The Holistic Approach to Absolute Values – Part IV

Quarter Wit, Quarter WisdomLast week, we looked at some absolute value questions involving inequalities. Today, we’ll continue this discussion by adding some more complications to our questions. Consider the question: What is the minimum value of the expression |x – 3| + |x + 1| + |x|? Technically, |x – 3| + |x + 1| + |x| is the sum of “the distance of x from 3,” “the distance of x from -1” and “the distance of x from 0.” To make solving such questions simpler, we’ll often use a parallel situation:

Imagine that there are 3 friends with houses at points -1, 0 and 3 in a straight line. They decide to meet at the point x.

  • |x – 3| will be the distance covered by the friend at 3 to reach x.
  • |x + 1| will be the distance covered by the friend at -1 to reach x.
  • |x| will be the distance covered by the friend at 0 to reach x.

So, the total distance the friends will cover to meet at x will be |x – 3| + |x + 1| + |x|.

Now we can choose to minimize this total distance, bring it to some particular value or make it more or less than some particular value.

If we want to minimize the total distance, we just make the friends meet at the second guy’s house, i.e. at the point 0. The friend at 3 and the friend at -1 need to travel 4 units total to meet anyway, so there’s no point in making the guy at 0 travel any distance at all. So the minimum total distance would be 4, which would then be the minimum value of |x – 3| + |x + 1| + |x|. This minimum value is given by the expression at x = 0.

With this in mind, when we move to the right or to the left of x = 0, the total distance will increase and, hence, the value of the expression |x – 3| + |x + 1| + |x| will also increase.

Thereafter, it is easy to solve for |x – 3| + |x + 1| + |x| = 10 or |x – 3| + |x + 1| + |x| < 10, etc., as seen in our previous post.

Today, let’s look at how to solve a more advanced GMAT question using the same logic:

For how many integer values of x, is |x – 5| + |x + 1| + |x|  + |x – 7| < 15?

(A) 0
(B) 2
(C) 4
(D) 6
(E) Infinite

In our parallel situation of friends and houses, we now have 4 friends with houses at points -1, 0, 5 and 7.

The friends at -1 and 7 are 8 units apart, so they will need to cover at least this total distance together to meet. It doesn’t matter where they meet between -1 and 7 (inclusive), they will need to cover exactly 8 units.

The friends at 0 and 5 will need to travel a minimum distance of 5 to meet. They can meet anywhere between 0 and 5 (inclusive) and the distance they will cover will still be 5.

So, all four friends can meet anywhere between 0 and 5 (inclusive) and the total distance covered will be 8 + 5 = 13. This would be the minimum total distance, and hence, the minimum value of the expression |x – 5| + |x + 1| + |x|  + |x – 7|.

When we move to the left of 0 or to the right of 5, the total distance covered will be more than 13. At any point between -1 and 7, the total distance covered by the friends at -1 and 7 will be only 8. When we move 1 unit to the left of 0 and reach -1, the total distance covered by the friends at 0 and 5 will be 1 + 6 = 7. So to meet at -1, the total distance traveled by all friends together will be 8 + 7 = 15.

Similarly, when we move 1 unit to the right of 5 and reach 6, the total distance covered by the four friends will be again 8 + 7 = 15. So at points x = -1 and x = 6, the value of the expression will be 15. Between these two points (excluding the points themselves), the value of the expression will be less than 15.

So now we know -1 < x < 6. With these parameters, x can take 6 integer values: 0, 1, 2, 3, 4, 5. Therefore, the answer is D.

Note that when we had 3 points on the number line, the minimum total distance was found at the second point. Now when we have 4 points on the number line, the minimum total distance has been found to be in the range between second and third points.

Let’s look at another question:

For how many integer values of x, is |2x – 5| + |x + 1| + |x| < 10?

(A) 1
(B) 2
(C) 4
(D) 5
(E) Infinite

|2x – 5| + |x + 1| + |x| < 10

2*|x – 5/2| + |x + 1| + |x| < 10

In this sum, now the distance from 5/2 is added twice.

In our parallel situation, this is equivalent to two friends living at 5/2, one living at 0 and one living at -1. Now note that the expression may not take the minimum value of x = 0 because there are 2 people who will need to travel from 5/2.

We have four friends in all, so we can expect to get a range in which we will get the minimum value of the expression. The second and third people are at 0 and 5/2, respectively.

The total distance at x = 0 will be 1 + 2*(5/2) = 6. The total distance at x = 5/2 will be 7/2 + 5/2 = 6.

So if we move to the left of 0 or to the right of 5/2, the total distance will increase. If we move 1 unit to the right of 5/2 and reach 7/2, the total distance covered by the four friends will be 9/2 + 7/2 + 2 = 10. If we move 1 unit to the left of 0 and reach -1, the total distance covered by the four friends will be 0 + 1 + 2*(7/2) = 8. Now all four friends are at -1. To cover a distance of another 2, they should move another 0.5 units to the left of -1 to reach -1.5.

Now the total distance covered by the four friends will be 0.5 + 1.5 + 2*4 = 10, so the total distance when x lies between the points -1.5 and 3.5 (excluding the points themselves) will be less than 10.

Now we know -1.5 < x < 3.5. With these parameters, x can take 5 integer values: -1, 0, 1, 2 and 3. Therefore, the answer is D.

Now use these concepts to solve the following question: For how many integer values of x, is |3x – 3| + |2x + 8| < 15?

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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!