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16 Dec 2021, 10:35
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B
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Difficulty:
55%
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Question Stats:
65% (02:09) correct
35%
(02:10)
wrong
based on 165
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10, 3, 12, 5, 6, \(x\)
What is the value of \(x\) in the above set of numbers?
(1) The median of the set is 2 less than the range of the set.
(2) the median of the numbers in the set is 7.
What is the value of \(x\) in the above set of numbers?
(1) The median of the set is 2 less than the range of the set.
(2) the median of the numbers in the set is 7.
19 Dec 2021, 00:43
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10, 3, 12, 5, 6, x
What is the value of in the above set of numbers?
(1) The median of the set is 2 less than the range of the set.
X = 5 range is 9
Median 5.5
X = 7
Median
6.5
Range 9
X= 8
Median 7
Range 9
X=13
Range 10
Median 8
Hence insufficient as we get two values for x.
(2) the median of the numbers in the set is 7.
So x would be 8 for median to be 7.
Sufficient.
Answer
B
Posted from my mobile device
What is the value of in the above set of numbers?
(1) The median of the set is 2 less than the range of the set.
X = 5 range is 9
Median 5.5
X = 7
Median
6.5
Range 9
X= 8
Median 7
Range 9
X=13
Range 10
Median 8
Hence insufficient as we get two values for x.
(2) the median of the numbers in the set is 7.
So x would be 8 for median to be 7.
Sufficient.
Answer
B
Posted from my mobile device
19 Dec 2021, 02:09
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If we arrange the set in ascending order -
_ 3 _ 5 _ 6 _ 10 _ 12 _
x can take any of places indicated by dash.
Another observation - the set consists of even number of terms, so the median would be the average of the middle two terms.
Statement 1
Case 1
Lets assume that x does not occupy the first or the last position. Therefore the available positions that x can take -
3 _ 5 _ 6 _ 10 _ 12
Range = 10 - 3 = 9
Median = 7
For median to be 7
x = 8
Case 2
Lets assume that x occupies the first position
Range = 12 - x
Median = 5.5
12 - x = 5.5 + 2
x = 4.5
This is not possible, hence disregard this case.
Case 3
Lets assume that x occupies the last position
Range = x - 3
Median = 8
12 - x = 8 + 2
x = 13
Hence we have two possible values of X.
Therefore statement 1 is not sufficient.
Statement 2
Median is 7
Now 7 is not present in the set. The possible position of the x can be between 6 and 10 (because we saw that median is 8 when 6 & 10 are in the center, so for median 7, x has to be less than 10 and greater than 6)
For median to be 7, x = 8
Hence sufficient.
IMO - B
_ 3 _ 5 _ 6 _ 10 _ 12 _
x can take any of places indicated by dash.
Another observation - the set consists of even number of terms, so the median would be the average of the middle two terms.
Statement 1
Case 1
Lets assume that x does not occupy the first or the last position. Therefore the available positions that x can take -
3 _ 5 _ 6 _ 10 _ 12
Range = 10 - 3 = 9
Median = 7
For median to be 7
x = 8
Case 2
Lets assume that x occupies the first position
Range = 12 - x
Median = 5.5
12 - x = 5.5 + 2
x = 4.5
This is not possible, hence disregard this case.
Case 3
Lets assume that x occupies the last position
Range = x - 3
Median = 8
12 - x = 8 + 2
x = 13
Hence we have two possible values of X.
Therefore statement 1 is not sufficient.
Statement 2
Median is 7
Now 7 is not present in the set. The possible position of the x can be between 6 and 10 (because we saw that median is 8 when 6 & 10 are in the center, so for median 7, x has to be less than 10 and greater than 6)
For median to be 7, x = 8
Hence sufficient.
IMO - B
