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The time taken by a man to cover 300 km on a scooter was 3/2 hrs more than the time taken by him during the return journey. If the speed in returning to 10 kmph more than the speed is going, find his speed in each direction
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The time taken by a man to cover 300 km on a scooter was 3/2 hrs more ...
Given:
- Distance covered = 300 km
- Let the speed during going be x kmph
- Speed during return journey = (x+10) kmph
- Time taken during going - time taken during return journey = 3/2 hours
To find:
- Speed in each direction
Solution:
Let's first find the time taken during going and return journey.
Time taken during going:
Distance = Speed × Time
300 = x × Time
Time = 300/x hours
Time taken during return journey:
Distance = Speed × Time
300 = (x+10) × Time
Time = 300/(x+10) hours
Given that time taken during going - time taken during return journey = 3/2 hours.
Therefore,
300/x - 300/(x+10) = 3/2
Let's simplify this equation by taking LCM.
300(x+10) - 300x = 3x(x+10)/2
3000 = 3x² + 30x
Dividing both sides by 3, we get:
1000 = x² + 10x
Rearranging, we get:
x² + 10x - 1000 = 0
Now, we can solve this quadratic equation to find the value of x.
Using the quadratic formula, we get:
x = (-b ± √(b²-4ac)) / 2a
where a = 1, b = 10 and c = -1000
x = (-10 ± √(10²-4×1×(-1000))) / 2×1
x = (-10 ± √(10400)) / 2
x = (-10 ± 102) / 2
Taking the positive value for x, we get:
x = (-10 + 102) / 2
x = 46 kmph
Therefore, the speed during going is 46 kmph and the speed during return journey is (46+10) = 56 kmph.
Final Answer:
- Speed during going = 46 kmph
- Speed during return journey = 56 kmph
- Distance covered = 300 km
- Let the speed during going be x kmph
- Speed during return journey = (x+10) kmph
- Time taken during going - time taken during return journey = 3/2 hours
To find:
- Speed in each direction
Solution:
Let's first find the time taken during going and return journey.
Time taken during going:
Distance = Speed × Time
300 = x × Time
Time = 300/x hours
Time taken during return journey:
Distance = Speed × Time
300 = (x+10) × Time
Time = 300/(x+10) hours
Given that time taken during going - time taken during return journey = 3/2 hours.
Therefore,
300/x - 300/(x+10) = 3/2
Let's simplify this equation by taking LCM.
300(x+10) - 300x = 3x(x+10)/2
3000 = 3x² + 30x
Dividing both sides by 3, we get:
1000 = x² + 10x
Rearranging, we get:
x² + 10x - 1000 = 0
Now, we can solve this quadratic equation to find the value of x.
Using the quadratic formula, we get:
x = (-b ± √(b²-4ac)) / 2a
where a = 1, b = 10 and c = -1000
x = (-10 ± √(10²-4×1×(-1000))) / 2×1
x = (-10 ± √(10400)) / 2
x = (-10 ± 102) / 2
Taking the positive value for x, we get:
x = (-10 + 102) / 2
x = 46 kmph
Therefore, the speed during going is 46 kmph and the speed during return journey is (46+10) = 56 kmph.
Final Answer:
- Speed during going = 46 kmph
- Speed during return journey = 56 kmph
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The time taken by a man to cover 300 km on a scooter was 3/2 hrs more than the time taken by him during the return journey. If the speed in returning to 10 kmph more than the speed is going, find his speed in each direction Related: Exercise 4.3 - Quadratic Equations, Class 10, Maths? for Class 10 2024 is part of Class 10 preparation. The Question and answers have been prepared according to the Class 10 exam syllabus. Information about The time taken by a man to cover 300 km on a scooter was 3/2 hrs more than the time taken by him during the return journey. If the speed in returning to 10 kmph more than the speed is going, find his speed in each direction Related: Exercise 4.3 - Quadratic Equations, Class 10, Maths? covers all topics & solutions for Class 10 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for The time taken by a man to cover 300 km on a scooter was 3/2 hrs more than the time taken by him during the return journey. If the speed in returning to 10 kmph more than the speed is going, find his speed in each direction Related: Exercise 4.3 - Quadratic Equations, Class 10, Maths?.
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