Question

Ten years ago, the age of a person’s mother was three times the age of her son. Ten years hence, the mother’s age will be two times the age of her son. Their present ages are​

Answer 1

$\sf\small\underline\purple{Let:-}$

$\tt{\implies The\: present\:age\:_{(mother)}=x}$

$\tt{\implies The\: present\:age\:_{(son)}=y}$

$\sf\small\underline\purple{To\: Find:-}$

$\tt{\implies The\: present\:age\:_{(both)}=?}$

$\sf\small\underline\purple{Solution:-}$

Question based on linear equation with two variables so, we will solve this question by setting up equation. In the Q given that we have to find present age of mother and his son as per the given clue in question.

$\sf\small\underline{Given\:in\:clue-(i):-}$

• Ten years ago, the age of a person’s mother was three times the age of her son]

$\rm{\implies Before\:10\: years\:_{(mother's\:age)}=Before\:10\: years\:_{(son's\:age)}}$

$\tt{\implies x-10=3(y-10)}$

$\tt{\implies x-10=3y-30}$

$\tt{\implies x-3y=-30+10}$

$\tt{\implies x-3y=-20----(I)}$

$\sf\small\underline{Given\:in\:clue-(ii):-}$

• Ten years hence, the mother’s age will be two times the age of her son]

$\rm{\implies After\:10\: years\:_{(mother's\:age)}=after\:10\: years\:_{(son's\:age)}}$

$\tt{\implies x+10=2(y+10)}$

$\tt{\implies x+10=2y-20}$

$\tt{\implies x-2y=20-10}$

$\tt{\implies x-2y=10----(II)}$

• By Substracting eq (I) and (ii) we get :-]

$\tt{\implies x-3y=-20}$

$\tt{\implies x-2y=10}$

• By solving we get here :-]

$\tt{\implies -y=-30}$

$\tt{\implies y=30}$

• Putting the value of y in eq (I):-]

$\tt{\implies x-3y=-20}$

$\tt{\implies x-3(30)=-20}$

$\tt{\implies x-90=-20}$

$\tt{\implies x=-20+90}$

$\tt{\implies x=70}$

$\sf\large{Hence,}$

$\sf\small\green{\implies The\: present\:age\:_{(mother)}=70\: years}$

$\sf\small\pink{\implies The\: present\:age\:_{(son)}=30\: years}$

Answer 2

Answer:

$\sf \green{\boxed{ \rightarrow \: The \: present \: age \: {of\:mother} = 70 \: years}}$

$\sf \green{\boxed{ \:\rightarrow\: The \: present \: age \: {of\:son} = 30 \: years}}$

Explanation :

Let us assumed that,

$\tt{ \odot\: the \: present \: age \: {(mother)} = x }$

$\tt{ \odot \: the \: present \: age \: {(son)} = y}$

To find :

$\tt { \odot\: the \: present \: age \: {(both)} = \: ?}$

Solution :

$\bf \bold{Given \: in \: clue \: (i) : - }$

Ten years ago, the age of a person’s mother was three times the age of her son.

$\rm{ \implies \: before \: 10 \: years \: {(mother \: age)} \: = before \: 10 \: years \: {(son \: age)}}$

$\:$

$\sf{ \implies \: x - 10 = 3(y - 10)}$

$\sf {\implies \:x - 10 = 3y - 30 }$

$\sf \implies{x -3y = 30 + 10 }$

$\sf{ \implies \: x - 3y = 20 ----(1) }$

$\sf \underline{Given \: in \: clue \: (ii) : - }$

Ten years hence, the mother’s age will be two times the age of her son.

$\rm{ \implies \: after \: 10 \: years \: {(mother \: age)} = after \: 10 \: years \: {(son \: age)}}$

$\sf{ \implies \: x + 10 = 2(y + 10)}$

$\sf{ \implies \: x + 10 = 2y - 20}$

$\sf{ \implies \: x - 2y = 20 - 10}$

$\sf{ \implies \: x - 2y = 10 ---- (2)}$

By Substracting eq (1) and (2) we get :-

$\sf{ \implies \: x - 3y = 20}$

$\sf{ \implies \: x - 2y = 10}$

By solving we get here :-

$\sf{ \implies \: - y = - 30}$

$\pink{\bf{ \implies \: y = 30}}$

Putting the value of y in eq (1):-

$\sf{ \implies \: x - 3y = - 20}$

$\sf{ \implies \: x - 3(30) = - 20}$

$\sf{ \implies \: x - 90 = - 20}$

$\sf{ \implies \: x = - 20 + 90}$

$\pink{\bf{{ \implies \: x = 70}}}$

$\bf {Hence,}$

$\sf \green{\boxed{ \rightarrow \: The \: present \: age \: {of\:mother} = 70 \: years}}$

$\sf \green{\boxed{ \rightarrow\: The \: present \: age \: {of\:son} = 30 \: years}}$