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Xander, Yolanda, and Zelda each have at least one hat. Zelda [#permalink]
07 Oct 2011, 21:23

1

This post received KUDOS

00:00

A

B

C

D

E

Difficulty:

85% (hard)

Question Stats:

54% (02:36) correct
46% (01:38) wrong based on 113 sessions

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.

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

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

So, I think answer should be E. Can you please help.

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.

Xander, Yolanda, and Zelda each have at least one hat. Zelda [#permalink]
23 Apr 2012, 09:37

Expert's post

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

So, I think answer should be E. Can you please help.

{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 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

So, I think answer should be E. Can you please help.

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.

Answer: C.

Hope it's clear.

Yes, It was slightly time consuming question. _________________

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

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

So, I think answer should be E. Can you please help.

{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...

gmatclubot

Re: Xander, Yolanda, and Zelda each have at least one hat. Zelda
[#permalink]
21 Aug 2014, 10:09

Back to hometown after a short trip to New Delhi for my visa appointment. Whoever tells you that the toughest part gets over once you get an admit is...