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Is the average (arithmetic mean) of x and y greater than 20?
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15 Dec 2014, 09:19
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Re: Is the average (arithmetic mean) of x and y greater than 20?
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15 Dec 2014, 10:24
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!



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Re: Is the average (arithmetic mean) of x and y greater than 20?
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16 Dec 2014, 23:50
Bunuel wrote: 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
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Re: Is the average (arithmetic mean) of x and y greater than 20?
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23 Dec 2014, 06:49



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Re: Is the average (arithmetic mean) of x and y greater than 20?
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08 Feb 2015, 03:29
Hello,
[1] is pretty much straight forward. [2] i guess also is not too complicated, but as I didn't see what I did I am posting it, just in case it was wrong..
So, we are looking for (x+y)/2 and whether it is more than 20. Substituting x for 3y in the equation above we get:
(3y+y)/2 = 4y/2 = 2y. We don't know what y is, so insufficient. Leading to ANS A.



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Re: Is the average (arithmetic mean) of x and y greater than 20?
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21 Jul 2016, 12:22
Bunuel wrote: 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.When I was solving this problem, I got statement one right, but when I got to statement 2, I got Y>10, and X>30, and assumed that X+Y must then be greater than 40. Am I missing something? Please help me clarify. Thank you!



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Re: Is the average (arithmetic mean) of x and y greater than 20?
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21 Jul 2016, 13:55



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Re: Is the average (arithmetic mean) of x and y greater than 20?
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21 Jul 2016, 14:24
Bunuel wrote: toby001 wrote: Bunuel wrote: 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.When I was solving this problem, I got statement one right, but when I got to statement 2, I got Y>10, and X>30, and assumed that X+Y must then be greater than 40. Am I missing something? Please help me clarify. Thank you! For (2) we only know that x = 3y. How did you get that y > 10 and x > 30 from that? I'm starting to feel like I did something stupid here, or I'm reading somewthing wrong. Anyway, Using the inequality and the eauation given in (2), I solved for X and Y.



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Re: Is the average (arithmetic mean) of x and y greater than 20?
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21 Jul 2016, 21:50
toby001 wrote: Bunuel wrote: toby001 wrote: 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 When I was solving this problem, I got statement one right, but when I got to statement 2, I got Y>10, and X>30, and assumed that X+Y must then be greater than 40. Am I missing something? Please help me clarify. Thank you! For (2) we only know that x = 3y. How did you get that y > 10 and x > 30 from that? I'm starting to feel like I did something stupid here, or I'm reading somewthing wrong. Anyway, Using the inequality and the eauation given in (2), I solved for X and Y. From the stem we have a question: Is the average (arithmetic mean) of x and y greater than 20? Is (x + y)/2 > 20? This is not given to be true, this is the question we need to answer.
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Re: Is the average (arithmetic mean) of x and y greater than 20?
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21 Dec 2016, 17:00
Nice one. Here is my solution to this one =>
For mean to be less than 20 => a+b<40
Statement 1> 2a+2b=2*48 Hence a+b=48 Sufficient => NO,a+b is not less than 40 Hence sufficient Statement 2>
Using test cases => 3,1 => YES 1000,2000=> NO
Hence not sufficient
Hence A
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Re: Is the average (arithmetic mean) of x and y greater than 20?
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