If x < y < z, z = ky, x = 0, and the average of the numbers x, ... ...
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Posted: 08/03/2012 19:45
I do not understand how the answer is not 3. I plugged in different numbers to where z is 3 times the median, for example 0<2<6 or 0<3<9 and the answer is still 3.
Contributor
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Posted: 08/03/2012 20:24
Solve the problem instead of plugging in.
Since y is the middle number, y = median
mean = (x + y + z)/3
(x + y + z)/3 = 3 * y
(0 + y + ky) = 9y, and so on.
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Posted: 09/25/2012 22:20
Not sure how you dropped y just cause it didnt equal zero?
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Posted: 09/25/2012 22:36
Dusty, what do you mean by "dropped the y"? See the post before yours for a quick explanation of the problem
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Posted: 12/04/2012 01:17
But with just a 0 given how can you get all these other numbers?
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Posted: 12/04/2012 02:31
Lauren, you are given other information too, such as the mean and median. Please follow the discussion thread.
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Posted: 12/04/2012 03:04
So x is 0, which is less than y
Y is less than z, which means z is more than zero,
Z is equal to k times y
Z is three times the median
Y is the median? So 3y is z
I don't understand how 8 appears. Can you explain?
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Posted: 12/04/2012 13:19
Lauren,
The median is the middle number, the x smaller than y smaller than z tells us y will be the median whatever values x, y and z become.
It states 'the average' = 3 times the median thus
( x + y + z ) / 3 = 3y
x=0 and z= ky thus
( 0 + y + ky ) / 3 = 3y >>> both sides times 3
( y + ky ) = 9y >>> both sides -1y
ky = 8y >>> divide both sides by y
k = 8
Niels
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Posted: 12/04/2012 12:12
Lauren, it does not say that z is 3 times the median. It says the mean of x, y and z is 3 times the median. Hence, (x+y+z)/3 = 3y.
Plug in x=0 and z=ky then you can solve for k.
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If x < y < z, z = ky, x = 0, and the average of the numbers x, ... ...
Posted: 05/08/2014 05:04
You say that Y(-8+k)=0 and y is not = 0
Why isn't y=0?
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Posted: 05/08/2014 11:53
Israel, can you please elude how you think the question will make sense if y=0, don't get me wrong the expression make sense but the question not ;)
Hope this helps,
Niels
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RE: You say that Y(-8+k)=0 and y is not = 0...
Posted: 05/12/2014 15:11
Israel, the problem clearly states that y>x, and x=0.
