / speed
= 140 / 14
= 10 secs
So the time taken to cover the entire length will be 10 secs.
1248.Two trains, 250 metres and 150 metres long respectively, are running on parallel lines. If they are running in the same directions, the faster train crosses the slower train in 40 seconds. If they are moving in the opposite direction they pass each other in eight seconds. What is the speed of the slower train?
12 sec
8 sec
108 seconds
252 seconds
Explanation:
Relative speed = (45 - 40) km/hr = 5 km/hr
= (5×5/18)m/sec
= 25/18 m/sec
Total distance covered = Sum of lengths of trains = (200 + 150) m = 350 m
∴ Time taken= (350×18/25)sec
= 252 seconds
= (5×5/18)m/sec
= 25/18 m/sec
Total distance covered = Sum of lengths of trains = (200 + 150) m = 350 m
∴ Time taken= (350×18/25)sec
= 252 seconds
1249.Two persons , Ram & Lakshman , who are at a distance of 100 km from each other, move towards each other from two places P and Q at speeds of 20 kmph and 25 kmph respectively. Lakshman reaches P, returns immediately and meets Ram at R, who started on the return journey to P immediately after reaching Q. What is the distance between Q and R?
92 kmph
72 kmph
33 1/3 km
25 km
Explanation:
Ram takes 100/20 = 5 hours to cover the distance from P to Q. By that time Lakshman covers
covers 5 *25 = 125 km
Lakshman covers 25 km more than the distance PQ . Now the distance between them = 75 km
Time taken by them to meet = Distance/ Relative speed = 75/(20+25) = 75/45 = 5/3 hrs.
Distance between Q and R is nothing but the distance covered by Ram in 5/3 hours = 20 *5/3 = 100/3 km or 33 1/3 km
covers 5 *25 = 125 km
Lakshman covers 25 km more than the distance PQ . Now the distance between them = 75 km
Time taken by them to meet = Distance/ Relative speed = 75/(20+25) = 75/45 = 5/3 hrs.
Distance between Q and R is nothing but the distance covered by Ram in 5/3 hours = 20 *5/3 = 100/3 km or 33 1/3 km
1250.With out any halt a train travels a certain distance with an average speed of 75 km ph , and with halts it covers the same distance at an average speed of 60 kmph. When it is traveling with halts, how many minutes/per hour does the train halt on an average ?
30km
27 1/3 km
48 min
12min
Explanation:
In a hour train travel 75km
Then in a min it will covered 60/75br> Therefore 60/75*60=48 min train travel in a hourbr> Hence it will stop for 12 minutes only(60-48).
Then in a min it will covered 60/75br> Therefore 60/75*60=48 min train travel in a hourbr> Hence it will stop for 12 minutes only(60-48).
1251.A lady starts from P towards Q and realizes that at a point R, if he walks 50 km further he will be at a point S, which is as far away from R as it is from Q. What is the distance between P and Q if the distance between P and R is half the distance from R to Q?( Assume that P, Q, R and S are all on the same straight line
15min
18min
150 km
200 km
Explanation:
P___50_____ R ___50_____S____50 ___Q
The above figure gives the locations of P, R, S &Q in relation to each other.
The above figure gives the locations of P, R, S &Q in relation to each other.
A and B participate in a 5000 m bicycle race which is being run on a circular track of 500 m. If the speed of A and B are 20 m/s and 10 m/s respectively, what is the distance covered by A when he passes B for the seventh time ?
Solution:
Total race distance = 5000 m
Track length = 500 m
Speed of A = 20 m/s; Speed of B = 10 m/s
For calculating the time at which they will meet for first time, relative speed is to be calculated.
Relative speed = 20 -10 = 10 m/s
Time taken by them to meet = 500 / (20-10) = 50 sec
Thus, they will meet after every 50 sec.
If they were to meet for 7th time, total time elapsed = 50 x 7 = 350 sec
Distance covered by A = 20 x 350 = 7000 m > 5000 m
By that time, A would have completed the race.
Ans: 7th meet is not possible.
Total race distance = 5000 m
Track length = 500 m
Speed of A = 20 m/s; Speed of B = 10 m/s
For calculating the time at which they will meet for first time, relative speed is to be calculated.
Relative speed = 20 -10 = 10 m/s
Time taken by them to meet = 500 / (20-10) = 50 sec
Thus, they will meet after every 50 sec.
If they were to meet for 7th time, total time elapsed = 50 x 7 = 350 sec
Distance covered by A = 20 x 350 = 7000 m > 5000 m
By that time, A would have completed the race.
Ans: 7th meet is not possible.
1254.Two buses A and B leave the same bus depot, A towards the North and B towads the East. The bus A travels at a speed of 5 km/hr more than that of the bus B. If after four hours the distance between the two buses is 100 km, find the speed of the bus A.
4000
situation is not possible
60 kmph
40 kmph
1255.A person traveled from his house to office at 30 kmph; then he was late to his office by 5 minutes. If he increases his speed by 10 kmph he would be early by 15 minutes to his office. What should be his speed so that he reaches his office on time ?
20 kmph
15 kmph
36 kmph
32 kmph
1346.Walking at the rate of 4kmph a man cover certain distance in 2hr 45 min. Running at a
speed of 16.5 kmph the man will cover the same distance in.
speed of 16.5 kmph the man will cover the same distance in.
12 min
25 min
40min
60 min
Explanation:
Distance = Speed * time
4*11/4 = 11km
New Speed = 16.5 kmph
Therefore time = D/S = 11/16.5 = 40min
Distance = Speed * time
4*11/4 = 11km
New Speed = 16.5 kmph
Therefore time = D/S = 11/16.5 = 40min
1348.Two boys starting from the same place walk at a rate of 5kmph and 5.5kmph
respectively. What time will they take to be 8.5km apart, if they walk in the same
direction?
respectively. What time will they take to be 8.5km apart, if they walk in the same
direction?
17 hrs
25 hrs
31 hrs
45 hrs
Explanation:
The relative speed of the boys = 5.5kmph – 5kmph
= 0.5 kmph
Distance between them is 8.5 km
Time= 8.5km / 0.5 kmph = 17 hrs
The relative speed of the boys = 5.5kmph – 5kmph
= 0.5 kmph
Distance between them is 8.5 km
Time= 8.5km / 0.5 kmph = 17 hrs
1354.A person walks at 5kmph for 6hr and at 4kmph for 12hr.The average speed is.
4 1/3
7 2/3
9 ½
8km
Explanation:
Avg speed = total distance/total time
= 5*6 + 4*12 / 18
=4 1/3 mph
Avg speed = total distance/total time
= 5*6 + 4*12 / 18
=4 1/3 mph
1355.A person covers a certain distance at 7kmph .How many meters does he cover in
2 minutes.
2 minutes.
14 kmph
7 kmph
2400m
1200m
Explanation:
no explanation
1356.If a man runs at 3m/s. How many km does he run in 1hr 40min
18 km
7 km
12 km
17 km
Explanation:
Speed of the man = 3*18/5 kmph = 54/5kmph
Distance covered in 5/3 hrs=54/5*5/3 = 18km
Speed of the man = 3*18/5 kmph = 54/5kmph
Distance covered in 5/3 hrs=54/5*5/3 = 18km
1358.Vikas can cover a distance in 1hr 24min by covering 2/3 of the distance at 4 kmph and
the rest at 5kmph.the total distance is?
the rest at 5kmph.the total distance is?
18 km
6 km
12 km
17 km
Explanation:
Let total distance be S
total time=1hr24min
A to T :: speed=4kmph
diistance=2/3S
T to S :: speed=5km
distance=1-2/3S=1/3S
21/15 hr=2/3 S/4 + 1/3s /5
84=14/3S*3
S=84*3/14*3 = 6km
Let total distance be S
total time=1hr24min
A to T :: speed=4kmph
diistance=2/3S
T to S :: speed=5km
distance=1-2/3S=1/3S
21/15 hr=2/3 S/4 + 1/3s /5
84=14/3S*3
S=84*3/14*3 = 6km
1359.walking at ¾ of his usual speed ,a man is late by 2 ½ hr.
the usual time is.
the usual time is.
4
7 ½
7
5
Explanation:
Usual speed = S
Usual time = T
Distance = D
New Speed is ¾ S
New time is 4/3 T
4/3 T – T = 5/2
T=15/2 = 7 ½
Usual speed = S
Usual time = T
Distance = D
New Speed is ¾ S
New time is 4/3 T
4/3 T – T = 5/2
T=15/2 = 7 ½
1362.A thief steals a car at 2.30pm and drives it at 60kmph. The theft is discovered at 3pm
and the owner sets off in another car at 75kmph when will he overtake the thie
and the owner sets off in another car at 75kmph when will he overtake the thie
6:00 PM
8:00 PM
9:00 PM
5:00 PM
Explanation:
Let the thief is overtaken x hrs after 2.30pm
distance covered by the thief in x hrs = distance covered by
the owner in x-1/2 hr
60x = 75 ( x- ½)
x= 5/2 hr
thief is overtaken at 2.30 pm + 2 ½ hr = 5 pm
Let the thief is overtaken x hrs after 2.30pm
distance covered by the thief in x hrs = distance covered by
the owner in x-1/2 hr
60x = 75 ( x- ½)
x= 5/2 hr
thief is overtaken at 2.30 pm + 2 ½ hr = 5 pm