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# If 0 < a < b, is c < 0? (1) cb < ca (2) a – b > c

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Re: If 0 < a < b, is c < 0? (1) cb < ca (2) a – b > c [#permalink]
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Bunuel wrote:
If 0 < a < b, is c < 0?

(1) cb < ca

(2) a – b > c

We are given that 0 < a < b and we need to determine whether c is negative.

Statement One Alone:

cb < ca

Since we know that a and b are positive and that b is greater than a, we see that c must be negative. For example, if a = 2 and b = 3, in order for cb < ca, c could be -1, such that -3 < -2. Statement one alone is sufficient to answer the question.

Statement Two Alone:

a – b > c

Since a is less than b and they are both positive, (a - b) must be negative; thus, we have:

Negative number > c

Since c is less than some negative number, c is also negative. Statement two is sufficient to answer the question.

Answer: D
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Re: If 0 < a < b, is c < 0? (1) cb < ca (2) a – b > c [#permalink]
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Bunuel wrote:
If 0 < a < b, is c < 0?

(1) cb < ca

(2) a – b > c

0 < a < b implies the following:
b-a = bigger - smaller = POSITIVE
a-b = smaller - bigger = NEGATIVE

Statement 1:
cb - ca < 0
c(b-a) < 0
Implication:
c and b-a have DIFFERENT SIGNS.
Since b-a is positive, c must be negative.
SUFFICIENT.

Statement 2:
c < a-b
c < negative
Thus, c must be negative.
SUFFICIENT.

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Re: If 0 < a < b, is c < 0? (1) cb < ca (2) a – b > c [#permalink]
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Re: If 0 < a < b, is c < 0? (1) cb < ca (2) a – b > c [#permalink]
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