2. 外语
  3. If x + y+z>0, is z> 1 ? (1) z>x + y+1 (2) x+y+1

If x + y+z>0, is z> 1 ? (1) z>x + y+1 (2) x+y+1

admin2022-10-18  49
问题 If x + y+z>0, is z> 1 ?
(1) z>x + y+1
(2) x+y+1<0
选项 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.
答案B
解析 Determine if z > 1 is true.
(1)     Given that z > x + y + 1, by adding z to both sides, it follows that 2z>x+y + z + 1. Also, x+y + z+1 > 1 because x + y + z > 0.Thus, 2z > 1 and z >1/2.  It is possible that z > 1 is true and it is possible that z > 1 is not true. For example, if z = 1.1 and x = y = 0, then x + y + z > 0 and z > x + y + 1 are both true, and z > 1 is true. However, if z = 1, x = -0.5 and y = -0.25,x+y+ z > 0 and z > x+ y+ 1 are both true, and z > 1 is not true; NOT sufficient.
(2)     Given that x + y + 1 < 0, it follows that 1 < -x - y. It is also given that x + y + z > 0, so z > -x -y or -x-y < z. Combining 1 < -x - y and -x - y < z gives 1 < z or z > 1; SUFFICIENT.
The correct answer is B;
statement 2 alone is sufficient.
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