Question

Difficulty: MediumPositive and Negative Number Properties

If aa, bb, and cc are non-zero real numbers such that ab<0\frac{a}{b} < 0, bc>0b c > 0, and ac>0a - c > 0, which of the following expressions MUST be negative?

  1. A
    abca b c
  2. B
    a+bc\frac{a + b}{c}
  3. ab2ca b^2 cAnswer
  4. D
    bca\frac{b - c}{a}
  5. E
    a2bca^2 b c

Answer

The expression ab2ca b^2 c must be negative.
From ab<0\frac{a}{b} < 0, aa and bb have opposite signs. From bc>0b c > 0, bb and cc have the same sign. Thus, aa and cc must have opposite signs. The inequality ac>0a - c > 0 implies a>ca > c, so aa must be positive (a>0a > 0) and cc must be negative (c<0c < 0). Since bb shares the sign of cc, bb is also negative (b<0b < 0). Evaluating ab2ca b^2 c: aa is positive, b2b^2 is strictly positive for any non-zero real number bb, and cc is negative. Therefore, ab2ca b^2 c is the product of two positive terms and one negative term, which MUST be negative.

Step-by-Step Solution

1
Determine the relative signs of aa, bb, and cc from the given inequalities.
ab<0\frac{a}{b} < 0 means aa and bb have opposite signs. bc>0b c > 0 means bb and cc have the same sign. Therefore, aa and cc must have opposite signs.
Quotients of numbers with opposite signs are negative, and products of numbers with the same sign are positive.
2
Use ac>0a - c > 0 to determine the explicit sign of each variable.
ac>0    a>ca - c > 0 \implies a > c. Since aa and cc have opposite signs and a>ca > c, aa must be positive (a>0a > 0) and cc must be negative (c<0c < 0). Since bb has the same sign as cc, bb must also be negative (b<0b < 0).
A positive number is always greater than a negative number.
3
Evaluate the sign of ab2ca b^2 c.
Since a>0a > 0, b2>0b^2 > 0 (as b0b \neq 0), and c<0c < 0, ab2c=(+)×(+)×()<0a b^2 c = (+) \times (+) \times (-) < 0.
The product of two positive real numbers and one negative real number is strictly negative.

Key Concept

Deducing signs of variables using properties of products, quotients, and inequalities.
Estimated Time:2m 0s
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