tonebeeze wrote:If a, b, c, and d are positive, is ac + bd > bc + ad?
1. c > d
2. b > a
OA = C
Rephrase the question:
ac + bd > bc + ad
ac - bc > ad - bd
c(a-b) > d(a-b).
Question rephrased: Is c(a-b) > d(a-b)?
Statement 1: c>d.
Let c=2, d = 1, and a-b = 20-10 = 10.
Is 2*10 > 1*10? Yes.
Let c=1, d=2, and a-b = 10-20 = -10.
Is 2(-10) > 1(-10)? No.
Since in the first case the answer is Yes, and in the second case the answer is No, insufficient.
Statement 2: b > a.
Let b=20 and a=10 so that a-b = -10.
Let c=1 and d=2.
Is 1(-10) > 2(-10)? Yes.
Let c=2 and d=1.
Is 2(-10) > 1(-10)? No.
Since in the first case the answer is Yes, and in the second case the answer is No, insufficient.
Statements 1 and 2 combined: c>d and b>a.
Thus, a-b < 0.
Since c>d, c(a-b) will be further below 0 than will be d(a-b).
Thus, we know that c(a-b) < d(a-b).
Sufficient.
The correct answer is
C.
Be careful when using division to simplify an expression.
Once the question has been rephrased as
Is c(a-b) > d(a-b)?, it's dangerous to divide by (a-b) because we don't know the value of (a-b).
If (a-b) = 0, then the quotient will be undefined.
If (a-b) < 0, then the direction of the inequality will have to change from > to <.
Since we don't know the value of (a-b), plugging in values is a safer approach for most test-takers.
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