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# Events & Promotions

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

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Current Student
Joined: 29 Jan 2005
Posts: 5244
Followers: 23

Kudos [?]: 175 [0], given: 0

Joan, Kylie, Lillian, and Miriam all celebrate their [#permalink]  18 Dec 2005, 01:25
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Question Stats:

0% (00:00) correct 0% (00:00) wrong based on 1 sessions
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
SVP
Joined: 24 Sep 2005
Posts: 1895
Followers: 11

Kudos [?]: 133 [1] , given: 0

Re: PS MGMAT 1-15 [#permalink]  18 Dec 2005, 04:32
1
KUDOS
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.
Re: PS MGMAT 1-15   [#permalink] 18 Dec 2005, 04:32
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