Solving Age Problems Part Two
Solving Age Problems Part Two
This hub is a sequel to my hub “Solving Age Problems.” In this hub I present additional four problems with their solution. I hope you will enjoy this.
Problem Number One :
Andr ewis nine year older than three timeshis son’s age. In fifteen years, the age of Andrew will be six years more than twice his son’s age. Find their present ages.
Solution :
Representation:
LetX= Son’s age
3X + 9= Andrew’s age
Their ages in 15 years or fifteen years from now
X + 15= Son’s age
3X + 9 + 15 = 3X + 24= Andrew’s age
Working Equation:
3X + 24= 2(X + 15) + 6
3X + 24= 2X + 30 + 6
3X – 2X= 36 – 24
X = 12= Son’s age
3X + 9 = 3(12) + 9= 45 = Andrew’s age
Problem Number Two :
Linda is two thirds as old as Francis. In five years, the sum of their ages will be fifty years. Find their present ages.
Solution :
Representation :
Let X = Francis age
2/3 X = Linda’s age
Their ages in five years
X + 5= Francis age
2/3 X +5= Linda’s age
Working Equation :
X + 5 + 2/3 X + 5= 50
X+ 2/3 X + 10 = 50
X + 2/3 X = 50 -10
( X + 2/3 X= 40) 3
3X + 2X=120
5X= 120
5X /5=120/5
X= 24= Francis age
2/3 X = 16= Linda’s age
Problem Number Three :
In eight years ,Alvin will be three times as old as he was eight year ago. How old is Alvin now ?
Solution :
Representation :
Let X = Alvin’s age
X + 8 = Alvin’s age in 8 years
X – 8+ Alvin’s age eight years ago
Working Equation :
X + 8 = 3 (X – 8)
X + 8= 3X -24
X -3X = -24 – 8
-2X= -32
-2x/-2= -32/-2
X = 16
Alvin is sixteen years old now.
Problem Number Four:
A man who is 42 years old has a daughter who is 12 years old. In how many years will the father be twice as old as his daughter ?
Solution:
Man’s present age= 42
Daughter’s present age= 12
LetX = In how many years will the father’s age be equal to twice his daughter’s age
Their ages in X years from now
42 + X= Father’s age
12 + X= Daughter’s age
Working equation :
42 + X= 2 (12 + X)
42 + X= 24 + 2X
X – 2X= 24 -42
(-X= - 18 ) -1
X = 18
In eighteen years, Father’s age will be twice his daughter’s age.