Quarter Wit, Quarter Wisdom: The Holistic Approach to Mods

Quarter Wit, Quarter WisdomI have a dream… A dream that one day, I will see my students making a bonfire of all their pens and pencils… that I will see them lost in thought in my Quant class, occasionally drawing lines and curves on a drawing sheet with colorful crayons… I will see them coming up with innovative logical solutions, just like that… But I know that no dream of mine is realized until and unless I keep my nose to the grindstone (I am not waiting with bated breath to achieve that elusive target weight.) So on this particular sleepless night, I will write a post with some more figures, figures that make complicated questions look like easy pickings. Let me explain using step by step approach.

A Complicated Question: If y = ||x – 5| – 10|, for how many values of x is y = 1? (I remember once someone said, “I think I would rather eat spinach than try such questions.”)

Easy Pickings:

Graph of y = x is a line passing through the center with slope 1.


Graph of y = |x| is as shown below. Modulus cannot be negative so all negative values of y are flipped to positive. (The red line shows the original position for reference.)

To get the graph of y = |x – 5|, shift the above graph 5 units to the right on the x axis. This is so because in the graph above, y = 0 when x = 0. But in the required graph, y should be 0 when x = 5. Hence the point at (0, 0) shifts to (5, 0). Since the slope of the line is 1, it makes an intercept of 5 on the y axis.

The graph of y = |x – 5| – 10 is just the above graph shifted down by 10 units because now y is 10 less than every previous value of y.

Now we need to take the modulus of the equation above to get y = ||x – 5| – 10|. Since a modulus is never negative, whatever part of the graph is below the x axis i.e. in quadrants III and IV, gets reflected above the x axis in quadrants I and II.

This is the graph we wanted. We see that the line y = 1 (shown in green below) intersects this graph at 4 points. So y = 1 when x = -6 or -4 or 14 or 16.

Put these values in the given equation if you want to cross check. Once you get the hang of it, you can arrive at this graph in under a minute! Such tricky questions can be elegantly handled using this approach. In fact, we can add many more levels of complexity and still easily arrive at our answer. For shakes, try out the graphs of y = |||x – 5| – 10| -5| and y = |||x| – 3| – x|!

Their answers are discussed here.

Now that you have lost your sleep, I think I will sleep easier!

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!

20 Responses

  1. Rakesh says:

    Thanks Krishma for this wonderful post.

    As we know that on GMAT we don’t have access to any graphing tools can you tell whether there is a simple way of finding the x and y intercepts of the final graph?

    Also, how do you calculate the x intercepts on the final graph when y = 1?

  2. Karishma says:

    Hey Rakesh,

    You do not need any graphing tools. Notice that the slope of the line is 1 which means that when y changes by some amount, x changes by the same amount e.g. in the third diagram when the graph shifts 5 units to right on the x axis, the line will intersect on the y axis at 5 as well.
    (Check out
    http://www.veritasprep.com/blog/2010/12/quarter-wit-quarter-wisdom-bagging-the-graphs/
    for explanation of slope)

    In the fourth diagram, when the lines move down by 10 units on the y axis, their x co-ordinates also change by 10 units and so on…

    We use the same concept to find the x coordinates in the final graph. Consider the point (15, 0). The line on the right intersects with y = 1. At that point we need to find the x co-ordinate. Since at y = 1, y coordinate has increased by 1 (from 0 it has become 1), x coordinate also increases by 1 and becomes 16. Similarly, the line on the left intersects with the line y = 1 at x = 14.
    You do the same thing for the point (-5, 0) to get all 4 points where the graph intersects with the line y = 1.

  3. jami says:

    Karishma, thank you so much. i have read your other articles as well and I must say they are very useful.

  4. Vicky says:

    Karishma..I am a big fan of your your blog. Its really helpful. Exactly what I had been looking for. I have a few queries on this one.

    1> How do we decide as to when the graph will move right/left and when will it it move up/down? I was clear till ||x-5|-10|. But I got confused with your second last question(|||x – 5| – 10| -5|), where you added “-5″ at the end. Will the graph move 5 units to the left on the X Axis or 5 units down the X axis at the end? Can you explain

    2> What is the solution of the last one – y = |||x| – 3| – x||?

  5. Karishma says:

    Think how did we go from y = |x – 5| to y = |x – 5| – 10? We moved down by 10 units because in this new equation, every value of y is 10 less than the previous value. Similarly, to go from ||x-5|-10| to ||x – 5| – 10| -5, we will just move the graph down by 5 units. Thereafter, we flip the part lying below the y axis to get the mod!

    Draw y = x, y = x – 5, y = 2x, y = 2(x-1), y = 2x – 1… etc to see how they move.

    As for the last one, it is quite an interesting question. I can send you the pdf if you mail me on kbansal@veritasprep.com. I don’t think I can attach a pdf here.

  6. Mary says:

    Hi Karishma

    For me it’s easier to solve without graphs.
    ||x-5|-10|=1

    |x-5|-10=1 and |x-5|-10=-1
    |x-5|=11 |x-5|=9
    x-5=11 and

  7. Mary says:

    x-5=-11
    x=16
    x=-6,
    then |x-5|=9

    x-5=9, x-5=-9
    x=14, x=-4
    So when y=1, X=16;-6;14;-4
    for 4 values of x is y = 1

  8. Karishma says:

    Hey Mary,

    It’s true for most people. They feel that the algebraic methods are easier. The reason is that the graphical approach has a learning curve since most of us do not learn to use graphs extensively in school. But I assure you – once you learn how to use graphs, you will never go back to the algebra…
    There is just something about the big picture approach; it makes you hate the ‘one tiny part at a time’ game. Anyway, it is an alternative approach. Use it only if you wish to.

    Best,
    Karishma

  9. Vishnu says:

    Hi karishma,

    The blogs are excellent and explain how to deal in case there is a question involving modulus and equalities. What about modulus and a inequality.

    for. e.g :

    Is |x| < 1?

    (1) |x + 1| = 2|x – 1|
    (2) |x – 3| ? 0

    How do we represent |x| < 1 on a graph in this case ?

    • Karishma says:

      Inequalities are no different from equations except in our perception. I have discussed mods & inequalities in these posts:
      http://www.veritasprep.com/blog/2011/01/quarter-wit-quarter-wisdom-do-what-dumbledore-did/
      http://www.veritasprep.com/blog/2012/07/quarter-wit-quarter-wisdom-some-inequalities-mods-and-sets/

      As for your question: Is |x| < 1?

      What does the question stem ask? Is the distance of x from 0 less than 1? That is, is x between -1 and 1?

      (1) |x + 1| = 2|x – 1|
      This implies that distance of x from -1 is twice the distance of x from 1. Draw it on the number line. You will see that there are two such points: one which is between -1 and 1 (divide the distance between -1 and 1 into 3 equal parts and x is 2 parts away from -1 i.e. x = 1/3. You don't really need to do this calculation. I gave it only for clarity.) and another which is to the right of 1 at a distance of 2 away from 1 (because distance between -1 and 1 is 2). This other point is 2 to the right of 1 so it must be x = 3. There will be no such point to the left of -1 since the distance of all points on the left of -1 from -1 will be less than the distance of these points from 1. With some practice, you can easily see this in a few seconds. So we have a value of x lying between -1 and 1 and one value which doesn't lie between -1 and 1. Not sufficient

      (2) |x – 3| ? 0
      This only tells us that x is not 3. Not sufficient.

      Taking both statements together, we can see that x can take only one value and that value lies between -1 and 1 (i.e. x = 1/3). Hence sufficient.

      Answer (C)

  10. vidya says:

    How do we solve for this?

    |x+3| – |4-x| = |8+x|

    How many solutions will this equation have?

  11. Dipankar says:

    How do we solve for this?
    |x+3| – |4-x| = |8+x|
    How many solutions will this equation have?

    I found no solutions to this. Can someone confirm if this is correct. My approach was as follows:
    Four possible scenarios -
    1. x>4
    2 -3<x<4
    3. -8<x<-3
    4 x<-8
    the equation had no roots for any values for x

  12. Karishma says:

    |x+3| – |4-x| = |8+x|
    |x+3| = |x-4| + |x+8| (re-adjust the terms)

    Now, plot the points on the number line. The distance of x from -3 should be equal to sum of distance of x from -8 and distance of x from 4. Notice that it is never possible (try out various points) and hence the equation has no roots.
    For details of this method, check out the links I suggested in my replies above.

    (Btw, I apologize for the delayed response. Somehow, my mail box did not deliver the notification in my Inbox!)

  13. kartic says:

    Hi Karishma,

    I have a question here.When you drew graph for y = |x – 5| – 10 ,how did you get the value of x as 15 and -5?

    Thanks.
    Kartic

    • Karishma says:

      The slope of the line is 1. A point on the graph which is 10 units above x axis is also 10 units to the right of (5, 0) i.e. points (6, 1), (7, 2), ….(15, 10) etc lie on the graph.
      When you pull the graph down by 10 units, these points will change to (5, -10), (6, -9), (7, -8) … (15, 0)
      Same logic for x = -5.

      Once you get a hang of graphs, you will be able to SEE it without even thinking about it.

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