Question

two years ago, a father wsa five times old as his son. two years later, his age will be 8 more than three times the age of his son. find the sum of the present ages of the father and son​

Answer 1

Answer:

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Answer 2

⇝ Given :-

• Two years ago, a father was five times old as his son.

• Two years later, his age will be 8 more than three times the age of his son.

⇝ To Find :-

• Sum of Their Present Ages.

⇝ Solution :-

Let,

• Present Age of Father = x years

• Present Age of Son = y years

★ Two Years Ago :

• Age of Father = (x - 2) years

• Age of Son = (y - 2) years

✏ According To Question :

$\large\red{ (\text x - 2) = 5(\text y - 2)}$

$:\longmapsto\text x - 2 = 5\text y - 10$

$:\longmapsto\text x - 5\text y = - 10 + 2$

$:\longmapsto \bf x - 5y = - 8 \: \: - - - (1)$

★ After Two Years :

• Age of Father = (x + 2) years

• Age of Son = (y + 2 ) years

✏ According To Question :

$\large \red{(\text x + 2) = 3(\text y + 2) + 8}$

$:\longmapsto\text x + 2 = 3\text y + 6 +8$

▪ Subtracting (1) From (2) :

$:\longmapsto2\text y = 20$

$:\longmapsto\text y = \cancel \dfrac{20}{2}$

$\purple{ \Large :\longmapsto  \underline {\boxed{{\bf y = 10} }}}$

▪ Putting Value of y in (2) :

$:\longmapsto\text x - 3 \times 10 = 12$

$:\longmapsto\text x - 30 = 12$

$:\longmapsto\text x =12 + 30$

$\purple{ \Large :\longmapsto  \underline {\boxed{{\bf x = 42} }}}$

Hence,

$\pink{\begin{cases} \bf Age \: of \: Father = 42 \: years \\ \\ \bf Age \: of \: Son = 10 \: years\end{cases}}$

So,

⇝ Sum of Their Ages = 42 + 10

⇝ Sum of Their Ages = 52 years