There is one more important concept in Modulus that I want to discuss. Once it is done, we will bid farewell to Mods (for the time being at least), I promise. The concept involves dealing with multiple Mod terms (I will explain in just a minute). Before I start with the discussion, let me point out that it is relevant only if you are looking for a 50/51 in Quant and if you are looking for a 50/51 in Quant, then it is definitely relevant (I remember seeing a mean Modulus question in my GMAT a while back). But remember, don’t waste time on advanced Modulus questions if you are uncomfortable with Number Properties or other such high-weightage topics. Only when you are above 48 consistently in Quant, should you spend time on the two posts titled ‘Holistic Approach to Mods’. That said, everyone is welcome to read the posts and elicit his/her own takeaways.

Let’s start.

Our problem for today is:

For what value of x, is |x – 3| + |x + 1| + |x| = 10?

The one good thing about GMAT is that it gives you five options. Here, the five options will be the possible values of x. So obviously, we will not waste time solving this question. We will just plug in the values and find out which value gives the sum 10. So let me write the complete question here:

For what value of x, is |x – 3| + |x + 1| + |x| = 10?

(A) 0

(B) 3

(C) -3

(D) 4

(E) -2

When you put x = 4, you get |4 – 3| + |4 + 1| + |4| = 10. So answer is (D).

When I made this question and was putting the options, obviously I had to solve the question to find the value of x (one option has to be the correct answer after all!) Using the method I will just discuss, I did it in my mind in under a minute! Curiosity piqued? I hope so.

Let us change the question a little to rope you in.

For how many values of x, is |x – 3| + |x + 1| + |x| = 10?

(A) 0

(B) 1

(C) 2

(D) 3

(E) Infinite

(Previously we discussed that Modulus represents distance. In this post, we will build upon that concept.)

In simple language, the question tells us that x is a point on the number line such that the sum of its distance from 3, -1 and 0 is 10. Let’s say, if x = 0,

Distance of 0 from 3 = 3

Distance of 0 from -1 = 1

Distance of 0 from 0 = 0

Sum of distance of 0 from 3, -1 and 0 is 3 + 1 + 0 = 4.

So we know that x is not equal to 0. Now let’s see what happens if x = 3.

Sum of distances of 3 from 3, -1 and 0 is 0 + 4 + 3 = 7. We need 3 more units of distance to make it 10. We need to make x go a little more to the right. Tell me, what happens when x goes 1 unit to the right? By how much will the distance increase? By 3 units! Because x will be 1 unit away from each of the 3 points. When x = 4, sum of distances of 4 from 3, -1 and 0 is 1 + 5 + 4 = 10.

So 4 is definitely one solution for x. What if we go further to the right? Every one unit further to right increases the distance by 3 units. So distance will keep increasing and will never be 10 again on this side of the number line.

Let’s go to the other side. What happens if x = -1? Sum of distances of -1 from 3, -1 and 0 is 4 + 0 + 1 = 5. To increase the distance, we need to go further to left. Remember, the same logic holds here – Every one step to left will increase the distance by 3 units. We need to increase the distance by 5 units. So we take 1 step to the left (reach -2) and then take 2/3^{rd} of a step to the left (reach – 2.667). So x = -2.667 is another solution. Now, every time we take another step to the left, the total distance will increase.

Therefore, there are only two solutions for x: 4 and -2.667

To review:

The Red line shows the region where the total distance of x from the 3 points is less than 10. The Blue lines show the region where the total distance of x from the 3 points is more than 10. The points -2.667 and 4 are the points where the total distance of x from the 3 points is 10.

Since there are 2 points where the total distance is 10, answer is (C).

A few things to ponder upon:

– I change the question to ‘For how many values of x, is |x – 3| + |x + 1| + |x| = 4?’ The answer now is 1. Why?

– I now change the question to ‘For how many values of x, is |x – 3| + 3|x + 1| + |x| = 4?’ The answer now is 0. Why?

– What happens if I change the question to ‘For how many values of x, is |2x – 3| + |x + 1| = 10?’

– What if I change it to ‘For how many values of x, is |x – 3| – |x + 1| + |3x| = 10?’

Thoughts on the points above are welcome. By the way, if a doubt arises somewhere, feel free to let me know and I will get back to you.

*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 in Detroit, Michigan, and regularly participates in content development projects such as this blog!*

## 11 thoughts on “Quarter Wit, Quarter Wisdom: The Holistic Approach to Mods – Part II”

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Another variant, <10

Hi Karishma,

I have just started following your posts and they are very useful.

I am unable to understand how for question |x-3| + 3|x+1| + |x| = 4 does the interval is 3. I mean if i take x = 0, the result is 6. if i calculate for x = 1, the result is 9. But shouldnt it be 1 for |x-3|, 3 for |x+1| and 1 for |x| means the distance should increment by 5 instead of 3.

Please help me out where am i failing to understand this calculation.

Thanks

Reply

Hey Rahul,

Here is the solution:

|x – 3| + 3|x + 1| + |x| = 4

Make the number line as discussed above. You have to find the point x such that sum of distance of x from 3, 0 and thrice of -1 (since we have 3|x + 1| ) is 4. Notice here that the distance between 3 and -1 itself is 4. To that, we will need to add the distance of x from 0. So the sum will be definitely more than 4. If you are thinking here that why don’t we choose x = 0 so that x’s distance from 0 is 0, then observe that we need thrice the distance from -1. So if x = 0, sum of distances = 3 + 3*1 + 0 = 6

Therefore, there is no such point where the sum of these distances is 4. Hence the equation has no solution.

If you are facing a problem in imagining mod as distance, I would request you to take a look at my previous mod posts on this blog. They will help you understand the logic better.

Thanks Karishma for the quick reply. I have read all your previous posts and climbing the ladder to be with you on your latest blog.

I certainly got the answer for the above question but the thing that wasnt clear to me was that why the distances between f(x)=0 and f(x) = 1 same. I solved a few more problems at my own end and the thing became clear to me. The distance is not necessarily same.

Thanks!

Hi Karishma

Can you please solve last two questions, it would be really helpful. Awaiting you reply. Regards

Question 1: |2x – 3| + |x + 1| = 10

2|x – 3/2| + |x + 1| = 10 (taking 2 out of the mod)

Now it is just like our previous questions.

You need to find x such that twice the distance from 1.5 plus the distance from -1 add up to 10. Notice that the distance between -1 and 1.5 is 2.5. So x will not lie between these two numbers.

There will be a value to the left of -1. When x = -1, the total distance covered = 2*2.5 = 5. We need to cover 5 more units to the left. So x will be 5/3 units to the left of -1 i.e. x = -8/3

There will be another value to the right of 1.5. When x = 1.5, the total distance covered = 2.5. So we need to cover another 7.5 to make the total distance = 10.Therefore, x must be 7.5/3 = 2.5 units to the right of 1.5 i.e. x = 4

Question 2: |x – 3| – |x + 1| + |3x| = 10

|x – 3| – |x + 1| + 3|x| = 10

Here there are 3 critical points 3, 0 and -1. We have to find x such that thrice the

distance from 0 plus the distance from 3 minus the distance from -1 adds up to 10.

If x = -1, total distance = 3*1 + 4 – 0 = 7

If x = 0, total distance = 3*0 + 3 – 1 = 2

If x = 3, total distance = 3*3 + 0 – 4 = 5

There must be a point to the left of -1 and one to the right of 3.

As you go to the left of -1, thrice the distance from 0 increases by 3 for every unit, distance from 3 increases by 1 for every unit and distance from -1 also increases by 1 for every unit. Therefore, the distance will keep increasing by 3 + 1 – 1 = 3 for every unit you go to the left. If you go one unit to the left i.e. if x = -2, total distance must be 10.

As you go to the right of 5, using the same concept as above, the distance will keep increasing by 3 for every unit you go to the right. Hence you must go 5/3 units to the right i.e. x = 3 + 5/3 = 14/3

Question 1: |2x – 3| + |x + 1| = 10

Thank you for answering the question. However, a couple of points are not clear to me yet. Probably a number line will be of some help here.

——(-1)——————-(3/2)

I understand when you say the distance between these two points is 2.5 (given that point x is either of them: 3/2 -(-1) = 2.5).

But then you say that the distance covered is 2*2.5 given that x = -1.5, that’s when I take a tumble.

The next point that’s not clear is when you divide 5 by 3. Since we only have two points so shouldn’t an increase by 1 point account for 2 units not 3?

Thanks.

The basic logic of mod is that it signifies the distance of x from a point.

|x| = 10

means the distance of x from 0 is 10. So x can be 10 or -10 since the distance of both these points from 0 is 10.

2|x| = 10

means twice the distance of x from 0 is 10. So the distance of x from 0 is 5 and hence x can be 5 or -5.

|2x – 3| = 2|x – 3/2|

This is twice the distance of x from 3/2.

So when you have the question: |2x – 3| + |x + 1| = 10

you need to find a point such that the sum of twice its distance from 3/2 and its distance from -1 is 10.

When x = -1, the distance from 3/2 becomes 2.5 and twice the distance from 3/2 becomes 5. We need to move x further to the left because we want the total distance to be 10. There is a distance of another 5 to be covered. Now every time we move one unit, we are moving away from -1 so we will sum the distance from -1 and twice the distance from 3/2. Hence every step will be counted three times. So we move 5/3 steps so that when it is counted thrice, we get a total distance of 5.

Try to understand it with the help of a number line and by using a simpler question first e.g.

2|x – 3| + |x – 6| = 9

Karishma,

I thought i understood this part ” By how much will the distance increase? By 3 units! Because x will be 1 unit away from each of the 3 points”

in the first question but then you still applied it in the second question

ie ” 2x – 3| + |x + 1| = 10″

I thought the first question increased by 3 points since there were 3 parts of the question ,then now in the later it 1 step will equal 2 units since the question has two parts ie “2x – 3| + |x + 1”

Or what did you mean in the above statement?

Hey i finally understood this part.thanks

Hi Karishma,

What you said above for Question 2: |x – 3| – |x + 1| + |3x| = 10:

“If x = -1, total distance = 3*1 + 4 – 0 = 7

If x = 0, total distance = 3*0 + 3 – 1 = 2

If x = 3, total distance = 3*3 + 0 – 4 = 5

There must be a point to the left of -1 and one to the right of 3.”

How can you conclude that “there must be a point to the left of -1 and one to the right of 3?” What if, when x 3, the distance will also decrease? Don’t you have to visualize the distance based on the given equation to reach your conclusion?