Lesson Age word problems

In January of the year 2000, my husband John was eleven times as old as my son William. In January of 2012, he will be three times as old as my son. How old was my son in January of 2000?
Of course, in "real life" you'd have walked up to my kid and and asked him, and he'd have proudly held up three grubby fingers, but that won't help you on your homework. Here's how you'd figure this out for class:
Let "J " stand for my husband John's age, and let "W " stand for William's age. Then J = 11W in the year 2000. In the year 2012, John and William will be twelve years older, so their ages will be J + 12 and W + 12. Also, John will be three times as old as William, so J + 12 = 3(W + 12). So now you've got two equations, each with two variables:
J = 11W
J + 12 = 3W + 36
If you know how to solve systems of equations, you can proceed with those techniques. Otherwise, use the first equation to simplify the second; since J = 11W, plug in "11W " for "J " in the second equation:
J + 12 = 3W + 36
(11W) + 12 = 3W + 36
11W – 3W = 36 – 12
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 at the beginning of the year 2000.
The important things here were to set up the variables, labelling them all clearly, and then to increment the variables by the required amount (in this case, by 12) to reflect the passage of time. Don't try to use the same expression to stand for two different things. If "J " stands for John's age in 2000, then "J " can not stand for his age in 2012. Make sure that you are very explicit about this when you set up your equations; write down the two sets of information (their ages at the first time, and their ages at the second time) as two distinct situations.
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 is 68. How old is each one now? Copyright © Elizabeth Stapel 2000-2007 All Rights Reserved
This problem 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. Let Miguel's present age be "m"; let his grandfather's age be "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; in other words:
g + 3 = 6(m – 1)
This gives you two equations with two variables:
m + g = 68
g + 3 = 6(m – 1)
Solving the first equation, you 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.) Plug this into the second equation:
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. Since m + g = 68, then m = 11, and Miguel is presently eleven years old.
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 in two years: ( 1/2 )(H + 2) = H/2 + 1
one-third of age three years ago: ( 1/3 )(H – 3) = H/3 – 1
The sum of these two numbers is twenty, so I'll add them and set this equal to 20:
H/2 + 1 + H/3 – 1 = 20
H/2 + H/3 = 20
3H + 2H = 120
5H = 120
H = 24
Heather is 24 years old.
Note: Remember that, when solving equations, you can always check your answer by plugging it 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, 13 + 7 = 20, so the solution works.
The following problem is an old one, but it keeps cropping up in various forms:
"Here lies Diophantus," the wonder behold . . .
Through art algebraic, the stone tells how old:
"God gave him his boyhood one-sixth of his life,
One twelfth more as youth while whiskers grew rife;
And then yet one-seventh ere marriage begun;
In five years there came a bouncing new son.
Alas, the dear child of master and sage
After attaining half the measure of his fathers life
chill fate took him.
After consoling his fate by this science of numbers
for four years, he ended his life."
Find Diophantus' age at death.
Let d be Diophantus' age at death. Then:
childhood: d/6
adolescence: d/12
bachelorhood: d/7
childless marriage: 5
age of child at death: d/2
life after child's death: 4
So his whole life had been divided into intervals which, when added together, give the sum of his life. That is:
.d/6 + d/12 + d/7 + 5 + d/2 + 4 = d
( 25/28 )d + 9 = d
9 = d – ( 25/28 )d
9 = ( 3/28 )d
84 = d
Diophantus lived to be 84 years old.
You can check the answer if you like, by plugging "84" into the original problem. If he lived to be 84, then one-sixth of his life is 14 years, one-twelfth of his life is 7 years (so he'd be 21, and he certainly should have a beard by this age), one-seventh of his life is 12 years (so he didn't marry until he was 33 years old), his child was born when he was 38, the boy died at 42 (when Diophantus was 80), and then Diophantus died four years later.
Always try to label your variables and expressions clearly, as this will go a long way toward helping you get your equations set up correctly. And remember that you can always check your answers (like I did on the last example above); this is an especially good idea on tests.