Five years ago,A’s age was four times the age of B.Five years hence A’s age will be twice of the age of B. Find their present ages

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1. Answer:

   Question:

   • Five years ago,A’s age was four times the age of B. 5 years hence, A’s age will be twice of the age of B. Find their present ages.

   Given:

   • Five years ago, A’s age was 4 times the age of B.

   • Five years hence A’s age will be twice of the age of B.

   To find:

   • Age of A

   • Age of B

   Let :

   • present age of A = x

   • present age of B = y

   Answer:

   Five years ago:

   • Age of A= x-5

   • Age of B=y-5

   As it’s Given 5 years ago A’s age was 4 times the age of B

   .°. Equation formed as follows:

   [tex]\sf{}x – 5 = 4(y – 5)[/tex]

   Now Let’s solve this equation:

   [tex]: \implies\sf{}x – 5 = 4y – 20[/tex]

   [tex]: \implies\sf{}x – 4y = – 15[/tex]

   [tex]: \implies \sf{}y = \dfrac{ – 15 – x}{ – 4} [/tex]

   [tex]: \implies \sf{}y = \dfrac{ 15 + x}{ 4} \small{… \green{⓵}}[/tex]

   Five years hence:

   • Age of A= x+5

   • Age of B = y+5

   As it’s given 5 years hence A’s age will be twice of the age of B.

   .°. Equation formed as follows:

   [tex] \sf{}x + 5 = 2(y + 5)[/tex]

   [tex]: \implies \sf{}x + 5 = 2y + 10[/tex]

   [tex]: \implies \sf{}x – 2y = 10 – 5[/tex]

   [tex]: \implies\sf{}x – 2y = 5\small{…\green{⓶}}[/tex]

   Now put value of y in Equation 2

   [tex]: \implies\sf{}x – 2 [\dfrac{ 15 + x}{ 4}] = 5[/tex]

   [tex]: \implies\sf{}x – \dfrac{ 30+2 x}{ 4}= 5[/tex]

   multiply each digit by 4

   [tex]: \implies \sf{}4x – (30 + 2x) = 20[/tex]

   [tex] : \implies\sf{}4x -30 – 2x = 20[/tex]

   [tex]: \implies\sf{}4x – 2x = 20 + 30[/tex]

   [tex] : \implies\sf{}2x = 50[/tex]

   [tex]: \implies\sf{}x = \dfrac{50}{2} [/tex]

   [tex]: \implies\sf{}x = \dfrac{ { \cancel{50}}^{ \: \: 25} }{ { \cancel2}^{ \: \: 1} } [/tex]

   [tex]: \implies \star \boxed{ \sf{}x = 25} \star[/tex]

   we got value of x i.e.25

   Put value of x in equation 1:

   [tex]: \implies \sf{}x – 2y = 5[/tex]

   [tex]: \implies \sf{}25 – 2y = 5[/tex]

   [tex]: \implies \sf{} – 2y = 5 – 25[/tex]

   [tex]: \implies \sf{} – 2y = – 20[/tex]

   [tex]: \implies \sf{} y = \dfrac{ – 20}{ – 2} [/tex]

   [tex]: \implies \sf{} y = \dfrac{ 20}{ 2} [/tex]

   [tex]: \implies \sf{} y = \dfrac{ { \cancel{20}}^{ \: 10} }{ { \cancel{2}}^{ \: 1} } [/tex]

   [tex]: \implies \star \boxed{ \sf{}y= 10} \star[/tex]

   As we have supposed A’s age as x

   .°. A’s age = 25

   As we have supposed B’s age as y

   .°. B’s age = 10

   Now Let’s Verify their ages:

   Put age of A and B in Equation 1:

   [tex]\sf{}25 – 5 = 4(10 – 5)[/tex]

   [tex]: \implies \sf{}20= 40 – 20[/tex]

   [tex]: \implies \sf{}20= 20[/tex]

   ☆Hence Verified ☆

   __________________________________________________

   And all we are done! ✔

   😀

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