Question

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A man walks a certain distance with certain speed. If he walks 1/2 km an hour faster, he takes 1 hour less. But, if he walks 1 km an hour slower, he takes 3 more hours. Find the distance covered by the man and his original rate of walking.​

Answer 1

Distance=Time×Speed

Hence,The speed of the man=4 km/hr

The time of walking=9 hrs.

The distance covered by man 4×9=36 km.

Answer 2

SOLUTION :—

Let the original speed y km/ hr and total time taken be x h

→ Distance = xy

[ Distance = time × speed]

→ (y + ½) (x - 1) = xy

→ xy - y + x/2 - ½ = xy

→ x/2 - y = ½ ………………(1)

[Distance covered in each case is same]

→ (y -1) (x + 3) = xy

→ xy + 3y -x -3 = xy

→ -x + 3y = 3……………..(2)

Multiply equation 1 by 3 ,

→ 3x/2 - 3y = 3/2…………(3)

On adding eq 2 and 3,

-x + 3y = 3

3x/2 - 3y = 3/2

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-x +3x/2 = 3/2 +3

→ (-2x + 3x)/2 = (3+6)/2

→ x/2 = 9/2

→ x = 9

On Putting x = 9 in eq 2,

→ -x + 3y = 3

→ -9 + 3y = 3

→ 3y = 3 +9

→ 3y = 12

→ y = 12/3

→ y = 4

Distance covered by man = xy = 9 × 4 = 36 km

Original speed of walking = y = 4 km/h

Hence,

The Distance covered by man is 36 km & Original speed of walking is 4 km/h.