{"id":35126,"date":"2016-11-21T12:36:27","date_gmt":"2016-11-21T19:36:27","guid":{"rendered":"https:\/\/gmatclub.com\/blog\/2016\/11\/absolute-value-inequalities\/"},"modified":"2016-11-21T12:36:27","modified_gmt":"2016-11-21T19:36:27","slug":"absolute-value-inequalities","status":"publish","type":"post","link":"https:\/\/gmatclub.com\/blog\/absolute-value-inequalities\/","title":{"rendered":"Absolute Value Inequalities"},"content":{"rendered":"

This is an extremely difficult question category that appears only infrequently on the GMAT. First, a couple of practice problems.<\/p>\n

 <\/p>\n

\"absolute<\/p>\n

1) The dark purple region on the number line above is shown in its entirety.\u00a0This region is delineated by which of the following inequalities?<\/p>\n

(A) 10 < |x + 10| < 80<\/p>\n

(B) 10 < |x \u2013 100| < 80<\/p>\n

(C) |x \u2013 20| < 70<\/p>\n

(D) |x \u2013 20| < | x \u2013 90|<\/p>\n

(E) |x \u2013 55| < 35<\/p>\n

 <\/p>\n

 <\/p>\n

2) If |x| < 20 and |x \u2013 8| > |x + 4|, which of the following expresses the allowable range for x?<\/p>\n

(A) \u201312 < x < 12<\/p>\n

(B) \u201320 < x < 2<\/p>\n

(C) \u201320 < x < \u201312 and 12 < x < 20<\/p>\n

(D) \u201320 < x < \u20138 and 4 < x < 20<\/p>\n

(E) \u201320 < x < \u20134 and 8 < x < 20<\/p>\n

 <\/p>\n

 <\/p>\n

3) If |(x \u2013 3)2<\/sup> + 2| < |x \u2013 7| , which of the following expresses the allowable range for x?<\/p>\n

(A) 1 < x < 4<\/p>\n

(B) 1 < x < 7<\/p>\n

(C) \u2013 1 < x < 4 and 7 < x<\/p>\n

(D) x < \u2013 1 and 4 < x < 7<\/p>\n

(E) \u2013 7 < x < 4 and 7 < x<\/p>\n

 <\/p>\n

Explanations to these will follow the discussion.<\/p>\n

A hard question type<\/h2>\n

I want to make clear that this question type is very tricky, something that might pose challenges even to students relatively strong in math.\u00a0I also want to make clear that we could send 50 people to take 50 separate GMATs, and it may be that not one of them would see such a question on their GMAT. This is a very infrequent question genre, and it would only be asked if you were acing the Quant sections and the CAT<\/a> were throwing everything including the kitchen sink at you!<\/p>\n

If you find what I say in this blog article helpful, that’s great: it may give you some insight into other, easier questions involving absolute values and\/or inequalities.\u00a0If you are still confused, don’t worry: you are rather unlikely to see this question type on test day.<\/p>\n

Thinking about absolute values<\/h2>\n

Fundamentally, the absolute value is about distance, and thinking about it geometrically is often the key to difficult absolute value questions<\/strong>. The na\u00efve understanding of absolute value is that it “makes things positive.”\u00a0While this is undeniably true, it is not always the most mathematically productive way to understand the concept.\u00a0It’s undeniably true that |+5| = +5 and that |-5| = +5.\u00a0 We could say that both are true because the absolute value “makes everything positive,” or we could say that these two numbers, +5 and \u20135, have a distance of 5 from zero on the number line.<\/p>\n

This geometric interpretation about distance becomes considerably more important when we are dealing with algebra.\u00a0The expression |x| is the distance from x to zero.\u00a0The expression |x \u2013 6| is the distance from x to +6.\u00a0The expression |x + 2| is the distance from x to \u2013 2: to understand this, remember that x + 2 = x \u2013 (\u2013 2).\u00a0In general, the expression |x \u2013 c| is the distance from any x to the fixed point x = c.<\/p>\n

This distance interpretation replaces a whole lot of complicated calculations.\u00a0 For example, if we had to solve the inequality |x \u2013 10| \u2264 |x \u2013 20|, that’s just saying that the distance from x to 20 is smaller than the distance from x to 10\u2014in other words, x is closer to 20 than to 10,<\/em> and because of the “or equal” part, it could be an equal distance also.\u00a0 Well, the point x = 15 is the only point on the number line that is equidistant from x = 10 and x = 20.\u00a0 If x has to be either an equal distance from 10 and 20, or closer to 20, then it has to be at the point x = 15 or to the right.\u00a0 Any point between 15 and 20 is closer to 20 than to 10, and any point to the right of 20 has to be closer to 20 than to 10.\u00a0 Thus, the solution is 15 \u2264 x.<\/p>\n

\"region-x-greater-than-or-equal-to-15\"<\/p>\n

What region is denoted by the inequality |x \u2013 25| > 10?\u00a0 Well, all we are saying is that we have to be more than 10 units away from the point x = 25.\u00a0 If we go 10 units to the left, we get to x = 15: we can’t be here, at a distance equal to 10, but we could be the left of x = 15, at a distance of more than 10.\u00a0 If we go 10 units to the right, we get to x = 35: we can’t be here, at a distance equal to 10, but we could be the right of x = 35, at a distance of more than 10.\u00a0 Thus, the solution includes the pair of regions x < 15 and 35 < x.<\/p>\n

\"more-than-ten-from-25\"<\/p>\n

Notice that x = 25 is the point of symmetry of this diagram, since everything is about distance from that point.<\/p>\n

Summary of absolute value inequalities<\/h2>\n

The perspective above may give you some insight into this problem.\u00a0 If you had any “aha” moments reading this, then you might want to give the practice problems another try before looking at the solutions below.\u00a0 The solutions may afford you a few more insights into problem solving.<\/p>\n

\"interlaced-five-petal-flowers\"<\/p>\n

Practice Problem Explanations<\/h2>\n

1) Step one: find the midpoint of the region.\u00a0 The midpoint, halfway between and 20 and 90, is 55.\u00a0 In other words, 20 and 90 have the same distance from 55, a distance of 35.\u00a0 These endpoints are not included, but the region includes all the points that have a distance from from x = 55 that is less than 35.\u00a0 Translating that into math, we get the following:<\/p>\n

|x \u2013 55| < 35<\/p>\n

Answer = (E)<\/strong><\/p>\n

 <\/p>\n

2) Some folks might think this involves a sophisticated calculation, but much of this can be done with simple spatial analysis.\u00a0 Look at the second inequality, the more complicated one: |x \u2013 8| > |x + 4|.\u00a0 All this says is that we are looking for points such that the distance to x = 8 is greater than the distance from x = \u20134; in other words, we want all the points that are closer to x = \u20134 and farther from x = 8.<\/p>\n

The midpoint between x = \u20134 and x = 8 is the point x = 2.\u00a0 This point is not included because it’s equidistant from both points, but everything to the left of this point on the number line is closer to x = \u20134 than it is to x = 8.\u00a0 That entire complicated inequality simplifies to x < 2.<\/p>\n

Combine that with the first inequality, |x| < 20, which in the negative realm means that x must be greater than \u201320.\u00a0 Thus, the allowed region is \u201320 < x < 2.<\/p>\n

Answer = (B)<\/strong><\/p>\n

 <\/p>\n

3) For this one, we need to be clever in a few ways.\u00a0 First of all, the expression inside the absolute value on the left is of the form [a square] + [a positive number].\u00a0 Anything squared is either zero or positive, because something squared can never be negative. When we add a positive number, we are guaranteed that it is always positive.\u00a0 Thus, the absolute values around it are entirely superfluous, because it’s always positive anyway.\u00a0 We can remove those absolute value signs on the left with changing the mathematical meaning of the statement one bit.<\/p>\n

\"te-3-first-part\"<\/p>\n

The other expression, (x \u2013 7), could be positive or negative, depending on the value of x, so it may equal +(x \u2013 7) or it may equal \u2013(x \u2013 7).\u00a0 We have to investigate either case.<\/p>\n

\"case-i-part-1\"<\/p>\n

Think about the graph of a parabola.<\/p>\n

\"case-i-parabola\"<\/p>\n

This is an upward opening parabola, one that goes up on both sides.\u00a0 As you may know,<\/p>\n

\"axis-of-symmetry-rule\"<\/p>\n

Furthermore, the vertex of the parabola is on this axis of symmetry.\u00a0 For this particular parabola, we get x = \u2013(\u20137)\/(2*1) = 7\/2 as the axis of symmetry.\u00a0 Plug this value in to find the height of the vertex.<\/p>\n

\"case-i-part-2\"<\/p>\n

We don’t have to solve that last expression.\u00a0 The fraction 49\/4 is between 12 and 13, so when this is subtracted from 18, we get something between 5 and 6.\u00a0 In other words, the vertex of this parabola starts above<\/em> the x-axis and continues up<\/em>.\u00a0 In other words, it is never negative, never less than zero.\u00a0 This means that the Case I inequality has no solutions<\/strong>.<\/p>\n

\"case-ii-part-1\"<\/p>\n

We can solve the equation to find the boundary points.\u00a0 We will have to factor the quadratic<\/a>.<\/p>\n

\"case-ii-part-2\"<\/p>\n

Think about the graph of this parabola.<\/p>\n

\"case-ii-parabola\"<\/p>\n

This is also an upward opening parabola, also one that goes up on both sides.\u00a0 It equals zero at x = 1 and x = 4, so it must have its vertex between them, and it must be negative between those two points and positive to the left of x = 1 and to the right of x = 4.\u00a0 Thus, the solution of the Case II inequality is 1 < x < 4<\/strong>.<\/p>\n

Since Case I had no solution, the Case II solution is the whole shebang.<\/p>\n

Answer = (A) <\/strong><\/p>\n

The post Absolute Value Inequalities<\/a> appeared first on Magoosh GMAT Blog<\/a>.<\/p>\n","protected":false},"excerpt":{"rendered":"

This is an extremely difficult question category that appears only infrequently on the GMAT. First, a couple of practice problems.   1) The dark purple region on the number line…<\/p>\n","protected":false},"author":133,"featured_media":0,"comment_status":"open","ping_status":"1","sticky":false,"template":"","format":"standard","meta":{"footnotes":""},"categories":[9,783,243,940],"tags":[],"class_list":["post-35126","post","type-post","status-publish","format-standard","hentry","category-gmat","category-magoosh-blog","category-blog","category-gmat-prep-gmat","entry"],"aioseo_notices":[],"_links":{"self":[{"href":"https:\/\/gmatclub.com\/blog\/wp-json\/wp\/v2\/posts\/35126","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/gmatclub.com\/blog\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/gmatclub.com\/blog\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/gmatclub.com\/blog\/wp-json\/wp\/v2\/users\/133"}],"replies":[{"embeddable":true,"href":"https:\/\/gmatclub.com\/blog\/wp-json\/wp\/v2\/comments?post=35126"}],"version-history":[{"count":0,"href":"https:\/\/gmatclub.com\/blog\/wp-json\/wp\/v2\/posts\/35126\/revisions"}],"wp:attachment":[{"href":"https:\/\/gmatclub.com\/blog\/wp-json\/wp\/v2\/media?parent=35126"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/gmatclub.com\/blog\/wp-json\/wp\/v2\/categories?post=35126"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/gmatclub.com\/blog\/wp-json\/wp\/v2\/tags?post=35126"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}