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A student travels to his school at a speed of 4 km/hour and reaches the school 15 minutes late. On traveling at a speed of 6 km/hour; he reaches the school 5 minutes early. At what speed must he travel to reach the school just in time?
- a)41/3 km/hr
- b)31/3 km/hr
- c)51/3 km/hr
- d)61/3 km/hr
Correct answer is option 'C'. Can you explain this answer?
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Verified Answer
A student travels to his school at a speed of 4 km/hour and reaches th...
Given, a student travels to his school at a speed of 4 km/hour and reaches the school 15 minutes late. On travelling at a speed of 6 km/hour; he reaches the school 5 minutes early.
Let the original speed and time be ‘s’ and ‘t’ respectively.
Distance = speed × time
Distance = s × t = 4 × (t + 15/60) = 6 × (t – 5/60)
Distance = st = 4t + 1 = 6t – 0.5
⇒ 4t + 1 = 6t – 0.5
⇒ 2t = 1.5
⇒ t = 0.75 hours
Now, st = 4t + 1
⇒ s = 4 + 1/0.75 = 16/3 = 5 1/3 km/hr
Most Upvoted Answer
A student travels to his school at a speed of 4 km/hour and reaches th...
To solve this problem, let's assume the distance between the student's home and school is 'd' km.
Let's calculate the time taken by the student to reach the school at a speed of 4 km/hour.
Time taken at 4 km/hour = distance / speed = d / 4
We know that the student reaches the school 15 minutes late, which is equal to 15/60 = 1/4 hours.
So, the actual time taken to reach the school is (d / 4) + 1/4.
Now, let's calculate the time taken by the student to reach the school at a speed of 6 km/hour.
Time taken at 6 km/hour = distance / speed = d / 6
We know that the student reaches the school 5 minutes early, which is equal to -5/60 = -1/12 hours.
So, the actual time taken to reach the school is (d / 6) - 1/12.
Since the student reaches the school just in time, the actual time taken at both speeds would be the same. Therefore, we can equate the two expressions for the actual time taken.
(d / 4) + 1/4 = (d / 6) - 1/12
Now, let's solve this equation to find the value of 'd'.
Multiplying the equation by 12 to eliminate the denominators, we get:
3d + 3 = 2d - 1
Simplifying the equation, we get:
3d - 2d = -1 - 3
d = -4
Since distance cannot be negative, we discard this solution.
Hence, there is no solution for this problem.
Therefore, the given options are incorrect and the correct answer cannot be determined.
Let's calculate the time taken by the student to reach the school at a speed of 4 km/hour.
Time taken at 4 km/hour = distance / speed = d / 4
We know that the student reaches the school 15 minutes late, which is equal to 15/60 = 1/4 hours.
So, the actual time taken to reach the school is (d / 4) + 1/4.
Now, let's calculate the time taken by the student to reach the school at a speed of 6 km/hour.
Time taken at 6 km/hour = distance / speed = d / 6
We know that the student reaches the school 5 minutes early, which is equal to -5/60 = -1/12 hours.
So, the actual time taken to reach the school is (d / 6) - 1/12.
Since the student reaches the school just in time, the actual time taken at both speeds would be the same. Therefore, we can equate the two expressions for the actual time taken.
(d / 4) + 1/4 = (d / 6) - 1/12
Now, let's solve this equation to find the value of 'd'.
Multiplying the equation by 12 to eliminate the denominators, we get:
3d + 3 = 2d - 1
Simplifying the equation, we get:
3d - 2d = -1 - 3
d = -4
Since distance cannot be negative, we discard this solution.
Hence, there is no solution for this problem.
Therefore, the given options are incorrect and the correct answer cannot be determined.
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A student travels to his school at a speed of 4 km/hour and reaches the school 15 minutes late. On traveling at a speed of 6 km/hour; he reaches the school 5 minutes early. At what speed must he travel to reach the school just in time?a)41/3 km/hrb)31/3 km/hrc)51/3 km/hrd)61/3 km/hrCorrect answer is option 'C'. Can you explain this answer?
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