CAT 2020 Set-3 | Question: 52

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 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 $

Answer 2

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.

 

Answer 3

Given information:
- Dick is thrice as old as Tom. Let Tom's age be x.
- Harry is twice as old as Dick.
- Dick's age is 1 year less than the average age of all three.

Step 1: Express Dick's age in terms of Tom's age (x).
Dick's age = 3x

Step 2: Express Harry's age in terms of Dick's age.
Harry's age = 2 × Dick's age = 2 × 3x = 6x

Step 3: Set up an equation for Dick's age being 1 year less than the average age of all three.
Dick's age = (Tom's age + Dick's age + Harry's age)/3 - 1
3x = (x + 3x + 6x)/3 - 1
3x = 10x/3 - 1
9x = 10x - 3
x = 3

Step 4: Find Harry's age by substituting x = 3 into the expression for Harry's age.
Harry's age = 6x = 6 × 3 = 18 years

Therefore, Harry's age, in years, is 18.