Mockingbird Book Gets a Sequel

To Kill a Mockingbird by Harper Lee occupies a special place in the history of American literature. Published out of the blue, written by a previously unknown author, it became an instant classic, was adapted into film in 1962, translated into more than 40 languages, sold more than 40 million copies and continues to sell about a million copies a year even now. It would have been amazing even by itself, but one fact makes it even more fascinating: Harper Lee retired to her native town of Monroeville, Ala immediately after the book’s release and never wrote again. And yet, now it seems that there will be another Mockingbird book, set for release this July. Or rather, it has been in existence this entire time. The second novel by Harper Lee, titled Go Set a Watchman, is set twenty years after the events of To Kill a Mockingbird and features a number of the same characters, but it is not a sequel per se. According to Harper Lee herself, it is a ‘parent’ to what so far was her first and only novel. Harper Lee wrote Go Set a Watchman in the mid-1950s, before To Kill a Mockingbird. In this book, Scout, the main heroine and protagonist of To Kill a Mockingbird, comes back to her native town to visit her aged father, Atticus Finch. Similarly to the first book, this work is mostly dedicated to depiction of racial tensions in a rural Southern town and father-daughter relationships. But it is not a follow-up – it is rather what To Kill a Mockingbird could have been. It was Harper Lee’s first novel, and it was rejected by the publisher; who, however, was fascinated with the flashbacks to Scout’s childhood and suggested that Ms. Lee should write a full-fledged novel about them. Thus was born To Kill a Mockingbird. But why now? Where has this book been for more than half a century? As it turns out, Harper Lee for a long time thought that the original manuscript was either lost or accidentally destroyed, and it was only recently rediscovered by Tonja Carter, her friend and lawyer. Ms. Lee reportedly wasn’t eager to publish the novel now, but was later persuaded by a small circle of friends that it was worthy of it. There is, however, a number of people who are skeptical about the quality of the book. Charles J. Shields, the author of biography of Harper Lee, believes that what we are going to get is a novel written by Ms. Lee without an editor’s participation when she was, for all practical purposes, an amateur. Another ethical consideration arises in concern of Ms. Lee’s involvement in the book’s publishing. She suffered a stroke in 2007 and has been living in an assisted-living facility ever since. Her sister Alice Lee who protected her from unwanted publicity for all these years, died last autumn. Ms. Lee didn’t make any public announcements concerning the new book personally. All this makes the sudden appearance of this manuscript right now look a little bit too convenient.

How to Go From a 48 to 51 in GMAT Quant - Part VI

Today’s post the next part in our â€Å"How to Go From 48 to 51 in Quant† series. Again, we will learn a technique that can be employed by the test-taker at an advanced stage of preparation   requiring  one to understand the situations in which one can use this simplifying technique. (Before you continue reading, be sure to check out parts I, II, III, IV, and V of this series.) We all love to use the plug-in method on GMAT Quant questions. We have an equation given, and if the answer choices are the possible values of x, we just plug in these values to find the one  that satisfies the equation. But what if the answer choices are all complicated values? What if it seems that five times the calculation (in the worst case) will be far more time consuming than actually solving the given equation? Then one is torn between using the favorite plug-in method and using algebra. Let’s take an example to review the methods we can use to solve the question and learn how to simplify the plug-in process by approximating the five available options: If |4x−4|=|2x+30|, which of the following could be a value of x? (A) –35/3 (B) −21/2 (C) −13/3 (D) 11/5   (E) 47/5 This question is an ideal candidate for the â€Å"plug-in† method. Here, you have the absolute value equation with the potential values of x given in the answer choices. The problem is that the values of x given are fractional. Of course, if we do plan to solve the equation rather than plug-in, we can still solve it using our holistic approach rather than pure algebra. Let’s take a look at that now, and later we will discuss the trick to making the answer choices easier for us to plug in. Method 1: |4x 4| = |2x + 30| 4 * |x 1| = 2 * |x + 15| 2 * |x 1| = |x + 15| This is how we rephrase the equation in our words: twice the distance of x from 1 should be equal to the distance of x from -15. ——————(-15) —————————————————(0)——(1)—————— There are two ways to find the value of x: Case 1: x could be between -15 and 1 such that the distance between them is split in the ratio 2:1. or Case 2: x could be to the right of 1 such that the distance between x and -15 is twice the distance between x and 1. Lets examine both of these cases in further detail: Case 1:  The distance from -15 to 1 is of 16 units   this can be split into  3 sections of 16/3 units each. So, the distance of x from 1 should be 16/3, which would make the distance of x from -15 two times 16/3, i.e. 32/3. So, x should be at a point 16/3 away from 1 toward the left. x = 1 16/3 = -13/3 This is one of our answer choices  and, hence, the correct answer. Normally, we would just move on to the next question at this point, but had we not found -13/3 in the answer options, we would have moved on to  Case 2: Case 2:  The distance between -15 and 1 is 16 units. x should be an additional 16 units to the right of 1, so the distance between x and 1 is 16 and the distance between x and -15 is two times 16, i.e. 32. This means that x should be 16 units to the right of 1, i.e. x = 17.  If you would not have found -13/3 in the answer choices, then you would have found 17. Now let’s move on to see how we can make the plug-in method work for us in this case by examining each answer choice we are given: Method 2: |4x 4| = |2x + 30| 2 * |x 1| = |x + 15| (A) -35/3 It is difficult to solve for x = -35/3 to see if both sides match. Instead, let’s solve for the closest integer, -12. 2 * |-12 1| = |-12 + 15| On the left-hand side, you will get 26, but on the right-hand side, you will get 3. These values are far away from each other, so x cannot be -35/3.   As the value of x approaches the point where the equation holds i.e. where the two sides are equal to each other the gap between the value of the two sides keeps reducing. With such a huge gap between the value of the two sides in this case, it is unlikely that a small adjustment of -35/3 from -12 will bring the two sides to be equal. (B) -21/2 For this answer choice, lets solve for the nearest integer, x = -10. 2 * |-10 1| = |-10 + 15| On the left-hand side, you will get 22; on the right-hand side, you will  get 5. Once again, these values are far away from each other and, hence, x will not be -21/2. (C) -13/3 For this answer choice, lets solve for x = -4. 2 * |-4 -1| = |-4 + 15| On the left-hand side, you will get 10; on the right-hand side, you will get 11. Here, there is a possibility that x can equal -13/3, as the two sides are so close to one another plug in the actual value of -13/3 and you will see that  the left-hand side of the equation does, in fact, equal the right-hand side. Therefore, C is the correct answer. Basically, we approximated the answer choices we were given  and shortlisted the one that gave us very close values. We checked for that and found that it is the answer. We can also solve this question using pure algebra (taking positive and negative signs of absolute values) but in my opinion, the holistic Method 1 is almost always better than that. Out of the two methods discussed above, you can pick the one you like better, but note that Method 2 does have  limited applications only if you are given the actual values of x, can you use  it. Method 1 is far more generic for absolute value questions. Getting ready to take the GMAT? We have  free online GMAT seminars  running all the time. And, be sure to follow us on  Facebook,  YouTube,  Google+, and  Twitter! Karishma, a Computer Engineer with a keen interest in alternative Mathematical approaches, has mentored students in the continents of Asia, Europe and North America. She teaches the  GMAT  for Veritas Prep and regularly participates in content development projects such as this blog!