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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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Difficulty:

15% (low)

Question Stats:

78% (01:39) correct
22% (00:28) wrong based on 144 sessions

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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This post received KUDOS

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.