Question

Two year ago a woman was 7 times as old as her daughter but in 3 year time she will be 3 year time as her

Answer 1

Step-by-step explanation:

♦ Let the present age of the daughter be 'x'.

♦ 5 years ago,

Daughter's age = x - 5

Mother's age = 7(x - 5) = 7x - 35

♦ At present,

Daughter's age = x - 5 + 5 = x

Mother's age = 7x - 35 + 5 = 7x - 30

♦ 5 years hence,

Daughter's age = x + 5

Mother's age = 7x - 30 + 5 = 7x - 25

♦ But according to the question, after 5 years, mother will be thrice daughter's age.

→→→ 7x - 25 = 3(x + 5)

=> 7x - 25 = 3x + 15

=> 7x - 3x = 15 + 25

=> 4x = 40

=> x = 40/4

=> x = 10

•°• Daughter's present age = x = 10 years.

•°• Mother's present age = 7x - 30 = 7(10) - 30 = 70 - 30 = 40 years.

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Answer 2

Answer:

woman=43/4 and daughter=5/4

Step-by-step explanation:

let woman be ' x' and daughter be 'y' then

two years ago= 7y-2

7y-2=x....(1)

3 years later= 3y+3

3y+3=x...(2)

subtracting both equations

3y+3=x

7y - 2=x

(-)   (+)   (-)

_______

-4y+5=0

or,4y=5

∴y=5/4

substituting the value of y in (1)

7×5/4 +2=x

35/4+2=x

∴x=43/4