In three more years, Miguel's grandfather will be six times as old as Miguel was last year. When Miguel's present age is added to his grandfather's present age, the total is68. How old is each one now? This exercise refers not only to their present ages, but also to both their ages last year and their ages in three years, so labelling will be very important. I will label Miguel's present age as "m" and his grandfather's present age as "g". Then m + g = 68. Miguel's age "last year" was m – 1. His grandfather's age "in three more years" will be g + 3. The grandfather's "age three years from now" is six times Miguel's "age last year" or, in math: g + 3 = 6(m – 1)

This gives me two equations with two variables:

m + g = 68

g + 3 = 6(m – 1)

Solving the first equation, I get m = 68 – g. (Note: It's okay to solve for "g = 68 – m", too. The problem will work out a bit differently in the middle, but the answer will be the same at the end.) I'll plug "68 – g" into the second equation in place of "m": g + 3 = 6m – 6

g + 3 = 6(68 – g) – 6

g + 3 = 408 – 6g – 6

g + 3 = 402 – 6g

g + 6g = 402 – 3

7g = 399

g = 57

Since "g" stands for the grandfather's current age, then the grandfather is 57 years old. Sincem + g = 68, then m = 11, and Miguel is presently eleven years old.

SOURCE: http://www.purplemath.com/modules/ageprobs.htm

❷AGE PROBLEM

One-half of Heather's age two years from now plus one-third of her age three years ago is twenty years. How old is she now? This problem refers to Heather's age two years in the future and three years in the past. So I'll pick a variable and label everything clearly: age now: H

age two years from now: H + 2

age three years ago: H – 3

Now I need certain fractions of these ages:

one-half of age two years from now: ( 1 2 )(H + 2) = H2 + 1 one-third of age three years ago: ( 13 )(H – 3) = H3 – 1 The sum of these two numbers is twenty, so I'll add them and set this equal to 20: H2 + 1 + H3 – 1 = 20

H2 + H3 = 20

3H + 2H = 120

5H = 120

H = 24

Heather is 24 years old.

Note: Remember that you can always check your answer to any "solving" exercise by plugging that answer back into the original problem. If Heather is 24 now, then she will be 26 in two years, half of which is 13, and she was 21 three years ago, a third of which is 7. Adding, I get 13 + 7 = 20, so the solution works.

SOURCE: http://www.purplemath.com/modules/ageprobs.htm

❸AGE PROBLEM

In January of the year 2000, I was one more than eleven times as old as my son William. In January of 2009, I was seven more than three times as old as him. How old was my son in January of 2000? First, name things and translate the English into math: Let "E " stand for my age in 2000, and let "W " stand for William's age. Then E = 11W + 1 in the year 2000 (from "eleven times as much, plus another one"). In the year 2009 (nine years after the year 2000), William and I will each be nine years older, so our ages will be E + 9 and W + 9. Also, I was seven more than three times as old as William was, so E + 9 = 3(W + 9) + 7 = 3W + 27 + 7 = 3W + 34. This gives you two equations, each having two variables: E = 11W + 1

E + 9 = 3W + 34

If you know how to solve systems of equations, you can proceed with those techniques. Otherwise, you can use the first equation to simplify the second: since E = 11W + 1, plug "11W + 1 " in for "E " in the second equation: E + 9 = 3W + 34

(11W + 1) + 9 = 3W + 34

11W – 3W = 34 – 9 – 1

8W = 24

W = 3

Remember that the problem did not ask for the value of the variable W; it asked for the age of a person. So the answer is: William was three years old in January of 2000.

SOURCE: http://www.purplemath.com/modules/ageprobs.htm

❹AGE PROBLEM

Alvin's age is three times Elga's age. The sum of their ages is " 40 ". What is Elga's age?

Let x be Alvin's age

Let y be Elga's age

x + y = 40

x = 3y

Replace x = 3y into x + y = 40

We get 3y + y = 40

4y = 40...