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

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by BTGmoderatorLU » Tue May 08, 2018 2:09 pm
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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?

$$A.\ \frac{T+2}{J}+1$$
$$B.\ JT+1$$
$$C.\ \frac{J}{T}+\frac{1}{T}$$
$$D.\ TJ-1$$
$$E.\ \frac{T+J}{J}$$

The OA is A.

I get the solution as follows,

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.

I would like to know how can I solve this PS question with an algebraic method. Please, can anyone explain an algebraic method to solve it? Thanks!
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by Jay@ManhattanReview » Tue May 08, 2018 8:48 pm
BTGmoderatorLU 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?

$$A.\ \frac{T+2}{J}+1$$
$$B.\ JT+1$$
$$C.\ \frac{J}{T}+\frac{1}{T}$$
$$D.\ TJ-1$$
$$E.\ \frac{T+J}{J}$$

The OA is A.

I get the solution as follows,

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.

I would like to know how can I solve this PS question with an algebraic method. Please, can anyone explain an algebraic method to solve it? Thanks!
The oldest child is T years old; the oldest would-be (T + 2) years old when the youngest child is born.

Every J years, Mr. and Mrs. Wiley have a child. Thus, (T + 2) must be divisible by J.

(T + 2)/J is the number of children except the oldest one.

Thus, the total number of children, incl. the oldest child = (T + 2)/J + 1

The correct answer: A

Hope this helps!

-Jay
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by GMATGuruNY » Wed May 09, 2018 2:26 am
BTGmoderatorLU 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?

$$A.\ \frac{T+2}{J}+1$$
$$B.\ JT+1$$
$$C.\ \frac{J}{T}+\frac{1}{T}$$
$$D.\ TJ-1$$
$$E.\ \frac{T+J}{J}$$
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 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 an EVENLY SPACED SET.
For any evenly spaced set:
Count = (biggest - smallest)/(increment) + 1.
The increment is the distance between one term and the next.

In the problem above:
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)
Increment = J.

Thus:
Number of children = (T+2 - 0)/J + 1 = (T+2)/J + 1.
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by Scott@TargetTestPrep » Thu May 10, 2018 4:51 pm
BTGmoderatorLU 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?

$$A.\ \frac{T+2}{J}+1$$
$$B.\ JT+1$$
$$C.\ \frac{J}{T}+\frac{1}{T}$$
$$D.\ TJ-1$$
$$E.\ \frac{T+J}{J}$$
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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