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emmak
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If x is the median of the set {9/2, 11/3, 28/9, 21/5, x}, x Updated on: 16 Feb 2013, 03:52
Originally posted by emmak on 15 Feb 2013, 12:01.
Last edited by Bunuel on 16 Feb 2013, 03:52, edited 1 time in total.
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C
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77% (01:45) correct 23% (02:24) wrong based on 203 sessions

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If x is the median of the set {9/2, 11/3, 28/9, 21/5, x}, x could be

A. 16/5
B. 17/5
C. 4
D. 30/7
E. 31/7
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C
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If x is the median of the set {9/2, 11/3, 28/9, 21/5, x}, x 15 Feb 2013, 19:20
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emmak
If x is the median of the set {9/2, 11/3, 28/9, 21/5, x}, x could be

16/5

17/5

4

30/7

31/7
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The median is the middle number once all the numbers are arranged in increasing/decreasing order.
We see that
11/3 = 3.something,
28/9 = 3.something
21/5 = 4.something
9/2 = 4.something

So x should greater than the smallest two numbers and smaller than the greatest two numbers. We can see that x = 4 is possible. (First look at the simplest option or the middle option since options are usually arranged in increasing/decreasing order)
Answer (C)
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If x is the median of the set {9/2, 11/3, 28/9, 21/5, x}, x 29 Mar 2013, 10:29
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VeritasPrepKarishma


The median is the middle number once all the numbers are arranged in increasing/decreasing order.
We see that
11/3 = 3.something,
28/9 = 3.something
21/5 = 4.something
9/2 = 4.something

So x should greater than the smallest two numbers and smaller than the greatest two numbers. We can see that x = 4 is possible. (First look at the simplest option or the middle option since options are usually arranged in increasing/decreasing order)
Answer (C)
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The median could be some number such as 4.01 or 3.99, in that we would have to test answer choices too ? Am I right ?
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If x is the median of the set {9/2, 11/3, 28/9, 21/5, x}, x 29 Mar 2013, 10:59
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emmak
If x is the median of the set {9/2, 11/3, 28/9, 21/5, x}, x could be

A. 16/5
B. 17/5
C. 4
D. 30/7
E. 31/7
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step 1

get all set members expressed in terms of a common denominator , i.e. 90

thus the set becomes (405/90 , 330/90 , 280/90 , 378/90 , 90x /90)

so 90x is the median of set (405,330,280,378)
reorder set (405 , 378,330,280) for 90x to be median of the set it has to be between 378 and 330

looking at number choices
d,e are excluded as 7 is no a factor of 90 , and then we are left with A,B,C

A - 16/5*90 = 288
B = 17/5*90 = 306
C = 90*4 = 360

Obviously then , C it is
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If x is the median of the set {9/2, 11/3, 28/9, 21/5, x}, x 30 Mar 2013, 03:40
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jainpiyushjain
VeritasPrepKarishma


The median is the middle number once all the numbers are arranged in increasing/decreasing order.
We see that
11/3 = 3.something,
28/9 = 3.something
21/5 = 4.something
9/2 = 4.something

So x should greater than the smallest two numbers and smaller than the greatest two numbers. We can see that x = 4 is possible. (First look at the simplest option or the middle option since options are usually arranged in increasing/decreasing order)
Answer (C)

The median could be some number such as 4.01 or 3.99, in that we would have to test answer choices too ? Am I right ?
Show more

Yes, x could be 3.99, 4.01 etc but x could be 4 too. The question says 'x could be', which means there are various possible values of x but only one value is given in the five options. Since we see that 4 is already there and x can certainly be 4, it means we don't need to test any other options because they cannot be the value of x. After all, PS questions have only one correct answer.
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If x is the median of the set {9/2, 11/3, 28/9, 21/5, x}, x 01 Apr 2013, 07:48
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emmak
If x is the median of the set {9/2, 11/3, 28/9, 21/5, x}, x could be

A. 16/5
B. 17/5
C. 4
D. 30/7
E. 31/7
Show more
\(9/2 = 4.5\)
\(11/3= 3.66\)
\(28/9 = 3.13\)
\(21/5 = 4.2\)

Since the set has 5 numbers the median has to be the middle value
Arranging the number in ascending order we get 3.13, 3.66, 4.2, 4.5. Now for x to be the median of the given set, x has to lie between 3.66 and 4.2.
Only one option satisfies the condition i.e x =4
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If x is the median of the set {9/2, 11/3, 28/9, 21/5, x}, x 03 Apr 2019, 12:07
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Convert to proper fractions and then to decimal equivalents:

9/2 = 4 R1/2 = 4.5
11/3 = 3 R2/3 = 3.66
28/9 = 3 R1/9 = 3.11
21/5 = 4 R1/5 = 4.2

So, x is between 3.66 and 4.2.

Check answers (one should be obvious at this point):

A) 16/5 = 3 R1/5 = 3.2, too low
B) 17/5 = 3.4, too low
C) 4 is between 3.6 and 4.2...
D) 30/7 = 4 R2/7 = 4.28, too high

Thus, it's C
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If x is the median of the set {9/2, 11/3, 28/9, 21/5, x}, x 18 Mar 2023, 11:41
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