Wiley Edge Time, Speed and Distance Quiz-4

The hands of the clock overlap approximately every 66 minutes. So between 5:00 AM and 5:00 PM the hands overlap : 720/66 = 10.90 ; so the hands have overlapped 10 times .

To find the speed of the car, we need to convert the distance and time units to a common format. Since the pedestrian was walking at 3.6 miles per hour (5.8 km/h), and the car remained visible for 10 minutes,
o the total distance the car traveled after crossing the pedestrian = (distance pedestrian walked)+0.9 miles = (3.6 * ⅙ ) + 0.9 = 0.6 + 0.9 = 1.5 miles.
Took time of 10 minutes to travel this distance. So , speed = 1.5 / (⅙) = 9 mph.

Since the helicopter's speed is five times(5x) that of the car(x), it will cover the same distance in one-fifth of the time.But he relative speed of car and helicopter is 4x .
Therefore, when the helicopter catches up with the car, it will have traveled one-fourth of the distance the car covered. If the car is 30 kilometers from the city, the helicopter will catch up with it after traveling one-fourth of 30 kilometers, which is 7.5 kilometers.
Thus, the helicopter catches up with the car when it is 30 kilometers + 7.5 kilometers = 37.5 kilometers from the city.

: In the first race, Kiran finishes 20 meters ahead of Rahul, which means Kiran covers the entire distance of 200 meters while Rahul covers 180 meters.
In the second race, Kiran starts 20 meters behind the starting line, so he has to cover 220 meters to reach the finishing line. Meanwhile, Rahul starts at the regular starting line and covers the full 200 meters.
Since both runners maintain their earlier speeds, Kiran covers an extra distance of 20 meters in the second race compared to Rahul.
However, since Kiran had finished 20 meters ahead of Rahul in the first race, they would reach the finishing line simultaneously in the second race.

The average speed of a cyclist can be calculated by dividing the total distance traveled by the total time taken.
In this case, the circumference of the circular track is given by 2 * π * radius = 2 * 3.14 * 50 = 314 meters. As the cyclist completes one revolution in 3 minutes, the total time taken is 3 minutes. Average Speed = Total Distance / Total Time = 314 meters / 3 minutes≈ 104.67 meters per minute. So, the approximate average speed of the cyclist is approximately 104.67 meters per minute.

Let's assume the speeds of the faster and slower planes are v1 and v2, respectively. When they move in opposite directions, the relative speed between them is the sum of their individual speeds, which is v1 + v2. Since they meet every 30 seconds and the circumference is 5 km, the relative speed can be calculated as 5 km / 30 s = 0.1667 km/s.
Therefore, v1 + v2 = 0.1667 km/s.
When they move in the same direction, the relative speed is the difference between their individual speeds, which is v1 - v2. Since one plane overtakes the other every 120 seconds, the relative speed can be calculated as 5 km / 120 s = 0.0417 km/s.
Therefore, v1 - v2 = 0.0417 km/s.
We now have a system of equations:
v1 + v2 = 0.1667
v1 - v2 = 0.0417
Solving this system of equations, we find v1 = 0.1033 km/s and v2 = 0.0634 km/s. The speed of the slower plane is v2, which is approximately 0.0634 km/s. Rounding it to two decimal places, we get 0.06 km/s.

To find the average speed, we need to calculate the total time taken for the journey and divide it by the total distance. Let's assume the total distance is 100 km.
The time taken to cover 40% of the distance at 15 km/h is (40/100) * (100/15) = 40/15 hours.
The time taken to cover 30% of the distance at 30 km/h is (30/100) * (100/30) = 1 hours.
The time taken to cover the remaining 30% of the distance at 20 km/h is (30/100) * (100/20) = 3/2 hours.The total time taken is (40/15) + (1) + (3/2) = (8/3) + (1) + (3/2) = 5.17 hours.
The average speed is calculated by dividing the total distance (100 km) by the total time taken (32/6 hours):
Average speed = 100 / 5.17 = 19.34 km/h.
Rounded to the nearest whole number, the average speed is 19 km/h.
Therefore, none of the given options accurately represents the average speed of 19 km/h.

Speed of first train = 60 km/h ; speed of 2nd train = 1.2*60=72 km/h; let the speed of the 3rd train be X km/h. Now, when the 3rd train departs, 1st and 2nd trains are 45 km and 54 km from the start respectively . let the time to over take 1st train be ‘t’ hours , so time to overtake 2nd train will be (t + 1) hr
So if we take the relative speed of 1st and 3rd trains we get : 60 - x km/h
So if we take the relative speed of 2nd and 3rd trains we get : 72 - x km/h
So the distance equations for the trains:
(60-x)*t = 45
(72-x)*(t+1)=54;
So on solving these two equations we get x=96 km/hr

We can observe that in the actual scenario, the snail covers an additional 3 cm in each minute compared to the previous one.

In the hypothetical scenario, where it covers 0.5 cm in the first minute and 5 cm more in each successive minute, the difference in the paths covered will increase by 4.5 cm each minute.
To find the time for which the difference falls between 5 cm and 20 cm, we need to calculate how many times the difference of 4.5 cm can fit within that range.

By dividing 15 cm (the difference between 5 cm and 20 cm) by 4.5 cm, we get approximately 3.33.

Since we're looking for a whole number of minutes, the snail would need to move for at least 4 minutes to have a difference greater than 5 cm but less than 20 cm.

Therefore, the correct answer is 6 minutes.

Let's denote the length of the shorter leg of the track as x km. According to the problem, the longer leg of the track is 2 km more than the shorter leg, so its length is (x + 2) km.
Hypotenuse^2 = x^2 + (x + 2)^2
= x^2 + x^2 + 4x + 4
= 2x^2 + 4x + 4
Now, let's calculate the time taken for the escort vehicle to cover each segment of the track:
Time taken on the side road = sqrt(2x^2 + 4x + 4) / 30 hours
Time taken on the first highway segment = x / 42 hours
Time taken on the second highway segment = (x + 2) / 42 hours
According to the problem, the time taken on the side road is equal to the combined time taken on the two highway segments:sqrt(2x^2 + 4x + 4) / 30 = x / 42 + (x + 2) / 42
To simplify the equation, we can eliminate the square root by squaring both sides:
(2x^2 + 4x + 4) / 900 = x^2 / 1764 + (x + 2)^2 / 1764
On solving the quadratic equation we will get x= 6 km; so the total length of the track will be : x+x+2 = 14 km.