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 Post subject: Age problem
 Post Posted: Thu Jul 17, 2008 10:07 am 
Joan, Kylie, Lillian, and Miriam all celebrate their birthdays today. Joan is 2 years younger than Kylie, Kylie is 3 years older than Lillian, and Miriam is one year older than Joan. Which of the following could be the combined age of all four women today?
51
52
53
54
55
------------------------------------------------------------------------------------------------------------------------------------------------
Please could someone explain this question to me. I can not figure out that the ages represent consecutive integers as it says in its explaination.


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 Post subject:
 Post Posted: Fri Jul 18, 2008 1:55 pm 
I think the answer is: 54.

If we see the age of all 4 w.r.t Kylie's age, then it will look like below:
L J M Kylie
K-3 K-2 K-1 K.

Adding all = 4K-6 = 2(2K-3). Therefore the sum of tthe ages should be an even number. There are 2 options - 52 and 54.
4k-6 = 52 => K=29/2, but Kylie is celeberating her birthday today, so K should be a integer. So this option discarded.

4k-6=54 ==> k=15. So the answer.


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 Post subject: Re: Age problem
 Post Posted: Sat Jul 19, 2008 4:28 am 
Offline
ManhattanGMAT Staff


Posts: 7146
Capthan wrote:
Joan, Kylie, Lillian, and Miriam all celebrate their birthdays today. Joan is 2 years younger than Kylie, Kylie is 3 years older than Lillian, and Miriam is one year older than Joan. Which of the following could be the combined age of all four women today?
51
52
53
54
55
------------------------------------------------------------------------------------------------------------------------------------------------
Please could someone explain this question to me. I can not figure out that the ages represent consecutive integers as it says in its explaination.


you could also just guess numbers, especially because it's pretty easy to estimate how big the ages should be: the ages are pretty close together, and the total of all four is fifty-something. therefore, (fifty-something)/4 is around thirteen or so, so let's just randomly guess that joan (the first one mentioned - no special reason) is 13.
then kylie is 15, lillian is 12, and miriam is 14.
sum = 54.
lucky guess!

if we were to have guessed joan = 12, then we'd get 12, 14, 11, 13, for a sum of 50; that'd be too small, so we'd just crank up our estimate by 1 year for each girl, and, voilĂ , 54.

--

still, you should be able to generate algebraic expressions such as the ones in the previous post at a moment's notice; the ones in this problem should be absolutely routine. if you're having trouble generating such expressions, consider borrowing an algebra book from the local library and working the problems that introduce algebraic word translations (usually in the chapters that are introducing algebraic expressions like these for the first time). good times!


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 Post subject:
 Post Posted: Sat Jul 19, 2008 7:25 am 
In its explanation the ages are consecutive integers, they can all be expressed in term of Lillian or L. I am having hard time to express them in term of L.

My translation of the problem looks like

Joan is 2 yrs younger then Kylie: J + 2=K

Kylie is 3 yrs older then Lillian: k-3=L

Miriam is on yr older then Joan: m-1=j

Could u explain how can I go from here to form an expression in term of Lillian?


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 Post subject: Visual Method
 Post Posted: Sun Jul 27, 2008 6:05 pm 
Offline
ManhattanGMAT Staff


Posts: 901
Location: St. Louis, MO
Here's how I would set up the problem: Draw a number line the scrapboard, and use dots or letters to represent the ages.

Here's an attempt to reproduce it:

"Joan is 2 years younger than Kylie" becomes <---|----|----(J)----|----(K)----|----|---> (note that | represents integers)

"Kylie is 3 years older than Lillian lets me put Lillian on the line: <---|----(L)----(J)----|----(K)----|----|--->

"Miriam is 1 year older than Joan" puts M one notch to the right of J: <---|----(L)----(J)----(M)----(K)----|----|--->

This method might serve you better than the more abstract algebraic proof.

_________________
Emily Sledge
Instructor
ManhattanGMAT


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