GMAT Question of the Day - Daily to your Mailbox; hard ones only

 It is currently 23 Jun 2018, 22:31

### GMAT Club Daily Prep

#### Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.

Customized
for You

we will pick new questions that match your level based on your Timer History

Track

every week, we’ll send you an estimated GMAT score based on your performance

Practice
Pays

we will pick new questions that match your level based on your Timer History

# Events & Promotions

###### Events & Promotions in June
Open Detailed Calendar

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

Author Message
TAGS:

### Hide Tags

Intern
Joined: 29 Mar 2011
Posts: 21
A town's oldest inhabitant is x years older than the sum of the ages [#permalink]

### Show Tags

Updated on: 01 Mar 2018, 21:41
7
00:00

Difficulty:

15% (low)

Question Stats:

79% (01:27) correct 21% (01:29) wrong based on 135 sessions

### HideShow timer Statistics

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 + 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.
Current Student
Joined: 26 May 2005
Posts: 531
Re: A town's oldest inhabitant is x years older than the sum of the ages [#permalink]

### Show Tags

17 Jul 2011, 06:22
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
Manager
Joined: 14 Apr 2011
Posts: 173
Re: A town's oldest inhabitant is x years older than the sum of the ages [#permalink]

### Show Tags

17 Jul 2011, 13:25
1
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
_________________

Looking for Kudos

Director
Joined: 01 Feb 2011
Posts: 686
Re: A town's oldest inhabitant is x years older than the sum of the ages [#permalink]

### Show Tags

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
Math Expert
Joined: 02 Sep 2009
Posts: 46297
Re: A town's oldest inhabitant is x years older than the sum of the ages [#permalink]

### Show Tags

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 + 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.
_________________
VP
Joined: 07 Dec 2014
Posts: 1020
Re: A town's oldest inhabitant is x years older than the sum of the ages [#permalink]

### Show Tags

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
Senior Manager
Status: love the club...
Joined: 24 Mar 2015
Posts: 275
Re: A town's oldest inhabitant is x years older than the sum of the ages [#permalink]

### Show Tags

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]

### Show Tags

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

DS Forum Moderator
Joined: 22 Aug 2013
Posts: 1209
Location: India
Re: A town's oldest inhabitant is x years older than the sum of the ages [#permalink]

### Show Tags

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

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
Re: A town's oldest inhabitant is x years older than the sum of the ages   [#permalink] 01 Mar 2018, 22:21
Display posts from previous: Sort by