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: Let the age of Tom be

\[ t \] Then the ages of Dick and Harry are: \[ \text{Dick} = 3t, \quad \text{Harry} = 6t \]

Step 2: Use the given condition

It is given that: 
Dick’s age = one less than the average of all three: \[ 3t = \frac{t + 3t + 6t}{3} - 1 \]

Step 3: Simplify the equation

Add the terms in the numerator: \[ 3t = \frac{10t}{3} - 1 \] Multiply both sides by 3: \[ 9t = 10t - 3 \] Solve for \( t \): \[ 9t - 10t = -3 \Rightarrow -t = -3 \Rightarrow t = 3 \]

Step 4: Age of Harry

\[ \text{Harry's age} = 6t = 6 \times 3 = \boxed{18} \]

Final Answer:

The age of Harry is: \[ \boxed{18 \text{ years}} \]

Answer 2

Let:

• D = Age of Dick
• T = Age of Tom
• H = Age of Harry

Step 1: Use the given relationships

From the problem: \[ \text{(i)} \quad D = 3T \Rightarrow T = \frac{D}{3} \] \[ \text{(ii)} \quad H = 2D \] \[ \text{(iii)} \quad D = \left( \frac{T + D + H}{3} \right) - 1 \Rightarrow D + 1 = \frac{T + D + H}{3} \]

Step 2: Substitute (i) and (ii) into (iii)

\[ D + 1 = \frac{ \frac{D}{3} + D + 2D }{3} \Rightarrow D + 1 = \frac{ \frac{D + 3D + 6D}{3} }{3} = \frac{ \frac{10D}{3} }{3} = \frac{10D}{9} \] Multiply both sides by 9: \[ 9(D + 1) = 10D \Rightarrow 9D + 9 = 10D \Rightarrow D = 9 \]

Step 3: Find Harry’s Age

Using \( H = 2D \): \[ H = 2 \times 9 = \boxed{18 \text{ years}} \]

Final Answer:

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