Math, asked by Charvigowda030, 11 months ago

A man has 2 daughters and one son. The sum of the ages of his two daughters and one son is equal to the age of the father. In 15 years, the sum of the ages of his children will be one and half times their fathers age then what is the father age now?

Answers

Answered by Anonymous
255

Answer

Age of father = f = 45 years.

Explanation

Let the age of father be 'f' years.

And the ages of his two daughters be 'x' and 'y' and age of his son be 'z'.

The sum of the ages of his two daughters and one son is equal to the age of the father.

\sf{\underline{\underline{As\: per \:given\: condition:}}}

\boxed{\bold{\sf{x+y+z\:=\:f}}}

\implies\:\sf{x+y+z\:=\:f} ....(1)

In 15 years, the sum of the ages of his children will be one and half times their fathers age.

In 15 years,

  • Age of his daughters = (x + 15) and (y + 15) years
  • Age of his son = (z + 15) years
  • Age of father = (f + 15) yeas

\sf{\underline{\underline{As\: per \:given\: condition:}}}

 \boxed{ \bold{\sf{x+15+y+15+z+15\:=\:1\frac{1}{2}{(f+15)}}}}

\implies\:\sf{x+15+y+15+z+15\:=\:1\frac{1}{2}{(f+15)}}

\implies\:\sf{x+15+y+15+z+15\:=\:\frac{3}{2}{(f+15)}}

\implies\:\sf{x+y+z+45\:=\:\frac{3}{2}{(f+15)}} ....(2)

Let us assume that x, y and z are equal to 'c'.

(As, all are children)

So, our another equation will be:

\boxed{\bold{\sf{c\:=\:f}}}

\implies\:\sf{x+y+z\:=\:c}

From (eq 1) it will be,

\implies\:\sf{c\:=\:f} ....(3)

Substitute value of (eq 3) in (eq 2)

\implies\:\sf{c+45\:=\:\frac{3}{2}{(f+15)}}

As c = f also; So,

\implies\:\sf{f+45\:=\:\frac{3}{2}{(f+15)}}

\implies\:\sf{2(f+45)\:=\:3(f+15)}

\implies\:\sf{2f+90\:=\:3f+45}

\implies\:\sf{2f-3f\:=\:45-90}

\implies\:\sf{-f\:=\:-45}

(negative sign throughout cancel, we left with)

\implies\:\sf{f\:=\:45}

•°• Age of father is 45 years.


Anonymous: Nice bro
Anonymous: thanku
Answered by EliteSoul
126

Answer:

{\underline{\boxed{\sf\green{Father's \: age = 45 \: years}}}}

Given:-

  • Sum of 2 daughters and 1 son = Father's age.
  • In 15 years, sum of ages of children = 1 & half times of father's age.

To find:-

  • Father's present age = ?

Let present ages of 2 daughters age and 1 son be a, b & c years respectively and father's age be x years.

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

\hookrightarrow\sf\green{a + b + c = x............(i)}

\rule{200}{1}

Secondly,

In 15 years, ages of children will be (a + 15) , (b + 15) & (c + 15) .Father's age will be (x + 15) years.

\hookrightarrow\sf (a + 15) + (b + 15) + (c + 15) = 1 \dfrac{1}{2} (x + 15) \\\\\hookrightarrow\sf a + 15 + b + 15 + c + 15 = \dfrac{3}{2} (x + 15) \\\\\hookrightarrow\sf (a + b + c) + 45 = \dfrac{3}{2} (x + 15) \\\\\sf \: \: \: [From\:(i)\:: \: (a + b + c) = x]\\\\\hookrightarrow\sf x + 45 = \dfrac{3(x + 15)}{2} \\\\\hookrightarrow\sf 2(x + 45) = 3(x + 15) \\\\\hookrightarrow\sf 2x + 90 = 3x + 45 \\\\\hookrightarrow\sf 3x - 2x = 90 - 45 \\\\\hookrightarrow\large{\boxed{\sf\pink{x = 45 \: years}}}

{\underline{\boxed{\therefore{\sf\blue{Father's \: present \: age = 45 \: years}}}}}


Anonymous: Good answer
Anonymous: Nice one
ShuchiRecites: Nice attempt
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