a boy is thirce as old as his brother.2 years ago,the product of their age is 20.find their present ages.
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a boy is thirce as old as his brother.2 years ago,the product of their age is 20.find their present ages.
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Given :
To Find :
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 :
[tex]\sf{x=3y\:\:(1)}[/tex]
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 :
[tex]\longrightarrow[/tex] [tex]\sf{(x-2) (y-2) =20}[/tex]
[tex]\longrightarrow[/tex] [tex]\sf{x(y-2) -2(y-2) =20}[/tex]
[tex]\longrightarrow[/tex] [tex]\sf{xy-2x-2y+4=20}[/tex]
From equation (1), x = 3y,
[tex]\longrightarrow[/tex] [tex]\sf{3y(y)-2(3y)-2y=20-4}[/tex]
[tex]\longrightarrow[/tex] [tex]\sf{3y^2\:-\:6y-2y=16}[/tex]
[tex]\longrightarrow[/tex] [tex]\sf{3y^2\:-\:8y-16=0}[/tex]
[tex]\longrightarrow[/tex] [tex]\sf{3y^2\:-\:12y\:+\:4y\:-\:16\:=\:0}[/tex]
[tex]\longrightarrow[/tex] [tex]\sf{3y(y-4) +4(y-4) =0}[/tex]
[tex]\longrightarrow[/tex] [tex]\sf{(y-4) \:\:\:(3y+4)=0}[/tex]
[tex]\longrightarrow[/tex] [tex]\sf{y-4=0\:\:or\:\:3y+4=0}[/tex]
[tex]\longrightarrow[/tex] [tex]\sf{y=4\:\:or\:\:3y=-4}[/tex]
[tex]\longrightarrow[/tex] [tex]\sf{y=4\:\:or\:\:y=\dfrac{-4}{3}}[/tex]
Since age cannot be negative.
•°• y = -4/3 is neglected.
Substitute, y = 4 in equation (1),
[tex]\longrightarrow[/tex] [tex]\sf{x=3y}[/tex]
[tex]\longrightarrow[/tex] [tex]\sf{x=3(4)}[/tex]
[tex]\longrightarrow[/tex] [tex]\sf{x=12}[/tex]
[tex]\large{\boxed{\sf{\red{Present\:age\:of\:boy\:=\:x\:=\:12\:years}}}}[/tex]
[tex]\large{\boxed{\sf{\purple{Present\:age\:of\:brother\:=\:y\:=\:4\:years}}}}[/tex]
[tex]\huge{\fbox{\fbox{\bigstar{\mathfrak{\red{Question:-}}}}}} [/tex]
a boy is thirce as old as his brother.2 years ago,the product of their age is 20.find their present ages.
[tex]\huge{\fbox{\fbox{\bigstar{\mathfrak{\red{Answer:-}}}}}} [/tex]
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
[tex]{\purple{\sf{x=3y\:\:(1)}x=3y(1)}}[/tex]___________( 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
[tex]\huge{\tt{\fbox{\fbox{\purple{Present ~age ~of ~boy =~12years}}}}} [/tex]
[tex]\huge{\tt{\fbox{\fbox{\red{Present ~age ~of ~brother =~4years}}}}} [/tex]