Q27 of 35 Page 3


A man walks a certain distance at a certain speed. Had he walked (1/2) km/hrfaster, he would have taken 3hours longer. Find the distance.


Let the original speed be x km/hr and time taken be y hrs.
Then, distance covered = xy km.
Speed = km/hr,
Time taken = (y - 1)hrs.
Distance = km.
Therefore xy =
xy = xy – x + ( y/2) – (1/2)
- x + (y/2) – (1/2) = 0
y - 2x - 1 = 0
y - 2x = 1 … (i)
New speed = (x - 1) km/hr

Time taken = (y + 3) hrs.
Therefore distance = (x - 1) (y + 3) km.
Therefore xy = (x - 1)(y + 3)
3x – y = 3 … (ii)
Adding (i) and (ii), we get x = 4
Substituting x = 4 in (i),

We get y =9
Therefore speed = 4 km/hr,

time taken = 9 hrs
Hence, distance= 4 × 9 km = 36 km.

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