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Max (x,y) is defined as the maximum of x and y, and min(x,y) [#permalink]
28 Feb 2012, 18:52

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Max (x,y) is defined as the maximum of x and y, and Min(x,y) is defined as the minimum of x and y. What is the average of Max(x,60) and Min(40,x) ?

(1) Min(x, 60) = x (2) Max(40, x) = x

guys my question is how to rephrase this question and consider the possible values of x i have considered the possible values of x if x>60 then max(x,60)=x and min(40,x)=40 if x<40 then max(x,60)=60 and min(40,x)=x if 40<x<60 then max(x,60)= 60 and min(40,x) =x

Re: Max (x,y) is defined as the maximum of x and y, and min(x,y) [#permalink]
28 Feb 2012, 22:32

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Expert's post

GMATD11 wrote:

Max (x,y) is defined as the maximum of x and y, and Min(x,y) is defined as the minimum of x and y. What is the average of Max(x,60) and Min(40,x) ?

(1) Min(x, 60) = x (2) Max(40, x) = x

guys my question is how to rephrase this question and consider the possible values of x i have considered the possible values of x if x>60 then max(x,60)=x and min(40,x)=40 if x<40 then max(x,60)=60 and min(40,x)=x if 40<x<60 then max(x,60)= 60 and min(40,x) =x

i have doubt in last two considered values.

First of all: max(x,y) and min(x,y) are just some functions defined as: max(x,y)=the maximum of x and y and min(x,y)=the minimum of x and y.

Max (x,y) is defined as the maximum of x and y, and Min(x,y) is defined as the minimum of x and y. What is the average of Max(x,60) and Min(40,x) ?

Question is: \(average=\frac{min(40,x)+max(x,60)}{2}=?\). Consider the following three cases:

If \(x<{40}\) then \(min(40,x)=x\), \(max(x,60)=60\) and \(average=\frac{x+60}{2}=?\); If \(40<x<60\) then \(min(40,x)=40\), \(max(x,60)=60\) and \(average=\frac{40+60}{2}=50\); If \(x>{60}\) then \(min(40,x)=40\), \(max(x,60)=x\) and \(average=\frac{40+x}{2}=?\).

(1) Min(x,60)=x --> just says that \(x<60\), so we have either the first or the second case. Not sufficient.

(2) Max(40,x)=x --> just says that \(x>40\), so we have either the second or the third case. Not sufficient.

(1)+(2) \(40<x<60\) so we have the second case: \(min(40,x)=40\), \(max(x,60)=60\) and \(average=\frac{40+60}{2}=50\). Sufficient.

Answer: C.

Hope it helps.

P.S. You could just plug some different values of x to get the three cases, for example: x=30<40, 40<(x=50)<60 and x=70>60. _________________

Re: Max (x,y) is defined as the maximum of x and y, and min(x,y) [#permalink]
19 Nov 2014, 07:12

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