If the average (arithmetic mean) of x, y, z, 5, and 7 is 8,
which of the value for x = 1?
I The median of the five numbers cannot be 5
II At least one of x, y and z is greater than 9
III The range of the five numbers is 2 or more
(A) I only
(B) II only
(C) III only
(D) I and III
(E) II and III
BY-- TAUSEEF INDIKAR from BANGALORE
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7 comments:
it can be B coz, (I) 5 may be(ex: y=2 and z=25) r may not(ex: y=13, z=14) be median (II) ya atleast one of x,y,and z is greater than 9
(III) x=1
answer is E
E
e - II and III are correct.
(x+y+z+5+7)/5 = 8
x+y+z = 28
one of the no's has to be greater than 9 in order to satisfy the above equation.
and the range of the no's is 2 or more
I believe its B for the following reasons:
1. for 5 to be the middle of the distribution, then the preceeding numbers must be 5 or less which will not give you the sum of 28 (x+y+z = 28)
2. at least one of x, y and z has to be more than 9 to give you sum of 28, so this point is correct
3. for the range to be 2, then the maximum number in the distribution is 9. in this case, even if x, y and z were 9, they can only give you a sum of 27 which is less than 28.
For these reasons, i believe b is the answer
B
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