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# Three children, Alice, Brian, and Chris have a total of

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Joined: 21 May 2011
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Three children, Alice, Brian, and Chris have a total of [#permalink]

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24 Jul 2011, 15:30
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Question Stats:

14% (00:00) correct 86% (00:28) wrong based on 15 sessions

Three children, Alice, Brian, and Chris have a total of $1.20 between them. Does Chris have the most money? (1) Alice has 35 cents. (2) Chris has 40 cents. Math Expert Joined: 02 Sep 2009 Posts: 38908 Followers: 7739 Kudos [?]: 106246 [1] , given: 11618 Re: Three children, Alice, Brian, and Chris have a total of [#permalink] ### Show Tags 28 Mar 2012, 00:59 1 This post received KUDOS Expert's post bschool83 wrote: Three children, Alice, Brian, and Chris have a total of$1.20 between them. Does Chris have the most money?

(1) Alice has 35 cents.

(2) Chris has 40 cents.

This is a poor quality question because of its ambiguous wording.

For statement (2) if all 3 children have 40 cents, does that mean that all of them have the most money or none of them have the most money? How are we supposed to treat ties?

So, I'd advice not to study this question at all.
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Re: Three children, Alice, Brian, and Chris have a total of [#permalink]

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29 Mar 2012, 05:16
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Expert's post
monu1101 wrote:
Answer should be C, what is the OA?

The explanation is in either of the case A or B, Chris may or may not have the most money.

Welcome to GMAt Club.

Please check this: three-children-alice-brian-and-chris-have-a-total-of-117635.html#p1066131 So as you can see it's a poorly designed question, and one can justify B as well as C for it, and that's exactly why this kind of question has zero chances of appearing on the real test.

By the way the OA for this question is B, though it's completely irrelevant since as discussed the question is quite ambiguous.

Hope it helps.
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Re: DS - 700 level - money [#permalink]

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24 Jul 2011, 20:01
bschool83 wrote:
Three children, Alice, Brian, and Chris have a total of \$1.20 between them. Does Chris have the most money?

(1) Alice has 35 cents.

(2) Chris has 40 cents.

$$A + B + C = 120$$, is $$C > A$$ and $$C>B$$?

1) $$A = 35$$, therefore $$B + C = 120 - 35 = 85$$.

It could be that $$C = 84$$ and $$A = 1$$, or that $$C = 1$$ and $$B = 84$$. $$C$$ is greater than both $$A$$ and $$B$$ in one scenario, but not in the other. Insufficient.

2) $$C = 40$$, therefore $$A + B = 120 - 40 = 80$$.

This means that the average of $$A$$ and $$B$$ is $$40$$, and either $$A = B = 40$$, or $$A > 40 > B$$, or $$B > 40 > A$$.

Either way, $$C$$ is NOT greater than both $$A$$ and $$B$$.

Sufficient.

[Reveal] Spoiler:
B
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Re: Three children, Alice, Brian, and Chris have a total of [#permalink]

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27 Mar 2012, 22:11
I have to disagree. In the case where each have 40c ie you have a set (40, 40, 40) they each individually have the most (ie the highest value = 40) and it so happens they each individually have the least, again 40.

Since this provides two cases, Brian having the most when they all share the most (similar to tied for 1st place - they are equally best) and Brian not having the most when any other values are chosen, one requires both (1) and (2) to determine if Chris does/doesn't have the most.

Clearly this is a definition debate around "most" and ties for most, and the question would likely (hopefully) be thrown out by the gmac folks!
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Re: Three children, Alice, Brian, and Chris have a total of [#permalink]

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29 Mar 2012, 03:58
Answer should be C, what is the OA?

The explanation is in either of the case A or B, Chris may or may not have the most money.
Re: Three children, Alice, Brian, and Chris have a total of   [#permalink] 29 Mar 2012, 03:58
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