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If xyz 0, is x (y + z) 0? (1) y + z = y + z (2) x + y = x + [#permalink]

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29 Mar 2009, 19:57

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A

B

C

D

E

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(N/A)

Question Stats:

90% (02:01) correct
10% (01:00) wrong based on 22 sessions

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If xyz ≠ 0, is x (y + z) ≥ 0? (1) │y + z│ = │y│ + │z│ (2) │x + y│ = │x│ + │y│ 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. _________________

Statement 1, insufficient since we don't know x. Statement 1, insufficient since we don't know z.

Taking both the statements together, it is xyz not equal to 0. So we can have multiple scenarios. 1 x, y and z are +ve - sufficient 2 x, y and z are -ve - sufficient 2. x is -ve while y and z are positive - not sufficient.

1) y and z have the same sign. So we have two possibilities, y<0 and z<0 OR y>0 and z>0

2) x and y have the same sign, So we have two possibilites, x<0 and y<0 OR x>0 and y>0

Combining, if we take y<0 and z<0 from (1) then from (2) we will get x<0 ( since from 2, if y<0 then x WILL be < 0). Hence, x,y,z all have the same sign. Similarly for the other case. Hence C.

If xyz ≠ 0, is x (y + z) ≥ 0? (1) │y + z│ = │y│ + │z│ (2) │x + y│ = │x│ + │y│ 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.

x (y + z) ≥ 0? Deducing the signs of x and (y+z) will be sufficient to answer the question.

(1) │y + z│ = │y│ + │z│

is possible only if both Y and Z are positive or 0. threfore (y+z) is positive or 0. but we dont know the sign of x. so not sufficient.

(2) │x + y│ = │x│ + │y│

For the above reasons, x is also positive.

Combining both we can say that x (y + z) ≥ 0. Therefore C.