题目信息
If x + y + z > 0, is z > 1 ?
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.
参考答案及共享解析
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已选答案：
正确答案：
B:Statement (2) ALONE is sufficient, but statement (1) alone is not sufficient.
Algebra Inequalities
Determine if z > 1 is true.
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 >
. 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.
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.
Determine if z > 1 is true.
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 >
. 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.
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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