Test: Speed, Time and Distance - 1 - UCAT MCQ

Let assume that length of train be x m
Length of the pole = 0 m
Length of platform = 500 m
According to the question:

Speed = 
= 
⇒ Speed = 
By cross multiplication
⇒ 68x = 20x + 10,000
⇒ 48x = 10,000
⇒ x = 10,000/48 m

Speed of the train = 
∴ Speed of the train = 37.5 km/hr
The correct option is 3 i.e. 37.5 km/hr

According to the question,
Time taken if he travels only by air = 4 - 2 = 2 hrs
Speed of man if he travels by air = 360/2 = 180 km/hr
He saved 4/5^th of time taken by train = 2 hrs
Time taken by train = 2 × (5/4) = 5/52 hours
Time taken by air = 4 - (5/2) = 3/2 hours
Distance covered by air  
⇒ 180 x  (3/2)
⇒ 90 × 3 = 270 km
∴ The distance he traveled by air is 270 km.

Speed of the stream = 1/2 (Downstream Speed – Upstream Speed)

Speed of the stream = 1/2 (36 - 18) = 9 km/hr

Speed of Boat in Still Water = 1/2 (Downstream Speed + Upstream Speed)

⇒ 1/2(36 + 18) = 27

Time to travel 54 km = {54/(27 - 9)} + {54/(27 + 9)}

⇒ (54/18) + (54/36) = 3 + 1.5 = 4.5

∴ The required time is 4.5 h.

Total distance to be covered = total length of both the trains

= 152. 5 + 157.5

= 310 m

Total time taken = 9.3 sec

Speed = distance/time

= (310/9.3) m/sec

= (310/9.3) × (18/5)

= 120 km/hr

∴ The combined speed of the two trains every hour would then be 120 km/hr

Speed = 1200/120 = 10 m/sec

Total distance = 1200 + 700 = 1900 m

Time = distance/speed = 1900/10 = 190 sec

∴ Time required to cross a platform is 190 sec.

Time taken by first man to cover journey = 2 PM – 6 AM = 8 hours

Time taken by another man to cover journey = 3 PM – 8 AM = 7 hours

Let total distance from P to Q be 56x km (LCM of 8 & 7)

⇒ Speed of first man = 7x km/hr

⇒ Speed of second man = 8x km/hr

⇒ Distance covered by first man in 2 hours = 14x km

⇒ Remaining distance = 56x – 14x = 42x km

⇒ Time taken to meet each other = 42x/ (7x + 8x) = 42/15 hrs

= 2 hrs 48 min

⇒ Time of meeting = 8:00 + 2:48 = 10:48 AM

Let the distance he travelled be D km and his usual speed be S km/h

According to the question

⇒ D/6 – D/S = 32/60 .....(1)

⇒ D/S – D/7 = 18/60 .....(2)

Adding equation (1) and (2), we get

⇒ D/6 – D/7 = (32 + 18)/60

⇒ (7D – 6D)/42 = 50/60

⇒ D/42 = 5/6

⇒ D = (42 × 5)/6

⇒ D = 35

Now,

Twice the distance (in km) of his destination from his starting point = 2D

⇒ (2 × 35) km

⇒ 70 km

∴ The required distance is70 km.

Let the speed of the boat in still water and the speed of the stream be x and y respectively.

200/(x - y) = 10 

⇒ x – y = 20

⇒ 140/(x + 10 + y) = 2

⇒ x + y = 60

⇒ x = 40 and y = 20

∴ The speed of the stream is 20 km/hr.

Let the length of the train be L

According to the question,

Total distance = 1500 m + L

Speed = 60(5/18)

⇒ 50/3 m/sec

Time = 2 × 60 = 120 sec

⇒ 1500 + L = (50/3)× 120

⇒ L = 2000 - 1500

⇒ L = 500 m

∴ The length of the train is 500 m.

Let the distance be d km.

We know that,

Distance = Speed x Time

⇒ d = 72 km