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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 06 Feb 2010, 12:17
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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

Please give the algebric solution.
[Reveal] Spoiler: OA

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Last edited by Bunuel on 22 Nov 2012, 04:41, edited 1 time in total.
Renamed the topic and edited the question.
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Re: Mr. and Mrs Wiley, VIC [#permalink] New post 06 Feb 2010, 12:35
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}

[Reveal] Spoiler: OA:
(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

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Re: Mr. and Mrs Wiley, VIC [#permalink] New post 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.
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Re: Mr. and Mrs Wiley, VIC [#permalink] New post 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.

Can you please explain. Thanks.

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Re: Mr. and Mrs Wiley, VIC [#permalink] New post 13 Feb 2010, 01:03
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
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Re: Mr. and Mrs Wiley, VIC [#permalink] New post 21 Nov 2012, 07:46
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}

[Reveal] Spoiler: OA:
(A)


Source: Manhattan Guide

Please give the algebric solution.


Bunuel - would really appreciate you providing your bit on solving the original problem above algebraically. The problem and various explanations remain confusing. Should we think of it as a progression or some other way? Please share your take. Thank you.
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Re: Mr. and Mrs Wiley, VIC [#permalink] New post 21 Nov 2012, 09:45
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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}

[Reveal] Spoiler: OA:
(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, VIC   [#permalink] 21 Nov 2012, 09:45
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