Question

five years ago A mother was 7 times as old as her daughter five years hence she will be 3 times as old as her daughter find their present ages

Answer 1

Let present age of daughter and her mother be X and Y year's.




Five year's ago age of daughter = ( X - 5 ).


Five years ago age of her mother = ( Y - 5 ).


According to the question,


Y - 5 = 7 ( X - 5 )



Y - 5 = 7X - 35



7X - Y = 35 - 5



7X - Y = 30 ----_-----(1)




Five years hence age of daughter = ( X + 5 ).



Five years hence age of mother = ( Y + 5 ).



According to the question,



Y + 5 = 3 ( X + 5 )


Y + 5 = 3X + 15



3X - Y = 5 - 15



3X - Y = -10 -----------(2)



From equation (1) , we get


7X - Y = 30



Y = ( 7X - 30 ) --------(3)

Putting the value of X in equation (2) ,


3X - Y = -10



3X - ( 7X - 30 ) = -10



3X - 7X + 30 = -10



-4X = -10 - 30


-4X = -40



X = 10




Putting the value of X in equation (3) ,


Y = ( 7X - 30 ) = ( 7 × 10 - 30 )



Y = 70 - 30 = 40 years.



Hence,



Present age of daughter = X = 10 year's.


and,

Present age of mother = Y = 40 years

Answer 2

нєуα

нєяє ιѕ тнє αиѕωєя

ℓєт тнє мσтнєя'ѕ αи∂ ∂αυgнтєя'ѕ αgє вє χ αи∂ у уяѕ яєѕρє¢тινєℓу.

fινє уяѕ αgσ,

мσтнєя'ѕ αgє = (χ - 5) уяѕ

∂αυgнтєя'ѕ αgє = (у - 5) уяѕ

єqυαтισи fσямє∂ α/q :-

(χ - 5) = 7(у - 5)

σи ѕσℓνιиg fυятнєя, ωє gєт

χ - 5 = 7у - 35

χ - 7у = -35 + 5

χ - 7у = -30

αfтєя 5 уяѕ,

мσтнєя'ѕ αgє = (χ + 5) уяѕ

∂αυgнтєя'ѕ αgє = (у + 5) уяѕ

єqυαтισи fσямє∂ α/q :-

(χ + 5) = 3(у + 5)

χ + 5 = 3у + 15

χ - 3у = 15 - 5

χ - 3у = 10

иσω, ωє нανє тωσ єqυαтισиѕ,

χ - 7у = -30

αи∂

χ - 3у = 10

ѕυвтяα¢т вσтн тнє єqυαтισиѕ, ωє gєт

-4у = -40

у = 10

ρυт тнιѕ ιи 1ѕт єqυαтισи

χ - 7(10) = -30

χ - 70 = -30

χ = 40

нєи¢є,

мσтнєя'ѕ αgє = 40 уяѕ

∂αυgнтєя'ѕ αgє = 10 уяѕ

нσρє тнιѕ нєℓρѕ.