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Mr. and Mrs. Wiley have a child every J years. Their oldest

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Mr. and Mrs. Wiley have a child every J years. Their oldest [#permalink] New post   19 Feb 2007, 12:40
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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
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Re: Mr. and Mrs. Wiley have a child every J years. Their oldest [#permalink] New post   21 Nov 2012, 10:45
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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) \(\frac{T+J}{J}\)

Show SpoilerOA:
(A)


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. )
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Re: Mr. and Mrs. Wiley have a child every J years. Their oldest [#permalink] New post   03 Nov 2011, 03:15
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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
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Re: Mr. and Mrs. Wiley have a child every J years. Their oldest [#permalink] New post   19 Feb 2007, 15:04
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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) TJ-1
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
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Re: Mr. and Mrs. Wiley have a child every J years. Their oldest [#permalink] New post   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) TJ-1
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



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 (x-1)J where x is the total number of kids.

T+2 = (x-1)J

Hence A. Not so QED.
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Re: Mr. and Mrs. Wiley have a child every J years. Their oldest [#permalink] New post   06 Feb 2010, 13:35
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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) \(\frac{T+J}{J}\)

Show SpoilerOA:
(A)


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
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Re: Mr. and Mrs. Wiley have a child every J years. Their oldest [#permalink] New post   06 Feb 2010, 16:05
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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.
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Re: Mr. and Mrs. Wiley have a child every J years. Their oldest [#permalink] New post   13 Feb 2010, 01:21
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.

Can you please explain. Thanks.

Cheers!
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Re: Mr. and Mrs. Wiley have a child every J years. Their oldest [#permalink] New post   13 Feb 2010, 02:03
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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
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Re: Mr. and Mrs. Wiley have a child every J years. Their oldest [#permalink] New post   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?
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Re: Mr. and Mrs. Wiley have a child every J years. Their oldest [#permalink] New post   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 = (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?
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Re: Mr. and Mrs. Wiley have a child every J years. Their oldest [#permalink] New post   04 Nov 2011, 00:18
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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
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Re: Mr. and Mrs. Wiley have a child every J years. Their oldest [#permalink] New post   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.
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Re: Mr. and Mrs. Wiley have a child every J years. Their oldest [#permalink] New post   10 May 2018, 10:40
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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
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Re: Mr. and Mrs. Wiley have a child every J years. Their oldest [#permalink] New post   27 Feb 2019, 09:23
Solved algebraic method but to ensure i add +1 used number picking method.
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Re: Mr. and Mrs. Wiley have a child every J years. Their oldest [#permalink] New post   31 May 2020, 02:17
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When you see all variables, it's better to test some values.
Let's assume they have kids after every 2 years. so J = 2.
and age of T = 10 years. So, they have had 6 kids till now. One at when T was 0 years old, then 2,4,6,8 and now 10.
After two years they will have another kid, so in total they will have 7 kids.

Now, let's move on to the options.

(A) \(\frac{T+2}{J} + 1 = \frac{12}{2} + 1 = 6+1 = 7 \). Keep it.

(B) \(JT + 1\) Very big. Don't even calculate. Eliminated.

(C) \(\frac{J}{T} + \frac{1}{T} = \frac{2}{12} + \frac{1}{12}\). Kids can't be in fractions. Eliminated

(D) \(TJ - 1\). Again very big. Eliminated.

(E) \(\frac{T+J}{J} = \frac{12}{2} = 6\). One less than what we need. Eliminated.

OA, A
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
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Re: Mr. and Mrs. Wiley have a child every J years. Their oldest [#permalink] New post   04 Jun 2020, 02:44
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


let J = 2 years
T = 10 years

currently, they have 6 children. After two years they will have 7 children.

the answer choice should give 7 when I substitute j for 2 and t for 10

\(\frac{T+2}{J} + 1\)

12/2 +1 = 7 hence correct.
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Re: Mr. and Mrs. Wiley have a child every J years. Their oldest [#permalink] New post   18 Oct 2022, 06:37
Hi,

I think there's come ambiguity wrt what the question says

I interpreted the question as "How many kids will they have in total (across their lifetime)"

I think we need to specify in the question with something such as "at that point in time"

Bunuel and other experts, thoughts?

Do let me know if it is simply an error I am making!

Thanks!
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Re: Mr. and Mrs. Wiley have a child every J years. Their oldest [#permalink] New post   29 Oct 2023, 06:47
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