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20 May 2015, 03:50
| FROM Veritas Prep Blog: The Reason Behind Absolute Value Questions on the GMAT |
 Even after working extensively on absolute value questions, sometimes students come up with “why?” i.e. why do we have to take positive and negative values? Why do we have to consider ranges etc. They know the process but they do not understand the reason they need to follow the process. So here today, in this post, we will try to explain the reason. You know how to solve an equation such as x + 2x = 4. Simple enough, right? Just add x with 2x to get 3x and separate out the x on one side. But what do you do when you have an equation with absolute values? How will you handle that equation? Say, you have |x| + 2x = 4. Is this your regular equation? No! You CANNOT say that x + 2x = 4 => 3x = 4 => x = 4/3. You have an absolute value and that complicates matters. You need to get rid of it to get a solution for x. How do you get rid of absolute values? The definition of absolute value helps us here: |x| = x if x >= 0 |x| = -x if x < 0 So you can substitute x for |x| to make it a regular equation but only if x is non negative. If x is negative, then you put -x instead of |x| to convert it into a simple equation. And that is the reason you need to take positive and negative values of what is inside the absolute value sign. Similarly, |x-5| = (x-5) if (x-5) >= 0 i.e. if x >= 5 |x-5| = -(x-5) if (x-5) < 0 i.e. if x < 5 Let’s go back to the previous example and see how we can get rid of the absolute value to make it a regular equation: Question 1: What is the value of x given |x| + 2x = 4? We don’t know whether x is positive or negative so we will look at what happens in both cases: Case 1: x is positive or 0 If x >= 0 then equation becomes x + 2x = 4 => x = 4/3 Our initial condition is that x is non negative. We get a positive solution on solving it and hence 4/3 is a valid solution. Case 2: x is negative If x < 0 then equation becomes -x + 2x = 4 => x = 4 Our initial condition is that x is negative. We get a positive solution on solving it and hence x = 4 is not a valid solution. Had we obtained a negative solution, it would have been valid. So there is only one solution x = 4/3. We hope the entire process makes more sense now. Let’s follow it up with a complex question from our algebra book. Question 2: If x and y are integers and y = |x+3| + |4-x|, does y equal 7? Statement 1: x < 4 Statement 2: x > -3 Solution: Now what do you do when you have y = |x+3| + |4-x|? How do you convert this into a regular equation? You don’t know whether whatever is in the absolute value sign is positive or negative. How will you get rid of the sign then? You will work on all the cases (messy algebra coming up!). Now, we see the same logic in this question: y = |x+3| + |4-x| |x+3| = (x+3) if (x+3) >= 0. In other words, if x >= -3 |x+3| = -(x+3) if (x+3) < 0. In other words, if x < -3 |4-x| = (4-x) if (4-x) >= 0. In other words, if x <= 4 |4-x| = -(4-x) if (4-x) < 0. In other words, if x > 4 So our absolute values behave differently when x < -3, between -3 and 4 and when x > 4. We say that -3 and 4 are our transition points. Case 1: When x < -3, |x+3| = -(x+3) and |4-x| = (4-x). So the equation becomes y = -(x+3) + (4-x) y = 1 – 2x For different values of x, y will take different values. Recall that x must be less than -3. Say x = -4, then y = 9. If x = -5, y = 11. Case 2: When -3 <= x <= 4, |x+3| = (x+3) and |4-x| = (4-x). So the equation becomes y = (x+3) + (4-x) y = 7 In this range, y will always be 7. Case 3: When x > 4, |x+3| = (x+3) and |4-x| = -(4-x) So the equation becomes y = (x+3) – (4-x) y = 2x – 1 For different values of x, y will take different values. Recall that x must be more than 4. Say x = 5, then y is 9. If x = 6, then y is 11. Note that y equals 7 only when x is between -3 and 4. Both statements together tell us that x is between -3 and 4. No statement alone gives us this information. Hence, using both statements, we know that y must be 7. Answer (C) 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! |
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