Problem Solving

Source:- "MGMAT"

If (a - b)c < 0, which of the following cannot be true?

(A) a < b
(B) c < 0
(C) |c| < 1
(D) ac > bc
(E) a2 - b2 > 0

Answer is D

But i am confused with option E. The explanation about E is, "If we factor this expression, we get (a + b)(a - b) < 0. This tells us that the expressions a + b and a - b have opposite signs, which is possible according to the question", confusing.

How can factoring the expression changes the sign from > (greater than) to < (lesser than) ? or Is this some rule of inequality?

I have no idea. Please reply.

Regards
Vinni

we have two options 1) (a - b)<0 and c>0, 2) (a-b)>0 and c<0
precisely these options 1) a<b and c>0 and 2) a>b, c<0

a) a < b is possible
b) c < 0 possible too
c) |c| < 1 translates into -1<c<1 which is possible (although we have specified that c cannot be 0, it's either positive or negative - this question is NOT MUST BE TRUE type; the question is asking about possibility)
d) ac > bc translates into ac-bc>0 and from the original questions (a-b)c<0 is equivalent to ac-bc<0 which is not possible
e) a^2 - b^2 > 0 translates into (a-b)(a+b)>0 or two options 1) (a-b)>0 and (a+b)>0, 2) (a-b)<0 and (a+b)<0. Both give 1) a>b and a>-b or a>|b|, 2) a<|b| which is also possible

Anurag@Gurome wrote:
No,factorization never changes the sign of the inequality.
What has been done is not correct.
(a^2 - b^2) > 0 implies that (a+b)(a-b) > 0.

Anurag thank you so much. Now i am relieved. This small doubt was making my brain puzzled.

Regards
Vinni.k

pemdas wrote: e) a^2 - b^2 > 0 translates into (a-b)(a+b)>0 or two options 1) (a-b)>0 and (a+b)>0, 2) (a-b)<0 and (a+b)<0. Both give 1) a>b and a>-b or a>|b|, 2) a<|b| which is also possible

How do you justify the above two possibilities?

algebra - the product of two positives or two negatives is positive. Do you need explanation for mod part?

First thing I did was multiply out.

So I get ac - bc < 0.

add bc to the other side and you get ac < bc. D is the opposite, so D.

vinni.k wrote:Source:- "MGMAT"

If (a - b)c < 0, which of the following cannot be true?

(A) a < b
(B) c < 0
(C) |c| < 1
(D) ac > bc
(E) a2 - b2 > 0


Regards
Vinni

Anurag answered Vinni's original question regarding the explanation, but I want to discuss a super-fast way to solve this particular problem.

It's often quicker to solve number property questions by focusing on the concepts underlying the question rather than by doing a whole lot of math. Let's approach this question with that in mind.

If (a - b)c < 0, which of the following cannot be true?

Since the product of two terms is negative, we know that one term must be negative and the other positive. In this question, our two terms are "a-b" and "c".

Upon scanning the choices, we note that answers A, B, C and E only focus on one of the two terms. Since each term could be positive or negative in a vacuum (i.e. without considering the other term), there's no possible way that any of these 4 choices could be correct.

(A small caveat - if A, B, C or E told us that one of the terms were equal to 0, then that would in fact be impossible and the correct answer to the question.)

Only D provides information about both terms; accordingly, D must be the correct answer to the question.

pemdas wrote:algebra - the product of two positives or two negatives is positive. Do you need explanation for mod part?

It's not the technicalities of the math I was interested in. My question was how do you say option "E" is possible based on the deductions you used to arrive on the mod part.

Thanks.

ah, you mean this

e) a^2 - b^2 > 0 translates into (a-b)(a+b)>0 or two options 1) (a-b)>0 and (a+b)>0, 2) (a-b)<0 and (a+b)<0. Both give 1) a>b and a>-b or a>|b|, 2) a<|b| which is also possible

by analyzing the question itself, here

we have two options 1) (a - b)<0 and c>0, 2) (a-b)>0 and c<0
precisely these options 1) a<b and c>0 and 2) a>b, c<0

i've deduced a<b and a>b as two options possible should satisfy

besides i've posted this

c) |c| < 1 translates into -1<c<1 which is possible (although we have specified that c cannot be 0, it's either positive or negative - this question is NOT MUST BE TRUE type; the question is asking about possibility)

so basically, we are left without 'c'-yes, BUT we are looking into possibility rather than MUST BE TRUE in all cases

have i cleared ur doubt?