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In an auditorium, 360 chairs are to be set up in a rectangular arrange

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In an auditorium, 360 chairs are to be set up in a rectangular arrange  [#permalink]

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New post 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  [#permalink]

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New post 22 Sep 2019, 00:12
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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  [#permalink]

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New post 22 Sep 2019, 00:21
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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  [#permalink]

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New post 12 Dec 2019, 08:26
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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  [#permalink]

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New post 08 Jan 2020, 19:41
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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  [#permalink]

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New post 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  [#permalink]

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New post 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  [#permalink]

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New post 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  [#permalink]

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New post 16 Mar 2020, 10:43
1
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  [#permalink]

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New post 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  [#permalink]

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New post 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   [#permalink] 20 Jun 2020, 06:19

In an auditorium, 360 chairs are to be set up in a rectangular arrange

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