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Bunuel
If \(a^8b^7c^6d^5>0\), which of the following products must be positive ?

I. ab

II. bc

III. bd


A. I only
B. II only
C. III only
D. I and II only
E. I, II, and III

Solution:

We are given: \(a^8b^7c^6d^5>0\)

From this we can infer that

    1. \(b\) and \(d\) are positive numbers because they have odd powers over them i.e., 7 and 5 respectively, and yet there product is positive or \(>0\).

    2. \(a\) and \(c\) on the other hand can be positive or negative. They have even powers over them. We cannot say.

Now upon looking at the available option we see that only III i.e., \(bd\) must always be positive because we know \(b,d>0\).

Hence the right answer is Option C.
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Bunuel
If \(a^8b^7c^6d^5>0\), which of the following products must be positive ?

I. ab

II. bc

III. bd


A. I only
B. II only
C. III only
D. I and II only
E. I, II, and III

Solution:

We are given: \(a^8b^7c^6d^5>0\)

From this we can infer that

    1. \(b\) and \(d\) are positive numbers because they have odd powers over them i.e., 7 and 5 respectively, and yet there product is positive or \(>0\).

    2. \(a\) and \(c\) on the other hand can be positive or negative. They have even powers over them. We cannot say.

Now upon looking at the available option we see that only III i.e., \(bd\) must always be positive because we know \(b,d>0\).

Hence the right answer is Option C.


The coloured portion is wrong.
\(b^7d^5>0\) does not mean that b and d are positive numbers. We can only say that both b and d have same sign.
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I solved it without calculating anything at all:

We know that a and c are greater than 0 no matter what - because their exponents are even.

So, it means that b and d have to help us maintain the inequality:

if b is <0 and d is >0 than we will have: positive*negative*positive*positive = negative (<0)
if b is >0 and d is <0 then we will have: positive*positive*positive*negative = negative (<0)

SO, b*d has to be positive in order to maintain the inequality: positive*positive*positive*positive = positive (>0)
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a^8 and c^6 are positive (even power), but the value of 'a' and 'c' can be either '+ve' or '-ve'. On the other side, the signs of 'b' and 'd' have to be the same for the entire product to be '+ve'. Hence bd must be positive.
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