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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 ...
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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
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