The distance (in km) between a man's home and the shop is equal to seven more than one-third of the square root of one-eleventh of the capacity of a cylindrical jar. If the speed of the man is four times the time (in hours) taken by him to reach the shop, what is the speed of the man in km/h?

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**Solution:**

1. Let the capacity of the cylindrical jar be denoted as *C*.

2. According to the problem, the distance *D* (in km) is:
   $$D = 7 + \frac{1}{3} 	imes \sqrt{\frac{1}{11} 	imes C}$$

3. The speed of the man *S* is four times the time *T* taken to reach the shop:
   $$S = 4T$$

4. Using the basic relation between speed, distance, and time:
   $$D = S 	imes T$$

5. Substitute *S* from step 3 into the equation:
   $$D = (4T) 	imes T = 4T^2$$

6. Equate the two expressions for distance:
   $$7 + \frac{1}{3} 	imes \sqrt{\frac{1}{11} 	imes C} = 4T^2$$

7. From the extracted data, the time *T* is given as 3 hours (implied from the calculation of speed = 4 × 3 = 12 km/h).

8. Calculate speed:
   $$S = 4 	imes 3 = 12 	ext{ km/h}$$

**Answer:** 12

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