Question

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Five years hence, father's age will be three times the age of his son. Five years ago, father was seven times as old as his son. Find their present ages.​

Answer 1

Step-by-step explanation:

Given :

Five years hence, father's age will be three times the age of his son.

Five years ago, father was seven times as old as his son.

To find :

Their present ages.

Solution :

Let the present ages of the Father and the Son be x and y years.

1st condition : [Hence/Future]

=> After 5 years man’s age = x + 5  

=> After 5 years ago son’s age = y + 5  

According to the question :

=> x + 5 = 3(y + 5)

=> x + 5 = 3y + 15

=> x - 3y = 10 -- (1)

2nd condition : [Ago/Past]

=> 5 years ago man’s age = x - 5

=> 5 years ago son’s age = y - 5  

According to the question :

=> x - 5 = 7(y - 5)

=> x - 7y = -30 -- (2)

Subtracting (2) from (1) :

=> x - 7y = -30

=> x - 3y = 10

• 4y = 40
• y = 10

Substitute the value of y in (1) :

=> x - 3y = 10

=> x - 3(10) = 10

=> x - 30 = 10

=> x = 40

Therefore :

Father's present age = 40 years.

Son’s present age = 10 years.

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Answer 2

Answer:

$\huge \huge \bf {\: \pmb{Question}}$

Five years hence, father's age will be three times the age of his son. Five years ago, father was seven times as old as his son. Find their present ages.

━━━━━━━━━━━━━━━━━━━━━

$\huge \huge \bf {\: \pmb{ \blue{Solution}}}$

let the present age of father be x

and, present age of son be y

According to the question :

• Five years hence, father's age will be three times the age of his son

so, adding 5 in both sides

$\dashrightarrow$ x + 5 = y + 5

it's says after 5 years the age of father become thrice of son's age

so multiplying with 3 in Father's age

$\dashrightarrow$ x + 5 = ( y + 5 ) 3

$\dashrightarrow$ x + 5 = 3y + 15

$\dashrightarrow$ x - 3y = 10 ----(1)

Five years ago ,

Five years ago, father was seven times as old as his son

so, subtracting 5 in both sides

$\dashrightarrow$ x - 5 = y - 5

it's says after 5 years the age of father become 7 times of son's age

so multiplying with 7 in Father's age

$\dashrightarrow$ x - 5 = ( y - 5 ) 7

$\dashrightarrow$ x - 5 = 7y - 35

$\dashrightarrow$ x - 7y = - 35 + 5

$\dashrightarrow$ x - 7y = - 30 ---(2)

Now subtracting both equations (1) & (2)

$\sf \dashrightarrow \: x - 7y = - 30 \\ \sf \: \: x \: - 3y = 10 \\ \: \: \tiny( - ) \: \: \: \: \: \: \: \: \: \: \: \: ( + ) \: \: \: \ \: \: \: \: \: \: \: ( - ) \\ \sf - - - - - - - - - \\ \sf \cancel- 4y = \cancel- 40$

$\dashrightarrow$ 4y = 40

$\dashrightarrow$ y = $\cancel\frac{40}{4}$

$\dashrightarrow$ y = 10

therefore son's present is 10 years

to find father's present age we need to put value of y in our first equation

$\dashrightarrow$ x - 3y = 10

$\dashrightarrow$ x - 3(10) = 10

$\dashrightarrow$ x - 30 = 10

$\dashrightarrow$ x = 10 + 30

$\dashrightarrow$ x = 40

therefore, father's present age is 40 years