Question

one year ago,a mother was 8 times as old as her daughter. At present, the mother's age is equal to the square of the daughter's age. find the present age of the mother​

Answer 1

Answer:

Let present age of daughter be x.

So 5 years ago her age =x−5

Her mother's age is 7(x−5)

Including 5 years from past to present and then to future =10

So 7x−35+10=3(x−5+10)

7x−25=3x+15

4x=40

x=10

so daughters age=10

mother's age=4×10=40.

Answer 2

Answer:

Presently mother is 49 years old and daughter is 7 years old.

Step-by-step explanation:

Let,

Present age of Daughter = x

Present age of Mother = x²

One year ago,

Age of Daughter = x - 1

Age of Mother = x² - 1

Also,

Mother was 8 times as old as her daughter.

$\longrightarrow$ x² - 1 = 8(x - 1)

$\longrightarrow$ x² - 1 = 8x - 8

$\longrightarrow$ x² - 8x - 1 + 8 = 0

$\longrightarrow$ x² - 8x + 7 = 0

$\longrightarrow$ x² - 7x - x + 7 = 0

$\longrightarrow$ x(x - 7) - 1(x - 7)

$\longrightarrow$ (x - 7)(x - 1)

$\longrightarrow$ x = 7 or x = 1

Daughter's present age = 7 years.

_______________________________

★ Mother's present age =

$\longrightarrow$ x²

$\longrightarrow$ (7)²

$\longrightarrow$ 49

Therefore, presently mother is 49 years old and daughter is 7 years old.