Dick is thrice as old as Tom and Harry is twice as old as Dick. If Dick's age is 1 year less than the average age of all three, then Harry's age, in years, is

The correct answer is 18.

To find Harry's age, let us set up the algebraic relationships between the ages of Tom, Dick, and Harry step-by-step.

Let the age of Tom be represented by T, the age of Dick by D, and the age of Harry by H.

From the problem statement, we are given the following relationships:
1. Dick is thrice as old as Tom:
$$D=3T$$
This can also be written in terms of Dick's age as:
$$T=\frac{D}{3}$$

2. Harry is twice as old as Dick:
$$H=2D$$

3. The average age of all three individuals (Tom, Dick, and Harry) is given by:
$$\text{Average Age}=\frac{T+D+H}{3}$$

We are told that Dick's age is 1 year less than this average age. We can write this relationship as:
$$D=\frac{T+D+H}{3}-1$$

Now, let us substitute the expressions for T and H in terms of D into the average age equation:
$$D=\frac{\frac{D}{3}+D+2D}{3}-1$$

Simplify the numerator inside the fraction:
$$\frac{D}{3}+D+2D=\frac{D}{3}+3D=\frac{D+9D}{3}=\frac{10D}{3}$$

Substitute this back into the equation:
$$D=\frac{\left(\frac{10D}{3}\right)}{3}-1$$
$$D=\frac{10D}{9}-1$$

To solve for D, rearrange the equation by moving the terms involving D to one side:
$$1=\frac{10D}{9}-D$$
$$1=\frac{10D-9D}{9}$$
$$1=\frac{D}{9}$$
Multiplying both sides by 9 gives:
$$D=9$$

Thus, Dick's age is 9 years.

Now we calculate Harry's age using the relationship:
$$H=2D$$
$$H=2\times 9=18$$

Harry's age is indeed 18 years.