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nkmungila
If \(\frac{a}{bc}\) is positive, which of the following must be true?

A. a + b + c > 0
B. bc > 0
C. a > 0 and bc > 0
D. a < 0 and bc < 0
E. abc > 0

For a/(bc) to be positive, either of two scenarios must occur: (1) a and bc are both positive, or (2) a and bc are both negative.

We can reject choices C and D since these two choices are an “either-or” and not a “must.”

Let’s analyze the values of a, b, c further. If bc is positive, then either b and c are positive or b and c are negative. Thus:

a > 0, b > 0, c > 0 or
a > 0, b < 0, c < 0

If bc is negative, then either b is positive and c is negative or b is negative and c is positive. Thus:

a < 0, b > 0, c < 0 or
a < 0, b < 0, c > 0

We see that in any of the four cases above, abc > 0.

Answer: E
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nkmungila
If \(\frac{a}{bc}\) is positive, which of the following must be true?

A. a + b + c > 0
B. bc > 0
C. a > 0 and bc > 0
D. a < 0 and bc < 0
E. abc > 0
With algebra, I eliminated answers, and chose E. But the manipulation of abstractions can make me wary.

So I chose numbers (mindful of Answer A, I kept absolute values close together)

Testing numbers was much easier.

Four iterations yield +2

1) \(\frac{4}{(2)(1)}\)

2) \(\frac{4}{(-2)(-1)}\)

3) \(\frac{-4}{(-2)(1)}\)

4) \(\frac{-4}{(2)(-1)}\)

C (per #s 1 and 2) and D (per #s 3 and 4) can be true. But each proves that the other is not a case of "must be true."

A is contradicted by case #4:
-4 + (-1) + 2 = -3

B is contradicted by #3 and #4:
(-2)(1) = -2 and (2)(-1) = -2

Remaining: E) abc > 0
All four cases: true.

Answer E
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Since a/(bc) is positive that means both numerator and denominator should have same sign.

If a = +ve , then bc = +ve => product of two numbers positive when 'b' and 'c' both are positive

If a = -ve , then bc = -ve => product of two numbers negative when 'b' and 'c' have opposite sign.

Possible cases:

=> [a = +, b = c = +]
=> [a = -, b = - , c = +]
=> [a = -, b = + , c = -]

Only option E satisfy this abc > 0

Answer E
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