Question

The sum of the ages of two sisters, Mary and Jane, is 24 years. Four years ago three times Mary’s age was 3 years more than twice Jane’s age

Answer 1

Answer:

Answer................

Answer 2

Answer :

• Present age of Mary = 11 years
• Present age of Jane = 13 years

$\underline{\frak{ \qquad Given : \qquad}}$

• The sum of the ages of two sisters, Mary and Jane, is 24 years.
• Four years ago three times Mary’s age was 3 years more than twice Jane’s age.

$\underline{\frak{\qquad To\:Find :\qquad}}$

$\bullet \: \: \textsf{Present age of Mary = ?}$

$\bullet \: \: \textsf{Present age of Jane = ?}$

$\underline{\frak{\qquad Solution :\qquad}}$

$\dag\:\:\textsf{Let Mary's age be \textbf{x years.}}$

$\dag\:\:\textsf{Then, Jane's age will be \textbf{24 - x years.}}$

$\large\bigstar \: \underline{\textsf{Four years ago :}}$

$\dag\:\:\textsf{Mary was \textbf{x - 4 years.}}$

$\dag\:\:\textsf{and Jane was 24 - x - 4 = \textbf{20 - x years.}}$

☯ $\underline{\boldsymbol{According\: to \:the\: Question\:now :}}$

$:\implies \sf 3(x - 4) = 2(20 - x) + 3$

$:\implies \sf 3x - 12= 40 - 2x + 3$

$:\implies \sf 3x - 12= 43 - 2x$

$:\implies \sf 3x + 2x= 43 + 12$

$:\implies \sf 5x= 55$

$:\implies \sf x= \dfrac{55}{5}$

$:\implies \underline{ \boxed{\sf x= 11 \: years} }\qquad \Bigg\lgroup \textsf{\textbf{Age of Mary}}\Bigg\rgroup$

☯ $\underline{\boldsymbol{Present\: age \:of\: Mary\:and\:Jane :}}$

$\bullet \: \: \textsf{Present age of Mary = x = \textbf{11 years}}.$

$\bullet \: \: \textsf{Present age of Jane = 24 - x = 24 - 11 = \textbf{13 years}}.$