Question

A mother is twice as old as her daughter. Six years ago their joint ages were 54. How old is mother now?

Answer 1

let the mother's age be x years and

daughter's age be y years

A.T.P.

x=2y

after 6 years

x+6+y+6=54

x+y+12=54

put x=2y in above equation

2y+y+12=54

3y=54-12=42

y=42/3

y=14

x=2y

x=2×14=28

•°•mother's age=28 years

daughter's age =14 years

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Answer 2

Solution :

Let the present age of Daughter's be R years

Let the present age of Mother's be 2R years.

A/q

$\underbrace{\bf{6\:years\:ago\::}}}}$

The age of Mother's will be (2R - 6) years.

The age of Daughter's will be (R - 6) years.

So;

$\longrightarrow\sf{(2R-6)+(R-6)=54}\\\\\longrightarrow\sf{2R-6+R-6=54}\\\\\longrightarrow\sf{2R+R-12=54}\\\\\longrightarrow\sf{3R=54+12}\\\\\longrightarrow\sf{3R=66}\\\\\longrightarrow\sf{R=\cancel{66/3}}\\\\\longrightarrow\bf{R=22\:years}$

Thus;

$\boxed{\sf{The\:present\:age\:of\:Mother's\:will\:2R=2(22)\:years=\boxed{\bf{44\:years}}}}}$