Question

2
Father's age was 4 times that of his son before 7 years. After 7 years his
age will be double than that of his son. Find their present ages.
[Ans. 35 years & 14 years]

​

Answer 1

$\bf{\huge{\underline{\boxed{\sf{\red{Answer:}}}}}}$

$\bf{\underline{\underline{\sf{\green{Given}}}}}$

• Father's age was 4 times that of his son before 7 years.
• After 7 years his
• age will be double than that of his son

$\bf{\underline{\underline{\sf{\green{To\:find}}}}}$

• Present age of Father
• Present age of Son

$\bf{\underline{\underline{\sf{\green{Solution}}}}}$

Let the present age of Father be x years.

Let the present age of Son be y years.

$\bf{\underline{\underline{\sf{\blue{As\:per\:first\:condition}}}}}$

• Father's age was 4 times that of his son before 7 years.

Ages 7 years before :-

Father's age = x - 7 years

Son's age = y - 7 years

Representing the given condition mathematically.

=> x - 7 = 4 ( y - 7)

=> x - 7 = 4y - 28

=> x - 4y = - 28 + 7

=> x - 4y = - 21 ----> 1

$\bf{\underline{\underline{\sf{\blue{As\:per\:second\:condition}}}}}$

• After 7 years his
• age will be double than that of his son

Ages after 7 years :-

Father's age = x + 7 years

Son's age = y + 7 years

Representing second condition mathematically.

=> x + 7 = 2 ( y + 7)

=> x + 7 = 2y + 14

=> x - 2y = 14 - 7

=> x - 2y = 7 ----> 2

Solve equations 1 and 2 simultaneously by elimination method.

Subtract equation 2 from equation 1,

.....+ x - 4y = - 21

- ( + x - 2y = + 7 )

---------------------------

- 2y = - 28

y = $\large\frac{-28}{-2}$

y = $\large\frac{28}{2}$

y = 14

Substitute y = 14 in equation 2,

=> x - 2y = 7

=> x - 2(14) = 7

=> x - 28 = 7

=> x = 7 + 28

=> x = 35

$\bf{\large{\underline{\boxed{\sf{\blue{Present\:age\:of\:Father\:=\:x\:=\:35\:years}}}}}}$

$\bf{\large{\underline{\boxed{\sf{\blue{Present\:age\:of\:Son\:=\:y\:=\:14\:years}}}}}}$

$\bf\huge\underline{Verification:}$

For first case :-

• Father's age was 4 times that of his son before 7 years.

Present age of Father = x = 35 years

Present age of Son = y =14 years

Ages before 7 years :-

Father = x - 7 = 35 - 7 = 28 years

Son = y - 7 = 14 - 7 = 7 years

=> x - 7 = 4 ( y - 7)

=> 28 = 4 ( 7)

=> 28 = 28

LHS = RHS.

For second case :-

Ages after 7 years :-

Father = x + 7 = 35 + 7 = 42 years

Son = y + 7 = 14 + 7 = 21 years

=> x + 7 = 2 ( y + 7)

=> 42 = 2 ( 21)

=> 42 = 42

LHS = RHS.

Hence verified.

Answer 2

here your answer........

explanation:

let present age of father be x year's

and his son be y year's.

7 year's ago father age be x-7 year's

and son age be y-7 year's.

According to question

(x-7)=4( y-7)

x-7=4y-28

x-4y+21=0................ (1)

and

after 7 year's father and his son ages

is x+7, y+7

ATQ,

x+7=2(y+7)

x+7=2y+14

x-2y-7=0..................... (2)

now substract equations (1)-(2)

x-4y+21-(x-2y-7) =0

x-4y+21-x+2y+7=0

-2y+28=0

-2y=-28

y=14

substitute you value in equation (1)

x-4(14) +21=0

x-56+21=0

x-35=0

x=35

therefore the present ages of

father is x=35 year's

and his son is y=14 year's.