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A motorcyclist left point A for point B. Two hours later, another motorcyclist left A for B and arrived at B at the same time as the first motorcyclist. Had both the motorcyclists started simultaneously from A and B travelling towards each other, they would have met in 80 minutes. How much time did it take the faster motorcyclist to travel from A to B?
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A motorcyclist left point A for point B. Two hours later, another moto...
Problem: A motorcyclist left point A for point B. Two hours later, another motorcyclist left A for B and arrived at B at the same time as the first motorcyclist. Had both the motorcyclists started simultaneously from A and B travelling towards each other, they would have met in 80 minutes. How much time did it take the faster motorcyclist to travel from A to B?
Solution:
Let's assume that the distance between A and B is d, and the speed of the slower motorcyclist is s1, while the speed of the faster motorcyclist is s2.
Case 1: When the two motorcyclists started at different times
- Let's assume that the first motorcyclist started at time t=0, then the second motorcyclist started at time t=2.
- Since they both arrived at point B at the same time, we can say that the time taken by the first motorcyclist to travel from A to B is t1, while the time taken by the second motorcyclist to travel from A to B is t2.
- Therefore, we can write the following equation: d/s1 = (d/s2) - 2
Case 2: When the two motorcyclists started at the same time
- In this case, we can use the formula: time = distance/speed
- Since the two motorcyclists are travelling towards each other, their relative speed is (s1 + s2).
- Therefore, the time taken for them to meet is: d/(s1+s2) = 80/60 = 4/3 hours = 80 minutes
Combining the two cases:
- We know that the time taken by the slower motorcyclist to travel from A to B is t1 = d/s1, and the time taken by the faster motorcyclist to travel from A to B is t2 = d/s2.
- From case 1, we have the equation: d/s1 = (d/s2) - 2
- We can substitute d/s1 = t1 and d/s2 = t2 in the above equation to get: t1 = t2 - 2
- From case 2, we have: d/(s1+s2) = 4/3
- We can substitute d = t2*s2 and s1 = s2 - k (where k is the speed difference between the two motorcyclists) in the above equation to get: t2 = 2k/5
- Substituting t2 in terms of k in the equation t1 = t2 - 2, we get: t1 = 2(k/5) - 2 = (2k-10)/5
- Since we know that t1 + t2 = 4/3, we can substitute t1 and t2 in terms of k in this equation to get: (2k-10)/5 + 2k/5 = 4/3
- Solving for k, we get k = 15 km/hr
- Therefore, the speed of the slower motorcyclist is s1 = s2 - k = s2 - 15 km/hr
- From case 1, we have the equation: d/s
Solution:
Let's assume that the distance between A and B is d, and the speed of the slower motorcyclist is s1, while the speed of the faster motorcyclist is s2.
Case 1: When the two motorcyclists started at different times
- Let's assume that the first motorcyclist started at time t=0, then the second motorcyclist started at time t=2.
- Since they both arrived at point B at the same time, we can say that the time taken by the first motorcyclist to travel from A to B is t1, while the time taken by the second motorcyclist to travel from A to B is t2.
- Therefore, we can write the following equation: d/s1 = (d/s2) - 2
Case 2: When the two motorcyclists started at the same time
- In this case, we can use the formula: time = distance/speed
- Since the two motorcyclists are travelling towards each other, their relative speed is (s1 + s2).
- Therefore, the time taken for them to meet is: d/(s1+s2) = 80/60 = 4/3 hours = 80 minutes
Combining the two cases:
- We know that the time taken by the slower motorcyclist to travel from A to B is t1 = d/s1, and the time taken by the faster motorcyclist to travel from A to B is t2 = d/s2.
- From case 1, we have the equation: d/s1 = (d/s2) - 2
- We can substitute d/s1 = t1 and d/s2 = t2 in the above equation to get: t1 = t2 - 2
- From case 2, we have: d/(s1+s2) = 4/3
- We can substitute d = t2*s2 and s1 = s2 - k (where k is the speed difference between the two motorcyclists) in the above equation to get: t2 = 2k/5
- Substituting t2 in terms of k in the equation t1 = t2 - 2, we get: t1 = 2(k/5) - 2 = (2k-10)/5
- Since we know that t1 + t2 = 4/3, we can substitute t1 and t2 in terms of k in this equation to get: (2k-10)/5 + 2k/5 = 4/3
- Solving for k, we get k = 15 km/hr
- Therefore, the speed of the slower motorcyclist is s1 = s2 - k = s2 - 15 km/hr
- From case 1, we have the equation: d/s
Community Answer
A motorcyclist left point A for point B. Two hours later, another moto...
2 hours is answer
in first case distance is same, so D = S1T = S2(T-2). From this we get S1 = S2(T-2)/T. Also when they are travelling towards each other. D = (S1 + S2)(80/60). So, S1 = 4S2/(3T-4). Now equate it with S1 = S2(T-2)/T. we will get T = 4 or T = 2/3. But T≠2/3 because time taken by fast motorcyclist is T -2 which will be negative if we take T = 2/3. therefore T = 4, and reqd time = T - 2 = 2hours
in first case distance is same, so D = S1T = S2(T-2). From this we get S1 = S2(T-2)/T. Also when they are travelling towards each other. D = (S1 + S2)(80/60). So, S1 = 4S2/(3T-4). Now equate it with S1 = S2(T-2)/T. we will get T = 4 or T = 2/3. But T≠2/3 because time taken by fast motorcyclist is T -2 which will be negative if we take T = 2/3. therefore T = 4, and reqd time = T - 2 = 2hours
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A motorcyclist left point A for point B. Two hours later, another motorcyclist left A for B and arrived at B at the same time as the first motorcyclist. Had both the motorcyclists started simultaneously from A and B travelling towards each other, they would have met in 80 minutes. How much time did it take the faster motorcyclist to travel from A to B? for CAT 2025 is part of CAT preparation. The Question and answers have been prepared according to the CAT exam syllabus. Information about A motorcyclist left point A for point B. Two hours later, another motorcyclist left A for B and arrived at B at the same time as the first motorcyclist. Had both the motorcyclists started simultaneously from A and B travelling towards each other, they would have met in 80 minutes. How much time did it take the faster motorcyclist to travel from A to B? covers all topics & solutions for CAT 2025 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A motorcyclist left point A for point B. Two hours later, another motorcyclist left A for B and arrived at B at the same time as the first motorcyclist. Had both the motorcyclists started simultaneously from A and B travelling towards each other, they would have met in 80 minutes. How much time did it take the faster motorcyclist to travel from A to B?.
A motorcyclist left point A for point B. Two hours later, another motorcyclist left A for B and arrived at B at the same time as the first motorcyclist. Had both the motorcyclists started simultaneously from A and B travelling towards each other, they would have met in 80 minutes. How much time did it take the faster motorcyclist to travel from A to B? for CAT 2025 is part of CAT preparation. The Question and answers have been prepared according to the CAT exam syllabus. Information about A motorcyclist left point A for point B. Two hours later, another motorcyclist left A for B and arrived at B at the same time as the first motorcyclist. Had both the motorcyclists started simultaneously from A and B travelling towards each other, they would have met in 80 minutes. How much time did it take the faster motorcyclist to travel from A to B? covers all topics & solutions for CAT 2025 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A motorcyclist left point A for point B. Two hours later, another motorcyclist left A for B and arrived at B at the same time as the first motorcyclist. Had both the motorcyclists started simultaneously from A and B travelling towards each other, they would have met in 80 minutes. How much time did it take the faster motorcyclist to travel from A to B?.
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A motorcyclist left point A for point B. Two hours later, another motorcyclist left A for B and arrived at B at the same time as the first motorcyclist. Had both the motorcyclists started simultaneously from A and B travelling towards each other, they would have met in 80 minutes. How much time did it take the faster motorcyclist to travel from A to B?, a detailed solution for A motorcyclist left point A for point B. Two hours later, another motorcyclist left A for B and arrived at B at the same time as the first motorcyclist. Had both the motorcyclists started simultaneously from A and B travelling towards each other, they would have met in 80 minutes. How much time did it take the faster motorcyclist to travel from A to B? has been provided alongside types of A motorcyclist left point A for point B. Two hours later, another motorcyclist left A for B and arrived at B at the same time as the first motorcyclist. Had both the motorcyclists started simultaneously from A and B travelling towards each other, they would have met in 80 minutes. How much time did it take the faster motorcyclist to travel from A to B? theory, EduRev gives you an
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