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 [This Question was asked as TITA]

• A. 20
• B. 17
• C. 16
• D. 18

Answer 1

The Correct Option is D

Let the present ages of Dick, Tom, and Harry be \(D, T,\) and \(H\) respectively. 

It is given that: 
Dick is 3 times as old as Tom: 

\[D = 3T \Rightarrow T = \frac{D}{3} \tag{i}\]

Harry is twice as old as Dick: 

\[H = 2D \tag{ii}\]

Dick’s age is one less than the average of their ages: 
 

\[D = \left( \frac{T + D + H}{3} \right) - 1 \Rightarrow D + 1 = \frac{T + D + H}{3} \tag{iii}\]

Now substitute equations (i) and (ii) into equation (iii): 
 

\[D + 1 = \frac{ \frac{D}{3} + D + 2D }{3}\]

Simplify the numerator: 
 

\[\frac{D}{3} + D + 2D = \frac{D + 3D + 6D}{3} = \frac{10D}{3}\]

 So the equation becomes: 
 

\[D + 1 = \frac{10D}{9}\]

Multiply both sides by 9: 

\[9D + 9 = 10D \Rightarrow 10D - 9D = 9 \Rightarrow D = 9 \tag{iv}\]

Substituting into equation (ii): 

\[H = 2D = 2 \times 9 = 18\]

Final Answer: \( \boxed{18} \) (Option D)

Answer 2

Let the age of Tom be \(t\). 

Then, the age of Dick is \(3t\), and the age of Harry is \(6t\).

It is given that: 
 

\[3t = \frac{t + 3t + 6t}{3} - 1\]

Simplify the numerator: 

\[t + 3t + 6t = 10t\]

 
So the equation becomes: 

\[3t = \frac{10t}{3} - 1\]

Multiply both sides by 3 to eliminate the denominator: 

\[9t = 10t - 3\]

Solving: 

\[9t - 10t = -3 \Rightarrow -t = -3 \Rightarrow t = 3\]

Therefore, the age of Harry is: 

\[6t = 6 \times 3 = 18\]

Final Answer: \( \boxed{18} \)