A bus left point X for point Y. Two hours later a car left point X for...
In this question consider the total distance as 100%. Hence the sum of their speeds will be 75% per hour. Checking option (c)
If the bus took 6 hours, it would cover 16.66% distance per hour and the car would cover 25% distance per hour. (as it takes 2 hours less than the bus.)
This gives an addition of only 41.66%. Hence, the answer is not correct.
Option (b) is the correct answer
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A bus left point X for point Y. Two hours later a car left point X for...
Given information:
- A bus left point X for point Y.
- Two hours later a car left point X for Y and arrived at Y at the same time as the bus.
- If the car and the bus left simultaneously from the opposite ends X and Y towards each other, they would meet 1.33 hours after the start.
To find: How much time did it take the bus to travel from X to Y?
Approach:
Let's assume that the distance between X and Y is 'd' km and the speed of the bus is 'b' km/h.
Using the formula: speed = distance/time, we can derive the following equations:
- When the bus travels from X to Y:
- speed of the bus = b km/h
- time taken by the bus = t hours (to be found)
- distance = d km
- Therefore, d = b * t
- When the car travels from X to Y:
- Let's assume the speed of the car is 'c' km/h.
- The car started 2 hours later than the bus, so it traveled for (t - 2) hours.
- Therefore, distance traveled by the car = c * (t - 2) km
- Also, we know that the car and the bus arrived at the same time, so the distance traveled by the car = distance traveled by the bus.
- Therefore, c * (t - 2) = b * t
- When the bus and the car start simultaneously from opposite ends of X and Y:
- The total distance to be covered = 2d km
- The combined speed of the bus and the car = b + c km/h
- Time taken to meet = 1.33 hours
- Therefore, 2d = (b + c) * 1.33
Now, we have three equations and three variables (d, b, and c). We can solve these equations to find the value of 't', which is the time taken by the bus to travel from X to Y.
Solution:
- From equation 1, d = b * t
- From equation 2, c * (t - 2) = b * t
- Simplifying, we get: c * t - 2c = b * t
- Rearranging, we get: t = 2c / (c + b)
- Substituting the value of 't' in equation 1, we get: d = 2bc / (c + b)
- Substituting the value of 'd' in equation 3, we get: 2 * 2bc / (c + b) = (b + c) * 1.33
- Simplifying, we get: 4bc = 1.33(c + b)^2
- Substituting the value of 't' in equation 2, we get: c * (2c / (c + b) - 2) = b * 2c / (c + b)
- Simplifying, we get: c^2 - 4bc + 4b^2 = 0
- Solving this quadratic equation, we get: c = 2b or c = 2b/3 (rejecting negative value)