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 Your Progress

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

Not interested in getting valuable practice questions and articles delivered to your email? No problem, unsubscribe here.

It appears that you are browsing the GMAT Club forum unregistered!

Signing up is free, quick, and confidential.
Join other 350,000 members and get the full benefits of GMAT Club

Registration gives you:

Tests

Take 11 tests and quizzes from GMAT Club and leading GMAT prep companies such as Manhattan GMAT,
Knewton, and others. All are free for GMAT Club members.

Applicant Stats

View detailed applicant stats such as GPA, GMAT score, work experience, location, application
status, and more

Books/Downloads

Download thousands of study notes,
question collections, GMAT Club’s
Grammar and Math books.
All are free!

Thank you for using the timer!
We noticed you are actually not timing your practice. Click the START button first next time you use the timer.
There are many benefits to timing your practice, including:

Mr. and Mrs. Wiley have a child every J years. Their oldest [#permalink]
19 Feb 2007, 11:40

1

This post was BOOKMARKED

00:00

A

B

C

D

E

Difficulty:

55% (hard)

Question Stats:

61% (02:40) correct
39% (01:38) wrong based on 237 sessions

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?

Re: Mr. and Mrs. Wiley have a child every J years. Their oldest [#permalink]
06 Feb 2010, 12:17

1

This post received KUDOS

5

This post was BOOKMARKED

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 OA|Please underline your SC questions while posting|Try posting the explanation along with your answer choice| |For CR refer Powerscore CR Bible|For SC refer Manhattan SC Guide|

~~Better Burn Out... Than Fade Away~~

Last edited by Bunuel on 22 Nov 2012, 04:41, edited 1 time in total.

Re: Mr. and Mrs. Wiley have a child every J years. Their oldest [#permalink]
06 Feb 2010, 12:35

1

This post received KUDOS

1

This post was BOOKMARKED

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?

Re: Mr. and Mrs. Wiley have a child every J years. Their oldest [#permalink]
13 Feb 2010, 01:03

1

This post received KUDOS

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)

Re: Mr. and Mrs. Wiley have a child every J years. Their oldest [#permalink]
03 Nov 2011, 02:15

1

This post received KUDOS

Expert's post

1

This post was BOOKMARKED

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 = (J-2) + 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.

Re-arranging, 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 _________________

Re: Mr. and Mrs. Wiley have a child every J years. Their oldest [#permalink]
03 Nov 2011, 23:18

1

This post received KUDOS

Expert's post

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 _________________

Re: Mr. and Mrs. Wiley have a child every J years. Their oldest [#permalink]
21 Nov 2012, 09:45

1

This post received KUDOS

Expert's post

2

This post was BOOKMARKED

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?

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. ) _________________

Re: Mr. and Mrs. Wiley have a child every J years. Their oldest [#permalink]
19 Feb 2007, 14:04

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?

Re: Mr. and Mrs. Wiley have a child every J years. Their oldest [#permalink]
19 Feb 2007, 17:56

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?

Re: Mr. and Mrs. Wiley have a child every J years. Their oldest [#permalink]
06 Feb 2010, 15:05

Their oldest kid is T y/o. From the question stem, their youngest kid alive is J-2 y/o. We need to know how many kids are there between T and J-2 spaced exactly J years apart. This is an arithmetic progression.

(J-2) + (n-1)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. _________________

Re: Mr. and Mrs. Wiley have a child every J years. Their oldest [#permalink]
13 Feb 2010, 00:21

BarneyStinson wrote:

Their oldest kid is T y/o. From the question stem, their youngest kid alive is J-2 y/o. We need to know how many kids are there between T and J-2 spaced exactly J years apart. This is an arithmetic progression.

(J-2) + (n-1)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 J-2 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.

Re: Mr. and Mrs. Wiley have a child every J years. Their oldest [#permalink]
02 Nov 2011, 10:08

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.

Re: Mr. and Mrs. Wiley have a child every J years. Their oldest [#permalink]
03 Nov 2011, 03:01

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 = (J-2) + 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.

Re-arranging, 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?

Re: Mr. and Mrs. Wiley have a child every J years. Their oldest [#permalink]
04 Nov 2011, 09:14

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.

Re: Mr. and Mrs. Wiley have a child every J years. Their oldest [#permalink]
31 Oct 2014, 22:33

Hello from the GMAT Club BumpBot!

Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).

Want to see all other topics I dig out? Follow me (click follow button on profile). You will receive a summary of all topics I bump in your profile area as well as via email. _________________

Re: Mr. and Mrs. Wiley have a child every J years. Their oldest [#permalink]
29 May 2015, 21:58

Hello from the GMAT Club BumpBot!

Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).

Want to see all other topics I dig out? Follow me (click follow button on profile). You will receive a summary of all topics I bump in your profile area as well as via email. _________________

Back to hometown after a short trip to New Delhi for my visa appointment. Whoever tells you that the toughest part gets over once you get an admit is...