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# Aunt Marge has 20 pieces of candy to disperse among her

Author Message
CEO
Joined: 21 Jan 2007
Posts: 2736

Kudos [?]: 1022 [0], given: 4

Location: New York City
Aunt Marge has 20 pieces of candy to disperse among her [#permalink]

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13 Nov 2007, 14:14
This topic is locked. If you want to discuss this question please re-post it in the respective forum.

Aunt Marge has 20 pieces of candy to disperse among her nephews and nieces.

R gets 2 more than K
B gets 6 less than M
M gets 2 more than R
K gets 2 more than B

How many does K get?

Kudos [?]: 1022 [0], given: 4

SVP
Joined: 05 Jul 2006
Posts: 1741

Kudos [?]: 418 [0], given: 49

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13 Nov 2007, 14:19
b m k r
x-4 x+2 x-2 x

4x-4 = 20

x = 4

k = 2

Kudos [?]: 418 [0], given: 49

Intern
Joined: 16 Apr 2007
Posts: 26

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

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13 Nov 2007, 17:40

R = K + 2
B = M - 6
M = R + 2
K = B + 2

So it will
B K R M

if B gets x, we can see that the difference is 2 between each
x + (x+2) + (x+4) + (x+6) = 20
Solve for x = 2

So K = x+2 = 4

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

VP
Joined: 09 Jul 2007
Posts: 1098

Kudos [?]: 139 [0], given: 0

Location: London
Re: easy algebra - aunt marge [#permalink]

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13 Nov 2007, 18:03
bmwhype2 wrote:
Aunt Marge has 20 pieces of candy to disperse among her nephews and nieces.

R gets 2 more than K
B gets 6 less than M
M gets 2 more than R
K gets 2 more than B

How many does K get?

R=k+2
b=m-6--> b=k-2
M=2+r--> m=4+k
k=2+b

r+b+k+m=20
2k+2+2k+2=20
4k=16
k=4

Kudos [?]: 139 [0], given: 0

Re: easy algebra - aunt marge   [#permalink] 13 Nov 2007, 18:03
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