Problem Solving

Hi

Can any one of you help me solving this problem using algebra

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 ?

The answers is (J-X+60)/3 .... But i am not able to arrive at the same answer . Need help !

Basic Algebra buddy.. check it line by line..

Let the age of a triplet be L
Age of the oldest guy is given as J

1. A town's oldest inhabitant is X years older than the sum of the ages of the Lee triplets (triplets is like twins.. in twins you have two children, triplets mean three children)

So this sentence in mathematical equation would be:
Oldest inhabitant = sum of age of triplets + X
J = 3L + X

so L = (J - X)/3

After 20 years the age of triplet will be: L + 20

= (J - X)/3 + 20

= (J - X + 60)/3

SIMPLY plug in numbers if algebra is not your strong suit...


make the triplets 5 years old each...their sum is therefore 15 year

X = 10....therefore J= 25...

so in 20 years, the triplets will = 25 years old...

now plug X and J into the answer and see what gives you 25...

(J - X + 60)/3

(25-10+60)/3 = 75/3 = 25..BINGO

There is something very strange about this problem

present condition
J - 3L = x as J is x years older than the sum of the three triplets

In 20 years time, obviously everyone's age will increase by 20 years that is the universal truth.
so in 20 years
J's age = J+20
Each lee's age = L+20 but since there are 3 Lee's so their total age after 20 years 3(L+20) = 3L+60

Now the difference between J's age and all lee's age after 20 years is still x ( Their age difference all always remain x, suppose I am 2 years older than my small brother , then 20( or 30 or 40 or any number ) years down the line , I will still be 2 years older than my brother. Because my brother has also grown similar number years as me

so J+20 - (3L + 60 ) = x
J+20-3L-60=x
J-40-x=3L
(J-40-x)/3= L (each lee's age in terms of J after 20 years)

so when we say that the answer is (J+60-x)/3 it seems that J has not grown in the 20 years that lee's have grown, is that possible?

This contradicts many similar sums given by GMAT in OG and Quant. reviews.
Can some expert please share his/her opinion on this.

Is my logic wrong? or are we trying to somehow arrive at the answer because we know the answer.

Please do share your views.

gmatter2012 wrote: "....................................................................................................
.........Now the difference between J's age and all lee's age after 20 years is still x ( Their age difference all always remain x, suppose I am 2 years older than my small brother , then 20( or 30 or 40 or any number ) years down the line , I will still be 2 years older than my brother. Because my brother has also grown similar number years as me .............."

well I got it to some extent,The portion above is not entirely true. This could have been true if there was one Lee. But there are 3 Lee's.
so the difference between 3 Lee's age and J's age after 20 years in not x but something else, what is that? can we find that out , is there a relationship?

suppose I am two years older than the sum of the ages of my 2 small brothers
so Suppose my age is 30 and my small brothers each are 14 so 30 -2(14) = 2 ...( first relation )

now 5 years down the line this relation will not hold

after 5 years my age 35

my brothers age each 14 + 5 = 19 so both the brothers age total =38

Now difference between my age and 2 bothers total age = 35-2(19)= -3
so as we can see the first relation does not hold after 5 years

similarly in J's case and Lee's case, the initial difference of X does not hold after 20 years.

so what could be the difference after 20 years?

can we solve the problem this way? or is it not possible this way and have to approach in some other ways.

I think if we can get x after 20 years ( age difference between J and 3 Lee's after 20 years ) we will arrive at the given OA.

Assumption :

Each triplet = Y years old.
Oldest man = 3Y+X = J --> Eq.1

We need :

Y+20

So, Rebuilding Eq.1 => Y=(J-X)/3

Y+20=(J-X)/3+20=(J-X+60)/3