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b< c < d && c> 0 ==> d>0 && (b<0 or b>0)

since c & d are greater than 0 ==> cd > 0 && cd > b
Thus, b - cd > 0 is not possible.

Ans.: C
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b can be 0 as we are not given that b,c,d are positive integers.

Thus only choice C fits.
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If \(b < c < d\) and \(c > 0\), which of the following cannot be true if b, c and d are integers?

\(A. bcd > 0\)

\(B. b + cd < 0\)

\(C. b – cd > 0\)

\(D. \frac{b}{cd} < 0\)

\(E. b^{3} cd < 0\)

We are given that b < c < d and c > 0, and that b, c, and d are integers.

Since c > 0, c is positive, and since d > c, d must also be positive. However, we are unsure of whether b is positive or negative. Let’s look at our answer choices to determine which MUST be true.

A. bcd > 0

If b is positive, then bcd is greater than zero. Answer choice A can be true.

B. b + cd < 0

If b is negative and its absolute value is greater than the product of cd, then b + cd is less than zero. For example, if b = -3, c = 1, and d = 2, we have -3 + (1)(2) = -1 < 0. Answer choice B can be true.

C. b - cd > 0

We can simplify answer choice C to read:

b > cd

If b is negative or zero, it certainly can’t be greater than cd, since cd is positive. If b is positive, it still can’t be greater than cd, since both c and d are integers greater than b. Thus, answer choice C can never be true.

Answer: C
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This question can be done by this approach.
The key point is that b can be greater than , equal to or less than zero.
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