44031.A train can travel 50% faster than a car. Both start from point A at the same time and reach point B 75 kms away from A at the same time. On the way, however, the train lost about 12.5 minutes while stopping at the stations. The speed of the car is:
100 kmph
110 kmph
120 kmph
130 kmph
Explanation:
Let speed of the car be x kmph.
Then, speed of the train =$\dfrac{150}{100}x=\left(\dfrac{3}{2}x\right)$kmph.
$\Rightarrow\dfrac{75}{x} -\dfrac{50}{(x} =\dfrac{5}{24}$
$\Rightarrow x=\left(\dfrac{25\times24}{5}\right)$
=120kmph
Let speed of the car be x kmph.
Then, speed of the train =$\dfrac{150}{100}x=\left(\dfrac{3}{2}x\right)$kmph.
$\Rightarrow\dfrac{75}{x} -\dfrac{50}{(x} =\dfrac{5}{24}$
$\Rightarrow x=\left(\dfrac{25\times24}{5}\right)$
=120kmph
44037.A farmer travelled a distance of 61 km in 9 hours. He travelled partly on foot @ 4 km/hr and partly on bicycle @ 9 km/hr. The distance travelled on foot is:
14 km
15 km
16 km
17 km
Explanation:
Let the distance travelled on foot be x km.
Then, distance travelled on bicycle = (61 -x) km.
So,$\dfrac{x}{4}+\dfrac{61-x}{9}=9$
=> 9x + 4(61 -x) = 9 x 36
=>5x = 80
=> x = 16 km.
Let the distance travelled on foot be x km.
Then, distance travelled on bicycle = (61 -x) km.
So,$\dfrac{x}{4}+\dfrac{61-x}{9}=9$
=> 9x + 4(61 -x) = 9 x 36
=>5x = 80
=> x = 16 km.
44044.If a person walks at 14 km/hr instead of 10 km/hr, he would have walked 20 km more. The actual distance travelled by him is:
50 km
56 km
70 km
80 km
Explanation:
Let the actual distance travelled be x km.
Then,
$\dfrac{x}{10}=\dfrac{x + 20}{14}$
=>14x = 10x + 200
=>4x = 200
=>x = 50 km.
Let the actual distance travelled be x km.
Then,
$\dfrac{x}{10}=\dfrac{x + 20}{14}$
=>14x = 10x + 200
=>4x = 200
=>x = 50 km.
44227.A man takes 5 hours 45 min in walking to a certain place and riding back. He would have gained 2 hours by riding both ways. The time he would take to walk both ways is
11 hrs
8 hrs 45 min
7 hrs 45 min
9 hrs 20 min
Explanation:
Given that time taken for riding both ways will be 2 hours lesser than the time needed for wakingonewayandridingback.
Therefore,
time needed for riding one way = time needed for waking oneway- 2 hours
Given that time taken in walking one way and riding back =5 hours 45 min
Hence, the time he would take to walk both ways
=5 hours 45 min + 2 hours
=7 hours 45 min
Therefore,
time needed for riding one way = time needed for waking oneway- 2 hours
Given that time taken in walking one way and riding back =5 hours 45 min
Hence, the time he would take to walk both ways
=5 hours 45 min + 2 hours
=7 hours 45 min
44228.A and B walk around a circular track. A and B walk at a speed of 2 rounds per hour and 3 rounds per hour respectively. If they start at 8 a.m. from the same point in opposite directions, how many times shall they cross each other before 9.30 a.m.?
5
6
7
8
Explanation:
Relative speed = Speed of A + Speed of B (∴ they walk in opposite directions)
=2+3=5 rounds per hour
Therefore, they cross each other 5 times in 1 hour and 2 times in 12 hour
Time duration from 8 a.m. to 9.30 a.m. =1.5 hour
Hence they cross each other 7 times before 9.30 a.m.
=2+3=5 rounds per hour
Therefore, they cross each other 5 times in 1 hour and 2 times in 12 hour
Time duration from 8 a.m. to 9.30 a.m. =1.5 hour
Hence they cross each other 7 times before 9.30 a.m.
44229.Two boys starts from the same place walking at the rate of 5 kmph and 5.5 kmph respectively in the same direction. What time willtheytaketo be 8.5 km apart?
17 hr
14 hr
12 hr
9 hr
Explanation:
relative speed =5.5−5=0.5 kmph (because they walk in the same direction)
distance =8.5 km
time=distance /speed=8.5/0.5=17 hr
distance =8.5 km
time=distance /speed=8.5/0.5=17 hr
44230.An athlete runs 200 metres race in 24 seconds. What is his speed?
30.5 km/hr
20 km/hr
35 km/hr
30 km/hr
Explanation:
speed=distance/time=200/24 m/s
=200/24×18/5 km/hr
=(40×3)/4 km/hr=30 km/hr
=200/24×18/5 km/hr
=(40×3)/4 km/hr=30 km/hr
44231. A car travels first 160 km at 64 km/hr and the next 160 km at 80 km/hr. What is the average speed for the first 320 km of the tour?
70.11 km/hr
77.11 km/hr
71.11 km/hr
74.44 km/hr
Explanation:
Average Speed =(2×64×80)/(64+80)
=(2×64×80)/144=(2×32×40)/36
=(2×32×10)/9=(64×10)/9=71.11 kmph
=(2×64×80)/144=(2×32×40)/36
=(2×32×10)/9=(64×10)/9=71.11 kmph
44239.A person has to cover a distance of 6 km in 45 minutes. If he covers one-half of the distance in two-thirds ofthetotaltime, to cover the remaining distance in the remaining time,whatshouldbe his speed in km/hr?
14 km/hr
12 km/hr
10 km/hr
8 km/hr
Explanation:
The person needs to cover 6 km in 45 minutes
Given that he covers one-half of the distance in two-thirds of the total time
⇒ he covers half of 6 km in two-thirds of 45 minutes
⇒ He covers 3 km in 30 minutes
Now he needs to cover the remaining 3 km in remaining 15 minutes
Distance =3 km
Time =15 minutes =1/4 hour
Required Speed =3/(1/4)=12 km/hr
Given that he covers one-half of the distance in two-thirds of the total time
⇒ he covers half of 6 km in two-thirds of 45 minutes
⇒ He covers 3 km in 30 minutes
Now he needs to cover the remaining 3 km in remaining 15 minutes
Distance =3 km
Time =15 minutes =1/4 hour
Required Speed =3/(1/4)=12 km/hr
44241.A man in a train notices that he can count 21 telephone posts in one minute. If they are known to be 50 metres apart, at what speed is the train travelling?
61 km/hr
56 km/hr
63 km/hr
60 km/hr
Explanation:
The man can count 21 telephone posts in one minute. Number of gaps between 21 posts is 20 and adjacent posts are 50 metres apart.
It means 20×50=1000 metres are covered in 1 minute.
distance =1000 m=1 km
time =1 min=1/60 hr
speed =1/(1/60)=60 km/hr
It means 20×50=1000 metres are covered in 1 minute.
distance =1000 m=1 km
time =1 min=1/60 hr
speed =1/(1/60)=60 km/hr
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