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

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

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07 Oct 2011, 22:23
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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
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Joined: 20 Dec 2010
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08 Oct 2011, 00:35
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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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Posts: 39588

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22 Apr 2012, 21:44
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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

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.
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Joined: 14 Oct 2011
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22 Apr 2012, 19:27
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

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23 Apr 2012, 10:17
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?
Math Expert
Joined: 02 Sep 2009
Posts: 39588
Xander, Yolanda, and Zelda each have at least one hat. Zelda [#permalink]

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23 Apr 2012, 10:37
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.
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GMAT 3: 790 Q51 V49
GPA: 3.3

### Show Tags

18 Oct 2013, 03:45
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

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

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21 Aug 2014, 05:52
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.

Intern
Joined: 07 Mar 2014
Posts: 4
Re: Xander, Yolanda, and Zelda each have at least one hat. Zelda [#permalink]

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21 Aug 2014, 11:09
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...
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Joined: 09 Sep 2013
Posts: 15912
Re: Xander, Yolanda, and Zelda each have at least one hat. Zelda [#permalink]

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12 Jan 2016, 04:28
Hello from the GMAT Club BumpBot!

Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).

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Re: Xander, Yolanda, and Zelda each have at least one hat. Zelda   [#permalink] 12 Jan 2016, 04:28
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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: 44
Xander, Yolanda, and Zelda each have at least one hat. Zelda [#permalink]

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07 Oct 2011, 22:23
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Difficulty:

85% (hard)

Question Stats:

51% (02:54) correct 49% (01:46) wrong based on 197 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
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Joined: 20 Dec 2010
Posts: 2010

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08 Oct 2011, 00:35
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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Posts: 39588

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22 Apr 2012, 21:44
1
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Expert's post
1
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BOOKMARKED
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

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.
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Joined: 14 Oct 2011
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22 Apr 2012, 19:27
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

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Posts: 298
Location: Kuwait
Schools: Sloan '16 (M)
GMAT 1: 710 Q49 V37
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### Show Tags

23 Apr 2012, 10:17
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?
Math Expert
Joined: 02 Sep 2009
Posts: 39588
Xander, Yolanda, and Zelda each have at least one hat. Zelda [#permalink]

### Show Tags

23 Apr 2012, 10:37
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.
_________________
Director
Status: Verbal Forum Moderator
Joined: 17 Apr 2013
Posts: 613
Location: India
GMAT 1: 710 Q50 V36
GMAT 2: 750 Q51 V41
GMAT 3: 790 Q51 V49
GPA: 3.3

### Show Tags

18 Oct 2013, 03:45
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

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.
_________________

Like my post Send me a Kudos It is a Good manner.
My Debrief: http://gmatclub.com/forum/how-to-score-750-and-750-i-moved-from-710-to-189016.html

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

### Show Tags

21 Aug 2014, 05:52
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.

Intern
Joined: 07 Mar 2014
Posts: 4
Re: Xander, Yolanda, and Zelda each have at least one hat. Zelda [#permalink]

### Show Tags

21 Aug 2014, 11:09
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...
GMAT Club Legend
Joined: 09 Sep 2013
Posts: 15912
Re: Xander, Yolanda, and Zelda each have at least one hat. Zelda [#permalink]

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12 Jan 2016, 04:28
Hello from the GMAT Club BumpBot!

Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).

Want to see all other topics I dig out? Follow me (click follow button on profile). You will receive a summary of all topics I bump in your profile area as well as via email.
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Re: Xander, Yolanda, and Zelda each have at least one hat. Zelda   [#permalink] 12 Jan 2016, 04:28
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