A father is 6 times older than his son. After 4 years, he will be only

1.	A father is 6 times older than his son. After 4 years, he will be only 4 times as old as his son. What are their present ages?
Answer» GIVEN : FATHER is 6 times older than his son. After 4 YEARS, father's age will be 4 times as old as son's age. To Find : Present age of Father Present age of Son SOLUTION : Let the present age of father be x years. Let the present age of son be y years. Case 1 : The age of Father, x is 6 times the age of son, y. Equation : $\longrightarrow$ $\sf{x=6y}$ $\sf{x-6y=0\:\:\:\:(1)}$ Case 2 : After 4 years, father's age, x will be 4 times the age of son. Age of father after 4 year, (x+4) years. Age of son after 4 year, (y+4) years. Equation : $\longrightarrow$ $\sf{(x+4)=4(y+4)}$ $\longrightarrow$ $\sf{x+4=4y+16}$ $\longrightarrow$ $\sf{x-4y=16-4}$ $\longrightarrow$ $\sf{x-4y=12\:\:\:(2)}$ Now, SUBTRACT equation (1) from (2), $\longrightarrow$ $\sf{x-6y-(x-4y) =0-12}$ $\longrightarrow$ $\sf{x-6y+x+4y=-12}$ $\longrightarrow$ $\sf{-6y+4y=-12}$ $\longrightarrow$ $\sf{-2y=-12}$ $\longrightarrow$ $\sf{y=\dfrac{-12}{-2}}$ $\longrightarrow$ $\sf{y=\dfrac{12}{2}}$ $\longrightarrow$ $\sf{y=6}$ Substitute, y = 6 in equation (1), $\longrightarrow$ $\sf{x-6y=0}$ $\longrightarrow$ $\sf{x-6(6)=0}$ $\longrightarrow$ $\sf{x-36=0}$ $\longrightarrow$ $\sf{x=36}$ $\large{\boxed{\bold{\purple{Present\:age\:of\:father\:=\:x\:=\:36\:years}}}}$ $\large{\boxed{\bold{\purple{Present\:age\:of\:son\:=\:y\:=\:6\:years}}}}$