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07 Jan 2015, 07:52
| FROM Veritas Prep Blog: Properties of Absolute Values on the GMAT |
 We have talked about quite a few concepts involving absolute value of x in our previous posts. But some absolute value questions involve two variables. Then do we need to consider the positive and negative values of both x and y? Certainly! But there are some properties of absolute value that could come in handy in such questions. Let’s take a look at them: (I) For all real x and y, |x + y| <= |x| + |y| (II) For all real x and y, |x – y| >= |x| – |y| We don’t need to learn them of course and there is no need to look at how to prove them either. All we need to do is understand them – why do they hold, when is the equality sign applicable and when can they be useful. Let’s look at both the properties one by one. (I) For all real x and y, |x + y| <= |x| + |y| The result of both the left hand side and the right hand side will be positive or zero. On the right hand side, the absolute values of x and y will always get added irrespective of the signs of x and y. On the left hand side, the absolute values of x and y might get added or subtracted depending on whether they have the same sign or different signs. Hence the result of the left hand side might be smaller than or equal to that of the right hand side. For which values of x and y will the equality hold and for which values will the inequality hold? Let’s think logically about it. The absolute values of x and y get added on the right hand side. We want the absolute values of x and y to get added on the left hand side too for the equality to hold. This will happen when x and y have the same sign. So the equality should hold when they have the same signs. For example, x = 4, y = 8: |4 + 8| = |4| + |8| = 12 OR x = -3, y = -4: |-3 -4| = |-3| + |-4| = 7 Also, when at least one of x and y is 0, the equality will hold. For example, x = 0, y = 8: |0 + 8| = |0| + |8| = 8 OR x = -3, y = 0: |-3 + 0| = |-3| + |0| = 3 What happens when x and y have opposite signs? On the left hand side, the absolute values of x and y get subtracted hence the left hand side will be smaller than the right hand side (where they still get added). That is when the inequality holds i.e. |x + y| < |x| + |y| For example, x = -4, y = 8: |-4 + 8| < |-4| + |8| 4 < 12 OR x = 3, y = -4: |3 -4| < |3| + |-4| 1 < 7 Let’s look at our second property now: (II) For all real x and y, |x – y| >= |x| – |y| Thinking on similar lines as above, we see that the right hand side of the inequality will always lead to subtraction of the absolute values of x and y whereas the left hand side could lead to addition or subtraction depending on the signs of x and y. The left hand side will always be positive whereas the right hand side could be negative too. So in any case, the left hand side will be either greater than or equal to the right hand side. When will the equality hold? When x and y have the same sign and x has greater (or equal) absolute value than y, both sides will yield a positive result which will be the difference between their absolute values For example, x = 9, y = 2; |9 – 2| = |9| – |2| = 7 OR x = -7, y = -3 |-7 – (-3)| = |-7| – |-3| = 4 Also when y is 0, the equality will hold. For example, x = 8, y = 0: |8 – 0| = |8| – |0| = 8 OR x = -3, y = 0: |-3 – 0| = |-3| – |0| = 3 What happens when x and y have the same sign but absolute value of y is greater than that of x? It is easy to see that in that case both sides have the same absolute value but the right hand side becomes negative. For example, x = -4, y = -9 |x – y| = |-4 – (-9)| = 5 |x| – |y| = |-4| – |-9| = -5 So even though the absolute values will be the same since we will get the difference of the absolute values of x and y on both sides, the right hand side will be negative. If we were to take further absolute value of the right hand side, the two will become equal i.e. the right hand side will become |(|x| – |y|)| = |-5| = 5 in our example above. In that case, the equality will hold again. Similarly, what happens when only x = 0? The right hand side becomes negative again so taking further absolute value will make both sides equal. For example, x = 0, y = -5 |x – y| = |0 – (-5)| = 5 |x| – |y| = |0| – |5| = -5 Taking further absolute value, |(|x| – |y|)| = |-5| = 5 So when we take further absolute value of the right hand side, this property becomes similar to property 1 above: |x – y| = |(|x| – |y|)| when x and y have the same sign or at least one of x and y is 0. Now let’s look at the inequality part of property 2. Whenever x and y have opposite signs, |x – y| > |x| – |y| On the left hand side, the absolute values will get added while on the right hand side, the absolute values will get subtracted. So the absolute value of the left hand side will always be greater than the absolute value of the right hand side. The left hand side will always be positive while the right hand side could be negative too. Hence even if we take the further absolute value of the right hand side, the inequality will hold: |x – y| > |(|x| – |y|)| when x and y have opposite signs For example, x = -4, y = 8: |-4 – 8| > |-4| – |8| 12 > -4 Taking further absolute value of the right hand side, we get |(|x| – |y|)| = |-4| = 4 Still, 12 > 4 i.e. |x – y| > |(|x| – |y|)| OR x = 3, y = -4: |3 –(-4)| > |3| – |-4| 7 > -1 Taking further absolute value of the right hand side, we get |(|x| – |y|)| = |-1| = 1 Still, 7 > 1 i.e. |x – y| > |(|x| – |y|)| Note that the inequality of the original property 2 also holds when x and y have the same sign but absolute value of y is greater than the absolute value of x since the right hand side becomes negative. It also holds when x is 0 but y is not. To sum it all neatly, (I) For all real x and y, |x + y| <= |x| + |y| |x + y| = |x| + |y| when (1) x and y have the same sign (2) at least one of x and y is 0. |x + y| < |x| + |y| when (1) x and y have opposite signs (II) For all real x and y, |x – y| >= |x| – |y| |x – y| = |x| – |y|when (1) x and y have the same sign and x has greater (or equal) absolute value than y (2) y is 0 |x – y| > |x| – |y| in all other cases (III) For all real x and y, |x – y| >= |(|x| – |y|)| |x – y| = |(|x| – |y|)| when (1) x and y have the same sign (2) at least one of x and y is 0. |x – y| > |(|x| – |y|)| when (1) x and y have opposite signs Note that property (III) matches property (I). There is another property we would like to discuss but let’s take it up next week along with some GMAT questions where we put these properties to use. 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! |
14 Jan 2015, 02:10
| FROM Veritas Prep Blog: An Official Question on Absolute Values |
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Now that we have discussed some important absolute value properties, let’s look at how they can help us in solving official questions. Knowing these basic properties can help us quickly analyze the question and arrive at the answer without getting stuck in analyzing different ranges, a cumbersome procedure. First we will look at a GMAT Prep question. Question 1: Is |m – n| > |m| – |n|? Statement 1: n < m Statement 2: mn < 0 Solution 1: Recall the property number 2 Property 2: For all real x and y, |x – y| >= |x| – |y| |x – y| = |x| – |y| when (1) x and y have the same sign and x has greater (or equal) absolute value than y (2) y is 0 |x – y| > |x| – |y| in all other cases So if m and n have the same sign with |m| >= |n|, equality will hold. Also, if n is 0, equality will hold. If we can prove that both these conditions are not met, then we can say that |m – n| is definitely greater than |m| – |n|. Statement 1: n < m We have no idea about the signs of m and n. Are they same? Are they opposite? We don’t know. Also n may or may not be 0. Hence we don’t know whether the equality will hold or the inequality. Statement 1 alone is not sufficient to answer the question. Statement 2: mn < 0 Since mn is negative, it means one of m and n is positive and the other is negative. This also implies that n is definitely not 0. So we know that m and n do not have the same sign and n is not 0. So under no condition will the equality hold. Hence |m – n| is definitely greater than |m| – |n|. Statement 2 alone is sufficient to answer the question. Answer (B) Let’s look at one more question now. Question 2: If xyz ≠ 0, is x(y + z) >= 0? Statement 1: |y + z| = |y| + |z| Statement 2: |x + y| = |x| + |y| Solution 2: xyz ≠ 0 implies that all, x, y and z, are non zero numbers. Question: Is x(y + z) >= 0? If we can prove that x(y + z) is not negative that is x and (y+z) do not have opposite signs, we can say that x(y + z) is positive or 0. Looking at the statements given, let’s review our property number 1: Property 1: For all real x and y, |x + y| <= |x| + |y| |x + y| = |x| + |y| when (1) x and y have the same sign (2) at least one of x and y is 0. |x + y| < |x| + |y| when (1) x and y have opposite signs The two statements give us equalities which means that the relevant part of the property is this: |x + y| = |x| + |y| when (1) x and y have the same sign (2) at least one of x and y is 0. We are also given in the question stem that x, y and z are not 0. Hence, given |x + y| = |x| + |y|, we can infer that x and y have the same sign. Statement 1: |y + z| = |y| + |z| This implies that y and z have the same signs. But we have no information about the sign of x hence this statement alone is not sufficient. Statement 2: |x + y| = |x| + |y| This implies that x and y have the same signs. But we have no information about the sign of z hence this statement alone is not sufficient. Using both statements together, we know that x, y and z have the same sign. Whatever is the sign of y and z, the same will be the sign of (y+z). Hence x and (y+z) have the same sign. This implies that x(y + z) cannot be negative. Hence we can answer our question with a definite ‘yes’. Answer (C). Mind you, both these questions can get time consuming (even though they aren’t really tough) if you don’t understand these properties well. You can certainly start your thinking from the scratch, arrive at the properties and then proceed or resort to more desperate measures such as number plugging but that is best avoided in DS questions. 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! |
14 Jan 2015, 02:07
| FROM Veritas Prep Blog: Properties of Absolute Values on the GMAT - Part II |
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We pick up this post from where we left the post of last week in which we looked at a few properties of absolute values in two variables. There is one more property that we would like to talk about today. Thereafter, we will look at a question based on some of these properties. (III) |x – y| = 0 implies x = y x and y could be positive/negative integer/fraction; if the absolute value of their difference is 0, it means x = y. They cannot have opposite signs while having the same absolute value. They must be equal. This also means that if and only if x = y, the absolute value of their difference will be 0. Mind you, this is different from ‘difference of their absolute values’ |x| – |y| = 0 implies that the absolute value of x is equal to the absolute value of y. So x and y could be equal or they could have opposite signs while having the same absolute value. Let’s now take up the question we were talking about. Question: Is |x + y| < |x| + |y|? Statement 1: | x | ≠ | y | Statement 2: | x – y | > | x + y | Solution: One of the properties we discussed last week was “For all real x and y, |x + y| <= |x| + |y| |x + y| = |x| + |y| when (1) x and y have the same sign (2) at least one of x and y is 0. |x + y| < |x| + |y| when (1) x and y have opposite signs” We discussed in detail the reason absolute values behave this way. So our question “Is |x + y| < |x| + |y|?” now becomes: Question: Do x and y have opposite signs? We do not care which one is greater – the one with the positive sign or the one with the negative sign. All we want to know is whether they have opposite signs (opposite sign also implies that neither one of x and y can be 0)? If we can answer this question definitively with a ‘Yes’ or a ‘No’, the statement will be sufficient to answer the question. Let’s go on to the statements now. Statement 1: | x | ≠ | y | This statement tells us that absolute value of x is not equal to absolute value of y. It doesn’t tell us anything about the signs of x and y and whether they are same or opposite. So this statement alone is not sufficient. Statement 2:| x – y | > | x + y | Let’s think along the same lines as last week – when will | x – y | be greater than | x + y |? When will the absolute value of subtraction of two numbers be greater than the absolute value of their addition? This will happen only when x and y have opposite signs. In that case, while subtracting, we would actually be adding the absolute values of the two and while adding, we would actually be subtracting the absolute values of the two. That is when the absolute value of the subtraction will be more than the absolute value of the addition. For Example: x = 3, y = -2 | x – y | = |3 – (-2)| = 5 | x + y | = |3 – 2| = 1 or x = -3, y = 2 | x – y | = |-3 – 2| = 5 | x + y | = |-3 + 2| = 1 If instead, x and y have the same sign, | x + y | will be greater than| x – y |. If at least one of x and y is 0, | x + y | will be equal to| x – y |. Since this statement tells us that | x – y | > | x + y |, it implies that x and y have opposite signs. So this statement alone is sufficient to answer the question with a ‘Yes’. Answer (B) Takeaway from this question: If x and y have the same signs, | x + y | >| x – y |. If x and y have opposite signs, | x + y | <| x – y |. If at least one of x and y is 0, | x + y | =| x – y |. You don’t need to ‘learn this up’. Understand the logic here. You can easily recreate it in the exam if need be. 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! |
13 Oct 2016, 00:50
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