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If the average (arithmetic mean) of x, y, z, 5, and 7 is 8, which of
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01 May 2021, 07:22
01 May 2021, 07:22
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If the average (arithmetic mean) of x, y, z, 5, and 7 is 8, which of the value for x = 1?
I The median of the five numbers cannot be 5
II At least one of x, y and z is greater than 9
III The range of the five numbers is 2 or more
(A) I only
(B) II only
(C) III only
(D) I and III
(E) II and III
Kudos for the right answer and explanation
I The median of the five numbers cannot be 5
II At least one of x, y and z is greater than 9
III The range of the five numbers is 2 or more
(A) I only
(B) II only
(C) III only
(D) I and III
(E) II and III
Kudos for the right answer and explanation
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If the average (arithmetic mean) of x, y, z, 5, and 7 is 8, which of
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01 May 2021, 09:32
01 May 2021, 09:32
1
Carcass wrote:
If the average (arithmetic mean) of x, y, z, 5, and 7 is 8, which of the value for x = 1?
I The median of the five numbers cannot be 5
II At least one of x, y and z is greater than 9
III The range of the five numbers is 2 or more
(A) I only
(B) II only
(C) III only
(D) I and III
(E) II and III
I The median of the five numbers cannot be 5
II At least one of x, y and z is greater than 9
III The range of the five numbers is 2 or more
(A) I only
(B) II only
(C) III only
(D) I and III
(E) II and III
\(\frac{x + y + z + 5 + 7}{5} = 8\)
x + y + z = 40 - 12 = 28
When, x = 1
y + z = 27
I. The median of the five numbers cannot be 5 - NO
x = 1, y = 2, z = 25
Terms in ascending order: 1, 2, 5, 7, 25, Median = 5
II. At least one of x, y and z is greater than 9
Since, they add up to 28, one of them has to be greater than 9
III The range of the five numbers is 2 or more
In order to minimize the range, let us maximize y and z
x = 1, y = 13, z = 14
Terms in ascending order: 1, 5, 7, 13, 14, Range = 13
Hence, option E
gmatclubot
If the average (arithmetic mean) of x, y, z, 5, and 7 is 8, which of [#permalink]
01 May 2021, 09:32
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