Which one is not true?

Question

If (a - b)c bc (E) a^2 - b^2 > 0 I can prove every answer choice except of E. E. (a-b)*(a+b)>0, ok we know that (a-b) and (a+b) have the same signs. Which means we have 2 cases: ((a>b) and (a>-b)) or ((a

sudish · Answer

Hi

I'm not sure how you proved option D. According to my analysis, D cannot be true, as shown below:

Given, (a-b)c < 0
This implies that either a-b < 0 OR c < 0 which means that (a-b) and c will always have opposite signs

Now consider the solution c < 0. This implies that (a-b) must be greater than 0
Therefore, a-b > 0
i.e. a > b
Multiplying both sides of the above equation by c (which we know is negative) gives
ac < bc --> Note that the sign is reversed because we have multiplied by a negative number
Clearly, this contradicts what is given in option D i.e ac > bc
hence Option D is the correct answer.

Note that when c > 0, we'll get the same answer i.e. ac < bc

Now, using the same logic, we can prove that option E is possible i.e. we know that when c < 0 then (a-b) > 0
i.e. a > b
Squaring both sides, we get a^2 > b^2
Subtracting b^2 on both sides, we get
a^2 - b^2 = 0
This is given in option E and hence cannot be the answer.

Hope this helps

Cheers!

SonyGmat · Answer

I proved that it is not TRUE.
 
You can't square if you don't know whether a and b are positive. If they are both positive ok. If they are both negative its a^2<b^2 and if we don't know the signs (which we don't) or if the signs are unclear (which they are) we can't square!

sbvgmat · Answer

Here (a-b).c<0.
Which means that c is clearly negative.now for the above condition to be satisfied a-b should positive.

case-I:

If a-b has to be positive lets take some arbitary values.

a=1,b=-2.c=-3

D)ac>bc
1*-3>-2*-3===>-3>6.
this cannot be true.

SonyGmat · Answer

It could also mean that c is positive and a-b is negative.

But your reasoning is correct, since you found some numbers that don't satisfy the inequality this is the correct answer.

KarishmaB · Answer

We know that one and only one of (a-b) and c is negative. Option E is very straight forward too since it doesn't bring c into the picture. It doesn't matter what you get from option E - whether you get that (a-b) is positive or you get that (a-b) is negative. You can always adjust the value of c to make exactly one of (a-b) and c negative. Only if it leads to (a-b) = 0, then you can say that it is not possible but you can see that that particular problem is not there. It is just like options (A), (B) and (C). It only talks about one of (a-b) and c. Option (A) tells you that (a-b) < 0. Then you can say that c could be positive which makes (a - b)c negative. So it can be true. Similarly, option (B) says that c is negative. Then you can say that (a-b) could be positive which makes (a - b)c negative. So it can be true. Option (C) says that 'c' could be positive or negative. So anyway you can adjust (a-b) to make (a-b)*c < 0. So it can be true too. If you still want to analyze (E), here goes: a^2 - b^2 > 0 (a-b)(a+b) > 0 So either both are positive or both are negative. So (a-b) could be positive or negative. We could adjust c in either case to make (a-b)*c negative. Hence a^2 - b^2 > 0 can be true.

SonyGmat · Answer

I didn't only understand how to solve the question but how to deal with similar questions from now on!

You are awesome! +1

Which one is not true?

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