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# Is ac + bd + bc + ad > 0?

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Joined: 02 Sep 2009
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27 Nov 2019, 01:03
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35% (medium)

Question Stats:

82% (01:58) correct 18% (02:11) wrong based on 44 sessions

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Is $$ac + bd + bc + ad > 0$$?

(1) $$ac^2 + bd^2 + bc^2 + ad^2 > 0$$

(2) $$a^2d + b^2c + a^2c + b^2d < 0$$

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27 Nov 2019, 01:24
1
Is $$ac + bd + bc + ad > 0$$?
or, $$a(c+d) + b(c+d) >0$$
or $$(a+b)(c+d) >0?$$

St1: (1) $$ac^2 + bd^2 + bc^2 + ad^2 > 0$$
$$c^2(a+b) +d^2(a+b) > 0$$
$$(c^2+d^2)(a+b) > 0$$
Since, square of a number is always non negative , hence $$(c^2+d^2)$$ is always non negative.
hence, (a+b)>0
but nothing can be said about (d+c)
Hence NOT SUFFICIENT

St2: $$a^2d + b^2c + a^2c + b^2d < 0$$
$$(a^2+b^2)(c+d)<0$$
Since square of a number is always non negative, hence $$(a^2+b^2)$$ is always non negative.
So $$(c+d)<0$$ but nothing can be said about $$(a+b).$$
Hence NOT SUFFICIENT

Combing St1 &2 , we get $$(a+b)>0$$ & $$(c+d)<0$$
Hence, $$(a+b)(c+d) <0$$

Bunuel wrote:
Is $$ac + bd + bc + ad > 0$$?

(1) $$ac^2 + bd^2 + bc^2 + ad^2 > 0$$

(2) $$a^2d + b^2c + a^2c + b^2d < 0$$

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Re: Is ac + bd + bc + ad > 0?  [#permalink]

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29 Nov 2019, 23:42
Bunuel wrote:
Is $$ac + bd + bc + ad > 0$$?

(1) $$ac^2 + bd^2 + bc^2 + ad^2 > 0$$

(2) $$a^2d + b^2c + a^2c + b^2d < 0$$

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Forget conventional ways of solving math questions. For DS problems, the VA (Variable Approach) method is the quickest and easiest way to find the answer without actually solving the problem. Remember that equal numbers of variables and independent equations ensure a solution.
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The first step of the VA (Variable Approach) method is to modify the original condition and the question. We then recheck the question. We should simplify conditions if necessary.

The question asks if $$ac+bd+bc+ad > 0$$ or $$(a+b)(c+d)>0$$.

Since we have 4 variables and 0 equations, E is most likely the answer. So, we should consider conditions 1) & 2) together first. After comparing the number of variables and the number of equations, we can save time by considering conditions 1) & 2) together first.

Conditions 1) & 2)
Condition 1) $$(a+b)(c^2+d^2)>0$$ tells $$a+b>0$$ since $$c^2+d^2>0$$.
Condition 2) $$(a^2+b^2)(c+d)>0$$ tells $$c+d>0$$ since $$a^2+b^2>0$$.
Then $$(a+b)(c+d) > 0$$ and the answer is 'yes'.

Since both conditions together yield a unique solution, they are sufficient.

Since this question is an inequality question (one of the key question areas), CMT (Common Mistake Type) 4(A) of the VA (Variable Approach) method tells us that we should also check answers A and B.

Condition 1)
Condition 1) $$(a+b)(c^2+d^2)>0$$ tells $$a+b>0$$ since $$c^2+d^2>0$$.
Since there is no information about $$c+d$$, condition 1) is not sufficient.

Condition 2)
Condition 1) $$(a^2+b^2)(c+d)>0$$ tells $$c+d>0$$ since $$a^2+b^2>0$$.
Since there is no information about $$a+b$$, condition 2) is not sufficient.

In cases where 3 or more additional equations are required, such as for original conditions with “3 variables”, or “4 variables and 1 equation”, or “5 variables and 2 equations”, conditions 1) and 2) usually supply only one additional equation. Therefore, there is an 80% chance that E is the answer, a 15% chance that C is the answer, and a 5% chance that the answer is A, B or D. Since E (i.e. conditions 1) & 2) are NOT sufficient, when taken together) is most likely to be the answer, it is generally most efficient to begin by checking the sufficiency of conditions 1) and 2), when taken together. Obviously, there may be occasions on which the answer is A, B, C or D.
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Re: Is ac + bd + bc + ad > 0?   [#permalink] 29 Nov 2019, 23:42
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