AAO Exam-CT 14: Quant (Speed Time and Distance)
AAO Exam-CT 14- Quant (Speed Time and Distance)
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20 min
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25 min
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40 min
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None of these
Given:
Delay in time = 40 min = 2/3 hrs
Ratio of new and usual speed = 3/7
Concept:
Time and speed are inversely proportional. If he walks at 3/7 of his usual speed, he would take 7/3 of the usual time to reach office.
Calculation:
Let the usual time to reach office = a hrs
∵ 7/3 of usual time - usual time = 2/3 hrs
⇒ 7a/3 - a = 2/3
⇒ (7a - 3a)/3 = 2/3
⇒ 4a/3 = 2/3
⇒ a = 2/4 hrs
⇒ a = 1/2 hrs = 30 min
∴ He takes 30 min to reach office while walking at his usual speed.
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58
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60
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64
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None of these
The correct answer is option 3.
Given: A man sitting in a train which is moving at 50 Kmph, observes that a goods train, travelling in the opposite direction, takes 9 seconds to pass him. The length of the goods train is 280 meters.
Calculation:
⇒ Let's assume the speed of the goods train = V kmph
⇒ When two objects move in the opposite direction then their speed is added.
Net speed of a man = 50 + V kmph
Length of goods train = 280 m = 0.28 km
Time = 9 second = 93600hour
Time=LengthSpeed
⇒ 93600=0.2850+V
⇒ 9(50 + V) = 0.28 x 3600
⇒ 450 + 9V = 28 x 36
⇒ 9V = 1008 - 450
⇒ V = 62 kmph
So, the speed of the goods train is 62 kmph.
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7 km
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9 km
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5 km
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8 km
Given:
Time taken to cover 5 km = 12 min late
Time taken to cover 8 km = 15 min early
Formula used:
Distance = speed × time
Calculation:
Let the distance between his college and his starting point = x km
According to question,
⇒ x/5 - x/8 = (12 + 15)/ 60
⇒ (8x - 5x)/40 = 9/20
⇒ x = (18 / 3) km
⇒ 6 km
∴ Required distance is 6 km
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75 km
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85 km
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80 km
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95 km
Given:
The ratio of upstream speed and speed of current = 7 : 1
Distance covered upstream in 5 hours = 70 km
Concept used:
Speed of Boat = Upstream speed + Speed of current
Downstream speed = Speed of boat + Speed of current
Calculation:
Upstream speed = (70/5) km/hr
⇒ 14 km/hr
Let the upstream speed and speed of current be 7x and x respectively
Now, 7x = 14 km/hr
⇒ x = 2 km/hr
Speed of boat = (14 + 2) km/hr
⇒ 16 km/hr
Downstream speed = (16 + 2) km/hr
⇒ 18 km/hr
Distance covered downstream in 5 hours = (18 × 5) km
⇒ 90 km
∴ Distance covered by car downstream in 5 hours is 90 km
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330 metre
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380 metres
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290 metres
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None of these
Given :
Time taken by train A at speed of 20 m/s to cross a platform = 30 seconds
Time taken by train B at speed of 10 m/s to cross a pole = 40 seconds
Length of train B = 150 + Length of train A
Concept used:
Length when crossing a platform = Length of Train + Length of Platform
Length when crossing a pole = Length of Train only
Calculation:
Let length of train A and platform be A and P respectively
Length of train B = (A + 150)
According to the question,
(A + P) = 30 × 20 = 600 ----(i)
And, (A + 150) = 40 × 10 = 400
⇒ A + 150 = 400
⇒ A = 250
Now, putting value of A in equation (i)
(250 + P) = 600
⇒ P = 350 m
∴ Length of platform is 350 m
6. The speed of three cars A, B and C are in the ratio 5 ∶ 6 ∶ 8 . What is the ratio between the time taken by these cars to travel the same distance?
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25 ∶ 12 ∶ 15
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4 ∶ 2 ∶ 15
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12 ∶ 21 ∶ 19
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None of these
Given:
Speed of three cars are in the ratio 5 ∶ 6 ∶ 8
Concept used:
Speed = 1/time ----(When distance is equal)
Calculation:
Let, the total distance be d
Speed of A car = 5x
Speed of B car = 6x
Speed of C car = 8x
Time taken to cover the distance by car A = d/5x
Time taken to cover the distance by car B = d/6x
Time taken to cover the distance by car C = d/8x
According to the question,
⇒ (d/5x) ∶ (d/6x) ∶ (d/8x)
⇒ (1/5) ∶ (1/6) ∶ (1/8)
⇒ 24 ∶ 20 ∶ 15
∴ Require time ratio = 24 ∶ 20 ∶ 15.
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800 km
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950 km
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620 km
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None of these
Given:
Uber car complete a certain distance in = 21 hours
One-third part of the distance cover by Uber car in 20 km/hr
Rest part of the distance cover by Uber car in 50 km/hr
Concept:
Speed = distance /time
Calculation:
Let, the distance = x km
According to the question,
⇒ [(x/3) × (1/20)] + [(2x/3) × (1/50)] = 21
⇒ (x/60) + (x/75) = 21
⇒ (9x/300) = 21
⇒ x = (21 × 300)/9
⇒ x = 700 km
∴ Total distance is 700 km.
8. Raktim covered a distance from village A to B on car at 28 km/hr. However he covered the distance from B to A on foot at 4 km/hr. What is his average speed in the whole journey?
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22 km/hr
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8 km/h
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10 km/h
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None of these
Given:
Speed to covered from village A to B by car = 28 km/hr
Speed to covered from village B to A by foot = 4 km/hr
Concept used:
Average Speed = (2 × s1 × s2)/(s1 + s2)
Calculation:
According to the question,
⇒ (2 × 28 × 4)/(28 + 4)
⇒ 224/32 = 7
∴ Average speed of Raktim is 7 km/hr
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30 minutes
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25 minutes
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10 minutes
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None of these
Given:
Excluding stoppages, the speed of an Ola cab = 75 km/h
Including stoppages, the speed of an Ola cab = 60 km/h
Concept used:
Speed = Distance/Time
Calculation:
Difference b/w the speed = (75 - 60) km/h = 15 km/hr
⇒ Time taken to cover the 15 km = (15/75) × 60 = 12 min
∴ Minutes does the cab stop per hour is 12 min.
10. Speed of boat in still water is 9 km/hr. Stream speed initially is 2 km/hr but it increases by 3 km/hr after every hour. Find the time after which boat will come back to the position where it started.(in hour)
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4(7/8)
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5(3/8)
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4(3/4)
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None of these
Given:-
Boat speed = 9 km/hr
Formula:-
Speed = Distance/Time
Upstream speed = u - v
Concept:-
Boat will come back only if it is travelling upstream as after some time stream speed will be more than boat speed and boat will start travelling in backwards direction.
Calculation:-
Distance travelled in first hour = (9 - 2) × 1 = 7 km
Distance travelled in second hour = (9 - 2 - 3) × 1 = 4 km
Distance travelled in third hour = (9 - 2 - 3 - 3) × 1 = 1 km
Distance travelled in fourth hour = (1 - 3) × 1 = - 2 km or 2 km in backward direction.
After 4 hours boat is at a distance of (7 + 4 + 1 - 2) = 10 km from starting point.
Distance in fifth hour = (- 2 - 3) × 1 = - 5 km
Now boat will be (10 - 5) = 5 km away
Time for next 5 km = 5/(- 5 - 3) = 5/8 hour (ignore negative sign for time)
So boat will be back in 5(5/8) hours.
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