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GMATD11
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Max (x,y) is defined as the maximum of x and y, and min(x,y) 28 Feb 2012, 19:52
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C
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25% (medium)

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76% (01:25) correct 24% (01:30) wrong based on 1271 sessions

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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


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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.

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C
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Bunuel
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Max (x,y) is defined as the maximum of x and y, and min(x,y) 28 Feb 2012, 23:32
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GMATD11
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.
Show more

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.
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Max (x,y) is defined as the maximum of x and y, and min(x,y) 14 Jun 2013, 04:04
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Bumping for review and further discussion*. Get a kudos point for an alternative solution!

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Max (x,y) is defined as the maximum of x and y, and min(x,y) 24 May 2016, 05:09
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GMATD11
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

Show more


Stem : To have a definite average (independent of x) : 40< x < 60

A)x < 60 : Says nothing about 40
B) x > 40 : Says nothing about 60

Combining together;

40 < x < 60 (the stem)

C
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Max (x,y) is defined as the maximum of x and y, and min(x,y) 18 Aug 2020, 01:16
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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

-----

We need to get: [Max(x,60) + Min(40,x)] / 2

(1) Min(x, 60) = x

This says x < 60, hence Max of (X, 60) = 60. We know nothing about Min (40,x) because we don't know if X > 40. NS

(2) Max(40, x) = x

This says X > 40, hence Min (40,x) = 40. We know nothing about Max of (X,60) because we don't know if X > 60 or X <60. NS

Combining the two,
[Max(x,60) + Min(40,x)] / 2
= (60 + 40) / 2
= 100/2
= 50
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Max (x,y) is defined as the maximum of x and y, and min(x,y) 25 Nov 2021, 11:03
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Bunuel
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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

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.
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very clear
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Max (x,y) is defined as the maximum of x and y, and min(x,y) 12 Jan 2023, 01:42
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Average=(x,60,x,40)/4
putting the values of x,60 i-e x
and x,40 i-e x
Average=(x+x)/2
Average= 2x/2
Average= x
Answer: C
Both together are sufficient and neither alone
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Max (x,y) is defined as the maximum of x and y, and min(x,y) 15 Feb 2024, 09:28
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