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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]

Updated On: Jan 15, 2026
  • 20
  • 17
  • 16
  • 18
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The Correct Option is D

Solution and Explanation

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

The given conditions are: 
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
 

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

Substitute equations (i) and (ii) into equation (iii): 
 

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

Simplify the numerator: 
 

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

 The equation becomes: 
 

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

Multiply both sides by 9: 

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

Substitute \(D=9\) into equation (ii): 

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

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

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