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15 Jan 2015, 07:40
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D
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
25%
(medium)
Question Stats:
77% (02:05) correct
23%
(02:23)
wrong
based on 141
sessions
History
Date
Time
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Not Attempted Yet
If x is a prime number, and x-1 is the median of the set {x-1, 3x+3, 2x-4}, then what is the average (arithmetic mean) of the set?
A. 2
B. 5/3
C. 3
D. 10/3
E. 14/3
A. 2
B. 5/3
C. 3
D. 10/3
E. 14/3
15 Jan 2015, 20:32
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X is Prime. So lets put the first prime value and find out of the values of the set.
X=2 ; Set { 1,9,0 } , clearly the value x-1 is the median of the set. and the average is 10/3, which is D but hold on. lets check further to double check.
X= 3 ; Set { 2,12,2}, Here the value f x-1 can be the median ok t the average is 16/3 , no such value in the given answers. So no
X=5 and bigger prime values. the value of x-1 would be lowest in the set and therefore it cannot be the median.
Therefore D is the correct answer.
X=2 ; Set { 1,9,0 } , clearly the value x-1 is the median of the set. and the average is 10/3, which is D but hold on. lets check further to double check.
X= 3 ; Set { 2,12,2}, Here the value f x-1 can be the median ok t the average is 16/3 , no such value in the given answers. So no
X=5 and bigger prime values. the value of x-1 would be lowest in the set and therefore it cannot be the median.
Therefore D is the correct answer.
Bunuel
15 Jan 2015, 21:33
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Answer = D. 10/3
Average \(= \frac{x-1 + 3x+3 + 2x-4}{3} = \frac{6x-2}{3}\)
Placing x = 2, mean \(= \frac{10}{3}\)
I wonder why its given that (x-1) is the median... sort of distraction?
Average \(= \frac{x-1 + 3x+3 + 2x-4}{3} = \frac{6x-2}{3}\)
Placing x = 2, mean \(= \frac{10}{3}\)
I wonder why its given that (x-1) is the median... sort of distraction?
