Question

Now the age of mother and daughter together is 78. The difference in their ages after 5 years is 36.What were their ages 3 years ago?​

Answer 1

let the present age of mother be = x years

and her daughter's age be = y years

According to question,

x + y = 78 -------------------1

after 5 years,

(x + 5) - (y + 5) = 36

x + 5 - y - 5 = 36

x - y = 36 -------------------2

equation ¹ - equation ²

(x + y) - (x - y) = 78 - 36

x + y - x + y = 42

2y = 42

y = 21

now substituting value of y in equation 1

x + y = 78

x + 21 = 78

x = 78 - 21

x = 57

hence, the present ages of mother and her daughter is 57 and 21 years respectively

mother's age three years ago

= 57 - 3 = 54

daughter's age three years ago

= 21 - 3 = 18

Answer 2

$\large{\underline{\rm{\green{\bf{Given:-}}}}}$

Total age of mother and daughter = 78 years

Difference of their ages after 5 years = 36

$\large{\underline{\rm{\green{\bf{To \: Find:-}}}}}$

The age of mother 3 years ago.

The age of daughter 3 years ago.

$\large{\underline{\rm{\green{\bf{Solution:-}}}}}$

Given, total age of mother and daughter is 78 years.

Difference of their ages after 5 years = 36

Let us consider the ages of mother and daughter to be x and y respectively

$\longrightarrow \sf x+y=78$

Equation (1),

$\implies \sf x=78-y \qquad ...(1)$

After 5 years ages would be,

$\implies \sf x + 5\: and\: y + 5$

According to the question,

$\implies \sf x + 5 - ( y + 5 ) =36$

$\implies \sf x + 5 - y - 5 =36$

$\implies \sf x - y =36$

Substituting the value of x from first equation,

$\implies \sf 78 -y - y =36$

$\implies \sf 2y = 42$

Finding the value of y,

$\implies \sf y=\dfrac{42}{2}$

$\implies \sf y=21$

$\sf x = 78 - 21 = 57$

Ages 3 years ago = x -3 and y - 3

$\implies \sf x - 3 = 57 - 3 =54$

$\implies \sf y-3 = 21-3 = 18$

x = 54

y = 18

Therefore, the age of mother is 54 and the daughter is 18 years.