Question

a boy is thirce as old as his brother.2 years ago,the product of their age is 20.find their present ages.​

Answer 1

Given :

• A boy is thirce as old as his brother.
• 2 years ago,the product of their age is 20.

To Find :

• Present age of Boy.
• Present age of brother.

Solution :

Let the present age of boy be x years.

Let the present age of brother be y years.

Case 1 :

Age of boy (x) is thrice the age of his brother (y).

Equation :

$\sf{x=3y\:\:(1)}$

Case 2 :

2 years ago, the product of x and y was 20.

Age of boy 2 years ago, (x -2) years.

Age of brother 2 years ago, (y-2) years.

Equation :

$\longrightarrow$ $\sf{(x-2) (y-2) =20}$

$\longrightarrow$ $\sf{x(y-2) -2(y-2) =20}$

$\longrightarrow$ $\sf{xy-2x-2y+4=20}$

From equation (1), x = 3y,

$\longrightarrow$ $\sf{3y(y)-2(3y)-2y=20-4}$

$\longrightarrow$ $\sf{3y^2\:-\:6y-2y=16}$

$\longrightarrow$ $\sf{3y^2\:-\:8y-16=0}$

$\longrightarrow$ $\sf{3y^2\:-\:12y\:+\:4y\:-\:16\:=\:0}$

$\longrightarrow$ $\sf{3y(y-4) +4(y-4) =0}$

$\longrightarrow$ $\sf{(y-4) \:\:\:(3y+4)=0}$

$\longrightarrow$ $\sf{y-4=0\:\:or\:\:3y+4=0}$

$\longrightarrow$ $\sf{y=4\:\:or\:\:3y=-4}$

$\longrightarrow$ $\sf{y=4\:\:or\:\:y=\dfrac{-4}{3}}$

Since age cannot be negative.

•°• y = -4/3 is neglected.

Substitute, y = 4 in equation (1),

$\longrightarrow$ $\sf{x=3y}$

$\longrightarrow$ $\sf{x=3(4)}$

$\longrightarrow$ $\sf{x=12}$

$\large{\boxed{\sf{\red{Present\:age\:of\:boy\:=\:x\:=\:12\:years}}}}$

$\large{\boxed{\sf{\purple{Present\:age\:of\:brother\:=\:y\:=\:4\:years}}}}$

Answer 2

$\huge{\fbox{\fbox{\bigstar{\mathfrak{\red{Question:-}}}}}}$

a boy is thirce as old as his brother.2 years ago,the product of their age is 20.find their present ages.

$\huge{\fbox{\fbox{\bigstar{\mathfrak{\red{Answer:-}}}}}}$

Given :

A boy is thirce as old as his brother.

2 years ago,the product of their age is 20.

To Find :

Present age of Boy.

Present age of brother.

Solution :

Let the present age of boy be X

Let the present age of brother be Y

${\purple{\sf{x=3y\:\:(1)}x=3y(1)}}$___________( eq.1)

2 years ago, the product of x and y was 20.

Age of boy 2 years ago, (x -2) years.

Age of brother 2 years ago, (y-2) years.

(x−2) (y−2) = 20

x ( y−2 ) − 2( y−2 ) = 20

xy − 2x − 2y + 4 = 20

From equation (1),

x = 3y,

3y(y) − 2(3y) − 2y = 20 − 4

3y² −6y − 2y =16

3y²2 −8y − 16 = 0

3y ² − 12y + 4y − 16 = 0

3y(y−4)+4(y−4)=0

(3y+4)=0

y−4=0

3y +4 =0

y= 4

3y = −4

y = 3/−4

Since age cannot be negative, y = -4/3 is neglect.

Substituting, y = 4 in equation (1),

x = 3 y

x = 3 (4)

x = 12

$\huge{\tt{\fbox{\fbox{\purple{Present ~age ~of ~boy =~12years}}}}}$

$\huge{\tt{\fbox{\fbox{\red{Present ~age ~of ~brother =~4years}}}}}$