Question

6. A father's age is 3 times the sum of ages of his two sons. Five years later he will be twice
the sum of ages of his two sons. Find the present age of the father.​

Answer 1

$\large \underline{ \red{\underline{ \sf Question}}} : - \\ \sf A \: Father's \: age \: is \: 3 \: times \: the \: sum \: of \: ages \\ \sf of \: his \: two \: sons \: .5 \: years \: later \: he \: will \: be \: \\ \sf twice \: the \: sum \: of \: ages \: of \: his \: two \: sons \: \\ \sf find \: the \: present \: age \: of \: father.$

$\large \underline{ \green{\underline{ \sf \: Answer \: }}} : - \: \\ \to \sf present \: age \: of \: father \: is \: 45 \: years \: \\ \\ \large \underline{ \red{\underline{ \sf To \: Find }}} \: - \: \\ \to \sf \: present \: age \: of \: father$

$\large \underline{ \pink{\underline{ \sf Explanation }}} \: : - \:$

According to the question :

→ Father's age is 3 times of the sum of his two sons, and after 5 years he will be 2 times of the sum of age of his two sons .

$\large \underline{ \blue{\underline{ \sf Solution }}} \: : - \:$

Let,

→ Age of two sons is "x" and "y" ,

according to the question ,

→ Father's age is 3 times of sum of the ages of both ,

hence,

→ present age of father is 3(x + y) ....eq.(1)

,now

also given that

→ after 5 years his age will be twice of the sum of them,

therefore,

→ after 5 years ,age of father will be - 3(x + y)+ 5 ,

and ,age of sons be - x + 5 and y + 5

it will be twice of the sum of age of sons ,

→ 3(x + y) + 5 = 2( x + 5 + y + 5 )

→ 3x + 3y + 5 = 2x + 10 + 2y + 10

→ 3x - 2x + 3y - 2x = 20 - 5

→ x + y = 15 ,

hence ,from eq.(1)

→ Present age of father is

→ 3(x + y)

→ 3(15)

→ 45 ,

•°• Present age of father is 45 years

Answer 2

Answer:

$\bold\red{Father's\:present\:age}$ = ${\boxed{\mathfrak\red{45\:years}}}$

Step-by-step explanation:

Let father's present age be x years.

Present age of both two of his sons be respectively y and z years.

${\underline{\underline{\bold{According\:to\:question:-}}}}$

$\Rightarrow\sf X = 3(Y + Z) ............(i)$

$\rule{300}{1}$

$\Rightarrow\sf (X + 5) = 2(Y + 5 + Z + 5)$

$\tt *From\: (i)\:we\:get:-$

$\Rightarrow\sf 3(Y + Z) + 5 = 2(Y + Z + 10) \\\\\\\sf \: \: \: [\because X = 3(Y + Z)] \\\\\\\Rightarrow\sf 3Y + 3Z + 5 = 2Y + 2Z + 20 \\\\\\\Rightarrow\sf 3Y + 3Z - 2Y - 2Z = 20 - 5 \\\\\\\Rightarrow\sf Y + Z = 15 .........(ii)$

$\tt *Adding\:value\:of\:(ii)\:in\:(i)\:we\:get:-$

$\sf \: \: \: \: [\because X = 3(Y + Z).........(i)]$

$\sf \: \: \: \: [\because Y + Z = 15..........(ii)]$

$\Rightarrow\sf X = 3(15) \:years$

$\Rightarrow{\boxed{\bold\green{X = 45\:years }}}$

$\therefore\bold{Present\:age\:of\:father = 45\:years}$