Speed Time and Distance MCQ Quiz - Objective Question with Answer for Speed Time and Distance - Download Free PDF

Given:

Downstream speed = x + 5

Upstream speed = 2x − 26

Still water speed = Distance ÷ Time = 42 ÷ 3.5 = 12 km/hr

Formula used:

Still water speed = (Downstream + Upstream) ÷ 2

Distance = Speed × Time

Calculations:

⇒ (x + 5 + 2x − 26)/2 = 12

⇒ (3x − 21)/2 = 12

⇒ 3x − 21 = 24

⇒ 3x = 45 ⇒ x = 15

⇒ Downstream speed = 15 + 5 = 20 km/hr

⇒ Time = x − 7 = 15 − 7 = 8 hrs

⇒ Distance = 20 × 8 = 160 km

∴ Distance covered downstream in (x − 7) hrs = 160 km

Given:

Upstream distance = 300 km

Downstream distance = 280 km

Speed of current = 14.28% of boat's speed in still water

Time difference = 3 hours (Upstream takes 3 hours more)

Let the speed of the boat in still water be x km/h:

Speed of current = 14.28% of x = (14.28 / 100) × x = x/7 km/h

Calculate upstream and downstream speeds:

Upstream speed = x - x/7 = 6x/7 km/h

Downstream speed = x + x/7 = 8x/7 km/h

Time upstream = Distance ÷ Speed = 300 ÷ (6x/7)

Time downstream = 280 ÷ (8x/7)

According to the problem:

Time upstream - Time downstream = 3

⇒ (300 / 6x/7) - (280 / 8x/7) = 3

⇒ (2100 / 6x) - (1960 / 8x) = 3

Multiply both sides by x:

⇒ (2100 / 6) - (1960 / 8) = 3x

2100 / 6 = 350

1960 / 8 = 245

⇒ 350 - 245 = 3x

⇒ 105 = 3x

⇒ x = 105 ÷ 3 = 35 km/h

Thus, the correct answer is 35 km/h.

Given:

Downstream speed = 22 km/hr

Upstream speed = 5.5 km/hr

Total time = 8 hours

Formula used:

Time = Distance / Speed

Total distance = Distance downstream + Distance upstream

Calculation:

Let the one-way distance be D km.

Time taken downstream = D / 22

Time taken upstream = D / 5.5

Total time = Time downstream + Time upstream

⇒ 8 = D / 22 + D / 5.5

⇒ 8 = (D / 22) + (4D / 22)

⇒ 8 = (5D / 22)

⇒ D = (8 × 22) / 5

⇒ D = 176 / 5

Total distance = 2 × D

⇒ Total distance = 2 × (176 / 5)

⇒ Total distance = 352 / 5 km

∴ The correct answer is option (3).

Given:

A man covers distances of 100 km, 408 km, and 1460 km at speeds of 20 km/hr, x km/hr, and 20 km/hr, respectively.

Average speed for the whole journey = 20 km/hr

Formula used:

Average speed = Total distance / Total time

Total time = (Distance₁ / Speed₁) + (Distance₂ / Speed₂) + (Distance₃ / Speed₃)

Calculations:

Total distance = 100 + 408 + 1460 = 1968 km

Total time = (100 / 20) + (408 / x) + (1460 / 20)

⇒ Total time = 5 + (408 / x) + 73

⇒ Total time = 78 + (408 / x)

Average speed = 20 = 1968 / (78 + (408 / x))

⇒ 20 × (78 + (408 / x)) = 1968

⇒ 1560 + (8160 / x) = 1968

⇒ (8160 / x) = 1968 - 1560

⇒ (8160 / x) = 408

⇒ x = (8160 / 408)

⇒ x = 20

∴ The correct answer is option (2).

Given:

Walking speed = (2)/(5) of normal speed

Delay = 18 minutes

Formula used:

Let usual time taken = t minutes

When speed is reduced, time taken becomes (t + 18) minutes

Calculation:

⇒ t + 18 = (5/2) × t

⇒ 2(t + 18) = 5t

⇒ 2t + 36 = 5t

⇒ 36 = 5t - 2t

⇒ 36 = 3t

⇒ t = 12 minutes

∴ The usual time taken by Abhijeet is 12 minutes.

Given

Length of first train (L1) = 400 m

Length of second train (L2) = 300 m

Speed of second train (S2) = 60 km/hr

Time taken to cross each other (T) = 15 s

Concept:

Relative speed when two objects move in opposite directions is the sum of their speeds.

Calculations:

Let the speed of the first train = x km/hr

Total length = 300 + 400

Time = 15 sec

According to the question:

700/15 = (60 + x) × 5/18

28 × 6 = 60 + x

x = 108 km/hr.

Therefore, the speed of the longer train is 108 km per hour.

Given:

Total track length = 1200 m

Speed of A = 2 m/s ; speed of B = 4 m/s

Speed of C = 6 m/s

Formula used:

Distance = relative speed × time

Calculation:

Relative speed of A and B = (4 - 2) = 2 m/s

Relative speed of B and C = (6 + 4) = 10 m/s

Relative speed of A and C = (6 + 2) = 8 m/s

Time taken by A and B = 1200/2 = 600 sec

Time taken by B and C = 1200/10 = 120 sec

Time taken by A and C  = 1200/8 = 150 sec

A, B and C will meet at = L.C.M {600,120, 150} = 600 sec = 600/60 = 10 minutes

∴ The correct answer is 10 minutes.

Given:

Speed is 60 km per hour,

Train passed through a 1.5 km long tunnel in two minutes

Formula used:

Distance = Speed × Time

Calculation:

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.

Concept used:

Speed × time = distance

Calculation:

In the 1^st 20 min the thief cover distance = 4 m,

20 min in hour = 20/60 hour

Let's assume that the speed of security guard = x m/hr, where x > 12

According to the question,

⇒ (x - 12) × 20/60 = 4

⇒ x - 12 = 12

⇒ x = 24

∴ The correct answer is 24 m/h

Given:

In a 1500 m race, Anil beats Bakul by 150 m and in the same race Bakul beats Charles by 75 m.

Concept used:

Time × Speed = Distance

Calculation:

According to the question,

Anil goes 1500m while Bakul goes (1500 - 150) i.e. 1350m.

Ratio of speed of Anil and Bakul = 1500 : 1350 = 10 : 9 = 200 : 180

According to the question,

Bakul goes 1500m while Charlie goes (1500 - 75) i.e. 1425m.

Ratio of speed of Bakul and Charlie = 1500 : 1425 = 20 : 19 = 180 : 171

So, the ratio of the speeds of Anil, Bakul and Charlie = 200 : 180 : 171

Let the speeds of Anil, Bakul and Charlie be 200k, 180k and 171k m/s respectively.

Time taken by Anil to finish the race = 1500/200k = 7.5/k seconds

Now, Anil beats Charlie by = (200 - 171)k ×7.5/k = 217.5m

∴ Anil beat Charlie by 217.5m.

Shortcut Trick

In a 1500 m race, Anil beats Bakul by 150 m

When Anil completes the race, Bakul covered (1500 - 150) = 1350 m

In a 1500 m race Charles is 75 m behind Bakul

So, in 1350 m race Charles is 75/1500 × 1350 = 67.5 m behind Bakul

So, Charles is (67.5 + 150) = 217.5 m behind from Anil in 1500 m race

∴ Anil beat Charlie by 217.5m.

Given:

A man travels from A to B at a speed of 36 km/hr in 74 minutes and he travels a distance from B to C with a speed of 45 km/hr in 111 minutes. 

Formula used:

Average speed = Total distance/Total time taken

Calculation:

Time taken = 74 min : 111 min   [given]

Ratio of Time taken = 2 : 3

Average Speed = 

Average Speed = 207/5

Average Speed = 41.4 km/hr

∴ The average speed of whole journey is 41.4 km/h

Given:

Geeta runs 5/2 times as fast as Babita 

Geeta gives a lead of 40 m to Babita 

Formula Used:

Distance = Speed × Time 

Calculation:

Let the speed of Babita be 2x

⇒ Speed of Geeta = (5/2) × 2x = 5x

Let the distance covered by Geeta be y meters

⇒ Distance covered by Babita = (y - 40) meters

As time is constant, distance is directly proportional to speed 

⇒  = 

⇒ 2y = 5y - 200 

⇒ y = 200/3 = 66.67m 

∴ The distance from the starting point where both of them will meet is 66.67 m

Given:

Speed of train 1 = 80 km/h

Speed of train 2 = 95 km/h

Difference in distance = 180 km

Concept:

Distance = Speed × Time 

Solution:

Let the time of travelling be 't' hours

Distance travelled by train with speed 80 km/h = 80 × t

Distance travelled by train with speed 95 km/h = 95 × t 

According to the question, 

Difference in distance travelled by two train  = 180 km 

⇒ 95 t - 80t = 180 

⇒ 15t =  180 

⇒ t = 12 hours 

Distance between stations = 80 × 12 + 95 × 12  = 960 + 1140 = 2100km

Hence, the distance between Bangalore and Chennai is 2100 km .

Concept used:

If upstream speed = U and downstream speed = D, then speed of boat = (U + D)/2

Calculation:

According to the question,

20/U + 44/D = 8  … (i)

15/U + 22/D = 5  … (ii)

Multiply by 2 the equation (ii) then subtract from 1 we get

20/U + 44/D = 8

30/U + 44/D = 10

- 10/U = - 2

⇒ U = 5 km/hr

Putting the value in equation (i), we get D = 11

So, the speed of boat = (U + D)/2 = (5 + 11)/2 = 8 km/hr

∴ The correct answer is 8 km/hr

Given:

Two trains are running on opposite tracks between stations A and B.

After crossing each other they take 4 hr and 9 hr respectively to reach their destination.

Speed of first train is 54 kmph.

Formula used:

After crossing each other, if time taken by 2 trains is T1 and T2 resp. then, S1/S2 = √T2/√T1

where, S1 and S2 are speeds of first and second train respectively

Calculation:

We have, T1 = 4hr, T2 = 9hr, S1 = 54 kmph

⇒ 54/ S2 = √[9/4] = 3/2

⇒ S2 = 54 × 2 × 1/3 = 36 kmph

⇒ Speed of second train = 36 kmph

Alternate Method

Let the speed of the second train be 'x' kmph

Also, time taken to cross each other = √(T1 × T2) = √(9 × 4) = 6 hrs 

Total distance = 54 × 6 + x × 6 = x × 9 + 54 × 4

⇒ 9x - 3x = 54 × (6 - 2)

⇒ 6x = 216

⇒ x = 36 kmph = Speed of second train