Using Equations in Using Equations in Solving Age ProblemSolving Age ProblemUsing Equations in Using Equations in

Solving Age ProblemSolving Age Probleme-math e-math

Elementary AlgebraElementary AlgebraBy: Karla Mae S. GlavezBy: Karla Mae S. Glavez

SPCPSPCP

• In dealing with age problems, it is important to keep in mind that the ages of different people change at the same rate. For example, after two years, all the people in the given problem are two years older than they were two years ago. Four years ago, all the people in the problem were four years younger.

• Also, it is easier if one makes a table showing the representation for current ages in the problem, “future” ages (a number of years from now), and “past” ages ( a number of years ago).

• If possible, represent the youngest present age by a single letter, then represent the other ages. This process is illustrated in the following examples.

example: Next. . .

example 1• EXAMPLE 1:• Alvin is now 21 years older than his son.

In 8 years, he will be twice as old as his son’s age. What are their present ages?

• SOLUTION:• READ: Reading the problem

thoroughly, we fin a relationship between the present ages of Alvin and his son and the relationship of their ages 8 years from now.

• REPRESENT: Using the relationship between the present ages, we have

• Let x = the son’s present age

• RELATE: Then, x + 21 = Alvin’s present age

•• After 8 years, each age is

increasen by 8: x + 8 and x + 29 as given

NOWFUTURE ( 8 YEARS FROM NOW)

son x x + 8

Alvin x + 21 x + 29

EQUATE: Using the vsecong relationship: In 8 years, Alvin’s age is twice his son’s age, we havex + 29 = 2 ( x + 8)

SOLVE: Manipulating the equation, we havex + 29 = 2(x + 8)x + 29 = 2x + 16

x = 13

• ANSWERS: x = 13 son’s agex + 21 = 13 + 21 = 34 Alvin’s age

• PROOF: Alvin’s age is 21 years more than his son’s age: 34 = 13 + 21.

In 8 years: son: 13 + 8 = 21

Alvin: 29 + 13 = 42 Alvin’s age is twice his son’s age: 42 = 2 (21)• 42 = 42

Example 2• The sum of Patrick’s age and Marko’s age

is 58. Eight years ago, Patrick was twice as old as Marko then. How old is Marko?

• SOLUTION:• READ: We find that we have a

relationship between their present ages and their ages 8 years ago.

• REPRESENT: Let x = Marko’s present age

• RELATE: Then 58 – x = Patrick’s present age

We subtract 8 years from each as shown in the table below.

NOW Past ( 8 Years Ago)

Mako x x - 8

Patrick 58 - x 50 - x

• EQUATE: The second relationship gives us our equation: Eight years ago, Patrick was twice as old as Marko then.

50 – x = 2(x – 8) SOLVE: Manipulating the equation, we

have: 50 – x = 2(x – 8) 50 – x = 2x – 16 3x = 66 x = 22• ANSWERS: x = 22 Marko’s age 58 – x = 36 Patrick’s age

• PROOF: The sum of their ages is 58:Marko’s age = 22

• Patrick’s age = 36

• Add58

• Eight years ago, Patrick was twice as old as Marko then:

• 36 – 8 = 2(22 – 8)• 28 = 2(14)• 28 = 28

Example 3• Aries is twice as old as Rico while Jake is 24

years younger than Aris. If half of Aries’ age six years ago was three less than one-half the sum of Rico’s age in four years and Jake’s present age, find the ages of each.

• SOLUTION:• Let x = Rico’s present age• Then2x = Aries’ present age• And 2x – 24 = Jake’s present

age

Now Past (6 Years Ago)

Future ( Years From Now)

Rico x x – 6 x + 4

Aries 2x 2x – 6 2x + 4

Jake 2x - 24 2x – 30 2x – 20

Half of Aries’ age 6 years ago was 3 les than one-half the sum of Rico’s age in 4 years and Jake’s present age:

½ (2x – 6) = ½ [(x + 4) + (2x – 24)] – 3 2x -6 = x + 4 + 2x – 24 -6

2x – 6 = 3x -26x = 20

• ANSWERS: x = 20 Rico’s age• 2x = 40 Aires’ age• 2x – 24 = 16 Jake’s age• PROOF: Arie’s is twice as old as Rico:

40 = 2(20)• 40 = 40• Jake’s is 24 years younger than

Aries: 16 = 40 -24• 16 = 16• Half of Aries’ age 6 years ago was 3

less than one-half the sum of Rico’s age in 4 years and Jakes present age:

• ½ (34) = ½ (24 + 16) -3• 17 = 20 – 3• 17 = 17

Example 4• Mrs. Canteno is 40 years old and her

eldest daughter is 12. When will the mother be twice as old as her eldest daughter?

• SOLUTION:• Let x = the number of years that mother

will be twice as old as her eldest daughter.

• Then40 + x = mother’s age in x years.• And 12 + x = daughter’s age in x years

Now FUTURE ( X YEARS FROM NOW)

Mother 40 40 + x

Daughter

12 12 + x

Mother will be twice as old as her eldest daughter:40 + x = 2(12 + x )40 + x = 24 + 2x

Proof:Mother’s age in 16 years: 40 + 16 = 56

Daughter’s age in 16 years: 12 + 16 = 28 56 = 2 (28)

56 = 56

EXERCISES:1.) Mercy is 17 years old now. Represent her age –

a. 2 years from now c. x years from now e. x + 7 years hence

b. 3 years ago d. 2x years ago2.) Melchor is m years old now. Represent his age –

a. 15 years from now c. a years ago e. 5x years from now

b. 11 years ago d. n + 5 years ago 3.) Mr. Matthew Matics is y + 5 years old now. Represent

his age –a. 3 years ago d. 5y years from now

b. 8 years from now e. 3y – 5 years from nowc. 2y + 3 years ago

4.) In 4 years, Mr5. Parabola will be 17 years old. Represent his age –

a. now c. 6 years from now e. a – 3 from nowb. 8 years ago d. x years ago

5.) Arielle will be 57 years old p years from now. Represent her age –a. now c. 4 years from now e. m years ago

b. 3 years ago d. m years from now 6.) Five years ago, Edwin was 27 years old. Represent his age –

a. at present c. 5 years from now e. q years from nowb. 3 years from now d. q years ago

7.) P years ago, Mr. Reyes was Q years old. Represent his age –a. at present c. 5 years ago e. P + 4 years ago

b. 2 years from now d. R years from now

ANSWERS1.) C.2.) C.3.) D.4.) B.5.) D.6.) C.7.) D.

Solve each of the following

1.) Mr. Reyes is 25 years older than his son. How old are the father and the son if Mr. Reyes is 5 more then thrice as old as his son?