Is a |b|? (1) 2^(a - b) 16 (2) |a b|

Question

Is  a > |b|?

(1) 2^{a-b} > 16 
(2) |a – b| < b

This is   Question 6    of     the e-GMAT Question Series on Absolute Value.

Provide your solution below. Kudos for participation. Happy Solving! :-D

Best Regards
The e-GMAT Team

https://e-gmat.com/blogs/wp-content/uploads/2018/08/Free-Session-Algebra-e1533292089856.jpg

EgmatQuantExpert · Accepted Answer

Official Explanation

Correct Answer: C

We are asked to find if a > |b|

Note that in order for the answer to be YES:

i) a should be positive, AND 
ii) a should be greater than the magnitude of b.

If we find that a is negative or that a is positive but its magnitude is lesser than that of b, then our answer will be NO.

Analysing Statement 1  
 2^{a-b} > 16 
 2^{a-b} > 2^4

This means, a – b > 4
That is, a > b

However, note that we cannot say for sure if a is positive or not. For example, a = -2, b = -6 also satisfy the inequality a – b > 4.

So, Statement 1 alone is not sufficient to answer the question.

Analysing Statement 2

|a - b | < b

For all possible values of a and b, |a-b| will be non-negative (that is, greater than or equal to zero)
So, if b > |a-b| (as given in Statement 2),
This means b is positive.

But in order to answer the question asked, we need to find if a is positive AND its magnitude is greater than that of b.

Since we are not able to answer this, St. 2 is clearly not sufficient.

Combining the two statements

From Statement 1, a > b
From Statement 2, b is positive.
Since b is positive, |b| = b . . . (1)

Substituting Equation 1 in the inequality from Statement 1:
a > |b|

Thus, the two statements together are sufficient to answer the question.

UJs · Answer

Is a > |b|? Stmt (1) 2^{a-b} > 16 a-b > 4 a >b+4 clearly not sufficient as we can have (a,b) pair as i) (-2,-6) and ii) (5,1) in i) a > |b| = no ; ii) a > |b| = yes Stmt (2) |a – b| 0 ; |a – b| = (a - b) when (a – b) < 0 ; ; |a – b| = -(a - b) case 1 : (a - b) < b a < 2b case 2 : -(a - b) < b -a < 0 a < 0 (bit confused here , not sure if we multiply both sides with -1, should the sign flip ? :?:

) but anyways with case 1, 2 we get two different solutions , case 1 ( say a=3, b=2 ; a < 2b true & a > |b| No ) not sufficient Note that b is non-negative as b > |some absolute value| combining both 1 & 2 a >b+4 ; b>=0 (note i am just using part of info from Stmt 2) as b is positive , a is going to be positive too , so a > b+4 ; implies a > b sufficient Ans C

justdoitxxxxx · Answer

C
statement 1, depending on whether a is positive or negative
statement 2, tell us that a is negative
so both together are sufficient.

reto · Answer

Im having troubles with absolute value, im never sure if I considered all the options. Let me try:

Statement 1:
\(2^{a-b} > 16\) can be rewritten as \(2^{a-b} > 2^4\). This means a-b must be > 4 in order to be true. Therefore a > 4 + b. However if B is negative, e.g. -10 and a is 3, the statement 1 is still true but it changes the answer to the question. Statement 1 = Insufficient.

Statement 2
Shows us that a-b 4 + b
a<0

Plug in: a = -10 and b -15 > Is \(a > |b|?\) > NO

Sufficient if you combine it.

EgmatQuantExpert · Answer

Dear  katzzzz

Please elaborate how you inferred from Statement 2 that a is negative.

Best Regards

Japinder

EgmatQuantExpert · Answer

Dear reto The key to getting comfortable with Absolute Value questions is: 1. Understanding what Absolute Value means in the first place. We should not try to solve questions just by applying formulae without understanding why we are doing what we are doing. For example, in an expression |something| < c, why do we FLIP the sign of inequality and put a - sign before c when the 'something' is negative? You should feel confident about the logic behind this. 2. When you practice Absolute Value questions, do them step-by-step. DO NOT try to skip steps or do calculations in the air. This leads to mistakes that could have been easily avoided if we were writing each step down. Keeping this in mind, would you want to give the red part in your solution another try?

Hope you found this discussion useful! Best Regards Japinder

UJs · Answer

one correction  katzzzz , 
if statement 2, tells you that a is negative ; then a > |b| should be sufficient ? absolute value of any |xyz| is non-negative (can be 0 or >0 )

BTW, there is one more piece of information in  stmt 2  (not much useful in this context , maybe) but good to know

|a–b| |absolute|

reto · Answer

Hello Japinder

Thanks for your comments. It's always nice to experience this spirit here on gmatclub.com (people helping each other). Im honest, yesterday I had to reread some theory about absolut value before I could start to solve your questions. Im still in the early stages of my preperation.

Okay, the red part:

a-b -b, which then translates into a>0

All in all Statement 2 tells us:
b > |a-b| this means, that b is 0 or positive right? That's what I should have discovered at first!
a<2b
a>0

Is it that what you were missing in my remarks? If I now combine everything:

From Statement 1
a>4+b

From Statement 2
a>0 this is useless in combination with St. 1
a<2b this is useless in combination with St. 1
The most important is to see that b is 0 or positive. Combined with statement 1 then it will be sufficient.

Any Remarks? Thx

Escalate · Answer

Hii
Here is my approach

Statement 1 gives a-b>4
a> b+4. Now putting values of b as +2 we get a=6
In this case ans to thw ques is yes
Now put b= -5 then a= -1.
In this case ans is no.

Statement 1 insufficient

Statement 2 |a-b| positive value

|a-b| is always positive

Now on a number line |a-b| means distance between a and b

We inferred b is positive
So following 2 cases are possible on number line
Case1
<----------0--------a------b-------->
In this case ans to the ques is No

Case 2
<---------0-----------b------a------>
In this case ans to the ques is Yes

Hence statement 2 is also insufficient

Combining 1+2
we are left with only the case 2 which I mentioned above on number line as statement 1 gives a>b hence case 1 is eliminated.
From case 2 on number line we get a definite Yes

Hence C

Regards
Abhishek

1991sehwag · Answer

Hi EGMAT ,

First of all , Thanks for this question ; Solving this helped me realizing the concepts of Inequalities

Please correct me if I am wrong

Is a>|b|?

OPTION A :
2^(a−b)>16
Solving :

2^(a-b) > 2^4

so , (a-b) > 4= a>b+4

So , this can take multiple values and as such is NOT SUFFICIENT

OPTION B :
|a–b| |some absolute value|

combining both 1 & 2

a >b+4 ;
b>=0

Ans C

We can solve using both the options together
:-D

Escalate · Answer

Here I guess it does not make much difference but still want to put up that we cannot say b is non negative in statement 2.

B has to be positive 
I agree |a-b| can be equal to zero when a is equal to b 
But we are given that b>|a-b| so even if RHS here is zero, b has to be greater than zero which follows b cannot be equal to zero and hence b cannot be inferred as non negative as this will include b=0.

EgmatQuantExpert · Answer

Dear UJs Yes, when we multiply both sides of an inequality with a negative number (like -1 here), the sign of inequality always flips. So, in Case 2, The inequality -a < 0 becomes a > 0 . . . (1) Now, what was the defining condition of Case 2? That a - b < 0 (that is why you wrote |a-b| = -(a-b) in this case) That is, a |b|' will be NO in this case. In Case 1, The defining condition was a - b >= 0 (that is why you wrote |a-b| = a-b in Case 1) That is, a > = b . . . (1') Then, by processing the inequality |a-b| |b|' will be YES (if a ≠ b) or NO (if a = b). Since we're not able to get a unique YES/NO answer from St. 2, this statement is not sufficient. Hope this helped!

Best Regards Japinder

EgmatQuantExpert · Answer

Dear reto 2 remarks, both pertaining to Statement 2 analysis: 1. Please look at the highlighted part. The correct statement would be: |a-b| can be positive or 0. Since b > |a-b|, this means b will be strictly greater than 0. 2. When you are considering different cases, always remember to also consider the defining conditions of those cases. Let me explain what I mean: Look at the pink line in the quote above. This is your case 1. What is the defining condition of Case 1? That a - b ≥ 0 (this is why you wrote |a-b| = a - b) So, a ≥ b So, you should combine the inequality you obtained from the pink line with this inequality to get: b ≤ a < 2b This can also be written as |b| ≤ a < 2b (|b| = b since b is positive) Looking at this inequality, you can say that the answer to 'Is a > |b|' is YES (if a ≠ b) or NO (if a = b) Now, look at the orange line in the quote above. You are absolutely right in the way you processed the negative scenario, but remember to also consider that the defining condition of this case is that a - b < 0, that is, a < b Therefore, the overall inequality that will result in this case will be: 0 < a < b. This can also be written as 0 < a < |b| (|b| = b since b is positive) Looking at this inequality, you can say that the answer to 'Is a > |b|' is NO Since we do not get a unique YES/NO answer from St. 2, it is not sufficient. Look at the blue part in the quote above. I hope that you are now able to see that the 2 cases of St. 2 felt useless in combination with St. 1 only because you did not take the defining conditions of the 2 cases into consideration.

Best Regards Japinder

PK32 · Answer

If we will put the value of a as 7 and b as 5 then also it satisfies the equation. so how we determine a as negative.

brandon7 · Answer

How does -2 and -6 satisfy the inequality? -2 - (-6) =4, 4 is not > 4.

Harshani · Answer

Could someone please elaborate why my chain of thoughts is incorrect for statement 2?

I was able to eliminate statement 1 as Insuffcieint. but I marked statement 2 as Sufficient for the following reasons

l a - b l b > a
Since B and A are both positives, the statement will always be No, a is not bigger than lbl

akshayk · Answer

Absolute Value of any number means, the distance of that number from zero without the sign => The value is positive. => |b| means the value will be positive. a > |b| => We need to be see if a > 0 and it's also greater than b. 1) 2^{a-b} > 16 => a > b + 4 b = 1, a = 5 => a > |b| b = -3, a = 1 => a < |b| Insufficient. 2) |a-b| 0 but we have no information about a. Insufficient. From S1 -> a > b + 4 From S2 -> b > 0 => a > |b| Sufficient. C is the answer.

niks18 · Answer

hi Harshani can you explain how you arrived at the highlighted part? As per your working you only got 02 i.e a<2b but a>b and if a=1 b=2 then a

NinetyFour · Answer

Here's my reasoning. If I made a mistake please let me know. The stem is asking if 1) a is positive and 2) a is greater than b. Statement 1) 16 = 2^4 . we can then set a-b > 4 . This is sadly inconclusive because a can be negative in this case and b can be a large positive number. A can also be positive and larger than b. For example: a = -2, b = -100. This gives us -2 - (-100) = 98 > 4. Here -2 < 100. However if we did a = 50 and b = 10. We get 50 > 10 -> a>b. Statement 2) You can create two equations here or you can plug in values. Let's take a = 50 and b = 100. 50-100 = 50 < 100 in this case a < b. However we can do a = 5 and b = 4 to get 1 < 4. In this case a > b. Combing both statements we can add the inequalities together. a-b>4 + b>a-b = a > 4 + a - b = b > 4. If b > 4 that means b MUST be positive. Going back to our original equation from statement 1 -> a-b > 4. If b is positive then that means a must be also positive and greater than b by a value of 4. If a = 5 then 5 -4 is not greater then 4. However, 9 - 4 > 4, so 9 > 4. Conclusion: A is always greater than B

Is a > |b|? (1) 2^(a - b) > 16 (2) |a b| < b

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