# AAO Exam-CT 14: Quant (Speed Time and Distance)

## AAO Exam-CT 14- Quant (Speed Time and Distance)

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1. By walking at 3/7 of his usual speed, a man reaches his office late by 40 min. Find the usual time taken by him to reach his office?
• 20 min
• 30 min
• 25 min
• 40 min
• 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.

2. 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 its speed is (in Kilometer per hour)
• 58
• 60
• 62
• 64
• 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$\frac{9}{3600}ℎ���$

Time=LengthSpeed$����=\frac{�����ℎ}{�����}$

⇒ 93600=0.2850+V$\frac{9}{3600}=\frac{0.28}{50+�}$

⇒ 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.

3. A student goes to college at the rate of 5 km/h and reaches 12 minutes late. If he travels at the speed of 8 km/h he is 15 minutes early. Find the distance between his college and his starting point.
• 7 km
• 9 km
• 5 km
• 6 km
• 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

4. The ratio between the upstream speed of boat and the speed of current is 7 : 1. If the boat can cover 70 km upstream in 5 hours, then find the distance it can cover running downstream in 5 hours
• 75 km
• 85 km
• 80 km
• 95 km
• 90 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

5. Train A running with the speed of 20 m/s crosses a platform in 30 seconds. Another train B whose length is 150 metre more than length of train A is running with the speed of 10 m/s crosses a pole in 40 seconds. Find the length of platform.
• 330 metre
• 380 metres
• 290 metres
• 350 metres
• 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?

• 25 ∶ 12 ∶ 15
• 24 ∶ 20 ∶ 15
• 4 ∶ 2 ∶ 15
• 12 ∶ 21 ∶ 19
• 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.

7. A Uber car can complete a certain distance in 21 hours, it covers one-third of the distance at 20 km/hr and rest at 50 km/hr. What is the total distance covered by the Uber car?
• 800 km
• 700 km
• 950 km
• 620 km
• 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?

• 22 km/hr
• 7 km/hr
• 8 km/h
• 10 km/h
• 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

9. Excluding stoppages, the speed of an Ola cab is 75 km/h, and including stoppages it is 60 km/hr. for how many minutes does the cab stop per hour?
• 30 minutes
• 12 minutes
• 25 minutes
• 10 minutes
• 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)

• 5(5/8)
• 4(7/8)
• 5(3/8)
• 4(3/4)
• 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.