Skip to main contentSkip to solution

Dick is thrice as old as Tom and Harry is twice as old as Dick. If Dick's age is 11 year less than the average age of all three, then Harry's age, in years, is

Entered answer:

Solution

✅ Correct Answer: 18

We need to set up equations based on the age relationships given.


Let Tom's age = x years

From the problem:

Dick is thrice as old as Tom → Dick's age = 3x years

Harry is twice as old as Dick → Harry's age = 2 × (Dick's age) = 2 × 3x = 6x years


The average age of all three people is:

Average=Tom’s age+Dick’s age+Harry’s age3=x+3x+6x3=10x3\text{Average} = \frac{\text{Tom's age} + \text{Dick's age} + \text{Harry's age}}{3} = \frac{x + 3x + 6x}{3} = \frac{10x}{3}


The problem states: "Dick's age is 1 year less than the average age of all three"

This means: Dick's age = Average age - 1

3x=10x3−13x = \frac{10x}{3} - 1


Multiply everything by 3 to eliminate the fraction:

3x×3=10x3×3−1×33x \times 3 = \frac{10x}{3} \times 3 - 1 \times 3

9x=10x−39x = 10x - 3

9x−10x=−39x - 10x = -3

−x=−3-x = -3

x=3x = 3


Since x = 3 and Harry's age = 6x:

Harry’s age=6×3=18 years\text{Harry's age} = 6 \times 3 = 18 \text{ years}


Therefore, Harry's age is 18 years.

Keyboard Shortcuts

  • Left arrow: Previous question
  • Right arrow: Next question
  • S key: Jump to solution
  • Q key: Jump to question