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Is the average (arithmetic mean) of x and y greater than 20?

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Bunuel
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Is the average (arithmetic mean) of x and y greater than 20? [#permalink] New post  15 Dec 2014, 09:19
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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
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Re: Is the average (arithmetic mean) of x and y greater than 20? [#permalink] New post  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!
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Re: Is the average (arithmetic mean) of x and y greater than 20? [#permalink] New post  16 Dec 2014, 23:50
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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? [#permalink] New post  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? [#permalink] New post  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? [#permalink] New post  21 Jul 2016, 13:55
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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?
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Re: Is the average (arithmetic mean) of x and y greater than 20? [#permalink] New post  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? [#permalink] New post  21 Jul 2016, 21:50
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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? [#permalink] New post  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? [#permalink]
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