A post has one-fifth of its length in the mud, one half in the water and the…

Concept: In a "whole split into fractional shares plus a known remainder" word problem, represent the unknown whole with one variable, write every described share as a fraction of that same variable, and set the sum of all the shares -- including the stated remainder -- equal to the whole. Solving that single linear equation for the variable gives the whole quantity.

Application:

1. Let the total length of the post be L (in feet).

2. The share in the mud is L/5 and the share in the water is L/2; the remaining share, out of the water, is given as 10 ft.

3. Since the three shares together make up the whole post: L = L/5 + L/2 + 10.

4. Move the fractional shares to the left side: L - (L/5 + L/2) = 10.

5. Convert to a common denominator of 10: L - (2L/10 + 5L/10) = 10, so L - 7L/10 = 10, so 3L/10 = 10.

6. Multiply both sides by 10 and divide by 3: L = 100/3 = 33 and 1/3 ft.

7. Convert the fractional foot to inches: 1/3 ft x 12 in/ft = 4 in, so L = 33 ft 4 in.

Cross-check: Substituting L = 100/3 ft back gives mud = 20/3 ft and water = 50/3 ft, a combined submerged share of 70/3 ft; the remainder out of the water is L - 70/3 = 100/3 - 70/3 = 30/3 = 10 ft, exactly matching the given remainder.

So the whole length of the post is 33 ft 4 in.