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# Xander, Yolanda, and Zelda each have at least one hat. Zelda

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Intern
Joined: 23 Aug 2011
Posts: 39

Kudos [?]: 36 [1], given: 4

Xander, Yolanda, and Zelda each have at least one hat. Zelda [#permalink]

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07 Oct 2011, 21:23
1
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Difficulty:

85% (hard)

Question Stats:

52% (01:53) correct 48% (01:44) wrong based on 203 sessions

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Xander, Yolanda, and Zelda each have at least one hat. Zelda has more hats than Yolanda, who has more than Xander. Together, the total number of hats the three people have is 12. How many hats does Yolanda have?

(1) Zelda has no more than 5 hats more than Xander.
(2) The product of the numbers of hats that Xander, Yolanda, and Zelda have is less than 36.
[Reveal] Spoiler: OA

Kudos [?]: 36 [1], given: 4

Math Forum Moderator
Joined: 20 Dec 2010
Posts: 1949

Kudos [?]: 2140 [1], given: 376

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07 Oct 2011, 23:35
1
KUDOS
kkalyan wrote:
Xander, Yolanda, and Zelda each have at least one hat. Zelda has more hats than Yolanda, who has more than Xander. Together, the total number of hats the three people have is 12. How many hats does Yolanda have?

1. Zelda has no more than 5 hats more than Xander.
2. The product of the numbers of hats that Xander, Yolanda, and Zelda have is less than 36.

kkalyan: Please tag the source correctly. It is a Manhattan GMAT question, not a GMAT Prep question. Also, please try to give a unique name for the subject.

MGMAT DS: Hats owned by Xander, Yolanda, and Zelda

Z>Y>X
Z+Y+X=12

1) Z-X<=5
1, 5, 6
2, 4, 6
Not Sufficient.

2) X.Y.Z<36
1, 5, 6
1, 2, 9
Not Sufficient.

Combined;
{1, 5, 6}
Y=5

Ans: "C"
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Kudos [?]: 2140 [1], given: 376

Math Expert
Joined: 02 Sep 2009
Posts: 42604

Kudos [?]: 135673 [1], given: 12706

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22 Apr 2012, 20:44
1
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Expert's post
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anushapolavarapu wrote:
Hi fluke,

what about the combination {1,4,7}. even this satisfies the 2 conditions right.
1+4+7=12(x<y<z) and
1*4*7=28 which is < 36

Welcome to GMAT Club. Below is an answer to your doubt.

The case when x=1, y=4, and z=7 does not satisfy the first statement, which says that $$z-x\leq{5}$$ (Zelda has no more than 5 hats more than Xander). Also notice that there are some cases missing for (1) and (2) in fluke's solution.

Complete solution:

Xander, Yolanda, and Zelda each have at least one hat. Zelda has more hats than Yolanda, who has more than Xander. Together, the total number of hats the three people have is 12. How many hats does Yolanda have?

Given: x<y<z and x+y+z=12. Question: y=?

Now, only following 7 cases are possible;

X-Y-Z
1-2-9
1-3-8
1-4-7
1-5-6
2-3-7
2-4-6
3-4-5

(1) Zelda has no more than 5 hats more than Xander --> $$z-x\leq{5}$$ --> first 3 cases are out and only following cases are left: {1, 5, 6}, {2, 3, 7}, {2, 4, 6}, and {3, 4, 5}. Not sufficient.

(2) The product of the numbers of hats that Xander, Yolanda, and Zelda have is less than 36 --> last 3 cases are out and only following cases are left: {1, 2, 9}, {1, 3, 8}, {1, 4, 7}, and {1, 5, 6}. Not sufficient.

(1)+(2) There is only one case common for (1) and (2): {1, 5, 6}, so z=6. Sufficient.

Hope it's clear.
_________________

Kudos [?]: 135673 [1], given: 12706

Intern
Joined: 14 Oct 2011
Posts: 1

Kudos [?]: [0], given: 19

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22 Apr 2012, 18:27
Hi fluke,

what about the combination {1,4,7}. even this satisfies the 2 conditions right.
1+4+7=12(x<y<z) and
1*4*7=28 which is < 36

Kudos [?]: [0], given: 19

Senior Manager
Joined: 01 Apr 2010
Posts: 296

Kudos [?]: 59 [0], given: 11

Location: Kuwait
Schools: Sloan '16 (M)
GMAT 1: 710 Q49 V37
GPA: 3.2
WE: Information Technology (Consulting)

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23 Apr 2012, 09:17
anushapolavarapu wrote:
Hi fluke,

what about the combination {1,4,7}. even this satisfies the 2 conditions right.
1+4+7=12(x<y<z) and
1*4*7=28 which is < 36

{1,4,7}. Z has more than 5 hats than X does. Again great explanation above.

Do you suggest during these type of questions to spend time writing out all combination possibilities at the start?

Kudos [?]: 59 [0], given: 11

Math Expert
Joined: 02 Sep 2009
Posts: 42604

Kudos [?]: 135673 [0], given: 12706

Xander, Yolanda, and Zelda each have at least one hat. Zelda [#permalink]

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23 Apr 2012, 09:37
aalba005 wrote:
anushapolavarapu wrote:
Hi fluke,

what about the combination {1,4,7}. even this satisfies the 2 conditions right.
1+4+7=12(x<y<z) and
1*4*7=28 which is < 36

{1,4,7}. Z has more than 5 hats than X does. Again great explanation above.

Do you suggest during these type of questions to spend time writing out all combination possibilities at the start?

It depends how many possible combinations are there. Luckily there are only 7 for this question, so it's not hard to write them all down. In this case everything will be in front of you so you won't miss any case while solving.
_________________

Kudos [?]: 135673 [0], given: 12706

Director
Status: Verbal Forum Moderator
Joined: 17 Apr 2013
Posts: 599

Kudos [?]: 649 [0], given: 298

Location: India
GMAT 1: 710 Q50 V36
GMAT 2: 750 Q51 V41
GMAT 3: 790 Q51 V49
GPA: 3.3

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18 Oct 2013, 02:45
Bunuel wrote:
anushapolavarapu wrote:
Hi fluke,

what about the combination {1,4,7}. even this satisfies the 2 conditions right.
1+4+7=12(x<y<z) and
1*4*7=28 which is < 36

Welcome to GMAT Club. Below is an answer to your doubt.

The case when x=1, y=4, and z=7 does not satisfy the first statement, which says that $$z-x\leq{5}$$ (Zelda has no more than 5 hats more than Xander). Also notice that there are some cases missing for (1) and (2) in fluke's solution.

Complete solution:

Xander, Yolanda, and Zelda each have at least one hat. Zelda has more hats than Yolanda, who has more than Xander. Together, the total number of hats the three people have is 12. How many hats does Yolanda have?

Given: x<y<z and x+y+z=12. Question: y=?

Now, only following 7 cases are possible;

X-Y-Z
1-2-9
1-3-8
1-4-7
1-5-6
2-3-7
2-4-6
3-4-5

(1) Zelda has no more than 5 hats more than Xander --> $$z-x\leq{5}$$ --> first 3 cases are out and only following cases are left: {1, 5, 6}, {2, 3, 7}, {2, 4, 6}, and {3, 4, 5}. Not sufficient.

(2) The product of the numbers of hats that Xander, Yolanda, and Zelda have is less than 36 --> last 3 cases are out and only following cases are left: {1, 2, 9}, {1, 3, 8}, {1, 4, 7}, and {1, 5, 6}. Not sufficient.

(1)+(2) There is only one case common for (1) and (2): {1, 5, 6}, so z=6. Sufficient.

Hope it's clear.

Yes, It was slightly time consuming question.
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Kudos [?]: 649 [0], given: 298

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Status: Do till 740 :)
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Re: Xander, Yolanda, and Zelda each have at least one hat. Zelda [#permalink]

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21 Aug 2014, 04:52
Hi Bunuel,
I took 3 minutes to solve this problem.I took time to list down all the values XYZ with both constraints ie , XYZ <36 and X+Y+Z=12. how can i speed up in solving such problems? Is there a better way and while listing I get tensed too as i might miss some cases.

Kudos [?]: 13 [0], given: 19

Intern
Joined: 07 Mar 2014
Posts: 4

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

Re: Xander, Yolanda, and Zelda each have at least one hat. Zelda [#permalink]

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21 Aug 2014, 10:09
Bunuel wrote:
aalba005 wrote:
anushapolavarapu wrote:
Hi fluke,

what about the combination {1,4,7}. even this satisfies the 2 conditions right.
1+4+7=12(x<y<z) and
1*4*7=28 which is < 36

{1,4,7}. Z has more than 5 hats than X does. Again great explanation above.

Do you suggest during these type of questions to spend time writing out all combination possibilities at the start?

It depends how many possible combinations are there. Luckily there are only 7 for this question, so it's not hard to write them all down. In this case everything will be in front of you so you won't miss any case while solving.

What do you suggest if we weren't able to write down all possible answers ?!
For this particular question...

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

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Joined: 09 Sep 2013
Posts: 14833

Kudos [?]: 287 [0], given: 0

Re: Xander, Yolanda, and Zelda each have at least one hat. Zelda [#permalink]

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12 Jan 2016, 03:28
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Re: Xander, Yolanda, and Zelda each have at least one hat. Zelda   [#permalink] 12 Jan 2016, 03:28
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# Xander, Yolanda, and Zelda each have at least one hat. Zelda

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