Question

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

Answer 1

Step 1: Define Variables

Let Tom's age be \( t \). Then Dick's age is \( 3t \) and Harry's age is \( 6t \).

Step 2: Apply Given Condition

Dick's age is one less than the average of the three ages: \( 3t = \frac{t + 3t + 6t}{3} - 1 \).

Step 3: Solve for \( t \)

Simplify the equation: \( 3t = \frac{10t}{3} - 1 \). Multiply by 3: \( 9t = 10t - 3 \). Rearrange: \( -t = -3 \), so \( t = 3 \).

Step 4: Calculate Harry's Age

Harry's age is \( 6t \). Substituting \( t = 3 \): \( 6 \times 3 = \boxed{18} \).

Final Answer:

Harry's age is: \[ \boxed{18 \text{ years}} \]