Usual Speed MCQ Quiz - Objective Question with Answer for Usual Speed - Download Free PDF
Last updated on May 21, 2025
Latest Usual Speed MCQ Objective Questions
Usual Speed Question 1:
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 1 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 2:
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 2 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 3:
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 3 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 4:
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 4 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 5:
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 5 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
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