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15 Dec 2014, 09:19
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A
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
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92% (00:57) correct
8%
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Is the average (arithmetic mean) of x and y greater than 20?
(1) The average (arithmetic mean) of 2x and 2y is 48.
(2) x = 3y
(1) The average (arithmetic mean) of 2x and 2y is 48.
(2) x = 3y
15 Dec 2014, 10:24
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x+y > 20?
statement 1: sufficient
(2x+2y) / 2 = 48. multiply both sides by 2 to remove the fraction, then divide both sides by 2 to remove coefficients of x and y to leave us with x+y=48. The average of x and y will always be 24 for any 2 values according to this statement.
statement 2: insufficient
x=1, y=3, average is 2, less than 20
x=100, y=300, average is 200, greater than 20
Answer A!
statement 1: sufficient
(2x+2y) / 2 = 48. multiply both sides by 2 to remove the fraction, then divide both sides by 2 to remove coefficients of x and y to leave us with x+y=48. The average of x and y will always be 24 for any 2 values according to this statement.
statement 2: insufficient
x=1, y=3, average is 2, less than 20
x=100, y=300, average is 200, greater than 20
Answer A!
16 Dec 2014, 23:50
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Bunuel
Tough and Tricky questions: Number Properties.
Is the average (arithmetic mean) of x and y greater than 20?
(1) The average (arithmetic mean) of 2x and 2y is 48.
(2) x = 3y
Kudos for a correct solution.
1) (2X+2Y)/2 = 48=> (x+y)/2 = 24 sufficient
2) Insufficient as the value cannot be calculated using the onformation given
Ans A
