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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.
(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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31 Jul 2008, 11:20
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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
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
31 Jul 2008, 11:30
Kudos
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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.
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
(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
