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11. A man covered a certain distance at some speed. Had he moved 3 kmph faster, he would have taken 40 minutes less. If he had moved 2 kmph slower, he would have taken 40 minutes more. The distance (in km) is:
Answer: Option D
Explanation:Let distance = x km and usual rate = y kmph.
Then, x/y - x/(y+3) = 40/60
2y(y + 3) = 9x ....(i)
And, x/(y-2) - x/y = 40/60
y(y - 2) = 3x ....(ii)
By Solving (i) by (ii), we get: x = 40
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12. Sanu travelled a distance of 20 kms at the speed of 5 km/hr, he reached 40 minutes late. If he had walked at the speed of 8 km/hr, how early from fixed time will he reach?
Answer: Option B
Explanation:Time taken to travel 20 kms at the speed of 5 kmph.
=> t = 20/5 = 4 hrs.
Actual time of reaching = (4 hrs - 40 mins) = 3 hrs 20 mins ......(1)
Time taken to travel 20 kms at the speed of 8 kmph.
=> t = 20/8 = 2.5 hrs. ......(2)
Difference between (1) and (2) => (3 hrs 20 mins - 2 hrs 30 mins)
(200 mins - 150 mins) = 50 mins.
Sanu would have reached 50 mins earlier.
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13. A person rides a bicycle round a circular path of radius 50 m. The radius of the wheel of the bicycle is 50 cm. The cycle comes to the starting point for the first time in 1 hour. What is the number of revolutions of the wheel in 15 minutes?
Answer: Option B
Explanation:Circumference of the circular path = 2/\50 = 100 /\ m
This means 100? m is the distance covered in 60 minutes
Therefore distance covered in 15 minutes = 1/4 x 100 /\ m = 25/\ m
Now distance covered in one revolution of the wheel = 2 /\(1/2) m = /\ m
Therefore number of revolutions in 15 minutes = 25/\ //\ = 25
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14. It takes 8 hours for a 600 km journey, if 120 km is done by train and the rest by car. It takes 20 minutes more, if 200 km is done by train and the rest by car. The ratio of the speed of the train to that of the cars is:
Answer: Option C
Explanation:Assume Here the speed of the train = x km/hr
and speed of car = y km/hr.
Given First journey information:
(120/x + 480/y) = 8 => (120/x + 480/y)/8 = 1
(15/x + 60/y) = 1 => (1/x + 4/y) = 1/15 ---------- (1)
Given Second journey information:
And, (200/x + 400/y) = 25 => (1/x + 2/y) = 25/200 ------- (2)
By Solving (1) and (2)
x = 60 and y = 80.
So Ratio of speeds = 60 : 80 = 3 : 4
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15. A thief steals a car at 2.30 p.m and drives it at 60 kmph. The theft is discovered at 3 p.m and the owner sets off in another car at 75 kmph. When will be overtake the thief.
Answer: Option D
Explanation:As Theft is discovered at 3:00pm but Thief stole the car at 2:30.
This means thief covered some distance in this 30 min gap.
Distance travelled by thief in 30 min = 60 * 1/2 = 30 km
Owner Discovered Car at 3:00pm
Now relative speed = (75-60)km/hr = 15km/hr
Time needed to travel 30km by the speed of 15km/hr.
Time at which owner meets thief = 30/15 = 2 hrs
So After 2 hrs (i.e,at 5:00 pm) the owner will catch/overtake the thief
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16. Excluding stoppages, the speed of a bus is 54 kmph and including stoppages, it is 45 kmph. For how many minutes does the bus stop per hour?
Answer: Option B
Explanation:Speed of the bus excluding stoppages =54 kmph
Speed of the bus including stoppages =45 kmph
Loss in speed when including stoppages =54−45=9 kmph
⇒ In 1 hour, bus covers 9 km less due to stoppages.
Hence, time in which the bus stops per hour
= Time taken to cover 9 km
=distancespeed=954 hour=16 hour =606 min=10 min
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17. If an object travels at five feet per second, how many feet does it travel in one hour?
Answer: Option E
Explanation:5 ft per second so in 1 minute 300 ft
In 60 minutes = 60 * 300= 18000 ft.
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18. Kartik rides a bicycle 8 km/hr and reaches his school 2.5 minutes late. The next day he increases his speed to 10 km/hr and reaches the school 5 minutes early. How far is the school from his house?
Answer: Option C
Explanation:Let the total distance be d km. We have speed = 8km/hr and 10km/hr.
We know, time (t) = distance (d)/speed (s).
= > t = d/8 (for the first day)
=> t = d/10 (for the second day)
Difference between the time taken on first day and second day is 2.5 + 5 = 7.5 mins.
Therefore, (d/8) - (d/10) = 7.5/60. (7.5 mins converted to hours)
=> (10d - 8d)/80 = 75/600
=> d = (75 x 80)/1200.
=> d = 5 kms.
Therefore, the distance between his school and house 5 km.
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19. 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:
Answer: Option C
Explanation:Let's Farmer travelled distance on foot = x km.
So distance travelled on bicycle = (61 -x) km.
Formula Used: Time = Distance/Speed
So, (x/4 + (61 -x)/9) = 9
=> 9x + 4(61 -x) = 9 x 36 => 9x + 244 -4x = 324
=> 5x = 80
x = 16 km.
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20. If I walk with 30 miles/hr i reach 1 hour before and if i walk with 20 miles/hr i reach 1 hour late. Find the distance between 2 points and the exact time of reaching destination is 11 am then find the speed with which it walks.
Answer: Option D
Explanation:the exact time is 11 am
then 30miles/ph reach it on 10am
20 miles/ph reach it on 12am
if they start walking in 6am
then 30m/ph-(10am-6am) 4h*30=120miles
20m/ph-(12am-6am) 6h*20=120miles
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