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01 Jan 2006, 13:49
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In a set of 11 consecutive integeres what is the smallest integer ?
a) if x is the smallest integer then (x+72)^1/3 =4
b) if x is the smallest and y is the largest in the set then 16x^-2=y^-2
a) if x is the smallest integer then (x+72)^1/3 =4
b) if x is the smallest and y is the largest in the set then 16x^-2=y^-2
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01 Jan 2006, 15:28
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D?
1) x = -8
2) 16/x^2-1/y^2 = 0
=> either 4/x-1/y =0 or 4/x+1/y =
Consider 4/x-1/y = 0
=> x = 4y ---(1)
Also since they are 11 consecutive integers y-x = 10 ---(2)
y-4y = 10
or y = -10/3 (NOT an integer) so x =4y cannot be TRUE
Hence 4/x+1/y = 0
or x = -4y
Using (1) y - (-4y) = 10 or y = 2
Hence x = -8
1) x = -8
2) 16/x^2-1/y^2 = 0
=> either 4/x-1/y =0 or 4/x+1/y =
Consider 4/x-1/y = 0
=> x = 4y ---(1)
Also since they are 11 consecutive integers y-x = 10 ---(2)
y-4y = 10
or y = -10/3 (NOT an integer) so x =4y cannot be TRUE
Hence 4/x+1/y = 0
or x = -4y
Using (1) y - (-4y) = 10 or y = 2
Hence x = -8
01 Jan 2006, 19:22
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go for D.
From stat.1
(x+72)^1/3 =4
or, (x+72)= 4*4*4=64
so x= -8, sufficient
from stat.2, 16/x^2=1/y^2
if i put the value of x as -8, (just as picking a number not from stat 1, a short cut) then y=2, which satisfy x y are smallest and largest of 11 consequtive integers. Other values of x/y will not satisfy 16/x^2=1/y^2
and therefore from stat2. x=-8
Sufficient.
From stat.1
(x+72)^1/3 =4
or, (x+72)= 4*4*4=64
so x= -8, sufficient
from stat.2, 16/x^2=1/y^2
if i put the value of x as -8, (just as picking a number not from stat 1, a short cut) then y=2, which satisfy x y are smallest and largest of 11 consequtive integers. Other values of x/y will not satisfy 16/x^2=1/y^2
and therefore from stat2. x=-8
Sufficient.
