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Of the three numbers, median is M and average is A, is

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Of the three numbers, median is M and average is A, is [#permalink]

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New post 31 Jul 2008, 10:41
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Of the three numbers, median is M and average is A, is M<A?
(1) M is less than the average of the largest and smallest number.
(2) The largest number is 250 greater than the average, and the smallest number is 150 less than the average.
A. Statement (1) ALONE is sufficient but Statement (2) ALONE is not sufficient.
B. Statement (2) ALONE is sufficient but Statement (1) ALONE is not sufficient.
C. BOTH Statements TOGETHER are sufficient, but NEITHER Statement alone is sufficient.
D. Each Statement ALONE is sufficient.
E. Statements (1) and (2) TOGETHER are NOT sufficient.

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Re: median , and mode [#permalink]

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New post 31 Jul 2008, 11:20
I think answer is E

1.For 1 3 9
Is M < A -> Yes
For 1 2 4
Is M < A -> No
Insufficient

2. Lets say A = 100
based on given condition -50 Y 350
Y could be anything
Insufficient

If you combine 1 & 2, still Insufficient

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Re: median , and mode [#permalink]

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New post 31 Jul 2008, 11:30
Of the three numbers, median is M and average is A, is M<A?
(1) M is less than the average of the largest and smallest number.
(2) The largest number is 250 greater than the average, and the smallest number is 150 less than the average.

S2. Let the three numbers be x,y,z in ascendign order; In this case y will be the median
Average = x+y+z/3 = A
z = A+250 and x = A -150

3A = A-150 + y + A+250
A = y + 100
Hence average is greater than the median A> M; so ACE are out BD in

S1. y < x+z/2 You can pick numbers here

10, 11, 29 M < A
6, 7, 18 M < A

IMO D

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Re: median , and mode [#permalink]

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New post 31 Jul 2008, 11:36
arjtryarjtry wrote:
Of the three numbers, median is M and average is A, is M<A?
(1) M is less than the average of the largest and smallest number.
(2) The largest number is 250 greater than the average, and the smallest number is 150 less than the average.
A. Statement (1) ALONE is sufficient but Statement (2) ALONE is not sufficient.
B. Statement (2) ALONE is sufficient but Statement (1) ALONE is not sufficient.
C. BOTH Statements TOGETHER are sufficient, but NEITHER Statement alone is sufficient.
D. Each Statement ALONE is sufficient.
E. Statements (1) and (2) TOGETHER are NOT sufficient.


say X and Y are smallest and largest numbers

A= (X+Y+M)/3
X+Y+M=3A
1) M<(X+Y)/2 --suffciient
X+Y>2M
X+Y+M>3M
3A>3M
A>M -->M<A
2) Y= A+250 -- Sufficient.
X= A-150
X+Y+M=3A --> 2A+100+M=3A
--> A=M+100
--> M<A

answer D.
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Re: median , and mode [#permalink]

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New post 03 Aug 2008, 07:17
arjtryarjtry wrote:
Of the three numbers, median is M and average is A, is M<A?
(1) M is less than the average of the largest and smallest number.
(2) The largest number is 250 greater than the average, and the smallest number is 150 less than the average.
A. Statement (1) ALONE is sufficient but Statement (2) ALONE is not sufficient.
B. Statement (2) ALONE is sufficient but Statement (1) ALONE is not sufficient.
C. BOTH Statements TOGETHER are sufficient, but NEITHER Statement alone is sufficient.
D. Each Statement ALONE is sufficient.
E. Statements (1) and (2) TOGETHER are NOT sufficient.


say numbers are x,m,y and m = median

a = average
3a= x+y+m now we know from(1) m< (x+y)/2 => x+y+m<3m => 31<3m => a<m => sufficient

(2) solving condn in (2) we get m=a-100 => m <a => sufficient

HENCE (D) is answer
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Re: median , and mode   [#permalink] 03 Aug 2008, 07:17
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Of the three numbers, median is M and average is A, is

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