Usual Speed MCQ Quiz - Objective Question with Answer for Usual Speed - Download Free PDF

Given:

Speed = (3/4) of normal speed

Delay = 45 minutes

Formula used:

Time = Distance / Speed

If time at normal speed = x minutes, then

Distance = Speed × Time = s × x

Calculation:

At reduced speed (3/4)s, time = x + 45

⇒ s × x = (3/4)s × (x + 45)

⇒ x = (3/4)(x + 45)

⇒ 4x = 3x + 135

⇒ x = 135

∴ Time at normal speed = 135 minutes

Given:

Zakir travels from City A to City B.

If Zakir drives his car at \(\frac{2}{3}\) of his normal speed, then he reaches City B 10 minutes late.

Find the time (in minutes) Zakir would have taken to travel from City A to City B if he drove at his normal speed.

Formula used:

Let normal time = t minutes.

Time taken at \(\frac{2}{3}\) speed = \(\frac{t}{\frac{2}{3}}\) = \(\frac{3t}{2}\)

Delay = Time at reduced speed - Normal time

Delay = 10 minutes

Calculation:

Delay = \(\frac{3t}{2} - t\)

⇒ \(\frac{3t}{2} - t = 10\)

⇒ \(\frac{3t}{2} - \frac{2t}{2} = 10\)

⇒ \(\frac{t}{2} = 10\)

⇒ t = 20

∴ The correct answer is option (1).

Given:

When Chetan drives at (3/4) of his normal speed, he reaches City B 33 minutes late.

Let the normal time taken to travel from City A to City B be T (in minutes).

Calculation:

If speed is reduced to (3/4) of the normal speed, time increases by 33 minutes:

\(\dfrac{\text{Distance}}{\text{(Normal Speed)}} + 33 = \dfrac{\text{Distance}}{\text{(3/4 × Normal Speed)}}\)

⇒ \(\dfrac{\text{Distance}}{\text{Normal Speed}} + 33 = \dfrac{4}{3} \times \dfrac{\text{Distance}}{\text{Normal Speed}}\)

⇒ \(\dfrac{T + 33}{T} = \dfrac{4}{3}\)

⇒ \(\dfrac{33}{T} = \dfrac{1}{3}\)

⇒ T = 99 minutes

∴ The correct answer is option (1).

Given:

Manoj travels from City A to City B. If Manoj drives his car at \(\dfrac{1}{5}\) of his normal speed, then he reaches City B 36 minutes late.

Formula used:

Let t be the time taken at normal speed.

Time taken at \(\dfrac{1}{5}\) of normal speed = 5t

Calculation:

5t - t = 36

⇒ 4t = 36

⇒ t = 9

∴ The correct answer is option 1.

Given:

Speed 1 = 54 km/hr → reaches 18 min early

Speed 2 = 36 km/hr → reaches 18 min late

Total time difference = 18 + 18 = 36 min = 36/60 = 3/5 hr

Formula used:

Distance = (S1 × S2 × Time Difference) / (S1 - S2)

Calculation:

⇒ Distance = (54 × 36 × 3/5) / (54 - 36)

⇒ Distance = (1944 × 3/5) / 18

⇒ Distance = (5832 / 5) / 18 = 5832 / (5 × 18) = 5832 / 90 = 64.8 km

Usual time = 64.8 / x

Also, 64.8 / 54 = 6/5 hr → 18 min early

⇒ Usual time = 6/5 + 3/10 = 15/10 = 3/2 hr

⇒ 64.8 / x = 3/2

⇒ x = 64.8 × 2 / 3 = 129.6 / 3 = 43.2 = 216/5

∴ Required speed = 216/5 km/hr

Calculation:

60% = 3/5

Let the speed of the man be 5x

60% of the speed = 5x × (3/5) = 3x

Ratio of speed of man before and after = 5x : 3x

As we know, Speed is inversely proportional to time.

Time ratio of man before and after = 3x : 5x

According to the question

5x – 3x = 36 min 

⇒ 2x = 36

⇒ x = 18min 

Required time = 3x = 3 × (18) = 54 mins.

∴ Option 4 is the correct answer.

Given:

Speed 1 = 54 km/hr → reaches 18 min early

Speed 2 = 36 km/hr → reaches 18 min late

Total time difference = 18 + 18 = 36 min = 36/60 = 3/5 hr

Formula used:

Distance = (S1 × S2 × Time Difference) / (S1 - S2)

Calculation:

⇒ Distance = (54 × 36 × 3/5) / (54 - 36)

⇒ Distance = (1944 × 3/5) / 18

⇒ Distance = (5832 / 5) / 18 = 5832 / (5 × 18) = 5832 / 90 = 64.8 km

Usual time = 64.8 / x

Also, 64.8 / 54 = 6/5 hr → 18 min early

⇒ Usual time = 6/5 + 3/10 = 15/10 = 3/2 hr

⇒ 64.8 / x = 3/2

⇒ x = 64.8 × 2 / 3 = 129.6 / 3 = 43.2 = 216/5

∴ Required speed = 216/5 km/hr

Given:

Speed = (3/4) of normal speed

Delay = 45 minutes

Formula used:

Time = Distance / Speed

If time at normal speed = x minutes, then

Distance = Speed × Time = s × x

Calculation:

At reduced speed (3/4)s, time = x + 45

⇒ s × x = (3/4)s × (x + 45)

⇒ x = (3/4)(x + 45)

⇒ 4x = 3x + 135

⇒ x = 135

∴ Time at normal speed = 135 minutes

Calculation:

60% = 3/5

Let the speed of the man be 5x

60% of the speed = 5x × (3/5) = 3x

Ratio of speed of man before and after = 5x : 3x

As we know, Speed is inversely proportional to time.

Time ratio of man before and after = 3x : 5x

According to the question

5x – 3x = 36 min 

⇒ 2x = 36

⇒ x = 18min 

Required time = 3x = 3 × (18) = 54 mins.

∴ Option 4 is the correct answer.

Given:

Manoj travels from City A to City B. If Manoj drives his car at \(\dfrac{1}{5}\) of his normal speed, then he reaches City B 36 minutes late.

Formula used:

Let t be the time taken at normal speed.

Time taken at \(\dfrac{1}{5}\) of normal speed = 5t

Calculation:

5t - t = 36

⇒ 4t = 36

⇒ t = 9

∴ The correct answer is option 1.

Given:

Speed 1 = 54 km/hr → reaches 18 min early

Speed 2 = 36 km/hr → reaches 18 min late

Total time difference = 18 + 18 = 36 min = 36/60 = 3/5 hr

Formula used:

Distance = (S1 × S2 × Time Difference) / (S1 - S2)

Calculation:

⇒ Distance = (54 × 36 × 3/5) / (54 - 36)

⇒ Distance = (1944 × 3/5) / 18

⇒ Distance = (5832 / 5) / 18 = 5832 / (5 × 18) = 5832 / 90 = 64.8 km

Usual time = 64.8 / x

Also, 64.8 / 54 = 6/5 hr → 18 min early

⇒ Usual time = 6/5 + 3/10 = 15/10 = 3/2 hr

⇒ 64.8 / x = 3/2

⇒ x = 64.8 × 2 / 3 = 129.6 / 3 = 43.2 = 216/5

∴ Required speed = 216/5 km/hr

Given:

Zakir travels from City A to City B.

If Zakir drives his car at \(\frac{2}{3}\) of his normal speed, then he reaches City B 10 minutes late.

Find the time (in minutes) Zakir would have taken to travel from City A to City B if he drove at his normal speed.

Formula used:

Let normal time = t minutes.

Time taken at \(\frac{2}{3}\) speed = \(\frac{t}{\frac{2}{3}}\) = \(\frac{3t}{2}\)

Delay = Time at reduced speed - Normal time

Delay = 10 minutes

Calculation:

Delay = \(\frac{3t}{2} - t\)

⇒ \(\frac{3t}{2} - t = 10\)

⇒ \(\frac{3t}{2} - \frac{2t}{2} = 10\)

⇒ \(\frac{t}{2} = 10\)

⇒ t = 20

∴ The correct answer is option (1).

Given:

When Chetan drives at (3/4) of his normal speed, he reaches City B 33 minutes late.

Let the normal time taken to travel from City A to City B be T (in minutes).

Calculation:

If speed is reduced to (3/4) of the normal speed, time increases by 33 minutes:

\(\dfrac{\text{Distance}}{\text{(Normal Speed)}} + 33 = \dfrac{\text{Distance}}{\text{(3/4 × Normal Speed)}}\)

⇒ \(\dfrac{\text{Distance}}{\text{Normal Speed}} + 33 = \dfrac{4}{3} \times \dfrac{\text{Distance}}{\text{Normal Speed}}\)

⇒ \(\dfrac{T + 33}{T} = \dfrac{4}{3}\)

⇒ \(\dfrac{33}{T} = \dfrac{1}{3}\)

⇒ T = 99 minutes

∴ The correct answer is option (1).

Given:

Speed = (3/4) of normal speed

Delay = 45 minutes

Formula used:

Time = Distance / Speed

If time at normal speed = x minutes, then

Distance = Speed × Time = s × x

Calculation:

At reduced speed (3/4)s, time = x + 45

⇒ s × x = (3/4)s × (x + 45)

⇒ x = (3/4)(x + 45)

⇒ 4x = 3x + 135

⇒ x = 135

∴ Time at normal speed = 135 minutes