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Word Problem

by nadib002 » Sun Apr 18, 2010 2:33 pm
A town's oldest inhabitant is x years older than the sum of the ages of the Lee triplets. If the oldest inhabitant is now J years old, how old will one of the triplets be in 20 years?

A. (J-50)/3

B. (3(J+20))/x

C. (J+x-50)/3

D. (J-x +60)/3

E. (J+x-20)/3

Please tell me the algebraic solution. I can get the answer by picking numbers.
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by liferocks » Sun Apr 18, 2010 5:27 pm
Let the sum of the ages of the Lee triplets is S.
according to the question
s=J-X
after 20 years the sum will be S+20*3=J-X+60
so,age of one of the triplets after 20 years will be (J-X+60)/3

Ans D
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by RJ43 » Mon Apr 19, 2010 11:32 am
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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by kewltot » Mon Apr 19, 2010 12:12 pm
A town's oldest inhabitant is x years older than the sum of the ages of the Lee triplets. If the oldest inhabitant is now J years old, how old will one of the triplets be in 20 years?


Sol:



oldest inhabitant age = 3(one of the Triplets age) + x

oldest inhabitant is now J years old

=> j = 3(one Triplets age) + x

=> one Triplets age = j-x/3

one of the triplets be in 20 years = (j-x/3) + 20 = j-x+60/3

OA is D
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