If |b|=2 and x = |a|*b, then which of the following is always true?

Question

If  |b|\geq 2  and  x = |a|*b , then which of the following is always true?

A)  a-x*b\geq 0 
B)  a+x*b\geq 0 
C)  a+x*b\leq 0 
D)  a-x*b\leq 0 
E) Both B and D

My specific question is that,when you substitute the value of  x in any of the given answer choice e.g. in the first answer choice

a- |a|b^2\geq 0  
Should again two values of  |a|  be considered  +a  and  -a  when  a>0  and when  a<0  respectively? Or Should it be taken as 
 a- a*b^2\geq 0  assuming the  |a|  is always going to be  + ?

KarishmaB · Accepted Answer

Given:   |b|\geq 2 
This implies that  b^2 \geq 4

Given:   x = |a|*b

|a| is either 0 or positive.

Assuming a is not 0,  |a|*b^2  is positive with greater absolute value than a because  b^2 >= 4

Let's consider the two expressions:

1.  a + xb = a + |a|*b^2

|a|*b^2  is positive with greater absolute value than a so irrespective of whether a is positive or negative,  a + |a|*b^2  will be positive.

2.  a - xb = a - |a|*b^2

Using the exact same analysis as above,
 |a|*b^2  is positive (so  -|a|*b^2  is negative) with greater absolute value than a so irrespective of whether a is positive or negative,  a - |a|*b^2  will be negative.

Hence, both (B) and (D) are correct.

gmat1220 · Answer

a could be zero. b could be +ve or -ve. At a = 0, x = 0. So a-xb is zero, a+xb is zero At b = -2. a=1 . x = -2. a+xb = 5 a -xb = -3 At b = 2. a=1. x= 2 a -xb = -3 a +xb = 5 I would have missed some test cases :-(

vjsharma25 · Answer

I don't think you need to plug in any values to check the inequality. 
I don't understand your approach.Can you please elaborate a bit more.

If you consider both the values for the first option then it will not be true always
if  a>0  the inequality will become
 a-a*b^2\geq 0  will be -ve (for some finite values of  a ,excluding  0 ).
and when  a<0 ,the inequality will become
 -a+a*b^2\geq 0  will be +ve (for some finite values of  a ,excluding  0 ).

So this will not be always true.Thats why I asked the question

gmat1220 · Answer

|b| >= 2. This means that b^2 >= 4. On the number line you will get two roots. (2,-2) both inclusive. Hence b >= 2 and b <=-2

x = |a| b
x can be +ve or -ve. since b can be +ve or -ve. 
Since b can be +ve or -ve. xb can be +ve or -ve.

Ultimately you will have to plugin "smart values" to determine the values of a-xb and a+xb.

This is definitely a tough question.

vjsharma25 · Answer

You are right about everything.But we have to look at the expression  x*b  which will become  |a|b^2  after plugging value of  x ,so  b^2  is always going to be +ve. It doesn't depend upon the value of  b  at all.

beyondgmatscore · Answer

Inequality is always  a - |a|*b^2

Also  |a|*b^2  is always positive irrespective of the sign of a or b (except when a=b=0, of course)

If a is negative than we are subtracting a positive term from a negative term, so it is always negative

If a is positive than we are subtracting from 'a' a positive term that is 4 times 'a', so again always negative.

Remember that sign of 'a' doesnt change the fact that the expression we are evaluating is still a - |a|*b^2

Therefore, I don't think plugging in is required in this case. We are told that  b^2 >=4 and  x = |a|* b

Since question is asking us about the signs of  a-b*x  and  a+b*x , we need to work scenarios for both of them

a-b*x  reduces to  a-b^2*|a|  Now  b^2 is >=4  and  |a| is always positive so we are always subtracting a higher number than 'a' from 'a' (irrespective of whether 'a' is negative or positive) so  a-b^2*|a| will always be <=0

a+b*x  reduces to  a+b^2*|a|  Now  b^2 is >=4  and  |a|  is always positive so we are always adding a higher number than 'a' to 'a' (irrespective of whether 'a' is negative or positive) so  a+b^2*|a| will always be >=0

Hence, E is correct option.

gmat1220 · Answer

That's an awesome interpretation ! Guys thank you so much ! beyondgmatscore I love your quant skills.

MBAhereIcome · Answer

a + x*b 
 = a + |a|b^2

as  b^2>=4  so,

= a + |a|4  ..... the value of this expression will always be +ve because 4|a|, a bigger +ve term, is added in a smaller term.
 a - |a|4  ..... this will always be -ve because 4|a|, a bigger +ve term, is subtracted from a smaller term.

therefore,

a + x*b >= 0 
 a - x*b <= 0

Bowtie · Answer

E. A could be 0, positive or negative. but the x=|a| * b tells you that x will be the same sign as b. so if b is negative x will be negative and likewise if b is positive so will x. Looking at the choices both B and D hold true.

MaheshGokhale · Answer

There are two expressions

a + x*b = a + |a|b^2 = a (1 +/- b^2)
when a<0 = a(1 - b^2) > 0
when a>=0 = a(1 + b^2) >=0

a - x*b = a - |a|b^2 = a (1 -/+ b^2)
when a<0 =a(1+b^2)< 0
when a>0 =a(1-b^2)< 0

So its must be E

hashjax · Answer

E
There are basically two equations to solve (considering B^2 >= 4)
?a - (+ab^2) .... always negative doesn't matter what ?a is...
?a + (+ab^2) .... always positive doesn't matter what ?a is... 
Thus B & D = E!

shapla · Answer

this question is really tough and i don't understand the explanation given above. Requesting Math expert to provide easy and logical explanation by dissecting every answer choices.

davedekoos · Answer

Hi Shapla,

The answer choices show two different expressions 
 a−x∗b 
 a+x∗b

and the question asks whether it is true that one of the expressions is always positive or always negative, or a combination of two options (answer choice E).

The question stem tells us that  |b|\geq{2} , and that  x=|a|*b . If we plug in that definition for x into each of the two expressions, then we get:

a-|a|*b*b  =  a-|a|*b^{2} 
 a+|a|*b*b  =  a+|a|*b^{2}

And since we know that  b^{2}\geq{4}  (because  |b|\geq{2} ), then we can plug in the value of 4 for  b^{2}  and the expressions become:

a-4*|a| 
 a+4*|a|

You might ask "But how can we assume the value of 4 for  b^{2} , when it could be anything greater than 4 too?"

If you look at the expressions, if we make  b^{2}  bigger than 4, it won't affect the  sign  of the expression. Note that  4*|a|  will always be positive, and always have a larger magnitude than  a . So, now when we look at the expressions, specifically the sign of the expressions, regardless of the sign of  a , we can see that:

a-4*|a|  will always be  \leq{0} 
 a+4*|a|  will always be  \geq{0}

Therefore the answer is E

I hope that cleared up your doubts.

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If |b|>=2 and x = |a|*b, then which of the following is always true?

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