A man can walk uphill at the rate of $2\frac{1}{2}$ $km/hr$ and down hill at the rate of $3\frac{1}{2}$ $km/hr$. If the total time required to walk a certain distance up the hill and return to the starting point is 4 hr. 36 min, then what is the distance walked up the hill by the man?

Uphill speed = $2\frac{1}{2}$ km/hr

Downhill speed = $3\frac{1}{2}$ km/hr

Total time = 4 hr 36 min

The distance going up equals the distance coming down (same path).

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Converting speeds to improper fractions:

Uphill speed = $2\frac{1}{2} = \frac{5}{2}$ km/hr

Downhill speed = $3\frac{1}{2} = \frac{7}{2}$ km/hr

Converting time to hours:

36 minutes = $\frac{36}{60}$ hours = $\frac{3}{5}$ hours

Total time = $4 + \frac{3}{5}$

$= \frac{20}{5} + \frac{3}{5}$

$= \frac{23}{5}$ hours

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Let $d$ = distance walked uphill (in km)

Using Time = Distance ÷ Speed:

Time to go uphill = $d \div \frac{5}{2}$

$= d \times \frac{2}{5}$

$= \frac{2d}{5}$ hours

Time to come downhill = $d \div \frac{7}{2}$

$= d \times \frac{2}{7}$

$= \frac{2d}{7}$ hours

Total time:

$\frac{2d}{5} + \frac{2d}{7} = \frac{23}{5}$

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Finding common denominator (LCM of 5 and 7 = 35):

$\frac{2d \times 7}{35} + \frac{2d \times 5}{35} = \frac{23}{5}$

$\frac{14d + 10d}{35} = \frac{23}{5}$

$\frac{24d}{35} = \frac{23}{5}$

Cross multiplying:

$24d \times 5 = 23 \times 35$

$120d = 805$

$d = \frac{805}{120}$

$= \frac{161}{24}$

$= 6\frac{17}{24}$ km

$\approx 6.7$ km

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The distance walked uphill is $6\frac{17}{24}$ km, which is approximately 6.7 km.