A perfect teaser!
Algebra
Level
pending
When the son will be as old as his father today, the sum of their ages then will be 126 . When the father was as old as his son is today, the sum of their ages then was 38 . Find their present ages.
Father: 50, Son 27
Father: 51, Son 31
Father: 52, Son 30
Father: 51, Son 29
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1 solution
Let the ages of father and son be denoted as 'x' and 'y' respectively.
And the difference of their ages be 'z' i.e. (x-y);
x+z+y+z=126
x+y+2z=126
Recall, z=(x-y)
:.x+y+2x-2y=126
3x-y=126.......eqn1
x=126/3+y/3
x=42+y/3.......eqn2
When the father was as old as his son;
x-z+y-z=38
x+y-2z=38
Recall, z=(x-y)
x+y-2x+2y=38
-x+3y=38
x-3y=-38........eqn3
x=3y-38.........eqn4
Equating eqn2 and eqn4 to solve for y: 42+y/3=3y-38
Find the LCD of LHS and cross multiply:
126+y=9y-114
y-9y=-114-126
-8y=-240
y=30 present age of son
substitute y=30 into eqn4;
x=3(30)-38
x=90-38
x=52 present age of father
PS: In 22 yrs father would be 74, and son 52 which sum up to 126. And 22 yrs ago, father was 30, son 8 which summed up 38 PS. This is a course problem and the solution isn't done by me.