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# Joan, Kylie, Lillian, and Miriam all celebrate their

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Joined: 29 Jan 2005
Posts: 5208

Kudos [?]: 434 [0], given: 0

Joan, Kylie, Lillian, and Miriam all celebrate their [#permalink]

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18 Dec 2005, 02:25
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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?

A. 51
B. 52
C. 53
D. 54
E. 55

How can we attack this problem (4 variables/3 equations) without backsolving

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SVP
Joined: 24 Sep 2005
Posts: 1884

Kudos [?]: 377 [1], given: 0

Re: PS MGMAT 1-15 [#permalink]

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18 Dec 2005, 05:32
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GMATT73 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?

A. 51
B. 52
C. 53
D. 54
E. 55

How can we attack this problem (4 variables/3 equations) without backsolving

We don't need to solve for the 4 variables coz the question asks "which...could be"
Let a, b, c and d be the ages of J, K, L and M respectively.
we have:
a+2= b
b-3=c ---> c= a-1
d-1= a ---> d= a+1

---> the combined age= a+b+c+d= a+ (a+2) + (a-1)+ ( a+1)= 4a+2----> the correct answer must be one which is divided by 4 has remainder of 2. Only 54 satisfies ---> D it is.

Kudos [?]: 377 [1], given: 0

Re: PS MGMAT 1-15   [#permalink] 18 Dec 2005, 05:32
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# Joan, Kylie, Lillian, and Miriam all celebrate their

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