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EIV page 123 problem number 4

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a.sarwari
 Post subject: EIV page 123 problem number 4
 Post Posted: Mon Oct 03, 2011 2:50 pm 
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Course Students


Posts: 6
4) A town's oldest inhabitant is x years older than the sum of the ages of the Lee triplets. If the oldest inhabitants is now J years old, how old will one of the triplets be in 20 years?

I would like to approach this using direct algebra. Please tell me what am I doing wrong?

1)J=3L+X
2) J+20=3(L+20)+X+20
J+20=3L+60+X+20
-20 -20
J=3L+60+x
L= (J-X-60)/3

In the book the correct answer is (J-x+60)/3. Please tell me what am I missing????

Thanks in advance.
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fazilot
 Post subject: Re: EIV page 123 problem number 4
 Post Posted: Thu Oct 13, 2011 2:35 pm 
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Students


Posts: 1
http://www.beatthegmat.com/word-problem-t56299.html
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a.sarwari
 Post subject: Re: EIV page 123 problem number 4
 Post Posted: Fri Oct 14, 2011 9:34 am 
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Course Students


Posts: 6
Thank you,

Here is the explanation through the link

T = Age of each Triplet

J = 3T + X

Since it wants to know the age of the Triplet in 20 years, we must solve for T. We get:

T = (J-X) / 3

Now we need to add the 20 years to it:

T = (J-X) / 3 + 20

This would normally be the answer, but we have to take it a step further to get a common denominator. We must convert the 20 into 60/3 to meet the requirement so we get:

T = (J-X) / 3 + 60/3 = (J-X+60) / 3 Answer D.
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StaceyKoprince
 Post subject: Re: EIV page 123 problem number 4
 Post Posted: Sun Nov 13, 2011 10:55 pm 
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ManhattanGMAT Staff


Posts: 6064
Location: San Francisco
Thanks for reposting the explanation. You didn't ask a new question, so I assume that the explanation worked for you? If not, let us know!

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Stacey Koprince
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