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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

Let the present ages of Dick, Tom and Harry be $D, T,$ and $H \; \text{years}$ respectively.

According to the question,

• $D = 3T \; \longrightarrow (1)$
• $H = 2D \; \longrightarrow (2)$

And, $D = \left( \frac{D+T+H}{3} \right) – 1$

$\Rightarrow D = \left( \frac{D+\frac{D}{3}+2D}{3} \right) – 1 \quad [ \because \text{From equation (1) and (2)}]$

$\Rightarrow D = \left( \frac{3D+D+6D}{9} \right) – 1$

$\Rightarrow 9D = 10D – 9$

$\Rightarrow \boxed{D = 9 \; \text{years}}$

Now, Harry present age $H = 2 \times 9$

$\Rightarrow H = 18 \; \text{years}.$

Correct Answer $: 18$

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Let us assume Tom age is $x$ then:

• Dick’s age= $3x$
• Harry’s age=$2.3x=6x$

Now according to the given question, Dick’s age is 1 year less than the average age of all three,

$\therefore 3x=\left (\frac{3x+6x+x}{3} \right )-1$

$\Rightarrow 3x=\left (\frac{10x}{3} \right )-1$

$\Rightarrow 1=\frac{10x}{3}-3x$

$\Rightarrow 1=\frac{10x-9x}{3}$

$\Rightarrow 1=\frac{x}{3}$

$\Rightarrow x=3$

So Tom age is $3$ year,

Harry’s age is $=6x=6*3=18$ year.

$\therefore$ Harry’s age is $18$ year.

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