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24 May 2020, 22:20
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In a list of 7 integers, one integer, denoted as x is unknown. The other six integers are 20, 4, 10, 4, 8, and 4. If the mean, median, and mode of these seven integers are arranged in increasing order, they form an arithmetic progression. The sum of all possible values of x is
A) 26
B) 32
C) 34
D) 38
E) 40
A) 26
B) 32
C) 34
D) 38
E) 40
26 May 2020, 23:49
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ARIEN3228
Let us arrange the known numbers in increasing order..
4, 4, 4, 8, 10, 20...x can be anywhere
So Mode will always be 4 irrespective of value of x...
The Median will depend on x.
Mean = \(\frac{50+x}{7}\)
Case I - \(x\leq 4\)
x, 4, 4, 4, 8, 10, 20
Mode = Median=4
Mean = \(\frac{50+x}{7}\)....
Mode, median, Mean = 4, 4, \(\frac{50+x}{7}\)....although x=-22 will give 4, 4, 4 but these are not in increasing order.
Case II - \(4<x<8\)
4, 4, 4, x, 8, 10, 20
Mode =4
Median = x = 5, 6, 7
Median = 5, Mean = \(\frac{50+5}{7}\)......Not an integer.
Median =6, Mean = \(\frac{50+6}{7}=8\)........ Mode, Median, Mean = > 4, 6, 8...an AP YES
Median = 7, Mean = \(\frac{50+7}{7}\)......Not an integer.
Case III - \(x\geq 8\)
4, 4, 4, 8, 10, 20...x anywhere above 8
Mode =4
Median = 8, so as per AP, 4, 8, 12, so Mean should be 12
Mean = \(\frac{50+x}{7}=12.....50+x=84...x=34\)......... Mode, Median, Mean = > 4, 8, 12...an AP YES
SO x can be 6 or 34....SUM =\( 6+34=40\)
E
General Discussion
27 May 2020, 01:02
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Mode = 4. Mean = (50+x)/7. Median depends on x.
No. of elements in list = 7. So median must be an element in the list. => median is an integer.
Since all 3 are in AP, therefore the mean value must also be an integer. And since any value of x<=4 will result in median being 4, the mode and median won't be in AP.
Thus, we'll consider values of x >= 5.
Since mean is an integer, next closest multiple of 7 is 56 => x=6=median. mean=8 =>(4,6,8) are in AP
Similarly checking for other multiples of 7, we get x=34. mode=4, median=8, mean=12. in AP. We don't need to check post this because the median value will always be 8. Mode and median are fixed => Mean will also be fixed. Thus, there can only be two such values of x (6, 34).
6+34=40
No. of elements in list = 7. So median must be an element in the list. => median is an integer.
Since all 3 are in AP, therefore the mean value must also be an integer. And since any value of x<=4 will result in median being 4, the mode and median won't be in AP.
Thus, we'll consider values of x >= 5.
Since mean is an integer, next closest multiple of 7 is 56 => x=6=median. mean=8 =>(4,6,8) are in AP
Similarly checking for other multiples of 7, we get x=34. mode=4, median=8, mean=12. in AP. We don't need to check post this because the median value will always be 8. Mode and median are fixed => Mean will also be fixed. Thus, there can only be two such values of x (6, 34).
6+34=40
