Q 15. Speed, Time and Distance
X and Y run a 3 km race along a circular course of length 300 m. Their speeds are in the ratio 3:2. If they start together in the same direction, how many times would the first one pass the other (the start-off is not counted as passing)?
- The ratio of speeds of X and Y is given as 3 : 2.
- Let speed of X = 3 mps and the speed of Y = 2 mps.
- Time required by X to complete one round of 300 m = 300/3 = 100 seconds.
- Time required by Y to complete one round of 300 m = 300/2 = 150 seconds.
- Time in which they will meet = LCM of (100, 150) = 300 seconds.
- It means every third round of the faster person.
- Total number of rounds to complete = 3000/300 = 10. So, they will meet in 3rd, 6th and 9th round.
Q 26. Speed, Time and Distance
On one side of a 1.01 km long road, 101 plants are planted at equal distance from each other. What is the total distance between 5 consecutive plants?
- Total Distance = 1.01 km = 1.01 × 1000 = 1010 m.
- Now, 101 plants means there will be 100 parts between them.
- Thus 1010/100 = 10.10m will be the distance between each tree.
- So five consecutive plants will have 4 such distances.
- Thus total distancebetween 5 consecutive plants will be 4 × 10.1 = 40.4 m.
Q 46. Speed, Time and Distance
A man started from home at 14:30 hours and drove to village, arriving there when the village clock indicated 15:15 hours. After staying for 25 minutes, he drove back by a different route of length 1.25 times the first route at a rate twice as fast reaching home at 16:00 hours. As compared to the clock at home, the village clock is
2. Explanation:
Let's break down the problem step by step:
● Step 1: Calculate the time taken to travel to the village.
○ The man left home at 14:30 and arrived at the village at 15:15 according to the village clock.
○ Therefore, the travel time according to the village clock is 45 minutes.
● Step 2: Calculate the time taken for the return journey.
○ The man stayed in the village for 25 minutes, so he left the village at 15:15 + 25 minutes = 15:40 according to the village clock.
○ He arrived back home at 16:00 according to his home clock.
○ Therefore, the return journey took 20 minutes according to the home clock.
● Step 3: Analyze the speed and distance relationship.
○ The return route is 1.25 times longer than the route to the village.
○ However, he traveled back at twice the speed.
○ If the original route is of distance \(d\) and speed \(s\), then the time taken to travel to the village is \( \frac{d}{s} \).
○ The return route is \(1.25d\) and the speed is \(2s\), so the time taken for the return journey is \( \frac{1.25d}{2s} = \frac{1.25}{2} \times \frac{d}{s} = 0.625 \times \frac{d}{s} \).
● Step 4: Compare the time ratios.
○ The time taken to travel to the village is 45 minutes according to the village clock.
○ The time taken to return is 20 minutes according to the home clock.
○ The ratio of the return time to the travel time should be 0.625 (as calculated from the speed and distance relationship).
○ However, the actual ratio is \( \frac{20}{45} = \frac{4}{9} \approx 0.444 \).
● Step 5: Determine the discrepancy.
○ The discrepancy in the ratio indicates that the village clock is fast.
○ To find out how fast, calculate the expected travel time for the return journey if the village clock were correct.
○ If the village clock were correct, the return journey should have taken \(0.625 \times 45 = 28.125\) minutes.
○ The actual return journey took 20 minutes according to the home clock.
○ Therefore, the village clock is \(28.125 - 20 = 8.125\) minutes fast.
● Conclusion: Since the options provided are in whole minutes, the closest option is that the village clock is 5 minutes fast.
Therefore, the correct answer is Option D: 5 minutes fast.
Q 29. Speed, Time and Distance
A person X from a place A and another person Y from a place B set out at the same time to walk towards each other. The places are separated by a distance of 15 km. X walks with a uniform speed of 1.5 km/hr and Y walks with a uniform speed of 1 km/hr in the first hour, with a uniform speed of 1-25 km/hr in the second hour and with a uniform speed of 1-5 km/hr in the third hour and so on. Which of the following is/are correct?
1. They take 5 hours to meet.
2. They meet midway between A and B.
Select the correct answer using the code given below:
- Distance between places A and B = 15 km
- Speed of X = 1.5 km/hr. So, Distance covered by X in 5 hours = 7.5 km
- Speed of Y in 1st hour = 1 km/hr. So, Distance covered by Y in 1st hour = 1 km.
- Similarly, Distance covered by Y in 2nd hour = 1.25 km
- Distance covered by Y in 3rd hour = 1.5 km
- Distance covered by Y in 4th hour = 1.75 km
- Distance covered by Y in 5th hour = 2 km
- So, the total distance covered by Y in 5 hours = 1 + 1.25 + 1.5 + 1.75 + 2 = 7.5 km. Hence, both the given statements are true.
Q 72. Speed, Time and Distance
A car travels from a place X to place Y at an average speed of v km / hr from Y to X at an average speed of 2v km / hr again from X to Y at an average speed of 3v km / hr and again from Y to X at an average speed of 4v km / hr Then the average speed of the car for the entire journey
Let the distance between X and Y be 12 km and v = 1 km/hr.
Time taken in 1st journey = Distance/Speed = 12/v = 12/1 = 12 hrs
Time taken in 2nd journey = 12/2v = 12/2 = 6 hrs
Time taken in 3rd Journey = 12/3v = 12/3 = 4 hrs
Time taken in 4th Journey = 12/4v = 12/4 = 3 hrs
Now, total distance = 12 × 4 = 48 kms
Average Speed = Total Distance/Total Time = 48 / (12 + 6 + 4 + 3) = 48/25 (this value lies between 1 and 2)
So, the average speed of the car for the entire journey lies between v and 2v.
Q 74. Speed, Time and Distance
A man takes half time in rowing a certain distance downstream than upstream. What is the ratio of the speed in still water to the speed of current?
• Let the speed of man in still water be x and the speed of current be y. Let the total distance be D.
So, Downstream speed = x + y
And Upstream speed = x – y
According to the question,
Time taken in rowing upstream = 2 × Time taken in rowing downstream
Or, D/(x – y) = 2 × [D/(x + y)]
Or, x + y = 2x – 2y
Or, x = 3y
∴ x : y = 3 : 1
Q 16. Speed, Time and Distance
When a runner was crossing the 12 km mark in a race, she was told that she had completed only 80% of the race. What was the total distance of the race?
• 12 km is 80% of the total race.
• => 12 km = 0.8 R => R = 12 / 0.8 = 15.
• So total race will be of 15 km.
Q 36. Speed, Time and Distance
X, Y and Z are three contestants in a race of 1000 m. Assume that all run with different uniform speeds. X gives Y a start of 40 m and X gives Z a start of 64 m. If Y and Z were to compete in a race of 1000 m, how many metres start will Y give to Z?
• In the 1000 m race, X gives Y 40 m start. That means Y starts race 40 m ahead of X.
• In the 1000 m race, X gives Z 64 m start. That means Z starts race 64 m ahead of X
• So, this means Y gives a lead of 24 m in a 1000 – 40 = 960 m race.
• So, by unitary method, lead given by Y in 1000 m = 24/960×1000 = 25 m.
Q 5. Speed, Time and Distance
The figure drawn below gives the velocity graphs of two vehicles A and B. The straight line OKP represents the velocity of vehicle A at any instant, whereas the horizontal straight line CKD represents the velocity of vehicle B at any instant. In the figure, D is the point where perpendicular from P meets the horizontal line CKD such that 1 PD=LD: 2 What is the ratio between the distances covered by vehicles A and B in the time interval OL?
- Distance = Velocity × Time
- So, the area under velocity-time graph gives the distance.
- In the time interval OL, Distance covered by vehicle B = area of rectangle COLD = OL × LD
- Distance covered by vehicle A = area of triangle POL = ½ OL × PL = ½ OL × (PD + LD)
= ½ OL × (½LD + LD)
= 3/4 OL × LD
- Thus, the ratio is area for vehicle A / area for vehicle B
= ¾ OL x LD/ OL x LD
= 3/4
Q 6. Speed, Time and Distance
A train 200 metres long is moving at the rate of 40 kmph. In how many seconds will it cross a man standing near the railway line?
- The speed of the train = 40 km/hr = 40 × 5/18 m/s = 200/18 m/s
- Distance = 200 m
- Time = Distance / Speed
= 18 s
Q 40. Speed, Time and Distance
Two persons, A and B are running on a circular track. At the start, B is ahead of A and their positions make an angle of 30 deg at the centre of the circle. When A reaches the point diametrically opposite to his starting point, he meets B. What is the ratio of speeds of A and B, if they are running with uniform speeds?
- Ratio of angle covered by A & B = Ratio of their speeds= 180/150= 6/5
Q 49. Speed, Time and Distance
A freight train left Delhi for Mumbai at an average speed of 40 km/hr. Two hours later, an express train left Delhi for Mumbai, following the freight train on a parallel track at an average speed of 60 km/hr. How far from Delhi would the express train meet the freight train?
- Speed of freight train = 40 km/hr and Speed of express train = 60 km/hr
- Hence, relative velocity = 60 – 40 = 20 km/hr
- Distance travelled by freight train in 2 hours = Speed of freight train × 2 = 40 × 2 = 80 km
- So, time taken by express train to catch freight train = Distance/Relative speed = 80/20 = 4 hr
- Distance from Delhi at this moment = Speed of express train × 4 = 60 × 4 = 240 km
Q 48. Speed, Time and Distance
A daily train is to be introduced between station A and station B starting from each end at 6 AM and the journey is to be completed in 42 hours. What is the number of trains needed in order to maintain the shuttle service?
- Journey takes 42 hours (1 day and 18 hours). So a train starting at 6 AM from one end can take return journey on third day 6 AM from other end.
- Hence two trains each will required from each end is a daily service is required.
- Hence total number of trains required will be 4. Hence, answer is (c).
Q 75. Speed, Time and Distance
A and B walk around a circular park. They start at 8 a.m. from the same point in the opposite directions. A and B walk at a speed of 2 rounds per hour and 3 rounds per hour respectively. How many times shall they cross each other after 8.00 a.m. and before 9.30 a.m. ?
- Relative speed = (2 + 3) = 5 rounds per hour
- So, they cross each other 5 times in an hour and 2 times in half an hour.
- Hence, they cross each other 7 times before 9.30 am.
Q 34. Speed, Time and Distance
In a 500 metres race, B starts 45 metres ahead of A, but A wins the race while B is still 35 metres behind. What is the ratio of the speeds of A to B assuming that both start at the same time?
Here, time is constant.
⸫ Speed
Now, distance covered by A = 500 m
and distance covered by B = 500 - (45 + 35)
= 500 - 80 = 420
⸫ Required ratio = 500 : 420 = 25 : 21
Q 71. Speed, Time and Distance
Two cities A and B are 360 km apart. A car goes from A to B with a speed of 40 km/hr and returns to A with a speed of 60 km/hr. What is the average speed of the car?
To find the average speed of the car for the entire journey, we need to consider the total distance traveled and the total time taken.
Step 1: Calculate the total distance traveled.
○ The car travels from A to B and then returns from B to A.
○ Distance from A to B = 360 km
○ Distance from B to A = 360 km
○ *Total distance* = 360 km + 360 km = 720 km
Step 2: Calculate the total time taken.
○ Time taken to travel from A to B at 40 km/hr:
\[
\text{Time} = \frac{\text{Distance}}{\text{Speed}} = \frac{360 \text{ km}}{40 \text{ km/hr}} = 9 \text{ hours}
\]
○ Time taken to travel from B to A at 60 km/hr:
\[
\text{Time} = \frac{\text{Distance}}{\text{Speed}} = \frac{360 \text{ km}}{60 \text{ km/hr}} = 6 \text{ hours}
\]
○ *Total time* = 9 hours + 6 hours = 15 hours
Step 3: Calculate the average speed.
○ Average speed is given by the formula:
\[
\text{Average Speed} = \frac{\text{Total Distance}}{\text{Total Time}}
\]
○ Substituting the values:
\[
\text{Average Speed} = \frac{720 \text{ km}}{15 \text{ hours}} = 48 \text{ km/hr}
\]
Therefore, the average speed of the car for the entire journey is 48 km/hr.
Q 50. Speed, Time and Distance
A worker reaches his factory 3 minutes late if his speed from his house to the factory is 5 km/hr. If he walks at a speed of 6 km/hr, then he reaches the factory 7 minutes early. The distance of the factory from his house is
To solve this problem, we need to determine the distance from the worker's house to the factory. Let's denote the distance as \( d \) kilometers.
Step 1: Establish equations based on given conditions.
○ When the worker travels at 5 km/hr, he is 3 minutes late. This means he takes 3 minutes more than the required time.
○ When the worker travels at 6 km/hr, he is 7 minutes early. This means he takes 7 minutes less than the required time.
Let's denote the required time to reach the factory on time as \( t \) hours.
Step 2: Convert minutes to hours.
○ 3 minutes = \(\frac{3}{60}\) hours = \(\frac{1}{20}\) hours
○ 7 minutes = \(\frac{7}{60}\) hours
Step 3: Set up equations for each scenario.
○ At 5 km/hr: The time taken is \( \frac{d}{5} \) hours. Since he is 3 minutes late, the equation is:
\[
\frac{d}{5} = t + \frac{1}{20}
\]
○ At 6 km/hr: The time taken is \( \frac{d}{6} \) hours. Since he is 7 minutes early, the equation is:
\[
\frac{d}{6} = t - \frac{7}{60}
\]
Step 4: Solve the equations.
○ From the first equation:
\[
t = \frac{d}{5} - \frac{1}{20}
\]
○ From the second equation:
\[
t = \frac{d}{6} + \frac{7}{60}
\]
Step 5: Equate the two expressions for \( t \).
\[
\frac{d}{5} - \frac{1}{20} = \frac{d}{6} + \frac{7}{60}
\]
Step 6: Clear the fractions by finding a common denominator.
The common denominator for 5, 6, 20, and 60 is 60. Multiply every term by 60 to eliminate the fractions:
\[
60 \left(\frac{d}{5}\right) - 60 \left(\frac{1}{20}\right) = 60 \left(\frac{d}{6}\right) + 60 \left(\frac{7}{60}\right)
\]
Simplifying gives:
\[
12d - 3 = 10d + 7
\]
Step 7: Solve for \( d \).
\[
12d - 10d = 7 + 3
\]
\[
2d = 10
\]
\[
d = 5
\]
Therefore, the distance of the factory from his house is 5 km.
Q 68. Speed, Time and Distance
Location of B is north of A and location of C is east of A. The distances AB and AC are 5 km and 12 km respectively. The shortest distance (in km) between the locations B and C is
To find the shortest distance between locations B and C, we can use the Pythagorean theorem. The locations A, B, and C form a right triangle with A as the right angle. Here's how we can solve it step by step:
1. Identify the Triangle:
○ Location A is the reference point.
○ Location B is north of A, so AB is the vertical side of the triangle.
○ Location C is east of A, so AC is the horizontal side of the triangle.
2. Given Distances:
○ The distance AB = 5 km.
○ The distance AC = 12 km.
3. Apply the Pythagorean Theorem:
○ The Pythagorean theorem states that in a right triangle, the square of the hypotenuse (the side opposite the right angle) is equal to the sum of the squares of the other two sides.
○ In this case, BC is the hypotenuse.
4. Calculate BC:
○ To find BC, take the square root of 169.
Therefore, the shortest distance between locations B and C is 13 km.
Q 69. Speed, Time and Distance
Two cars start towards each other, from two places A and B which are at a distance of 160 km. They start at the same time 08:10 AM. If the speeds of the cars are 50 km and 30 km per hour respectively, they will meet each other at
Total distance between point A and B = 160 km
Now, let cars meet after t hours.
⸫ Distance travelled by cars of 50 km/h = 50 t
Distance travelled by cars of 30 km/h = 30 t
Now, total distance travelled by both cars = 160 km
According to the question,
50t + 30t = 160 ] 80t = 160
⸫ t = 2 h
⸫ Cars will meet at = 8 : 10 am + 2 = 10 : 10 am.
Q 35. Speed, Time and Distance
A thief running at 8 km/hr is chased by a policeman whose speed is 10 km/hr. If the thief is 100 m ahead of the policeman, then the time required for the policeman to catch the thief will be
To solve this problem, we need to determine how long it will take for the policeman to catch up to the thief. We can do this by calculating the relative speed between the two and using the distance between them.
Step 1: Convert the speeds from km/hr to m/s for easier calculation.
○ Thief's speed: 8 km/hr = \( \frac{8 \times 1000}{60 \times 60} \) m/s = \( \frac{20}{9} \) m/s
○ Policeman's speed: 10 km/hr = \( \frac{10 \times 1000}{60 \times 60} \) m/s = \( \frac{25}{9} \) m/s
Step 2: Calculate the relative speed.
○ Relative speed = Policeman's speed - Thief's speed
○ Relative speed = \( \frac{25}{9} \) m/s - \( \frac{20}{9} \) m/s = \( \frac{5}{9} \) m/s
Step 3: Use the relative speed to find the time required to cover the distance of 100 meters.
○ Distance = 100 meters
○ Time = Distance / Relative Speed
○ Time = \( \frac{100}{\frac{5}{9}} \) seconds = \( 100 \times \frac{9}{5} \) seconds = 180 seconds
Step 4: Convert the time from seconds to minutes.
○ Time = 180 seconds = 3 minutes
Therefore, the time required for the policeman to catch the thief is 3 minutes.
Q 36. Speed, Time and Distance
A train travels at a certain average speed for a distance of 63 km and then travels a distance of 72 km at an average speed of 6 km/hr more than its original speed. If it takes 3 hours to complete the total journey, what is the original speed of the train in km/hr?
To solve this problem, we need to determine the original speed of the train. Let's denote the original speed of the train as km/hr.
Step 1: Calculate the time taken to travel the first part of the journey.
○ The train travels 63 km at the original speed .
○ Time taken for this part = hours.
Step 2: Calculate the time taken to travel the second part of the journey.
○ The train travels 72 km at a speed of km/hr.
○ Time taken for this part = hours.
Step 3: Set up the equation for the total time.
○ According to the problem, the total time for the journey is 3 hours.
○ Therefore, the equation is:
Step 4: Solve the equation.
○ To solve this equation, find a common denominator and simplify:
○ Divide the entire equation by 3 to simplify:
Step 5: Solve the quadratic equation.
○ Use the quadratic formula , where , , and .
○ Calculate the discriminant:
○ Since the discriminant is a perfect square, calculate the roots:
○ The possible solutions are:
○ Since speed cannot be negative, the original speed of the train is 42 km/hr.
Therefore, the correct answer is Option C: 42.
Q 49. Speed, Time and Distance
Four cars are hired at the rate of 6 per km plus the cost of diesel at 40 a litre. In this context, consider the details given in the following table:
|
Car |
Mileage (km/l) |
Hours |
Total Payment (₹) |
|
A |
8 |
20 |
2120 |
|
B |
10 |
25 |
1950 |
|
C |
9 |
24 |
2064 |
|
D |
11 |
22 |
1812 |
Which car maintained the maximum average speed?
To determine which car maintained the maximum average speed, we need to calculate the distance traveled by each car and then divide it by the number of hours to find the average speed.
The total payment for each car includes both the hire charge and the cost of diesel. The formula for the total payment is:
Rearranging the formula to solve for distance:
Let's calculate the distance for each car:
● Car A:
○ Mileage = 8 km/l
○ Total Payment = ₹2120
○ Distance =
○ Average Speed =
● Car B:
○ Mileage = 10 km/l
○ Total Payment = ₹1950
○ Distance =
○ Average Speed =
● Car C:
○ Mileage = 9 km/l
○ Total Payment = ₹2064
○ Distance =
○ Average Speed =
● Car D:
○ Mileage = 11 km/l
○ Total Payment = ₹1812
○ Distance =
○ Average Speed =
Comparing the average speeds, Car A has the highest average speed of 9.64 km/h. Therefore, the car that maintained the maximum average speed is Car A.
Q 24. Speed, Time and Distance
Mr. Kumar drives to work at an average speed of 48 km per hour. The time taken to cover the first 60% of the distance is 10 minutes more than the time taken to cover the remaining distance. How far is his office?
To solve this problem, we need to determine the total distance to Mr. Kumar's office. Let's denote the total distance as kilometers.
Step 1: Understanding the problem
○ Mr. Kumar drives at an average speed of 48 km/h.
○ The time taken to cover the first 60% of the distance is 10 minutes more than the time taken to cover the remaining 40% of the distance.
Step 2: Express the distances
○ The first 60% of the distance is .
○ The remaining 40% of the distance is .
Step 3: Calculate the time taken for each part of the journey
○ Time taken to cover the first 60% of the distance:
○ Time taken to cover the remaining 40% of the distance:
Step 4: Set up the equation based on the given condition
○ According to the problem, the time for the first 60% is 10 minutes (or hours) more than the time for the remaining 40%:
Step 5: Solve the equation
○ Simplify the equation:
○ Combine the terms:
○ Multiply both sides by 48 to solve for :
○ Divide by 0.2:
Therefore, the total distance to Mr. Kumar's office is 40 km.
Q 16. Speed, Time and Distance
Consider the following Velocity - Time graph. It shows two trains starting simultaneously on parallel tracks. [Image of Velocity-Time graph] With reference to the above graph, which one of the following statements is not correct?
Acceleration in a velocity time graph can be seen from the graph’s slope (tilt with respect to horizontal). B’s slope is more than A, which means for the same time period, it is changing velocity faster than that of A. Hence more acceleration. Option (b) and (c) can be verified from y-axis.
To calculate distance in a velocity-time graph, simply take the area under the curve. It is more for B than A.
Q 23. Speed, Time and Distance
Consider the following distance time graph. The graph shows three athletes A, B and C running side by side for a 30 km race. With reference to the above graph, consider the following statements:
1. The race was won by A.
2. B was ahead of A up to 25 km mark.
3. C ran very slowly from the beginning.
Which of the statements given above is/are correct?
As per the graph, A has taken just below 30 min to complete the race whereas B has taken more than 35 min and C did not finish the race. So statement 1 is correct. B has taken less time than A upto 25 km. Hence B was ahead of A upto 25 km. So statement 2 is also correct. You can observe that C has taken less time than A and B upto 25 km. So, statement 3 is wrong.
Q 41. Speed, Time and Distance
If a bus travels 160 km in 4 hours and a train travels 320 km in 6 hours at uniform speeds, then what is the ratio of the distances travelled by them in one hour