prachich1987 wrote:If in number 0.xyz, x,y and z stand respectively for the first, second, and third digit to the right of the decimal point, is 0.xyz>(2/3)
1) x+y>13
2) x+z>14
Statement 1:
Tells us that x≥5 (because the greatest possible value of y is 9, and if x=4 and y=9, then x+y = 4+9 = 13, and statement 1 indicates that x+y>13.)
If x=5, y=9, and z=1, is .591>.666? No.
If x=6, y=8, and z=1, is .681>.666? Yes.
Since the answer can be both no and yes, insufficient.
Statement 2:
Tells us that x≥6 (because the greatest possible value of z is 9, and if x=5 and z=9, then x+y = 5+9 = 14, and statement 2 indicates that x+z>14.)
If x=6, y=1, and z=9, is .619>.666? No.
If x=6, y=8, and z=9, is .689>.666? Yes.
Since the answer can be both no and yes, insufficient.
Statements 1 and 2 together:
Statement 2 tells us that x≥6.
Since statement 1 indicates that x+y>13, if x=6, then y≥8.
Since statement 2 indicates that x+z>14, if x=6, then z=9.
Thus, the smallest possible value of .xyz is .689.
Since .689>.666, we know that .xyz>.666.
Sufficient.
The correct answer is
C.
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