Question

(vi) Five years hence, the age of Jacob will be three times that of his son. Five years
ago, Jacob’s age was seven times that of his son. What are their present ages?

Answer 1

Let the present age of Jacob and his son be x and y respectively.

Case l : After five years ago of Jacob = ( x + 5 ),

After five years the age of his son = ( y + 5 ).

According to question,

x + 5 = 3 ( y + 5 )

⇒ x + 5 = 3y + 15

⇒ x - 3y = 15 - 5

⇒ x - 3y = 10 ...........( 1 )

Case 2: Five years ago Jacob's age = x - 5, and his son's = y - 5 . Then,

According to question,

x - 5 = 7 ( y - 5 )

⇒ x - 5 = 7y - 35

⇒ x = 7y - 30 ............( 2 )

☆Putting x = 7y - 30 from ( 2 ) in ( 1 ), we get

7y - 30 - 3y = 10

⇒ 4y - 30 = 10

⇒ 4y = 40

⇒ y = 10

☆ Putting y = 10 in ( 1 ), we get

x - 3y = 10

⇒ x - 3 × 10 = 10

⇒ x - 30 = 10

⇒ x = 40

Hence, age of Jacob is 40 years and age of his son is 10 years.

Answer 2

✯✯ QUESTION ✯✯

Five years hence, the age of Jacob will be three times that of his son. Five years ago, Jacob’s age was seven times that of his son. What are their present ages?

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✰✰ ANSWER ✰✰

$\longmapsto\tt{Let\:the\:age\:of\:Jacob=x}$

$\longmapsto\tt{Let\:the\:age\:of\:Son=y}$

Now, Five years later : -

$\longmapsto\tt{x+5=3(y+5)}$

$\longmapsto\tt{x+5=3y+15}$

$\longmapsto\tt{x-3y=10}$

$\longmapsto\tt{x=3y+10}$------(1)

And ,

$\longmapsto\tt{x-5=7(y-5)}$

$\longmapsto\tt{x-5=7y-35}$

$\longmapsto\tt{x-7y=-3}$------(2)

Putting 1 in 2 : -

$\longmapsto\tt{3y+10-7y=-30}$

$\longmapsto\tt{-4y=-30-10}$

$\longmapsto\tt{-4y=-40}$

$\longmapsto\tt{y=\dfrac{-40}{-4}}$

$\red\longmapsto\:\large\underline{\boxed{\bf\green{y}\orange{=}\purple{10}}}$

$\longmapsto\tt{Present\:Age\:of\:son=y}$

$\longmapsto\tt{\large{\boxed{\bold{\bold{\purple{\sf{10\:years}}}}}}}$

$\longmapsto\tt{Present\:Age\:of\:Father=3y+10}$

$\longmapsto\tt{3(10)+10}$

$\longmapsto\tt{\large{\boxed{\bold{\bold{\green{\sf{40\:years}}}}}}}$