Question

Two years ago a woman was 7 times as old as her daughter but in 3 years time she would be 4 times as old as the girl. How old are they now?

Answer 1

Answer:

37 year and 5 year

Step-by-step explanation:

Let the present age of daughter be a + 2 year.

Before 2 year -

Age of daughter = a + 2 - 2 = a

Age of mother = 7*a = 7a

After 3 years:

Age of daughter = a + 2 + 3 = a + 5 year

Age of mother = 7a + 2 + 3 = 7a + 5 year

According to question -

= > age of mother after 3 year = 4 * age of daughter

= > 7a + 5 = 4( a + 5 )

= > 7a + 5 = 4a + 20

= > 7a - 4a = 20 - 5

= > 3a = 16

= > a = 5

Hence, the present age of -

Mother = 7a + 2 = 7(5) + 2 = 37 year

Daughter = a + 2 = 5 + 2 = 7 year

Answer 2

Answer:

let x be woman and y be daugther

then :

first equation

x-2=7(y-2)

x-2=7y-14

x-7y= -14+2

x-7y=-12---(1)

second equation

x+3=4(y+3)

x+3=4y+12

x-4y=12-3

x-4y=9---(2)

Subtract (1) from (2)

x-4y=9

x-7y= -12

---------------

3y = 21

y =21/3

y = 7---(3)

substitute (3) in (1)

x-7 (7) = -12

x-49 = -12

x = 49 - 12

x = 37

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