Question

The present ages of a,b and c are in proportions 4:7:9. eight years ago, the sum of their ages was 56. what are their present ages (in years)?

Answer 1

$\underline { \pink { At \: present :} }$

$Ratio \:of \: present \:ages \:of \: a,b \:and \:c \\= 4 : 7 : 9$

$Let \:present \:age \: of \: \red{a }= 4x \: years$

$present \:age \: of \: \red{b }= 7x \: years$

$present \:age \: of \: \red{c }= 9x \: years$

$\underline { \pink { Eight \: years \:ago :} }$

$Age \: of \: \red{a }= (4x -8 )\: years$

$Age \: of \: \red{b }= (7x-8) \: years$

$Age \: of \: \red{c }= (9x -8)\: years$

/* According to the problem given */

$Sum \: of \:the \:ages = 56 \: years$

$\implies (4x-8)+(7x-8)+(9x-8) = 56$

$\implies 20x - 24 = 56$

$\implies 20x = 24 + 56$

$\implies 20x = 80$

$\implies x = \frac{80}{20} = 4$

Therefore.,

$present \:age \: of \: \red{a }= 4x \: years\\= 4 \times 4 \\= 16 \:years$

$present \:age \: of \: \red{b }= 7x\\ = 7\times 4\\ = 28 \: years$

$present \:age \: of \: \red{c }= 9x \: years \\= 9\times 4 \\= 36 \: years$

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