Quarter Wit, Quarter Wisdom: Bagging the Graphs - Part III

Quarter Wit, Quarter WisdomIf you haven’t read parts I and II of this topic, I strongly suggest you read them first: Bagging the Graphs – Part I, Bagging the Graphs – Part II

While introducing this concept in Part I of Graphs, I had mentioned: “the one thing that I would suggest to increase speed in Co-ordinate Geometry and Algebra is Graphs”

Did you wonder why I included “Algebra” here? If yes, then this post will answer your question. In part 2, I gave an example of a Geometry question that can be easily solved using Graphs. In this post, I will take up an Algebra question for which you can do the same.

In part I of Graphs, I had also mentioned “Learn how to draw a line from its equation in under ten seconds and you shall solve the related question in under a minute.” After this post, you won’t have to take my word for it!

Before I begin, let me also add here that this Data Sufficiency question is not a question I created. So don’t think I made it to conveniently suit my needs. It is a question someone asked me on a GMAT forum and was created by some third party. I chose it to demonstrate the beauty of graphs to you.

Question: If x and y are positive, is 4x > 3y?

Statements:

  1. x > y – x
  2. x/y < 1

Let us look at the question stem first: x and y are positive, ‘is 4x > 3y?’ or rephrase it as ‘is 4x – 3y > 0?’

Let us draw 4x – 3y = 0. Then we can figure out which region represents 4x – 3y > 0. When x = 0, y = 0 so the line passes through (0, 0). The slope of the line is 4/3. This is what it looks like:

The line has divided the graph into two regions: 4x – 3y < 0 and 4x – 3y > 0. Let us check in which region, the point (3, 0) lies. (This is an arbitrary choice. You can check for any point.) When you put x = 3 and y = 0 in 4x – 3y > 0, you get 12 > 0 which is true. Hence the point (3, 0) lies in 4x – 3y > 0 region. Since x and y are positive, we are only concerned with quadrant I. The shaded Green region is where 4x – 3y > 0. So the question boils down to: “Does the point (x, y) lie in the shaded Green region for all values of x and y?”

Statement 1: x > y – x or 2x – y > 0

Draw 2x – y = 0. When x= 0, y = 0 so this line passes through the center. The slope of the line is 2. The slope of this line, 2, is greater than 4/3, the slope of the line drawn above. Hence, this line will be steeper than the line drawn above. Check for point (3, 0) again to find whether it lies in region 2x – y > 0. Putting x = 3 and y = 0, we get  6 > 0 which is true so the region 2x – y > 0 includes the point (3, 0) and is as shown below:



The Red shaded region here includes all the points of the Green shaded region above plus some more. Hence all points of Red region may not lie in the Green region. Therefore, if values of x and y satisfy 2x – y > 0, they may or may not satisfy 4x – 3y > 0. Hence statement 1 alone is not sufficient.

Statement 2: x/y < 1

Since x and y are positive, we can multiply both sides of the inequality by y to get x < y or x – y < 0. Draw x – y = 0. When x= 0, y = 0 so this line passes through the center. The slope of the line is 1. The slope of this line, 1, is less than 4/3, the slope of the line in the question. Hence, this line will be less steep than the line in the question stem. Check for point (3, 0) again to find whether it lies in region x – y < 0. Putting x = 3 and y = 0, we get 3 < 0 which is not true so the region x – y < 0 does not include the point (3, 0) and is as shown below:

Note: Our concern is limited to first quadrant since x and y are both positive.

The Blue shaded region here includes some points of the Green shaded region above plus some more. Hence all points of Blue region may not lie in the Green region. Therefore, if values of x and y satisfy x – y < 0, they may or may not satisfy 4x – 3y > 0. Hence statement 2 alone is not sufficient.

Taking both statements together, x and y will have values that overlap in the Red and the Blue region as shown in the graph below. Some of these values will lie in the Green shaded region above, some will not.

So even if we take both statements together, they are not sufficient. Answer (E).

I hope you have come to appreciate the wide range of applicability of graphs. Next time, I will introduce a graphical way of working with Modulus and Inequalities.

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!

7 Responses

  1. Mike says:

    Hi,
    I think, we can solve this problem faster by using algebra.

    Stmt 1:
    x > y – x
    2x > y
    4x > 2y

    Not enough information.

    Stmt 2.
    x/y < 1
    x < y
    4x< 4y

    Again, not enough information.

    Stmt 1 and 2 combined : 4y2y
    Again, not enough information.

    What say thou?

  2. Mike says:

    Correction:
    Stmt 1 and 2 combined : 4y< 4x<2y or 2y<2x<y
    Again, not enough information.

  3. Mike says:

    Oops Correction 2:
    Stmt 1 and 2 combined : 4y> 4x>2y or 2y>2x>y
    Again, not enough information.

  4. Karishma says:

    Yes, most people like to use Algebra to solve such questions. You can use whatever method you like as long as you get the correct answer within a reasonable frame of time. Through this blog, I aim to provide some alternative approaches that have worked very well for me and some others. I strongly believe that once a person is good at drawing graphs, he/she can save a lot of time and reduce the scope of errors (especially in Data Sufficiency questions). Thereafter, working with equations/inequalities feels exhausting and time consuming. But that’s just my opinion.

  5. Tanay says:

    Hi karishma,

    A basic question may be, but its not allowing me to understand the concept. Please help me.

    Initially you considered the point as (3,0) which is arbitrarily chosen. But what if we considered a point like (2,4) . Then that point would lie in 4x-3y <0 but in first quadrant. And then for the equation 2x-y = 0 ; If I substitute this value (2,4) it becomes zero. So the point exists on that line .How would I decide the direction of the region in such a case.

    • Karishma says:

      Pick points which you know will lie in the region in which you are interested. From the figure, I see that (3, 0) does not lie on the line. That is why I tried it. You could try (10, 1) or something – whatever you know will not lie on the line. If you pick a point which lies on the line, it will not help you in determining what you want to know.

  6. Vaishno says:

    Karishma, shouldn’t the green line in both graphs above have the equation 4x-3y=0 instead of 4x-y=0?

    I followed almost everything you wrote until this point – The Red shaded region here includes all the points of the Green shaded region above plus some more. Hence all points of Red region may not lie in the Green region. Therefore, if values of x and y satisfy 2x – y > 0, they may or may not satisfy 4x – 3y > 0. Hence statement 1 alone is not sufficient.

    Please shed some light on the above mentioned point, since I am really interested in learning this approach. I must admit that after following your approach of using number line to solve inequalities with Mods, I feel very confident and get the questions right in less than a minute. So thank you for that.

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