Question

eight years ago, a mother's age 11 times that of her son. the sum of their present ages is 40 years. find their present ages
find using only one variable

Answer 1

Answer:

assume father’s age= x, son’s age= y

x+y=40 ………………………(I)

8 years ago,

x-8=11(y-8)

or, x-11y=-80…………….(II)

subtract (II) from (I) gives

x+y-(x-11y)=40-(-80)

or, 12y=120 => y=10 and x=30

Step-by-step explanation:

Answer 2

Answer:

Let the present age of son be x.

Age of son eight years ago = x-8

Age of mother eight years ago = 11(x-8) = 11x-88

Present age of mother = 11x-88+8 = 11x-80

According to Question,

x+11x-80 = 40

=> 12x = 40+80

=> 12x = 120

=> x = 10

Hence, present age of son = 10 years

Present age of mother = 30 years