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# A town's oldest inhabitant is x years older than the sum of the ages

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Intern
Joined: 29 Mar 2011
Posts: 16
A town's oldest inhabitant is x years older than the sum of the ages  [#permalink]

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Updated on: 01 Mar 2018, 21:41
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Difficulty:

15% (low)

Question Stats:

82% (02:04) correct 18% (02:07) wrong based on 128 sessions

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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?

A. $$\frac{J - 50}{3}$$

B. $$\frac{3(J + 20)}{x}$$

C. $$\frac{J + x - 50}{3}$$

D. $$\frac{J - x + 60}{3}$$

E. $$\frac{J + x - 20}{3}$$

(J-X+60)/3 .... But i was trying to solve it using algebra and got a wrong solution. I get this answer if i plug in numbers but i am trying to find the algebraic solution.

(J-X-40)/3 is my answers.

J = X + L + L + L is the initial situation
After 20 years
J + 20 = X + L + L + L + 60 ...(20 years for each triplet so 60 years totally).
(J - X - 40 ) / 3 = L is my answer.

What wrong am i doing ? Since the age asked is after 20 years i also consider adding 20 years to J .

Regards,
Mustu

Originally posted by mustu on 17 Jul 2011, 06:12.
Last edited by Bunuel on 01 Mar 2018, 21:41, edited 2 times in total.
Renamed the topic, edited the question and added the OA.
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Joined: 26 May 2005
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Re: A town's oldest inhabitant is x years older than the sum of the ages  [#permalink]

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17 Jul 2011, 06:22
2
here it goes:

Oldest inhabitant = sum of age of triplets + X
J = 3L + X so L = (J - X)/3

After 20 years = L + 20

= (J - X)/3 + 20

= (J - X + 60)/3

hope this helps
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Joined: 14 Apr 2011
Posts: 163
Re: A town's oldest inhabitant is x years older than the sum of the ages  [#permalink]

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17 Jul 2011, 13:25
2
hi mustu,

the initial eq is J=X+3L, your later equation i.e after 20 yrs J+20 = X+3L +60 is wrong as this is not what is stated. the relationship between J and L is only valid now and not after 20 yrs. therefore, you need to extract L from the initial eq and add 20 to it.

or you can do it as : J=X+Sum of triplets => sum of triplets = (J-X) and after 20 years: sum of triplets = J-X+60 => age of one triplet =(J-X+60)/3
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Re: A town's oldest inhabitant is x years older than the sum of the ages  [#permalink]

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17 Jul 2011, 14:14
Current In 20 yrs

Oldest Inhabitant J = 3L +X 3L+X+20

Lee Triplet 1 L L+20

Lee Triplet 2 L L+20

Lee Triplet 3 L L+20

we are asked to find out triplets age in 20 yrs from now = L+20

from current we have J = 3L+X = > L = (J-X)/3

=> L+20 = (J-X)/3 +20 = (J-X+60)/3
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Posts: 58327
Re: A town's oldest inhabitant is x years older than the sum of the ages  [#permalink]

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20 Nov 2014, 07:18
mustu wrote:
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?

A. (J - 50)/3
B. 3(J + 20)/x
C. (J + x - 50)/3
D. (J - x + 60)/3
E. (J + x - 20)/3

(J-X+60)/3 .... But i was trying to solve it using algebra and got a wrong solution. I get this answer if i plug in numbers but i am trying to find the algebraic solution.

(J-X-40)/3 is my answers.

J = X + L + L + L is the initial situation
After 20 years
J + 20 = X + L + L + L + 60 ...(20 years for each triplet so 60 years totally).
(J - X - 40 ) / 3 = L is my answer.

What wrong am i doing ? Since the age asked is after 20 years i also consider adding 20 years to J .

Regards,
Mustu

Check other Age Problems HERE.
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Joined: 07 Dec 2014
Posts: 1222
Re: A town's oldest inhabitant is x years older than the sum of the ages  [#permalink]

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18 Jun 2016, 10:32
1
t=age of one triplet
3t=J-x
t=(J-x)/3
t+20=(J-x)/3+20
t+20=(J-x+60)/3
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Re: A town's oldest inhabitant is x years older than the sum of the ages  [#permalink]

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20 Oct 2017, 14:08
[quote="mustu"]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?

A. (J - 50)/3
B. 3(J + 20)/x
C. (J + x - 50)/3
D. (J - x + 60)/3
E. (J + x - 20)/3

think simply

J - 3L = x

or L = (J - x)/3

in 20 years age of each "L" will be
(J - x)/3 + 20

which implies
(J - x + 60)/3

thanks
Intern
Joined: 30 Nov 2017
Posts: 1
Re: A town's oldest inhabitant is x years older than the sum of the ages  [#permalink]

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01 Mar 2018, 21:15
Hi, after reading all these I see the logic in keeping the "sum of the triplets" in mind..... but isn't that already accounted for in the original J = x+3L formula? As long as you're adding 20 to each side, aren't you already counting for each of the 3 "L"s? (ie --- J+20 = x+3(L+20) )

I'm running into the same problem as the OP and these explanations only half a little sense - I know 20x3 = 60... that's literally the only sensical party I understand from the jump of if L = (J -x)/3 , then in 20 years, that will be J-x + 60/3.... Especially since that 3 there SHOULD be counting (at least in my mind) for all the triplets....

Can someone break this down to an elementary level of how 20 years later becomes +60 in the equation?

My final formula was L+20 = ((J+20)-x)/3
leading to... L = (J-40-x)/3

Thank you in advance!
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Re: A town's oldest inhabitant is x years older than the sum of the ages  [#permalink]

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01 Mar 2018, 22:21
santisd1 wrote:
Hi, after reading all these I see the logic in keeping the "sum of the triplets" in mind..... but isn't that already accounted for in the original J = x+3L formula? As long as you're adding 20 to each side, aren't you already counting for each of the 3 "L"s? (ie --- J+20 = x+3(L+20) )

I'm running into the same problem as the OP and these explanations only half a little sense - I know 20x3 = 60... that's literally the only sensical party I understand from the jump of if L = (J -x)/3 , then in 20 years, that will be J-x + 60/3.... Especially since that 3 there SHOULD be counting (at least in my mind) for all the triplets....

Can someone break this down to an elementary level of how 20 years later becomes +60 in the equation?

My final formula was L+20 = ((J+20)-x)/3
leading to... L = (J-40-x)/3

Thank you in advance!

Hi

First of all you need to be clear what is the meaning of 'L' in your solution. Are you taking L to be age of each triplet? OR are you taking L to be the 'sum of ages of 3 triplets'?

So lets say we take L to mean age of each of Lee's triplets. Then sum of their ages right now = L + L + L = 3L. So now we have J = x + 3L or 3L = J - x or L = (J - x)/3

Or we can say that age of each triplet right now = (J - x)/3.
This means age of each triplet after 20 years = (J - x)/3 + 20
Now this (J-x)/3 + 20 can also be written as: (J-x)/3 + 60/3 (since 20 = 60/3)
So this becomes = (J - x + 60) / 3
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Re: A town's oldest inhabitant is x years older than the sum of the ages  [#permalink]

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08 Sep 2019, 21:36
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Re: A town's oldest inhabitant is x years older than the sum of the ages   [#permalink] 08 Sep 2019, 21:36
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