Age

Algebra Level 3

Today, Jim Higgins and his three sons are together a hundred years old. Moreover, Jim is twice as old as his oldest son who, in turn, is twice as old as his second son who, in turn, is twice as old as the youngest. How old is the oldest son of the Higgins family? Give your answer as months.

The answer is 320.

2 solutions

A Former Brilliant Member
May 25, 2017

let $J,A,B$ and $C$ be the present ages of Jim, oldest son, second son and youngest son, respectively.

$J+A+B+C=100$ $\color{#D61F06}(1)$

$J=2A$ $\color{#D61F06}(2)$

$A=2B$ $\implies$ $B=\dfrac{A}{2}$ $\color{#D61F06}(3)$

$B=2C$ $\implies$ $C=\dfrac{B}{2}$ $\implies$ $C=\dfrac{A}{4}$ $\color{#D61F06}(4)$

Substitute $\color{#D61F06}(2)$ , $\color{#D61F06}(3)$ and $\color{#D61F06}(4)$ in $\color{#D61F06}(1)$ .

$2A+A+\dfrac{A}{2}+\dfrac{A}{4}=100$

$3.75A=100$

$A=\dfrac{80}{3}~years$ or $320~months$

If x is the age of the youngest, the second youngest would be 2x, then 4x, and Jim is 8x. Add them together and we get 15x=100 years=1200 months, so x=80. We need 4x, the age of the biggest son, which is 320.

József Inczefi - 4 years ago

Rocco Dalto
May 29, 2017

Let $J$ be Jim higgins age.

$O_{1}$ be oldest son's age.

$O_{2}$ be second oldest son's age.

$O_{3}$ be youngest son's age.

$J + O_{1} + O_{2} + O_{3} = 100$

$J = 2 O_{1}, \:, O_{1} = 2 O_{2}, \:, O_{2} = 2 O_{3} \implies J = 8 O_{3}, \: O_{1} = 4 O_{3} \implies$

$15 O_{3} = 100 \implies O_{3} = \dfrac{100}{15} \implies O_{1} = 4 O_{3} = \dfrac{400}{15} = 26 + \dfrac{2}{3}$ years old.

In months we obtain:

$O_{1} = (26 + \dfrac{2}{3}) * 12 = 312 + 8 = \boxed{320}$ .

Age

The answer is 320.

2 solutions

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