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In an auditorium, 360 chairs are to be set up in a rectangular arrange
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21 Sep 2019, 13:51
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In an auditorium, 360 chairs are to be set up in a rectangular arrangement with x rows of exactly y chairs each. If the only other restriction is that 10 < x < 25, how many different rectangular arrangements are possible? A. Four B. Five C. Six D. Eight E. Nine PS66602.01
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Re: In an auditorium, 360 chairs are to be set up in a rectangular arrange
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22 Sep 2019, 00:12
360 will have ; 2^3*3^2*5^1 ; 24 factors possible integers for array x*y where x has restrictions 10<x<25 ; 12,15,18,20,24 ; 5 options IMO B gmatt1476 wrote: In an auditorium, 360 chairs are to be set up in a rectangular arrangement with x rows of exactly y chairs each. If the only other restriction is that 10 < x < 25, how many different rectangular arrangements are possible?
A. Four B. Five C. Six D. Eight E. Nine
PS66602.01



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Re: In an auditorium, 360 chairs are to be set up in a rectangular arrange
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22 Sep 2019, 00:21
gmatt1476 wrote: In an auditorium, 360 chairs are to be set up in a rectangular arrangement with x rows of exactly y chairs each. If the only other restriction is that 10 < x < 25, how many different rectangular arrangements are possible?
A. Four B. Five C. Six D. Eight E. Nine
PS66602.01 x and y are positive integers such that 360 = xy. The rephrased question: How many factors does 360 have between 10 and 25, exclusive? Since \(360=2^3\cdot 3^2\cdot 5\), we can quickly list the suitable factors. \(3\cdot5=15\) \(2^2\cdot5=20\) \(2\cdot3^2=18\) \(2^2\cdot3=12\) \(2^3\cdot3=24\) Answer: B
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Re: In an auditorium, 360 chairs are to be set up in a rectangular arrange
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12 Dec 2019, 08:26
gmatt1476 wrote: In an auditorium, 360 chairs are to be set up in a rectangular arrangement with x rows of exactly y chairs each. If the only other restriction is that 10 < x < 25, how many different rectangular arrangements are possible?
A. Four B. Five C. Six D. Eight E. Nine
PS66602.01 From the given information, the TOTAL number of chairs = xy This means: xy = 360 Since x and y must be POSITIVE INTEGERS, there is a finite number of possibilities. To help us list the pairs of values with a product of 360, let's find the prime factorization of 360 360 = (2)(2)(2)(3)(3)(5) When we consider the fact that 10 < x < 25, the possibilities are: x = 12 & y = 30 x = 15 & y = 24 x = 18 & y = 20 x = 20 & y = 18 x = 24 & y = 15 There are five such possibilities Answer: B Cheers, Brent
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Re: In an auditorium, 360 chairs are to be set up in a rectangular arrange
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08 Jan 2020, 19:41
gmatt1476 wrote: In an auditorium, 360 chairs are to be set up in a rectangular arrangement with x rows of exactly y chairs each. If the only other restriction is that 10 < x < 25, how many different rectangular arrangements are possible?
A. Four B. Five C. Six D. Eight E. Nine
PS66602.01 We need to find integers x and y such that xy = 360 and 10 < x < 25. Since 360 = 12 * 30 = 15 * 24 = 18 * 20 = 20 * 18 = 24 * 15 We see that x can be 12, 15, 18, 20, or 24. Therefore, there are 5 different arrangements. Answer: B
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In an auditorium, 360 chairs are to be set up in a rectangular arrange
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Updated on: 13 Mar 2020, 07:14
GMATPrepNow wrote: gmatt1476 wrote: In an auditorium, 360 chairs are to be set up in a rectangular arrangement with x rows of exactly y chairs each. If the only other restriction is that 10 < x < 25, how many different rectangular arrangements are possible?
A. Four B. Five C. Six D. Eight E. Nine
PS66602.01 From the given information, the TOTAL number of chairs = xy This means: xy = 360 Since x and y must be POSITIVE INTEGERS, there is a finite number of possibilities. To help us list the pairs of values with a product of 360, let's find the prime factorization of 360 360 = (2)(2)(2)(3)(3)(5) When we consider the fact that 10 < x < 25, the possibilities are: x = 12 & y = 30 x = 15 & y = 24 x = 18 & y = 20 x = 20 & y = 18 x = 24 & y = 15 There are five such possibilities Answer: B Cheers, Brent ScottTargetTestPrep This question is simple but took a bit of time in order to avoid missing any values of x. While I know how to solve it instantly by taking factor pairs, it did take me 3 min for me to double check to make sure I didn't miss any value of x in that range. Is there a quick to find all values of x that satisfy the condition 10 < x < 25 since we can have (3+1)(2+1)(1+1) = 24 values of x? Thanks!
Originally posted by dabaobao on 13 Mar 2020, 05:49.
Last edited by dabaobao on 13 Mar 2020, 07:14, edited 1 time in total.



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Re: In an auditorium, 360 chairs are to be set up in a rectangular arrange
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13 Mar 2020, 06:20
gmatt1476 wrote: In an auditorium, 360 chairs are to be set up in a rectangular arrangement with x rows of exactly y chairs each. If the only other restriction is that 10 < x < 25, how many different rectangular arrangements are possible?
A. Four B. Five C. Six D. Eight E. Nine
PS66602.01 Possible values of (X , Y) are  1. ( 12 , 30) 2. ( 15 , 24) 3. ( 18 , 20) 4. ( 20, 18 ) 5. (24 , 15 ) Hence, 5 arrangements are possible, correct Answer must be (B)
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Re: In an auditorium, 360 chairs are to be set up in a rectangular arrange
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13 Mar 2020, 15:35
X rows by Y chairs = X*Y X*Y = 360
Prime factors of 360 = 2*2*2*3*3*5 X values allowed 10<x<25 = 11 through 24 values of X that you can build with the prime factors of 360 12, 15, 18, 20, 24
5 values of x



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Re: In an auditorium, 360 chairs are to be set up in a rectangular arrange
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16 Mar 2020, 10:43
dabaobao wrote: GMATPrepNow wrote: gmatt1476 wrote: In an auditorium, 360 chairs are to be set up in a rectangular arrangement with x rows of exactly y chairs each. If the only other restriction is that 10 < x < 25, how many different rectangular arrangements are possible?
A. Four B. Five C. Six D. Eight E. Nine
PS66602.01 From the given information, the TOTAL number of chairs = xy This means: xy = 360 Since x and y must be POSITIVE INTEGERS, there is a finite number of possibilities. To help us list the pairs of values with a product of 360, let's find the prime factorization of 360 360 = (2)(2)(2)(3)(3)(5) When we consider the fact that 10 < x < 25, the possibilities are: x = 12 & y = 30 x = 15 & y = 24 x = 18 & y = 20 x = 20 & y = 18 x = 24 & y = 15 There are five such possibilities Answer: B Cheers, Brent ScottTargetTestPrep This question is simple but took a bit of time in order to avoid missing any values of x. While I know how to solve it instantly by taking factor pairs, it did take me 3 min for me to double check to make sure I didn't miss any value of x in that range. Is there a quick to find all values of x that satisfy the condition 10 < x < 25 since we can have (3+1)(2+1)(1+1) = 24 values of x? Thanks! Since you know there are (3+1)(2+1)(1+1) = 24 possible values of x, I understand you already factored 360 = 2^3 * 3^2 * 5. First of all, you definitely don't need to consider all 24 possibilities for x since we are only interested in the values of x that satisfy 10 < x < 25. In order to make sure we are not missing any factor pairs, we will find all values of x in the interval 10 < x < 25 that are a factor of 360. We already have the prime factorization 360 = 2^3 * 3^2 * 5; therefore we can immediately eliminate the primes in the interval 10 < x < 25 which do not appear in the prime factorization of 360. Namely, we eliminate the values x = 11, 13, 17, 19 and 23. Next, we can eliminate any multiples of 7 and 11 as well (since 360 cannot be divisible by any multiple of 7 or 11). Thus, the values x = 14, x = 21 and x = 22 are also eliminated. We are left with x = 12, 15, 16, 18, 20, 24. At this point, we can simply check these six values and determine that 360 is divisible by all except 16 (too few factors of 2 in 360 to be divisible by 16). This is the fastest way I can come up with which will guarantee that you won't miss any factor pairs.
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Re: In an auditorium, 360 chairs are to be set up in a rectangular arrange
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19 Jun 2020, 01:57
We need to find integers x and y such that xy = 360 and 10 < x < 25.
360 = \(2^3 3^2 5\)
Any value of x that contains a prime factor that is not 2, 3, or 5 will not satisfy the condition of xy = 360. Any value of x that contains a higher order prime than those in 360 will also not satisfy the condition of xy = 360. For example, 16 = \(2^4\) and hence is not a value of x. 14 = 2 x 7 and hence is not a value of x.
x = 11, 12, 13, 14, 15, 16, 17, 18, 19, 20, 21, 22, 23, 24
Number of rectangular arrangements = 5 > (B)
Hope it helps.



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Re: In an auditorium, 360 chairs are to be set up in a rectangular arrange
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20 Jun 2020, 06:19
Values of x which divide 360 > 12,15,18,20,24
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Re: In an auditorium, 360 chairs are to be set up in a rectangular arrange
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