Mar 23 07:00 AM PDT  09:00 AM PDT Christina scored 760 by having clear (ability) milestones and a trackable plan to achieve the same. Attend this webinar to learn how to build trackable milestones that leverage your strengths to help you get to your target GMAT score. Mar 27 03:00 PM PDT  04:00 PM PDT Join a free live webinar and learn the winning strategy for a 700+ score on GMAT & the perfect application. Save your spot today! Wednesday, March 27th at 3 pm PST Mar 29 10:00 PM PDT  11:00 PM PDT Right now, their GMAT prep, GRE prep, and MBA admissions consulting services are up to $1,100 off. GMAT (Save up to $261): SPRINGEXTRAGMAT GRE Prep (Save up to $149): SPRINGEXTRAGRE MBA (Save up to $1,240): SPRINGEXTRAMBA
Author 
Message 
TAGS:

Hide Tags

Manager
Joined: 14 Apr 2003
Posts: 82

Mr. and Mrs. Wiley have a child every J years. Their oldest
[#permalink]
Show Tags
19 Feb 2007, 12:40
Question Stats:
67% (02:01) correct 33% (02:08) wrong based on 548 sessions
HideShow timer Statistics
Mr. and Mrs. Wiley have a child every J years. Their oldest child is now T years old. If they have a child 2 years from now, how many children will they have in total? (A) \(\frac{T+2}{J} + 1\) (B) \(JT + 1\) (C) \(\frac{J}{T} + \frac{1}{T}\) (D) \(TJ  1\) (E) \(\frac{T+J}{J}\) Source: Manhattan Guide
Official Answer and Stats are available only to registered users. Register/ Login.




Veritas Prep GMAT Instructor
Joined: 16 Oct 2010
Posts: 8998
Location: Pune, India

Re: Mr. and Mrs. Wiley have a child every J years. Their oldest
[#permalink]
Show Tags
21 Nov 2012, 10:45
jeeteshsingh wrote: Need the solution using Algebra.... Mr. & Mrs Wiley have a child every J years. Their oldest child is now T years old. If they have a child 2 years from now, how many children will they have in total? (A) \(\frac{T+2}{J} + 1\) (B) \(JT + 1\) (C) \(\frac{J}{T} + \frac{1}{T}\) (D) \(TJ  1\) (E) \(\frac{T+J}{J}\) Source: Manhattan Guide Please give the algebric solution.Think of it as an Arithmetic Progression where every subsequent term (child) has a difference of J yrs from the previous term (child). 1st child, 2nd child, 3rd child, ....... nth child (to be born after 2 yrs) What is the difference between first and last terms (children)? (T + 2) yrs What is the common difference (age difference between two consecutive kids)? J yrs What is the number of terms (children)? (T + 2)/J + 1 (Number of terms of an AP is n = (Last term  First term)/Common Difference + 1. )
_________________
Karishma Veritas Prep GMAT Instructor
Learn more about how Veritas Prep can help you achieve a great GMAT score by checking out their GMAT Prep Options >




Senior Manager
Joined: 22 Dec 2009
Posts: 292

Re: Mr. and Mrs. Wiley have a child every J years. Their oldest
[#permalink]
Show Tags
Updated on: 22 Nov 2012, 05:41
Mr. and Mrs. Wiley have a child every J years. Their oldest child is now T years old. If they have a child 2 years from now, how many children will they have in total? (A) \(\frac{T+2}{J} + 1\) (B) \(JT + 1\) (C) \(\frac{J}{T} + \frac{1}{T}\) (D) \(TJ  1\) (E) \(\frac{T+J}{J}\) Source: Manhattan Guide Please give the algebric solution.
_________________
Cheers! JT........... If u like my post..... payback in Kudos!!
Do not post questions with OAPlease underline your SC questions while postingTry posting the explanation along with your answer choice For CR refer Powerscore CR BibleFor SC refer Manhattan SC Guide
~~Better Burn Out... Than Fade Away~~
Originally posted by jeeteshsingh on 06 Feb 2010, 13:17.
Last edited by Bunuel on 22 Nov 2012, 05:41, edited 1 time in total.
Renamed the topic and edited the question.




Intern
Joined: 04 Apr 2006
Posts: 35

Re: Mr. and Mrs. Wiley have a child every J years. Their oldest
[#permalink]
Show Tags
19 Feb 2007, 15:04
jainvineet wrote: Mr. and Mrs. Wiley have a child every J years. Their oldest child is now T years old. If they have a child 2 years from now, how many children will they have in total?
1) (T+2)/J + 1 2) JT + 1 3) J/T + 1/T 4) TJ1 5) (J+T)/J
My answer is (T+2J) /J OA is 1.
Try with numbers:
They have a child every 3 years (J=3)
child 1 is 1 year old
ch 2 4y
ch 3 7y
ch 4 10y (oldest)
T=10
They have 4 children now
(10+2)/3 = (T+2)/J
After 2 years they will have (T+2)/J +1



Manager
Joined: 14 Apr 2003
Posts: 82

Re: Mr. and Mrs. Wiley have a child every J years. Their oldest
[#permalink]
Show Tags
19 Feb 2007, 18:56
jvujuc wrote: jainvineet wrote: Mr. and Mrs. Wiley have a child every J years. Their oldest child is now T years old. If they have a child 2 years from now, how many children will they have in total?
1) (T+2)/J + 1 2) JT + 1 3) J/T + 1/T 4) TJ1 5) (J+T)/J
My answer is (T+2J) /J OA is 1. Try with numbers: They have a child every 3 years (J=3) child 1 is 1 year old ch 2 4y ch 3 7y ch 4 10y (oldest) T=10 They have 4 children now(10+2)/3 = (T+2)/JAfter 2 years they will have (T+2)/J +1
Please show me without number substituion.
Ok finally I am able to understand this question.
Age of the oldest kid at the time of birth of latest sibling is (x1)J where x is the total number of kids.
T+2 = (x1)J
Hence A. Not so QED.



Senior Manager
Joined: 25 Jun 2009
Posts: 276

Re: Mr. and Mrs. Wiley have a child every J years. Their oldest
[#permalink]
Show Tags
06 Feb 2010, 13:35
jeeteshsingh wrote: Need the solution using Algebra.... Mr. & Mrs Wiley have a child every J years. Their oldest child is now T years old. If they have a child 2 years from now, how many children will they have in total? (A) \(\frac{T+2}{J} + 1\) (B) \(JT + 1\) (C) \(\frac{J}{T} + \frac{1}{T}\) (D) \(TJ  1\) (E) \(\frac{T+J}{J}\) Source: Manhattan Guide Please give the algebric solution.Lets take it this way, They had first child today, after J years they will have 2nd child, and again after J years they will have 3rd child.. And now according to the Q stem in T + 2 years they will have J kids Hence number kids in T + 2 years = T+2 /J We will have to add 1 (their first/oldest Kid) Hence Total number of kids = (T+2 / J) + 1 cheers



Senior Manager
Joined: 21 Jul 2009
Posts: 322
Schools: LBS, INSEAD, IMD, ISB  Anything with just 1 yr program.

Re: Mr. and Mrs. Wiley have a child every J years. Their oldest
[#permalink]
Show Tags
06 Feb 2010, 16:05
Their oldest kid is T y/o. From the question stem, their youngest kid alive is J2 y/o. We need to know how many kids are there between T and J2 spaced exactly J years apart. This is an arithmetic progression. (J2) + (n1)J = T, inclusive. This gives n = (T+2)/J. Question stem also states that they are going to have a baby after 2 yrs from now, so we have to add that kid to n to get the total number of kids the Wiley family will have. Therefore, answer is A.
_________________
I am AWESOME and it's gonna be LEGENDARY!!!



Intern
Joined: 17 Nov 2009
Posts: 31
Schools: University of Toronto, Mcgill, Queens

Re: Mr. and Mrs. Wiley have a child every J years. Their oldest
[#permalink]
Show Tags
13 Feb 2010, 01:21
BarneyStinson wrote: Their oldest kid is T y/o. From the question stem, their youngest kid alive is J2 y/o. We need to know how many kids are there between T and J2 spaced exactly J years apart. This is an arithmetic progression.
(J2) + (n1)J = T, inclusive. This gives n = (T+2)/J. Question stem also states that they are going to have a baby after 2 yrs from now, so we have to add that kid to n to get the total number of kids the Wiley family will have. Therefore, answer is A. Hi Barney, Your solution makes sense as its simple AP you applied, but I failed to comprehend that how you get J2 as first element. As Q says that they'll have kids after every J years and their oldest kid is T years old and the next kid will be due after two years from now but we don't know that how many years back first kid was born....may be I didn't get rightly. Can you please explain. Thanks. Cheers!
_________________
Action is the foundational key to all success.



CEO
Status: Nothing comes easy: neither do I want.
Joined: 12 Oct 2009
Posts: 2573
Location: Malaysia
Concentration: Technology, Entrepreneurship
GMAT 1: 670 Q49 V31 GMAT 2: 710 Q50 V35

Re: Mr. and Mrs. Wiley have a child every J years. Their oldest
[#permalink]
Show Tags
13 Feb 2010, 02:03
Ans A their first child was born after J years... thus 1 child > j years => thus after another J years his age = J thus his age is J > after 2J years and 2j after 3j years his present age is T which is after T years. thus total time after 2years will be T+2 since after every J year they have a child after T+2 they will have \(\frac{(T+2)}{J}\) + 1 ( +1 is for the oldest) thus A
_________________
Fight for your dreams :For all those who fear from Verbal lets give it a fight
Money Saved is the Money Earned
Jo Bole So Nihaal , Sat Shri Akaal
Support GMAT Club by putting a GMAT Club badge on your blog/Facebook
GMAT Club Premium Membership  big benefits and savings
Gmat test review : http://gmatclub.com/forum/670to710alongjourneywithoutdestinationstillhappy141642.html



Manager
Joined: 08 Jun 2011
Posts: 81

Re: Mr. and Mrs. Wiley have a child every J years. Their oldest
[#permalink]
Show Tags
02 Nov 2011, 11:08
Sorry for pumping this old one up but I seem to never get it working when J =5 and T =20.
If they have a child every 5 years and their oldest is 20 years then this means that they have 1 child at year 0 2 children at year 5 3 children at year 10 4 children at year 15 5 children at year 20 They have 5 total children right now. After two years, they should have a total of 6.
If I plug in numbers, none of the choices fit.
The answer provided is (A) but if I plug in T + 2 /J + 1 I end up having a non integer.
20 + 2 /5 +1 = 4.5
Any help?



Veritas Prep GMAT Instructor
Joined: 16 Oct 2010
Posts: 8998
Location: Pune, India

Re: Mr. and Mrs. Wiley have a child every J years. Their oldest
[#permalink]
Show Tags
03 Nov 2011, 03:15
Lstadt wrote: Sorry for pumping this old one up but I seem to never get it working when J =5 and T =20.
If they have a child every 5 years and their oldest is 20 years then this means that they have 1 child at year 0 2 children at year 5 3 children at year 10 4 children at year 15 5 children at year 20 They have 5 total children right now. After two years, they should have a total of 6.
If I plug in numbers, none of the choices fit.
The answer provided is (A) but if I plug in T + 2 /J + 1 I end up having a non integer.
20 + 2 /5 +1 = 4.5
Any help? The numbers you chose are incorrect. If J = 5, T cannot be 20 because they are going to have another child in 2 yrs. Hence T can be 23 or 28 etc but not 20. If T = 23 and J = 5, T + 2 /J + 1 = (23 + 2)/5 + 1 = 6 which is correct Think of it this way: After 2 yrs, there will be another child. If they have a child every J years, it means (J  2) yrs have passed since they had their last child. Let's say T = (J2) + nJ n is the number of periods of J years that have passed since T was born. In each one of these periods, one child must have been born. Rearranging, n = (T+2)/J  1 so total number of children the couple will have after 2 yrs is n+2 (+2 to account for T and for the child that will be born 2 yrs from now) Total number of children = (T+2)/J  1 + 2 = (T+2)/J + 1
_________________
Karishma Veritas Prep GMAT Instructor
Learn more about how Veritas Prep can help you achieve a great GMAT score by checking out their GMAT Prep Options >



Manager
Joined: 08 Jun 2011
Posts: 81

Re: Mr. and Mrs. Wiley have a child every J years. Their oldest
[#permalink]
Show Tags
03 Nov 2011, 04:01
VeritasPrepKarishma wrote: Lstadt wrote: Sorry for pumping this old one up but I seem to never get it working when J =5 and T =20.
If they have a child every 5 years and their oldest is 20 years then this means that they have 1 child at year 0 2 children at year 5 3 children at year 10 4 children at year 15 5 children at year 20 They have 5 total children right now. After two years, they should have a total of 6.
If I plug in numbers, none of the choices fit.
The answer provided is (A) but if I plug in T + 2 /J + 1 I end up having a non integer.
20 + 2 /5 +1 = 4.5
Any help? The numbers you chose are incorrect. If J = 5, T cannot be 20 because they are going to have another child in 2 yrs. Hence T can be 23 or 28 etc but not 20. If T = 23 and J = 5, T + 2 /J + 1 = (23 + 2)/5 + 1 = 6 which is correct Think of it this way: After 2 yrs, there will be another child. If they have a child every J years, it means (J  2) yrs have passed since they had their last child. Let's say T = (J2) + nJ n is the number of periods of J years that have passed since T was born. In each one of these periods, one child must have been born. Rearranging, n = (T+2)/J  1 so total number of children the couple will have after 2 yrs is n+2 (+2 to account for T and for the child that will be born 2 yrs from now) Total number of children = (T+2)/J  1 + 2 = (T+2)/J + 1 Thank you very much. Please bear with me for a minute. I am still confused as how landed at 23 or 28. Did you choose these numbers to make sure that they end up being divisible by J? Since we must have an integer as the number of children?



Veritas Prep GMAT Instructor
Joined: 16 Oct 2010
Posts: 8998
Location: Pune, India

Re: Mr. and Mrs. Wiley have a child every J years. Their oldest
[#permalink]
Show Tags
04 Nov 2011, 00:18
Lstadt wrote: Thank you very much.
Please bear with me for a minute. I am still confused as how landed at 23 or 28. Did you choose these numbers to make sure that they end up being divisible by J? Since we must have an integer as the number of children? You chose J as 5. I only picked some values for T that would work with J = 5 How did I do that? After 2 yrs, there will be another child. It means 3 yrs have passed since their last child (since J = 5). If T was born in year 0, there would have been a child every 5 years. Right now, we are 3 yrs more than some multiple of 5. So T could be 3 yrs old (no child born after T)/8 yrs old (1 child born after T)/13 yrs old (2 children born after T) etc
_________________
Karishma Veritas Prep GMAT Instructor
Learn more about how Veritas Prep can help you achieve a great GMAT score by checking out their GMAT Prep Options >



Manager
Joined: 08 Jun 2011
Posts: 81

Re: Mr. and Mrs. Wiley have a child every J years. Their oldest
[#permalink]
Show Tags
04 Nov 2011, 10:14
VeritasPrepKarishma wrote: Lstadt wrote: Thank you very much.
Please bear with me for a minute. I am still confused as how landed at 23 or 28. Did you choose these numbers to make sure that they end up being divisible by J? Since we must have an integer as the number of children? You chose J as 5. I only picked some values for T that would work with J = 5 How did I do that? After 2 yrs, there will be another child. It means 3 yrs have passed since their last child (since J = 5). If T was born in year 0, there would have been a child every 5 years. Right now, we are 3 yrs more than some multiple of 5. So T could be 3 yrs old (no child born after T)/8 yrs old (1 child born after T)/13 yrs old (2 children born after T) etc Thank you so much. I now understand it. To me, it seems as though the wording of the problem is quite obscure because when the question says "They will have a child after two years, it didn't specify whether these two years are included in J or they are just some arbitrary years. That is, they suddenly decided to change their plan and up had a child two years after their last one.



Target Test Prep Representative
Status: Founder & CEO
Affiliations: Target Test Prep
Joined: 14 Oct 2015
Posts: 5381
Location: United States (CA)

Re: Mr. and Mrs. Wiley have a child every J years. Their oldest
[#permalink]
Show Tags
10 May 2018, 10:40
jainvineet wrote: Mr. and Mrs. Wiley have a child every J years. Their oldest child is now T years old. If they have a child 2 years from now, how many children will they have in total?
(A) \(\frac{T+2}{J} + 1\)
(B) \(JT + 1\)
(C) \(\frac{J}{T} + \frac{1}{T}\)
(D) \(TJ  1\)
(E) \(\frac{T+J}{J}\)
Source: Manhattan Guide We can let the oldest child be 4 years old and they have a child every 3 years. Thus, T = 4 and J = 3. In this case, 2 years from now, they will have 3 children all together. That is because the oldest child was born 4 years ago, another was born 1 year ago and a new baby will be born 2 years from now. We see that we can obtain the number of children, 3, by adding 1 to (4 + 2)/3. Since 4 is really T and the denominator 3 is really J, the number of children they will have 2 years from now is [(T + 2)/J] + 1. Answer: A
_________________
5star rated online GMAT quant self study course See why Target Test Prep is the top rated GMAT quant course on GMAT Club. Read Our Reviews



Intern
Joined: 24 Jan 2019
Posts: 17
Location: India
GPA: 3.94

Re: Mr. and Mrs. Wiley have a child every J years. Their oldest
[#permalink]
Show Tags
27 Feb 2019, 09:23
Solved algebraic method but to ensure i add +1 used number picking method.
_________________
***Please give Kudos. Kudos help win legendary GMAT club tests***




Re: Mr. and Mrs. Wiley have a child every J years. Their oldest
[#permalink]
27 Feb 2019, 09:23






