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# S98-13

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Math Expert
Joined: 02 Sep 2009
Posts: 55271

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16 Sep 2014, 01:52
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Difficulty:

65% (hard)

Question Stats:

55% (02:39) correct 45% (02:24) wrong based on 51 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.

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Math Expert
Joined: 02 Sep 2009
Posts: 55271

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16 Sep 2014, 01:52
Official Solution:

Write $$x$$ for the number of hats Xander has, $$y$$ for the number of hats Yolanda has, and $$z$$ for the number of hats Zelda has. From the question stem, we know that $$x \lt y \lt z$$ and that $$x + y + z = 12$$. Moreover, since each person has at least one hat, and people can only have integer numbers of hats, we know that $$x$$, $$y$$, and $$z$$ are all positive integers. With this number of constraints, we should go ahead and list scenarios that fit all the constraints. Start with $$x$$ and $$y$$ as low as possible, then adjust from there, keeping the order, keeping the sum at 12, and ensuring that no two integers are the same.
Scenario $$x$$ $$y$$ $$z$$ (a) 1 2 9 (b) 1 3 8 (c) 1 4 7 (d) 1 5 6 (e) 2 3 7 (f) 2 4 6 (g) 3 4 5
These are the only seven scenarios that work. As a reminder, we are looking for the value of $$y$$. Now, we turn to the statements.

Statement (1): INSUFFICIENT. We are told that $$z - x$$ is less than or equal to 5. This rules out scenarios (a) through (c), but the last four scenarios still work. Thus, $$y$$ could be 3, 4, or 5.

Statement (2): INSUFFICIENT. We are told that $$xyz$$ is less than 36. We work out this product for the seven scenarios:

(a) 18

(b) 24

(c) 28

(d) 30

(e) 42

(f) 48

(g) 60

We can rule out scenarios (e) through (g), but (a) through (d) still work. Thus, $$y$$ could be 2, 3, 4, or 5.

Statements (1) and (2) together: SUFFICIENT. Only scenario (d) survives the constraints of the two statements. Thus, we know that $$y$$ is 5.

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Joined: 20 Apr 2014
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26 Jul 2016, 20:13
is there any shorter way to solve such question ?
Intern
Joined: 11 Oct 2016
Posts: 1

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21 Oct 2016, 23:12
I think this is a high-quality question and I agree with explanation. There seems to be a bug in the average times shown in the Question Stats tab.

Average time for inccorect response shows 1246 minutes :D
Intern
Joined: 28 Dec 2017
Posts: 1

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02 Jan 2018, 19:24
hatemnag wrote:
is there any shorter way to solve such question ?

I did it this way:

(2) is obviously not enough to solve the question since all combos where x is 1 will work so you can immediately look at (1). Make note that combos where x is bigger than 1 won't work though.

(1) this means z <= 5 + x, this is also clearly not enough to figure out the question on its own

With both then you know that x must be 1 because of (2). Now you have z <= 6 because of (1). Now it should be apparent that only 1 way will work to add the numbers to 12 (since if z were 5 then 1+4+5 < 12 so z must be 6)
Re: S98-13   [#permalink] 02 Jan 2018, 19:24
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# S98-13

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