Usual Speed MCQ Quiz - Objective Question with Answer for Usual Speed - Download Free PDF
Last updated on Jun 11, 2025
Latest Usual Speed MCQ Objective Questions
Usual Speed Question 1:
A man started 10 minutes late from his home to his office. He travelled at a speed of \(1\frac{1}{4}\) times of his usual speed and reached his office in time. Then the time taken by the man to reach his office at his usual speed is (in minutes)
Answer (Detailed Solution Below)
Usual Speed Question 1 Detailed Solution
Given:
The man started 10 minutes late, traveled at \(1\frac{1}{4}\) (i.e 5/4) times his usual speed, and reached his office in time.
Formula used:
Time = Distance / Speed
Calculation:
Let the usual time taken to reach the office be T minutes.
At 5/4 times the speed, the new time taken = T × (4/5).
Since he reached on time despite being 10 minutes late:
T × (4/5) + 10 = T
⇒ T × (4/5) + 10 = T
⇒ T - 4T/5 = 10
⇒ T/5 = 10
⇒ T = 50
∴ The correct answer is option (1).
Usual Speed Question 2:
Kamal travels from City A to City B. If Kamal drives his car at 3/4 of his normal speed, then he reaches City B 45 minutes late.
Find the time (in minutes) that Kamal would have taken to travel from City A to City B if he drove at his normal speed.
Answer (Detailed Solution Below)
Usual Speed Question 2 Detailed Solution
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
Usual Speed Question 3:
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) that Zakir would have taken to travel from City A to City B if he drove at his normal speed.
Answer (Detailed Solution Below)
Usual Speed Question 3 Detailed Solution
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).
Usual Speed Question 4:
Chetan travels from City A to City B. If Chetan drives his car at \(\frac{3}{4}\) of his normal speed, then he reaches City B 33 minutes late. Find the time (in minutes) that Chetan would have taken to travel from City A to City B if he drove at his normal speed.
Answer (Detailed Solution Below)
Usual Speed Question 4 Detailed Solution
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).
Usual Speed Question 5:
Manoj travels from City A to City B. If Manoj drives his car at \(\frac{1}{5}\) of his normal speed, then he reaches City B 36 minutes late. Find the time (in minutes) that Manoj would have taken to travel from City A to City B if he drove at his normal speed.
Answer (Detailed Solution Below)
Usual Speed Question 5 Detailed Solution
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.
Top Usual Speed MCQ Objective Questions
If Ram covers a certain journey at 60% of his usual speed, he reaches the destination late by 36 min. His usual time (in min) to reach the destination is:
Answer (Detailed Solution Below)
Usual Speed Question 6 Detailed Solution
Download Solution PDFCalculation:
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.
A person goes to his office at the speed of 54 km/hr and reaches 18 minutes earlier. If he goes at the speed 36 km/hr, then he reaches 18 minutes late. What should be the speed to reach on the usual time?
Answer (Detailed Solution Below)
Usual Speed Question 7 Detailed Solution
Download Solution PDFGiven:
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
Kamal travels from City A to City B. If Kamal drives his car at 3/4 of his normal speed, then he reaches City B 45 minutes late.
Find the time (in minutes) that Kamal would have taken to travel from City A to City B if he drove at his normal speed.
Answer (Detailed Solution Below)
Usual Speed Question 8 Detailed Solution
Download Solution PDFGiven:
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
Usual Speed Question 9:
If Ram covers a certain journey at 60% of his usual speed, he reaches the destination late by 36 min. His usual time (in min) to reach the destination is:
Answer (Detailed Solution Below)
Usual Speed Question 9 Detailed Solution
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.
Usual Speed Question 10:
Manoj travels from City A to City B. If Manoj drives his car at \(\frac{1}{5}\) of his normal speed, then he reaches City B 36 minutes late. Find the time (in minutes) that Manoj would have taken to travel from City A to City B if he drove at his normal speed.
Answer (Detailed Solution Below)
Usual Speed Question 10 Detailed Solution
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.
Usual Speed Question 11:
A person goes to his office at the speed of 54 km/hr and reaches 18 minutes earlier. If he goes at the speed 36 km/hr, then he reaches 18 minutes late. What should be the speed to reach on the usual time?
Answer (Detailed Solution Below)
Usual Speed Question 11 Detailed Solution
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
Usual Speed Question 12:
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) that Zakir would have taken to travel from City A to City B if he drove at his normal speed.
Answer (Detailed Solution Below)
Usual Speed Question 12 Detailed Solution
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).
Usual Speed Question 13:
Chetan travels from City A to City B. If Chetan drives his car at \(\frac{3}{4}\) of his normal speed, then he reaches City B 33 minutes late. Find the time (in minutes) that Chetan would have taken to travel from City A to City B if he drove at his normal speed.
Answer (Detailed Solution Below)
Usual Speed Question 13 Detailed Solution
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).
Usual Speed Question 14:
Kamal travels from City A to City B. If Kamal drives his car at 3/4 of his normal speed, then he reaches City B 45 minutes late.
Find the time (in minutes) that Kamal would have taken to travel from City A to City B if he drove at his normal speed.
Answer (Detailed Solution Below)
Usual Speed Question 14 Detailed Solution
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