A lady said, "If you reverse my age digits, you'll get my husband's age. He is older, and the difference between their ages is one-eleventh of their sum." What is the man's age?

Let the tens digit of the husband's age be $x$ and the units digit be $y$.
Since the husband is older, his age is $10x+y$ and the wife's age is $10y+x$, where $x>y$.

According to the given condition:

$$\text{Difference}=\frac{1}{11}\times \text{Sum}$$

$$(10x+y)-(10y+x)=\frac{1}{11}[(10x+y)+(10y+x)]$$

$$9x-9y=\frac{1}{11}(11x+11y)$$

$$9(x-y)=x+y$$

$$9x-9y=x+y$$

$$8x=10y⟹\frac{x}{y}=\frac{5}{4}$$

Since $x$ and $y$ must be single digits, we find $x=5$ and $y=4$.

* Husband's age = $10(5)+4=54\text{ years}$
* Wife's age = $10(4)+5=45\text{ years}$