Question

5 years ago A mother was 7 times as old as her daughter 5 year hands she will be 3 times as old as her daughter find their present age

Answer 1

Let daughter's and mother's present ages be x and y respectively.
As,5 years ago mother's age was 7 times her daughter's age, so,mother's will be y-5=7(x-5) 

Similarly,5 years from now the mother's age will be y+5=3(x+5)
By solving both the equations we get x=10 & y=40.
So the present ages of mother and her daughter are 40 & 10 years respectively.

Answer 2

✬ Mother age = 40 years ✬

✬ Daughter age = 10 years ✬

Step-by-step explanation:

Given:

• Five years ago mother was 7 times as old as her daughter.
• 5 years hence mother will be 3 times as old her daughter.

To Find:

• What is the present ages of both?

Solution: Let daughter age be x and mother age be y years respectively. Therefore,

A/q

[ 5 years ago mother was 7 times as old as her daughter ]

➛ Daughter's age = (x – 5) years.

➛ Mother's age = (y – 5) years.

Equation will be :- (y – 5) = 7(x – 5)

$\implies{\rm }$ y – 5 = 7x – 35

$\implies{\rm }$ y – 7x = – 35 + 5

$\implies{\rm }$ y – 7x = – 30

$\implies{\rm }$ y = –30 + 7x.....(1)

A/q

[ 5 years hence mother will be 3 times as old her daughter ]

➛ Daughter's age = (x + 5) years

➛ Mother's age = (y + 5) years

Equation will be :- (y + 5) = 3(x + 5)

$\implies{\rm }$ y + 5 = 3x + 15

$\implies{\rm }$ y – 3x = 15 – 5

$\implies{\rm }$ –30 + 7x – 3x = 10

$\implies{\rm }$ 4x = 10 + 30

$\implies{\rm }$ 4x = 40

$\implies{\rm }$ x = 40/4

$\implies{\rm }$ x = 10

So,

➼ Daughter's present age = x = 10 years.

➼ Mother's present age = y = –30 + 7x

=> y = –30 + 7(10)

=> y = –30 + 70 = 40 years.