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What if the oldest child was 6 years old?

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What if the oldest child was 6 years old?

by rrobiinn » Mon Jun 11, 2012 9:22 pm
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)(T+2)/J +1
(B)JT+1
(C)J/T+1/T
(D)TJ-1

I found the answer explanation from Manhattan GMAT Book in the following way:

6. If the Wileys have a child every 2 years and the oldest child is "12", they
have 7 children, ages 0 (just born), 2, 4, 6, 8, 10, and 12. If they have another
child 2 years from now, they will have 8 children. Test each answer choice to
find the one that yields the Target value of 8.

(A) (T + 2)/J + 1 = 12 + 2 + 1 = 8 CORRECT

* What if we supposed that the oldest child was 6 years old? I don't get why they supposed that the child was 12 years old. If we would have picked up any other number than "12", the answer would be different.
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by mathbyvemuri » Mon Jun 11, 2012 10:08 pm
It's a problem on Arithmetic progressions.
If the first term is A and last term is B in an increasing AP with common difference d,
then the number of terms = {(B-A)/d}+1
Just consider a small example:
2,4,6,8,10
here common diff = 2
first term = 2
last term = 10
number of terms = (10-2)/2 + 1 = 5
Now, come to the problem:
The youngest child will born 2 yrs from now and the eldest child is T yrs old now.
So after two years, the eldest child will be T+2 yrs old, the youngest child will be 0 yrs old and ages of all other children span in between these two (0 and T+2).
If we convert ages of the children in to a series, that looks like:
0,J,2J...(T+2)
Number of children = (T+2-0)/J +1 = (T+2)/J +1
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by GMATGuruNY » Tue Jun 12, 2012 1:31 am
Mr. and Mrs. Wiley have a child every J years. Their oldest child is now Tyears old. If
they have a child 2 years from now, how many children will they-have in total?
(A) {(T+2)/J} +1 (B)JT+1 (C)J/T + T/J (D)TJ-1 (E) (J+T)/J
Let J=2.
Let 2000 = year in which the first child is born.
Let now = 2006.
Since a child is born every 2 years, there are now 4 children -- born in 2000, 2002, 2004 and 2006 -- and the oldest age is T=6.
When a child is born 2 years from now, the number of children = 5. This is our target.

Now we plug J=2 and T=6 into the answers to see which yields our target of 5.

Only answer choice A works:
(T+2)/J + 1 = (6+2)/2 + 1 = 5.

The correct answer is A.

To solve algebraically:

Since the ages are all J years apart, they constitute a set of evenly spaced integers.

Number of evenly spaced integers = (biggest - smallest)/increment + 1.

Here, the increment -- the distance between successive values in the set -- is J.
Biggest = T+2 (the age of the oldest child 2 years from now)
Smallest = 0 (the age of the last child at its birth 2 years from now)
Number of children = (T+2 - 0)/J + 1 = (T+2)/J + 1.
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