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a.sarwari
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EIV page 123 problem number 4

by a.sarwari Mon Oct 03, 2011 2:50 pm

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.
fazilot
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Re: EIV page 123 problem number 4

by fazilot Thu Oct 13, 2011 2:35 pm

a.sarwari
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Re: EIV page 123 problem number 4

by a.sarwari Fri Oct 14, 2011 9:34 am

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.
StaceyKoprince
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Re: EIV page 123 problem number 4

by StaceyKoprince Sun Nov 13, 2011 10:55 pm

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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