## Questions Regarding the "Age" Word Problems lesson

Simple patterns, variables, the order of operations, simplification, evaluation, linear equations and graphs, etc.
leah0723
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### Questions Regarding the "Age" Word Problems lesson

For simplicity, I am just going to copy and paste. My questions are in bold:
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) = 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 How is H/2 + H/3 = 20 reduced to 3H + 2H = 120?
3H + 2H = 120
5H = 120
H = 24

Heather is 24 years old.

Now here is where I am totally lost, on page two of the "age" word problem lesson:
"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.

My first task is to "translate" the poetry from the headstone into practical terms:

"Boyhood" can stand for pre-adolscent childhood; he spent one-sixth of his life in this period.
"Youth while whiskers grew" can stand for pubescence (the teenage years, into young adulthood); he spent one-twelfth of his life in this period.
"Ere marriage began" can stand for "unmarried adulthood" or "bachelorhood"; he spent one-seventh of his life in this period.
He had five years between the wedding and the time his first child was born.
Tragically, this child died young, living only half as long as his father eventually would; looked at the other way, half of Diophantus' life was spent while his child was alive.
Diophantus died four years after burying his child.
I will let d stand for Diophantus' age at death. Then:

childhood: d/6
bachelorhood: d/7
childless marriage: 5
age of child at death: d/2
life after child's death: 4

His whole life had been divided into intervals which, when added together, give the sum of his life. So I'll add the lengths of those periods, set their sum equal to his (as-yet unknown) total age, and solve:

d/6 + d/12 + d/7 + 5 + d/2 + 4 = d
( 25/28 )d + 9 = d How does d/6 + d/12 +d/7 + d/2 = (25/28)d?
9 = d – ( 25/28 )d How does d - (25/28)d become (3/28)d? I understand 3 is the difference between 25 and 28 but how was it reduced in this manner?
9 = ( 3/28 )d Finally, if I'm assuming correctly, how does 9/(3/28) = 84?
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.

stapel_eliz
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H/2 + 1 + H/3 – 1 = 20
H/2 + H/3 = 20 How is H/2 + H/3 = 20 reduced to 3H + 2H = 120?
What is the Least Common Multiple (or Lowest Common Denominator) of 2 and 3? Then what is the common denominator of the two fractions on the left-hand side? Then by what will you be multiplying each side, to clear the denominators? What is the resulting linear equation?
d/6 + d/12 + d/7 + 5 + d/2 + 4 = d
( 25/28 )d + 9 = d How does d/6 + d/12 +d/7 + d/2 = (25/28)d?
What is the Least Common Multiple of 2, 6, 12, and 7? Then what is the common denominator of the fractions? Once you have the common denominator, what are the numerators? What is the result, upon adding these numerators?
9 = d – ( 25/28 )d How does d - (25/28)d become (3/28)d?

What is the result, when subtracting 25/28 from 28/28?

9 = ( 3/28 )d Finally, if I'm assuming correctly, how does 9/(3/28) = 84?

What is the result, when multiplying 9/1 by 28/3?

All of your questions relation to working with fractions. To learn how to work with fractions (simplifying, multiplying and dividing, adding and subtracting, converting the common denominators, etc), try here.