Question

14 years ago the age of the mother was 4 times the age of her Daughter the present age of the mother is 2 times the age of her Daughter will be 4 years hence what are their present age​

Answer 1

Step-by-step explanation:

Let the present ages of the mother and daughter be x and y respectively.

X= mother

Y= daughter

So the first equation (10 years ago ) is as follows :

x-10=4(y-10)

So, x-10=4y-40

x=4y-30 …1

Now coming to the second equation :

x+10=2(y+10)

So,x+10=2y+20

x=2y+10 …2

From 1 and 2

2y+10=4y-30

2y=40 y=20

Therefore the daughter’s age is 20

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Answer 2

thus, the present age mother be x = 58years, and the present age daughter be y = 25years.

Step-by-step explanation:

let the present age mother be x and the present age daughter be y.

A. T. Q

(x-14) = 4(y-14)

or, x = 4y - 56 +14

or, x = 4y - 42 ......... 1

again

x = 2(y+4)

or, x = 2y + 8..........2

substituting the value of x from equation(1) in (2) we get,

4y -42 = 2y +8

or, 4y-2y = 8+42

or, 2y =50

or, y =50/2

or, y = 25years

put the value of x from equation(2)

x = 2×25 + 8

so, x =50+8

i.e x =58

thus, the present age mother be x = 58years, and the present age daughter be y = 25years.