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

Let us take the speed of the boat and river to be \( b \) and \( x \) respectively and let the distance be \( d \).

As per the condition in the question, \( \frac{d}{x + b} \) is the normal time taken. \( \frac{d}{2x + b} \) is the special time taken i.e. if the speed gets doubled.

Since the time gets reduced by 75%, \( \frac{1}{4} \left[ \frac{d}{x + b} + \frac{d}{x - b} \right] = \left[ \frac{d}{2x + b} + \frac{d}{2x - b} \right] \)

We have to find the ratio of the speed of the motor boat to the speed of the river = \( \frac{x}{b} \)

Let us divide throughout by \( b \) and assume \( \frac{x}{b} = k \)

\( \frac{1}{k + 1} + \frac{1}{k - 1} = 4\left[\frac{1}{2k + 1} + \frac{1}{2k - 1}\right] \)

Solving for \( k \), we get: \( 2k \times (k^2 - 1) = 16k \times \left(k^2 - 1\right) \)

Hence, the ratio of the original speed of the motor boat to the speed of the river is equal to: \( \frac{\sqrt{7}}{2} \)

Train P crosses a pole in 17 seconds. The length of Train P is equal to the distance it covers in 17 seconds.

Speed of Train P = 144 km/h

Convert speed to m/s:

144 × 5/18 = 40 m/s

Length of Train P = Speed × Time

Length of Train P = 40 × 17 = 680 m

Train P and Train Q are moving in opposite directions. Their relative speed is the sum of their individual speeds.

Speed of Train P = 144 km/h = 40 m/s

Speed of Train Q = 72 km/h

Convert speed to m/s:

72 × 5/18 = 20 m/s

Relative speed = 40 + 20 = 60 m/s 

Train P takes 22.8 seconds to cross Train Q. The combined length of the two trains is equal to the distance covered at the relative speed in 22.8 seconds.

Combined Length = Relative Speed × Time = 60 × 22.8 = 1368 m

The combined length of Trains P and Q is 1368 m, and the length of Train P is 680 m. Therefore, the length of Train Q is:

Length of Train Q = Combined Length Length of Train P = 1368 - 680 = 688 m

Thus, the length of Train Q is 688 meters.

Let speed of boat is x km /hr.

so, ATQ, [72 / (x - 3) ] - [ 72 / (x + 3)] 2

or, x² - 9 = 36 × 6 = 216

or, x² = 225

or, x = 15

so, speed of boat is 15 km /hr.

upstream speed is 15 - 3 = 12 km /hr 

Train cover 420 km distance at 140 km /hr.

so, Time taken = 420 /140 = 3 hour.

Time travel by train at p km/hr speed is 6.5 - 3 = 3.5

Distance travel at p km/hr is 420 km.

so, p = 420 / 3.5 = 120 km /hr

so, time taken to cover 600 km distance at p km /hr is  = 600 / 120 = 5 hour.

Here, the distance traveled by Shreya is constant in both the cases.

So, \(20 \times t = 25 \times (t - 1) = D\)

\( \Rightarrow 20t = 25t - 25\)

\( \Rightarrow 5t = 25\)

\( \Rightarrow t = 5 \text{ hrs. So, Distance travelled} = 20 \times 5 = 100 \text{ Km.}\)

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:

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.

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:

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 = \(\frac{{36\ \times\ 2\ +\ 45\ \times\ 3}}{{2\ +\ 3}}\)

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 

⇒ \(\frac{2x}{5x}\) = \(\frac{y-40}{y}\)

⇒ 2y = 5y - 200 

⇒ y = 200/3 = 66.67m 

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

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:-

Train₁= 152.5m

Train₂= 157.5m

Time = 9.3 sec

Calculation:-

⇒ 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.

Alternate Method When two trains are moving in opposite direction-

Let the speed of ine is 'v' and the second is 'u'

∴ Combined speed = v + u

Total distance = 152.5 + 157.5

= 310 m

∴ Combined speed = Total distance/total time

⇒ (v + u) = 310/9.3

⇒ (v + u) = 33.33 m/s

⇒ (v + u) = 33.33 × (18/5)

⇒ (v + u) = 120 km/hr

Given,

Sathish completes 900 m race.

Kiran covers = 900 – 270 = 630 m

Rahul covers = 900 – 340 = 560 m

⇒ Ratio of their speed = 630/560

When Kiran covers 900 m race  then

⇒ Rahul would cover = 900 × 560/630 = 800 m