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# The ULTIMATE age problem!

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Senior Manager
Joined: 12 May 2010
Posts: 288
Location: United Kingdom
Concentration: Entrepreneurship, Technology
GMAT Date: 10-22-2011
GPA: 3
WE: Information Technology (Internet and New Media)
Followers: 4

Kudos [?]: 51 [0], given: 12

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02 Oct 2011, 05:30
from a fantastic resource for mastering age problems: http://www.purplemath.com/modules/ageprobs.htm

"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.

a) 62
b) 73
c) 84
d) 89
e) 102
Senior Manager
Joined: 12 May 2010
Posts: 288
Location: United Kingdom
Concentration: Entrepreneurship, Technology
GMAT Date: 10-22-2011
GPA: 3
WE: Information Technology (Internet and New Media)
Followers: 4

Kudos [?]: 51 [0], given: 12

Re: The ULTIMATE age problem! [#permalink]

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02 Oct 2011, 05:31

"
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: Copyright © Elizabeth Stapel 2000-2011 All Rights Reserved
.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.
"
Manager
Joined: 03 Mar 2011
Posts: 90
Location: United States
Schools: Erasmus (S)
GMAT 1: 730 Q51 V37
GPA: 3.9
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Kudos [?]: 125 [0], given: 12

Re: The ULTIMATE age problem! [#permalink]

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02 Oct 2011, 05:42
Very famous problem, but it's pretty anyway. And it obviously will be good drill for anyone having problems with word PS on GMAT.

Great thanks!
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Re: The ULTIMATE age problem!   [#permalink] 02 Oct 2011, 05:42
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