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Intern
Joined: 01 Jun 2010
Posts: 23
Location: United States
Schools: Harvard Business School (HBS) - Class of 2014
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04 Mar 2012, 15:19
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While reviewing advanced quant and doing some questions on the forum, I realiized that my understanding of inequalities isnt as good as I thought. Here are few things I ran into that I am curious about...

1) Is |a| + |b| > |a + b| ?

Usually I break abs value questions into the possible cases when i see something like this: 4 > |a + b|. I rewrite with 4 as a pos and neg solution. With |b + c| > |a + b| i do the all positive case and the switch the signes on one side case. However, with |a| + |b| > |a + b| or even this |a| + |b| > 4, Im not sure how to break it down...

2) Not sure how to spell this out in a forum post, but in advanced gmat quant they use a numer line approach where you take the quadratic values that set the equation equal to zeros and use them to divide the number line. You then put a +/- for the different segments. This seems cleaner than listing and testing number cases. Is there a disadvantage to is I am not seeing?

3) Two parter. A) is there a way to factor a + a^-1? This comes from DS question #4 in chater #4 of advanced quant (given below) and B) what else did I do wrong on the question ? (the one below).

Is a + a^-1 > 2?

1) a>0
2) a<1

I manipulated to (a+1)/a > 2 then took pos and neg cases since we dont know the sign of a from the base a>1 or a<1.
Statement 1 s then not enough, and Statement 2 matches one of the cases i found from the base so I picked B. OA is C. Clearly I did something wrong...
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09 Mar 2012, 17:35
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Hi, there. I'm happy to help a bit with this.

QUESTION #1

Yes, it's true that, as long as a=/0 and b=/0, then |a + b| =< |a| + |b|. That's an actual math rule. First of all, notice it's not a "less than" sign, but a "less than or equal to" sign --- that's very important. This idea is closely related to something Geometry known as the Triangle Inequality. The basic logic is ---if a and b are both positive, or both negative, then the two sides will be the same, but if a and b have opposite sign, they will get subtracted on the left side, while their absolute values will be added on the right side. Example |(-3) + 5| = |2| = 2, but |(-3)| + |5| = 3 + 5 = 8. Does that make sense?

Caveat: this is always true, but not necessarily all that useful in problem-solving. If you have a DS question asking explicitly about comparing |a + b| to |a| + |b|, then this formula would be relevant, but in a general absolute value question, you are much better off enumerating cases, as you say you do anyway.

QUESTION #2

Yes, using a number line to mark +/- cases is the standard and most widely used method of approaching quadratic inequalities. It is a 100% valid and legitimate approach. If this method makes sense to you, there is no disadvantage to using it every single time.

QUESTION #3

OK, a DS question.

Prompt: Is a + a^-1 > 2?

Here's an algebraic approach. The big idea is --- when you have an x + 1/x, that is, an x to the +1 and an x to the -1, those are exponents that differ by two --- that means, in hidden form, you are dealing with a quadratic. That's what I will bring out in the solution below.

Subtract 2 from both sides:

a - 2 + 1/a > 0

Factor out an 1/a

(1/a)*(a^2 - 2a + 1) > 0 we have the square of a binomial!

(1/a)*(a - 1)^2 > 0

So, of these two factors, (1/a) is positive when a > 0 and negative when a < 0. By contrast, (a - 1)^2 is positive for all real number a EXCEPT when a = 1 --- that makes the expression equal to zero.

Statement #1: a > 0
The expression will be positive for all values of a except at a = 0, in which case the left side equals zero, and the inequality is false. So, a continuous infinity of possible values of a would make the prompt question true, and only one would make it false, but that's enough to say we can't answer the prompt definitively. Statement #1 by itself is insufficient.

Statement #2: a <1
The good news is --- the dreaded value of a = 1 is avoided, so the (a - 1)^2 is always positive. BUT, now a can be less than zero, and if a is negative, then 1/a is also negative, and that would make the inequality false. Again, we don't have a definitive answer. Statement #2 by itself is insufficient.

Statements #1 & #2 combined:
Now, we know 0 < a < 1. We avoid the dreaded a = 1 value, so (a - 1)^2 is always positive, and because a has to be greater than zero, we know that 1/a is also always positive. Therefore, the product of those two terms is always positive, and the inequality is always true. Combined, the statements are sufficient.

[Reveal] Spoiler:

I realize this is probably a very different approach from the one you took to solve this DS. If you'd like to post your work, I can help you figure out where you went wrong and what mistakes to avoid in the future.

Here's another related GMAT DS question, for more practice.

http://gmat.magoosh.com/questions/999

The question at that link, when you submit your answer, will be followed by a page with a complete video explanation.

Does everything I've said this make sense? Please let me know if you have any questions on what I've said, or if I can help you in any other way.

Mike
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Mike McGarry
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11 Mar 2012, 16:52
mikemcgarry wrote:
Hi, there. I'm happy to help a bit with this.

QUESTION #1

Caveat: this is always true, but not necessarily all that useful in problem-solving. If you have a DS question asking explicitly about comparing |a + b| to |a| + |b|, then this formula would be relevant, but in a general absolute value question, you are much better off enumerating cases, as you say you do anyway.

QUESTION #3

OK, a DS question.

Prompt: Is a + a^-1 > 2?

Here's an algebraic approach. The big idea is --- when you have an x + 1/x, that is, an x to the +1 and an x to the -1, those are exponents that differ by two --- that means, in hidden form, you are dealing with a quadratic. That's what I will bring out in the solution below.

Subtract 2 from both sides:

a - 2 + 1/a > 0

Factor out an 1/a

(1/a)*(a^2 - 2a + 1) > 0 we have the square of a binomial!

(1/a)*(a - 1)^2 > 0

So, of these two factors, (1/a) is positive when a > 0 and negative when a < 0. By contrast, (a - 1)^2 is positive for all real number a EXCEPT when a = 1 --- that makes the expression equal to zero.

[Reveal] Spoiler:

I realize this is probably a very different approach from the one you took to solve this DS. If you'd like to post your work, I can help you figure out where you went wrong and what mistakes to avoid in the future.

Here's another related GMAT DS question, for more practice.

http://gmat.magoosh.com/questions/999

The question at that link, when you submit your answer, will be followed by a page with a complete video explanation.

Does everything I've said this make sense? Please let me know if you have any questions on what I've said, or if I can help you in any other way.

Mike

Thanks for the thoughful reply, Mike.

1) Thanks for the insight here. Is the cleanest way to test an abs value problem with multiple variables to just plug in numbers and check? It seems tough algebraically...

2) I'll keep using the line then. Thanks,

3) Your feedback on recognizing this as a quadratic helped a ton. I think the problem I keep running into with this type of problem is the rule set for switching the inequality. I think I am confusing inequalities with abs values and those without. Now that I am factoring properly, here is how I solved it. Instead of factoring out the 1/a as you did, I multiplied the a through and made equations for the positive and negative cases (here is where confusion sometimes sets in) and got a^2-2a>1 and a^2-2a<1. From there I just solved as you did using the number line concept.

When putting together the inequalities for positive and negative cases, if there are no abs values, you multiple through the same as with the positive case, but just switch the sign, right? And with abs values, you solve once for the postive case and for the negagtive you slap a minus on everything that was in the brackets, flippling the sign only if you multiple the other side by a negative, correct? It seems like my brain fluctuates begin "getting it" ad "not". This type of problem seems to be my only weakness in quant...

Thanks again for the help! Kudos!
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25 May 2012, 04:11
Hi,

For the below expression:
|a| + |b| > |a + b|
You may consider squaring both the sides:
LHS = $$a^2 + b^2 + 2|a||b|$$
RHS = $$a^2 + b^2 + 2ab$$

Now based on sign of a & b, $$|ab| >= ab$$

Thus, LHS > RHS.
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