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

Let the age of Tom be \(t\).
Ages of Dick and Harry are \(3t\) and \(6t\) respectively. 
Given, \(3t = \frac{t+3t+6t}{3}-1 ⇒ 9t \Rightarrow10t-3 ⇒ t=3\)
Age of Harry =\(6×3 = 18\)

Answer 2

Suppose the present ages of Dick, Tom and Harry be D, T and H respectively.
Now, 3T = D
\(T=\frac{D}{3}\) ….. (i)
H = 2D ……. (ii)
\(D=[\frac{(T+D+H)}{3}]-1\)
D + 1 = \([\frac{(T+D+H)}{3}]\) …… (iii)
Now, by substituting values of eqn (i) and (ii) into eqn (iii), we get :
D + 1 = \([\frac{(\frac{D}{3})+D+2D}{3}]\)
By this , D = 9 ….. (iv)
So, by putting the value of D in eqn (iv), we get the Harry's age as :
H = 2 × 9
= 18 years