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SajjadAhmad
If \(abcd ≠ 0\), and if \(0 < c < b < a < 1\), is it true that \(\frac{a^4bc}{d^2} < 1\)?

(1) \(a = \sqrt{d}\)

(2) \(d > 0\)

Since d^2 is automatically positive, we can multiply the inequality in the question by d^2 on both sides, so our question becomes:

Is a^4 b c < d^2 ?

If a = √d, we can substitute for a in the question:

Is (√d)^4 bc < d^2 ?
Is d^2 bc < d^2 ?
Is bc < 1 ?

and since 0 < c < b < 1, it must be true that bc < 1, and Statement 1 is sufficient.

The sign of d is irrelevant, since we're squaring d anyway in the question, so Statement 2 isn't useful and the answer is A.
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SajjadAhmad
If \(abcd ≠ 0\), and if \(0 < c < b < a < 1\), is it true that \(\frac{a^4bc}{d^2} < 1\)?

(1) \(a = \sqrt{d}\)

(2) \(d > 0\)

S1: a = root d can also be written as a^4 = d^2. Therefore these two terms cancel out in the above equation and the new equation is bc<1. Since we know that b and c are between 0 and 1, the product will always be less than 1. SUFFICIENT.

S2: d^2 could be smaller, larger, or equal to the numerator, which will produce different answers. NOT SUFFICIENT.

ANSWER: A
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