Circle and Inscribed Regular Polygon Relations

Quarter Wit, Quarter WisdomAs promised last week, let’s figure out the relations between the sides of various inscribed regular polygons and the radius of the circle.

We will start with the simplest regular polygon – an equilateral triangle. We will use what we already know about triangles to arrive at the required relations.

Look at the figure given below. AB, BC and AC are sides (of length ‘a’) of the equilateral triangle. OA, OB and OC are radii (of length ‘r’) of the circle.

The interior angles of an equilateral triangle are 60 degrees each. Therefore, angle OBD is 30 degrees (since ABC is an equilateral triangle, BO will bisect angle ABD). So, triangle BOD is a 30-60-90 triangle.

As discussed in your geometry book, the ratio of sides in a 30-60-90 triangle is 1:?3:2 therefore, a/2 : r = ?3:2 or a:r = ?3:1

Side of the triangle = ?3 * Radius of the circle

You don’t have to learn up this result. You can derive it if needed. Note that you can derive it using many other methods. Another method that easily comes to mind is using the altitude AD. Altitude AD of an equilateral triangle is given by (?3/2)*a. The circum center is at a distance 2/3rd of the altitude so AO (radius) = (2/3)* (?3/2)*a = a/?3

Or side of the triangle = ?3 * radius of the circle

Let’s look at a square now.

AB is the side of the square and AO and BO are the radii of the circle. Each interior angle of a square is 90 degrees so half of that angle will be 45 degrees. Therefore, ABO is a 45-45-90 triangle. We know that the ratio of sides in a 45-45-90 triangle is 1:1:?2.

r:a = 1: ?2

Side of the square = ?2*Radius of the circle

Again, no need to learn up the result. Also, there are many methods of arriving at the relation. Another one is using the diagonal of the square. The diagonal of a square is ?2 times the side of the square. The radius of the circle is half the diagonal. So the side is the square is ?2*radius of the circle.

The case of a pentagon is more complicated since it needs the working knowledge of trigonometry which is beyond GMAT scope so we will not delve into it.

We will look at a hexagon though.

Notice that the interior angle of a regular hexagon is 120 degrees so half of that will be 60 degrees. Therefore, both angles OAB and OBA will be 60 degrees each. This means that triangle OAB is an equilateral triangle with all angles 60 degrees and all sides equal. Hence,

Side of the regular hexagon = Radius of the circle.

The higher order regular polygons and more complicated and we will not take them up. We will discuss a circle inscribed in a polygon next week.

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!

Inscribing Polygons and Circles

Quarter Wit, Quarter WisdomLast week we looked at regular and irregular polygons.  Today, let’s try to understand how questions involving one figure inscribed in another are done.  The most common example of a figure inscribed in another is a polygon inscribed in a circle or a circle inscribed in a polygon. Let’s see the various ways in which this can be done.

To inscribe a polygon in a circle, the polygon is placed inside the circle so that all the vertices of the polygon lie on the circumference of the circle.

There are a few points about inscribing a polygon in a circle that you need to keep in mind:

–  Every triangle has a circumcircle so all triangles can be inscribed in a circle.

–  All regular polygons can also be inscribed in a circle.

–  Also, all convex quadrilaterals whose opposite angles sum up to 180 degrees can be inscribed in a circle.

There are also a few points about inscribing a circle in a polygon that you need to keep in mind:

–  All triangles have an inscribed circle (called incircle). When a circle is inscribed in a triangle, all sides of the triangle must be tangent to the circle.

–  All regular polygons have an inscribed circle.

–  Most other polygons do not have an inscribed circle

A simple official question will help us see the relevance of these points:

Question: Which of the figures below can be inscribed in a circle?

(A) I only
(B) III only
(C) I & III only
(D) II & III only
(E) I, II & III

Solution:

I think it will suffice to say that the answer is (C).

Next week, we will look at the relations between the sides of these polygons and the radii of the circles.

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!

Regular Polygons and the Irregular Ones

Quarter Wit, Quarter WisdomContinuing our Geometry journey, let’s discuss polygons today. Some years back, I used to often get confused in the polygon sum-of-the-interior-angles formula if I had to recall it after a gap of some months because I had seen two variations of it:

Sum of interior angles of a polygon = (n – 2)*180

Sum of interior angles of a polygon = (2n – 4)*90

Now, I don’t want you to judge me. Of course, in the second formula, 2 has been removed from 180 and multiplied to the first factor. It is quite simple so why would anyone get tricked here, you wonder? The problem was that after a few months, I would somehow remember (2n – 4) and 180. So I was mixing up the two and I wasn’t sure of the logic behind this formula. That is until I came across the simple explanation of this formula in our Veritas Prep Geometry book (the one which explains how you can divide every polygon with n sides into (n – 2) triangles and hence get the sum of (n – 2)*180). Now it made perfect sense! I couldn’t believe that I had not come across that explanation before and had just learned up (well, tried to!) the formula blindly. So now I ensure that all my students understand every formula that I teach them.

Usually, we are given a regular polygon and we need to find the measure of interior angles or the number of sides. But what if we are given a polygon instead, not a regular polygon. Does this formula still apply? We wouldn’t know if we didn’t understand how the formula came into being. But since we know that we obtain the formula by dividing the polygon into (n-2) triangles, we know that the sum of all interior angles of a triangle is 180 irrespective of the kind of triangle. So it doesn’t matter whether the polygon is regular or not. The sum of all interior angles will still be (n-2)*180.

Let’s look at a question to see the application of this formula in irregular polygon scenario.

Question: The measures of the interior angles in a polygon are consecutive odd integers. The largest angle measures 153 degrees. How many sides does this polygon have?

A) 8
B) 9
C) 10
D) 11
E) 12

Solution:

The interior angles are: 153, 151, 149, 147 … and so on.

Now there are two ways to approach this question – one which is straight forward but uses algebra so is time consuming, another which makes you think but doesn’t take much time. You can guess which one we are going to focus on! But before we do that let’s take a quick look at the algebraic solution too.

Method 1: Algebra

Sum of interior angles of this polygon = 153 + 151 + 149 + … (153 – 2(n-1)) = (n – 2)*180

If there are n sides, there are n interior angles. The second largest angle will be 153 – 2*1. The third largest will be 153 – 2*2. The smallest will be 153 – 2*(n-1). This is an arithmetic progression.

Sum of all terms = [(First term + Last term)/2] * n = [(153 + 153 – 2(n-1))/2] * n

Equating, we get [(153 + 153 – 2(n-1))/2] * n = (n – 2)*180

Solving this you get, n = 10

But let’s figure out a solution without going through this painful calculation.

Method 2: Capitalize on what you know

Angles of the polygon: 153, 151, 149, 147, 145, 143, 141, … , (153 – 2(n-1))

The average of these angles must be equal to the measure of each interior angle of a regular polygon with n sides since the sum of all angles is the same in both the cases.

Measure of each interior angle of n sided regular polygon = Sum of all angles / n = (n-2)*180/n

Using the options:

Measure of each interior angle of 8 sided regular polygon = 180*6/8 = 135 degrees

Measure of each interior angle of 9 sided regular polygon = 180*7/9 = 140 degrees

Measure of each interior angle of 10 sided regular polygon = 180*8/10 = 144 degrees

Measure of each interior angle of 11 sided regular polygon = 180*9/11 = 147 degrees apprx
and so on…

Notice that the average of the given angles can be 144 if there are 10 angles.

The average cannot be higher than 144  i.e. 147 since that will give us only 7 sides (153, 151, 149, 147, 145, 143, 141 – the average is 147 is this case). But the regular polygon with interior angle measure of 147 has 11 sides. Similarly, the average cannot be less than 144 i.e. 140 either because that will give us many more sides than the required 9.

Hence, the polygon must have 10 sides.

Answer (C).

Interesting, eh? Well, it will be when you understand method 2 well and can do it intuitively!

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!

Geometry Diagrams for DS Questions

Quarter Wit, Quarter WisdomLet’s go back to geometry now. We will discuss how to use diagrams to solve DS questions today. Though we discussed a DS question in a previous geometry post, we didn’t discuss how the thought process used for a DS question is different from the thought process used for a PS question. To find whether a statement is sufficient to answer the question, you should try to prove that it is not sufficient. Try to make two cases which answer the question differently using the give information. If there are two or more different answers possible, it means the given information is not enough. Let’s discuss this with the help of an official question.

Question: A circle and a line lie in the XY plane. The circle is centered at the origin and has a radius 1. Does the line intersect the circle?

Statement I: The x-Intercept of the line is greater than 2
Statement II: The slope of the line is -1/5

Solution:

We are given that there is a circle and a line on the XY plane. The line can lie anywhere – it may or may not intersect the circle. The circle has radius 1 so it intersects the x axis at (1, 0) and (-1, 0).

Let’s look at the information given in the two statements:

Statement I: The x-Intercept of the line is greater than 2.

If x intercept > 2, the line can be any of the following (and can be drawn in many more ways)

We found two cases – one in which the line intersects the circle and another in which it doesn’t. The line could have different slopes and different x intercepts (as long as it is greater than 2) to get different cases. Hence we see that this information alone is not sufficient to answer the question.

Statement II: The slope of the line is -1/5.

If slope of the line is -1/5, the line can be drawn in any of the following ways (and many more).

Again, we found two cases – one in which the line intersects the circle and another in which it doesn’t. The slope of the line stays the same but you can move it up or down to get the two different cases (and different x intercepts). Hence we see that this information alone is not sufficient to answer the question.

Using both together: Now the line has a defined slope = -1/5 but it has no defined x intercept. To get x intercept greater than 2, all we need is that y intercept must be greater than 2/5. If you are wondering how we arrived at this, recall from an earlier post:

Slope of a line = – (y intercept)/(x intercept) = – 1/5

y intercept = (1/5) * x intercept

Since x intercept must be greater than 2, y intercept will be greater than 2/5.

If the y intercept is much more than 2/5, it will not intersect the circle. If the y intercept is a little more than 2/5, the line will intersect the circle (as shown by the two diagrams in Fig 2). In one case, it will intersect the circle, in the other case, it will not. So both statements together are not sufficient.

Answer (E)

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!

 

Or Just Use Inequalities!

Quarter Wit, Quarter WisdomIf you are wondering about the absurd title of this post, just take a look at last week’s title. It will make much more sense thereafter. This post is a continuation of last week’s post where we discussed number plugging. Today, as per students’ request, we will look at the inequalities approach to the same official question. You will need to go through our inequalities post to understand the method we will use here.

Recall that, given a < b, (x – a)(x – b) < 0 gives us the range a < x < b and (x – a)(x – b) > 0 gives us the range x < a or x > b.

Question: If x is positive, which of the following could be the correct ordering of 1/x, 2x and x^2?

(I) x^2 < 2x < 1/x

(II) x^2 < 1/x < 2x

(III) 2x < x^2 < 1/x

(A) none
(B) I only
(C) III only
(D) I and II
(E) I, II and III

Solution:  The question has three complex inequalities. We will take each in turn. Note that each inequality consists of two more inequalities. We will split the complex inequality into two simpler inequalities e.g. x^2 < 2x < 1/x gives us x^2 < 2x and 2x < 1/x. Next we will find the range of values of x which satisfy each of these two inequalities and we will see if the two ranges have an overlap i.e. whether there are any values of x which satisfy both these simpler inequalities. If there are, it means there are values of x which satisfy the entire complex inequality too. Things will become clearer once we start working on it so hold on.

Let’s look at each inequality in turn. We start with the first one:

(I) x^2 < 2x < 1/x

We split it into two inequalities:

(i) x^2 < 2x

We can rewrite x^2 < 2x as x^2 – 2x < 0 or x(x – 2) < 0.

We know the range of x for such inequalities can be easily found using the curve on the number line. This will give us 0 < x < 2.

(ii) 2x < 1/x

It can be rewritten as x^2 – 1/2 < 0 (Note that since x must be positive, we can easily multiply both sides of the inequality with x)

This gives us the range -1/?2 < x < 1/?2 (which is 0 < x < 1/?2 since x must be positive).

Is there a region of overlap in these two ranges i.e. can both inequalities hold simultaneously for some values of x? Yes, they can hold for 0 < x < 1/?2. Hence, x^2 < 2x < 1/x will be true for the range 0 < x < 1/?2. So this could be the correct ordering. Let’s go on to the next complex inequality.

(II) x^2 < 1/x < 2x

Again, let’s break up the inequality into two parts:

(i) x^2 < 1/x

x^1 < 1/x is rewritten as x^3 – 1 < 0 which gives us x < 1.

(ii) 1/x < 2x

1/x < 2x is rewritten as x^2 – 1/2 > 0 which gives us x < -1/?2 (not possible since x must be positive) or x > 1/?2

Can both x < 1 and x > 1/?2 hold simultaneously? Sure! For 1/?2 < x < 1, both inequalities will hold and hence x^2 < 1/x < 2x will be true. So this could be the correct ordering too.

(III) 2x < x^2 < 1/x

The inequalities here are:

(i)  2x < x^2

2x < x^2 can be rewritten as x(x – 2) > 0 which gives us x < 0 (not possible) or x > 2.

(ii) x^2 < 1/x

x^2 < 1/x gives us x^3 – 1 < 0 i.e. x < 1

Can x be less than 1 and greater than 2 simultaneously? No. Therefore, 2x < x^2 < 1/x cannot be the correct ordering.

Answer (D)

Is this method simpler?

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!

Plug Using Transition Points

Quarter Wit, Quarter WisdomLet’s take a break from Geometry today and discuss the concept of transition points. This is especially useful in questions where you are tempted to plug in values. A question often asked is: how do I know which values to plug and how do I know that I have covered the entire range in the 3-4 values I have tried? What transition points do is that they give you the ranges in which the relationships differ. All you have to do is try one value from each range. If you do, you would have figured out all the different relationships that can hold. We will discuss this concept using a GMAT Prep question. You can solve it using our discussion on inequalities too. But if number plugging is what comes first to your mind in this question, then it will be a good idea to get the transition points.

Let’s begin:

If x is positive, which of the following could be correct ordering of 1/x, 2x and x^2?

(I) x^2 < 2x < 1/x

(II) x^2 < 1/x < 2x

(III) 2x < x^2 < 1/x

(A) none
(B) I only
(C) III only
(D) I and II
(E) I, II and III

Solution:

Notice that the question says “could be correct ordering”. This means that for different values of x, different orderings could hold. We need to find the one (or two or three) which will not hold in any case. So what do we do? We cannot try every value that x can take so how do we know for sure that one or more of these relations cannot hold? What if we try 4-5 values and only one relation holds for all of those values? Can we say for sure that the other two relations will not hold for any value of x? No, we cannot since we haven’t tried all values of x. So there are two options you have in this case:

1. Use logic to figure out which relations can hold and which cannot. This you can do using inequalities (but we will not discuss that today).

2. You can figure out the ranges in which the relationships are different and then try one value from each range. This is our transition points concept which we will discuss today.

Let’s discuss the second option in more detail.

First of all, we are just dealing with positives so there is less to worry about. That’s good.

To picture the relationship between two functions, we first need to figure out the points where they are equal.

x^2 = 2x

x^2 – 2x = 0

x = 0 or 2

x cannot be 0 since x must be positive so this equation holds when x = 2

So x =2 is the transition point of their relation. x^2 is less than 2x when x is less than 2 and it will be greater than 2x when x is greater than 2.

Let’s try to figure out the relation between 1/x and 2x now.

1/x = 2x

x = 1/sqrt(2)

Since 1/x is less than 2x when x is greater than 1/?2, it will be more than 2x when x is less than 1/sqrt(2).

Move on to the relation between 1/x and x^2.

1/x = x^2

x^3 = 1 (notice that since x must be positive, we can easily multiply/divide by x without any complications)

x = 1

So you have got three transition points: 1/sqrt(2), 1 and 2.

Now all you need to do is try a number from each of these ranges:

(i) x < 1/sqrt(2)

(ii) 1/sqrt(2) < x < 1

(iii) 1 < x < 2

(iv) x > 2

If a relation doesn’t hold in any of these ranges, it will not hold for any value of x.

(i) For x < 1/sqrt(2), put x = a little more than 0 (e.g. 0.01)

1/x = 100, 2x = 0.02, x^2 = 0.0001

We get x^2 < 2x < 1/x is possible. So (I) is possible

(ii) For 1/sqrt(2) < x < 1, put x = a little less than 1 (e.g. 0.99)

1/x = slightly more than 1, 2x = slightly less than 2, x^2 = slightly less than 1

We get x^2 < 1/x < 2x is possible. So (II) is also possible.

(iii) For 1 < x < 2, put x = 3/2

1/x = 2/3, 2x = 3, x^2 = 9/4 = 2.25

We get 1/x < x^2 < 2x is possible.

(iv) For x > 2, put x = 3

1/x = 1/3, 2x = 6, x^2 = 9

We get 1/x < 2x < x^2 is possible.

We see that for no positive value of x is the third relation possible. We have covered all different ranges of values of x.

Answer (D)

Try using inequalities instead of number plugging to see if solving the question becomes easier in that case.

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!

Diagrams of Geometry – Part II

Quarter Wit, Quarter WisdomLast week, we discussed how drawing extreme diagrams can help solve Geometry questions. Today we will see how to solve another Geometry question by making diagrams. The diagram can help you understand exactly what it is that you need to do; doing it will be quite straightforward.

Question: If 10, 12 and ‘x’ are sides of an acute angled triangle, how many integer values of ‘x’ are possible?

 

(A) 7
(B) 12
(C) 9
(D) 13
(E) 11

Solution: The question is very interesting. It asks you for an acute triangle i.e. a triangle with all angles less than 90 degrees. It’s a little hard to wrap your head around it, isn’t it? We know that the third side of a triangle can take many values. Right from a little more than the difference of the other two sides to a little less than the sum of the other two sides (Since we know that the sum of any two sides of a triangle is always greater than the third side). So x can be anything from a little more than 2 to a little less than 22. But how do we find out the values for which all the angles will be less than 90?

We want no obtuse or right angles. An obtuse angled triangle has one angle more than 90. So the thought here is that before one of the angles reaches 90, find out all the values that x can take.

Look at the figure given above. The value of x in the first figure is very small – slightly more than 2 – minimum required to make a triangle. There is an obtuse angle in that triangle. We keep making x bigger and bigger and the angle keeps becoming smaller till it reaches 90 (Fig III). We use Pythagorean theorem to get the value of x in that case:

x = ?(12^2 – 10^2)
x = ?44 which is 6.something
x should be greater than 6.something because the angle cannot be 90.

We further keep increasing x and all the angles are acute now. We reach Fig V where we hit another right triangle. We use Pythagorean theorem again to get the value of x (the hypotenuse) in this case:

x = ?(12^2 + 10^2)
x = ?244 which is 15.something
x should be less than 15.something so that the angle is not 90.

Further on, in Fig VI, we obtain an obtuse angle again.

We only need integral values of x so values that x can take range from 7 to 15 which is 9 values.

Answer (C).

Note: We made two angles 90 and found the values of x in between those two angles. The third angle cannot be 90 because that will make 10 the hypotenuse but hypotenuse is always the greatest side.

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!

Diagrams of Geometry – Part I

Quarter Wit, Quarter WisdomLet’s continue with geometry today. We would like to discuss how drawing extreme diagrams can help you solve questions. Most GMAT questions are quite intuitive and hence our non-traditional methods are perfect for them. They are not typical MATH problems per se; instead, they are logical puzzles. If you can prove why some things will not work, it means whatever is left will work.

Let me explain with the help of an official Data Sufficiency question.

 

Question:

In the figure above, is the area of the triangle ABC equal to the area of the triangle ADB?

Statement 1: (AC)^2=2(AD)^2

Statement 2: ?ABC is isosceles.

Solution:

When presented with this question, people see right triangles and jump to Pythagorean theorem, isosceles triangles and then wage a war on AC, AB, CB and AD relations. Well, that is our traditional approach. But what do we do if making equations and solving for relations isn’t our style?

We make diagrams and figure out the relations! One thing that is apparent the moment we read statement 1 is that the figure is not to scale. From the figure it looks as if AD is greater than or at best, equal to AC. That itself is an indication that if you draw the figure on your own, you could see something that will make this question very simple. The question setter doesn’t want to show you that and hence he made the distorted figure.

Anyway, let’s first analyze the question. Then we will look at the statements.

We need to compare areas of ABC and ADB. Both are right angled triangles.

Area of ABC = (1/2) * AC * BC

Area of ADB = (1/2) * AD * AB

We need to figure out whether these two are the same.

Think about it this way – we are given a triangle ABC with a particular area. So the length of AD must be defined. If AD is very small, (shown by the dotted lines in the diagram given below) the area of ADB will be very close to 0. If AD is very large, the area will be much larger than the area of ABC. So for only one value of AD, the area of DAB will be equal to the area of ABC.

We need to figure out whether for the given relations, the triangles have equal area.

Statement 1: (AC)^2=2(AD)^2
This gives us AD = AC/?2. Let’s draw AC and AD such that AD is somewhat shorter than AC. Now can we say that the areas of the two triangles are the same? No. The area of ABC is decided by AC and BC both not just AC. We can vary the length of BC to see that the relation between AC and AD is not enough to say whether the areas will be the same (see the diagrams given below).

So this statement alone is not sufficient.

(2) ?ABC is isosceles.
This means that AB = BC. Notice that the triangle is right angled so the hypotenuse must be the largest side. If ABC is isosceles, it means that the two legs of the triangle must be equal. Hence sides of ABC must be in the ratio 1:1:?2 = AC:BC:AB. Since we only need to consider relative length of the sides, let’s say that AC = 1, BC = 1 and AB = ?2 or some multiple thereof.

We have no idea about the length of AD so this statement alone is also not sufficient.

Let’s consider both statements together now:

AD = AC/?2 = 1/?2 (Since AC = 1)

Area of ABC = (1/2) * AC * BC = (1/2) * 1 * 1 = 1/2

Area of ADB = (1/2) * AD * AB = (1/2) *  (1/?2 ) * ?2 = 1/2

Both triangles have the same area. Sufficient!

Answer (C)

Now compare this approach with your Pythagorean approach. Is this simpler?

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!

Graphs of Geometry – Part III

Quarter Wit, Quarter WisdomThis week, we will further build up on what we have discussed in the past two weeks. You will need to sum up everything we discussed last week in a few seconds and arrive at a conclusion and then, move on and solve the question on the basis of that conclusion. We will take you through the ‘summing up’ and ‘getting a feel for it’ process step by step so that it’s intuitive to you next time you come across this concept.

Question: A certain square is to be drawn on a coordinate plane such that all the coordinates of its vertices are integers. One of the vertices must be at the origin, and the area of the square must be 25. How many different squares can be drawn?

(A) 4
(B) 6
(C) 8
(D) 10
(E) 12

Solution:

Since the question tells us that there are many different ways to draw the square, let’s draw it in one particular way. We will have something with which to proceed.

One vertex must be at (0, 0). Area must be 25 which means the side of the square must be 5. The easiest way to draw such a square would be the blue square shown in the figure.

How will we get many different squares? By turning the square around (0, 0) as shown (Push one side of the square – notice the vertex with the green dot at (0, 5) moving clockwise). Now the problem is that we need the coordinates of all the vertices to be integers. How do we assure that? One vertex is at (0, 0) and will always stay at (0, 0) so we don’t need to worry about that. What about the rest of the three vertices? When you turn the square, at any one position, they may or may not have integer coordinates – and we have three of them to worry about!

This is where our last two posts come in. First let’s establish that we only need to focus on one vertex – the other two will follow. Look at the blue square. Each one of its vertices is at integer coordinates: (0, 0), (0, 5) – the green dot vertex, (5, 5) and (5, 0). Let’s say, when you push the side joining the center and the green dot, the green dot moves by 1 unit down and 3 units to the right. What happens to the side perpendicular to this side (the one joining the green and the yellow dots)? Let’s think about it.

Pay attention to the diagram. When we turn the square, the coordinates of all three vertices change. The change is similar in all three vertices.

Hope it’s intuitive that if one vertex takes integral steps, all vertices will move in a similar fashion. The diagram shown above is just to reinforce this point. So if our first square has all integral coordinates and we move one vertex such that its coordinates remain integers, all the other vertices will follow suit.

Let’s catch hold of the green dot vertex with coordinates (0, 5). We want the x and y coordinates to stay integers such that

x^2 + y^2 = 5^2 (since the length of the side must stay 5)

As discussed above, x = 0 and y = 5 satisfies this. When x = 1, will y be integral? No.

What about when x = 2? No.

When x = 3? Yes, y will be 4.

When x = 4? Yes, y will be 3.

When x = 5? Yes, y will be 0.

So you have two pairs of numbers 0, 5 and 3, 4 (and their variations) which satisfy this equation. (Hope it reminded you of Pythagorean theorem and you jumped to this conclusion right away!)

Notice that coordinates can be negative too so 0, -5 and -3, -4 will also work. So how many total squares do we get?

So the green dot can take any of the following coordinates:

(0, 5), (0, -5)

(5, 0), (-5, 0)

(3, 4), (3, -4)

(4, 3), (4, -3)

(-3, 4), (-3, -4)

(-4, 3), (-4, -3)

A total of 12 values.

Alternatively, you have 3 different squares keeping the green dot in the first quadrant and when rotated, they will give you one square in each of the three other quadrants too.

 

Total 12 such squares.

The correct answer is (E).

If it seems difficult, that is because it is – 750 level.

But keep practicing. Most of it will become intuitive once you get the hang of it.

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!

Graphs of Geometry – Part II

Quarter Wit, Quarter WisdomLet’s pick up from where we left last week. We had discussed a coordinate geometry concept using clock faces and had left you with a tough question. Today we will see how you can solve that question using the concepts discussed last week.

You might wonder whether we can expect such a question in actual GMAT. The question we discussed in the last post was an official question and we could solve it easily using this concept. Of course there are many other ways of solving it but this is simplest (or trickiest depending on how you look at it), and it certainly is the fastest, no two ways about it! It is a very logical big-picture approach and people who get Q50-51 often use such methods. The question we will discuss today can also be solved in other ways but we will use the last week’s ‘turning minute hand 90 degrees’ approach.

Question: On the coordinate plane (6, 2) and (0, 6) are the endpoints of the diagonal of a square. What are the coordinates of the point on the corner of the square which is closest to the origin?

(A) (0, 1)
(B) (1, 0)
(C) (1, 1)
(D) (2, 0)
(E) (2, 2)

Solution:

First rule of coordinate geometry – draw what you can.

So we make the xy axis and plot the given points, (6, 2) and (0, 6) on it. Let’s say the square is denoted by points ABCD. Say, A is (6, 2) and C is (0, 6). We see that AC is a sloping line. Its two end points are two vertices of the square. We need to find the other two vertices of the square. One of them will lie closest to the origin. The other two vertices will be the end points of the diagonal BD. BD will be perpendicular to AC at the mid-point of AC since a square’s diagonals bisect each other and are perpendicular. So the question is, how do we obtain the end points, B and D? Let’s try to figure out what information we need to draw BD. BD must pass through the mid-point of AC.

How will we obtain the mid-point of AC? By averaging x and y co-ordinates of the points A and C:

x coordinate of mid-point is (0 + 6)/2 = 3

y coordinate of mid-point is (6 + 2)/2 = 4

So BD must pass through (3, 4). When AC turns by 90 degrees, with point (3, 4) as the axis, we get the diagonal BD. So how do the coordinates of AC change when it turns by 90 degrees? Go back to last week’s post and look at the clock face again.

Think of a horizontal line PQ passing through (3, 4).

P coordinate will be given by (0, 4) and Q coordinate will be given by (6, 4) since length of P to mid point is 3 and length of mid-point to Q will also be 3. P shifts up by 2 units to give the point A and Q shifts down by 2 units to give the point C.

Now rotate PQ by 90 degrees and you get RS. We know the coordinates of a line perpendicular to PQ. R will be (3, 1) and S will be (3, 7). This is because R and S will have the same x coordinates as the mid-point (3, 4) and S is 3 units above the mid-point and R is 3 units below the mid-point. We are assuming that you can intuitively see these values on the graph. If not, it may be too soon to spend time on this post.

Now, can we obtain the diagonal BD using RS as reference? If you move S two units to the right, you will get point B (just like A was obtained by moving P two units up) and if you move R two units to the left, you will get point D. Notice that we are using PQ and RS as reference lines. It is easy to calculate vertical/horizontal distances. So B will be given by (5, 7) and D will be given by (1, 1).

The closest co-ordinate to (0, 0) is (1, 1).

Answer (C).

Take a few minutes to review the logic discussed here. The ability to ‘see’ such symmetry makes GMAT Quant very simple.

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!

 

Graphs of Geometry – Part I

Let’s start with geometry today. It has some very interesting and intuitive concepts. We will discuss one of them today. It’s surprising how a little bit of imagination can go a long way in helping you solve questions. Let’s discuss the concept first. We will look at a question later.

Imagine a clock face. Think of the minute hand on 10. Ignore the hour hand for our discussion today. Say, the length of the minute hand is 2 cm. Its distance from the vertical and horizontal axis is shown in the diagram below (using the green and the red dotted lines). Let’s say the minute hand moves to 1. Can you say something about the lengths of the dotted black and dotted blue lines?

Isn’t it apparent that when the minute hand moved by 90 degrees, the dotted green line became the dotted black line and the dotted red line became the dotted blue line. So can we say that the dotted black line is ?3 cm in length and the dotted blue line is 1 cm in length? The same thing will happen when the minute hand goes to 4 and to 7. We don’t think there is much explanation needed here, right? The diagram makes it all clear.

Let’s look at the clock from coordinate geometry perspective. Let’s say the center of the clock is the origin (0, 0). What are the x and y coordinates of the “10’o clock point” i.e. the tip of the minute hand before it moves to 1? Notice that the x coordinate will be -?3 (since the point is in the second quadrant, x coordinate will be negative) and y coordinate will be 1.

What are the x and y coordinates of the “1’o clock point” i.e. the tip of the minute hand after it moves to 1. Notice that the absolute values of x coordinate and y coordinate have switched because the hand has turned 90 degrees. The x coordinate is 1 now and the y coordinate is ?3. Since it is the first quadrant, both the coordinates will be positive.

Now, think, what will be the coordinates of the “4’o clock point”, “7’o clock point”? What about the “11’o clock point”, “12’o clock point” etc?

Using this concept, we can solve a very tough GMAT Prep question in a few seconds.

Question 1:

In the figure above, points P and Q lie on the circle with center O. What is the value of s?

(A) 1/2

(B) 1

(C) ?2

(D) ?3

(E) 1/?2

Solution: You might be tempted to think on the lines of ‘slope of a line’ or ’30-60-90’ triangle (because of the presence of ?3) etc. But we should be able to arrive at the answer without using any of those.

Point P is (-?3, 1). O is the center of the circle at (0, 0). When OP is turned 90 degrees to give OQ, the x and y co-ordinates get interchanged. Also both x and y co-ordinates will be positive in the first quadrant. Hence s, the x co-ordinate of Q will be 1 (and y co-ordinate of Q will be ?3).

Answer (B)

The question doesn’t seem difficult now (after understanding the concept); actually, it is a 700+ level.

Try another question using the same concept:

Question 2: On the coordinate plane (6, 2) and (0, 6) are the endpoints of the diagonal of a square. What are the coordinates of the point on the corner of the square which is closest to the origin?

(A) (0, 1)

(B) (1, 0)

(C) (1, 1)

(D) (2, 0)

(E) (2, 2)

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!

Quarter Wit, Quarter Wisdom: Stuck in Assumptions Again

Quarter Wit, Quarter WisdomThere is a particular issue in assumption questions that I would like to discuss today. We discussed in our previous posts that assumptions are ‘necessary missing premises’. Many students get stuck between two options in assumption questions. The correct option is the necessary premise. The incorrect one is often a sufficient premise. Due to the sufficiency, they believe that that particular option is a stronger assumption. But the point to remember is that an assumption is only necessary for the conclusion to be true. It may not actually lead to the conclusion beyond a reasonable doubt. You only have to answer what has been asked (which is an assumption), not what you think is better to make the conclusion true.

Let me explain this with an example:

Question: Exports of United States cotton will rise considerably during this year. The reason for the rise is that the falling value of the dollar will make it cheaper for cloth manufacturers in Japan and Western Europe to buy American cotton than to get it from any other source.

Which of the following is an assumption made in drawing the conclusion above?

(A) Factory output of cloth products in Japan and Western Europe will increase sharply during this year.
(B) The quality of the cotton produced in the United States would be adequate for the purposes of Japanese and Western European cloth manufacturers.
(C) Cloth manufacturers in Japan and Western Europe would prefer to use cotton produced in the United States if cost were not a factor.
(D) Demand for cloth products made in Japan and Western Europe will not increase sharply during this year.
(E) Production of cotton by United States companies will not increase sharply during this year.

Solution:

First, let’s analyze the given argument:

Premises:
– Dollar is falling.
– It will be cheaper for cloth manufacturers in Japan and Western Europe to buy American cotton than to get it from any other source.

Conclusion:
– Exports of United States cotton will rise considerably during this year.

The conclusion links ‘sale of cotton’ to ‘cost of cotton’. It says that since the cost of American cotton will be lower than the cost of cotton from any other source, American cotton will sell. We are assuming here that the American cotton is adequate in all other qualities that the cloth manufacturers look for while buying cotton or that lower cost is all that matters. We are assuming that lowest cost will automatically lead to sale.

Let’s look at each of the options now:

(A)   Factory output of cloth products in Japan and Western Europe will increase sharply during this year.

Notice that we don’t NEED the factory output to increase. Even if it stays the same or in fact, even if it falls, as long as the manufacturers find American cotton suitable, the cotton exports of US could rise. Hence this is not the assumption.

(B)   The quality of the cotton produced in the United States would be adequate for the purposes of Japanese and Western European cloth manufacturers.

This option says that the quality is adequate and hence this is an assumption. Notice that it is necessary for our conclusion. If the quality is not adequate, no matter what the cost, US cotton sale may not increase. Hence, answer is (B).

(C)   Cloth manufacturers in Japan and Western Europe would prefer to use cotton produced in the United States if cost were not a factor.

This is the tricky non-correct option! Many people will swing between (B) and (C) for a while and then choose (C).  This option says that Japanese and Europeans prefer to use US cotton if cost does not matter. Do we NEED this to be true? No. It is good if it is true because it means that if cost of US cotton goes down, US cotton will sell more (hence, it is sufficient for the conclusion to be true – assuming all else stays constant). But do we NEED them to prefer US cotton? No. It is not necessary for our conclusion to be true. Even if the manufacturers don’t particularly prefer US cotton, US cotton exports could still increase if the price is the lowest.

Beware of this difference between ‘necessary’ and ‘sufficient’ conditions. Remember that assumptions are NECESSARY conditions; they don’t need to be sufficient. We end up incorrectly choosing sufficient conditions because they seem to be all encompassing and hence more attractive for our conclusion. If the sufficient condition is satisfied, then the conclusion has to be true. But mind you, that is not what the question is asking you. The question is looking for only a necessary condition, not a sufficient one. Also notice that sufficient conditions may not be necessary.

(D)   Demand for cloth made in Japan and Western Europe will not increase sharply during this year.

This is incorrect. We are not assuming that the demand for their cloth will not increase.

(E)    Production of cotton by United States companies will not increase sharply during this year.

It doesn’t matter what happens to the production of cotton in US. All we care about is that exports should rise.

The correct answer is (B). 

Hope you will be careful next time when you come across such a question.

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!

Quarter Wit, Quarter Wisdom: And Now, Evading Formulas!

Quarter Wit, Quarter WisdomToday, we again pay homage to the lazy bum within each one of us in our QWQW series. If you are wondering what we mean by ‘again’, check out our last two posts of the QWQW series. We have been discussing how to avoid calculations. Today let’s learn why it is advisable to avoid learning formulas too!

You really don’t need to know many formulas for GMAT – just the basic ones e.g. Distance = Speed*Time, Work = Rate*Time (which are actually the same if you look at them closely) etc. If a Time-Distance-Speed question pertains to GMAT, rest assured it can be solved using just the formula given above and that too, within 1-2 mins. Then, do you need to learn the many formulas that people claim speed up question solving? No! In fact, the more specific the formula, the more constraints it has. It can be used in only particular circumstances and hence when the situation differs even a little bit from the ideal, you could end up using the formula incorrectly. Therefore, we recommend our students to stay away from the umpteen, less generic formulas until and unless they have already used them extensively. Let’s discuss this point with an example:

Question: A and B start from Opladen and Cologne respectively at the same time and travel towards each other at constant speeds along the same route. After meeting at a point between Opladen and Cologne, A and B proceed to their destinations of Cologne and Opladen respectively. A reaches Cologne 40 minutes after the two meet and B reaches Opladen 90 minutes after their meeting. How long did A take to cover the distance between Opladen and Cologne?

(A) 1 hour
(B) 1 hour 10 minutes
(C) 2 hours 30 minutes
(D) 1 hour 40 minutes
(E) 2 hours 10 minutes

Solution: People often like to use a formula for this situation. Let’s quickly discuss that first.

If two objects A and B start simultaneously from opposite points and, after meeting, reach their destinations in ‘a’ and ‘b’ hours respectively (i.e. A takes ‘a hrs’ to travel from the meeting point to his destination and B takes ‘b hrs’ to travel from the meeting point to his destination), then the ratio of their speeds is given by:

Sa/Sb = sqrt(b/a)

i.e. Ratio of speeds is given by the square root of the inverse ratio of time taken.

Sa/Sb = sqrt(90/40) = 3/2

This gives us that the ratio of the speed of A : speed of B as 3:2. We know that time taken is inversely proportional to speed. If ratio of speed of A and B is 3:2, the time taken to travel the same distance will be in the ratio 2:3. Therefore, since B takes 90 mins to travel from the meeting point to Opladen, A must have taken 60 (= 90*2/3) mins to travel from Opladen to the meeting point

So time taken by A to travel from Opladen to Cologne must be 60 + 40 mins = 1 hr 40 mins

Now let’s see how we can solve the question without using the formula.

Think of the point in time when they meet:

A starts from Opladen and B from Cologne simultaneously. After some time, say t mins of travel, they meet. Since A covers the entire distance of Opladen to Cologne in (t + 40) mins and B covers it in (t + 90) mins, A is certainly faster than B and hence the Meeting point is closer to Cologne.

Now think, what information do we have? We know the time taken by A and B to reach their respective destinations from the meeting point. We also know that they both traveled the same distance i.e. the distance between Opladen and Cologne. So let’s try to link distance with time taken. We know that ‘Distance’ varies directly with ‘Time taken’. (Check out this post if you don’t know what we are talking about here.)

Distance between Opladen and Meeting point /Distance between Meeting point and Cologne = Time taken to go from Opladen to Meeting point/Time taken to go from Meeting point to Cologne = t/40 (in case of A) = 90/t (in case of B)
t = 60 mins

So A takes 60 mins + 40 mins = 1 hr 40 mins to cover the entire distance.

Answer (D)

We could easily solve the question without using any specific formula. So stick to your basics and kick those little grey cells to get to the answer!

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!

Quarter Wit, Quarter Wisdom: Evading Calculations Part II

Quarter Wit, Quarter WisdomLast week we discussed how to solve equations with the variable in the denominator. We also said that the technique generally works for PS questions but you need to be careful while working on DS questions. Today, let’s look at the reason behind the caveat.

Say, the question stem of a DS question asks you to find the value of n, the number of people in the room. Statement 1 of the question gives you the following equation:

60/(n – 5) – 60/n = 2

We can easily figure out that a value of n that satisfies this equation is 15. Now, is that enough to say that statement 1 is sufficient alone? No! It could be a trap! The equation, when manipulated, gives us a quadratic. It is important to find out whether the second solution of the quadratic works for us. When n is the number of people, it must be positive. So one extra step that we should take is re-arrange the equation to get the quadratic. If the constant term i.e. the product of the roots is negative, it means one root is positive and one is negative. Since we have already found the positive root, it is the only answer and hence we can say that the statement 1 is sufficient alone.

60/(n – 5) – 60/n = 2

60*n – 60*(n – 5) = 2*n*(n – 5)

n^2 – 5n – 150 = 0

The constant term, -150, is negative so the product of the roots must be negative. This means one root must be negative and the other must be positive. Since we have already found the positive root i.e. the number of people in the room, we can say that statement 1 is sufficient alone.

Let’s look at an example where we could fall in the trap.

Say statement 1 gives us an equation which looks like this:

60/(n +5) – 10/(n – 5) = 2

As discussed last week, we will easily see that n = 10 satisfies this equation. So should we move on now and say that statement 1 is sufficient alone? No, not so fast! Let’s try to manipulate the equation to get the quadratic.

60/(n +5) – 10/(n – 5) = 2

60*(n – 5) – 10*(n + 5) = 2*(n – 5)(n + 5)

n^2 – 25n + 150 = 0

n = 10 or 15

So actually, there are two values of n that satisfy this equation. In PS questions, since we have a single answer, there would be only one solution so once you get one, you are done. In DS questions, you need to be certain that only one value satisfies. There is a possibility that both values satisfy your constraints in which case your answer would change.

Therefore, it may not be necessary to solve the equation for the PS question, but it is certainly necessary to solve it for DS. That’s counter intuitive, isn’t it? We hope you understand the reason.

Another related trap in DS questions: Statement 1 gives you a quadratic and asks you for the value of x (no constraints that x must be an integer or positive number etc). You know that it is a quadratic and it will give you two values of x so you say that statement 1 is not sufficient alone and move on. But hold it! What if both the roots of the equation are same? It may not apparent to you when you look at the equation. When you solve it, you realize that the roots are the same. Hence, ensure that you solve the equation in DS questions before you decide on the sufficiency.

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!

Quarter Wit, Quarter Wisdom: Evading Calculations!

Quarter Wit, Quarter WisdomWe have discussed before how GMAT is not a calculation intensive exam. Whenever you land on an equation which looks something like this: 60/(n – 5) – 60/n = 2, you probably think that we don’t know what we are talking about! You obviously need to cross multiply, make a quadratic and finally, solve the quadratic to get the value of n. Actually, you usually don’t need to do any of that for GMAT questions. You have an important leverage – the options. Even if the options don’t directly give you the values of n or n-5, you can use the knowledge that every GMAT question is do-able in 2 mins and that the numbers fit in beautifully well.

Let’ see whether we can get a value of n which satisfies this equation without going the whole nine yards. We will not use any options and will try to rely on our knowledge that GMAT questions don’t take much time.

60/(n – 5) – 60/n = 2

So, the difference between the two terms of the left hand side is 2. Try to look for values of n which give us simple numbers i.e. try to plug in values which are factors of the numerator.
Say, if n = 10, you get 60/5 – 60/10 = 12 – 6 = 6. The difference between them is much more than 2. 60/n and 60/(n – 5) need to be much closer to each other so that the difference between them is 2. The two terms should be smaller to bring them closer together. So increase the value of n.

Put n = 15 since it is the next number such that (15 – 5 =) 10 as well as 15 divide 60 completely. You get 60/10 – 60/15 = 6 – 4 = 2. It satisfies and you know that a value that n can take is 15. Usually, you will get a solution within 2-3 iterations. This is enough for a PS question. Notice that this equation gives us a quadratic so be careful while working on DS questions. You might need to manipulate the equation a little to figure out whether the other root is a possible solution as well. Anyway, today we will focus on the application of such equations in PS questions only. Let’s take a question now to understand the concept properly:

Question: Machine A takes 2 more hours than machine B to make 20 widgets. If working together, the machines can make 25 widgets in 3 hours, how long will it take machine A to make 40 widgets?

(A) 5
(B) 6
(C) 8
(D) 10
(E) 12

Solution: We need to find the time taken by machine A to make 40 widgets. It will be best to take the time taken by machine A to make 40 widgets as the variable x. Then, when we get the value of x, we will not need to perform any other calculations on it and hence the scope of making an error will reduce. Also, value of x will be one of the options and hence plugging in to check will be easy.

Machine A takes x hrs to make 40 widgets.

Rate of work done by machine A = Work done/Time taken = 40/x

Machine B take 2 hrs less than machine A to make 20 widgets hence it will take 4 hrs less than machine B to make 40 widgets. Think of it this way: Break down the 40 widgets job into two 20 widget jobs. For each job, machine B will take 2 hrs less than machine A so it will take 4 hrs less than machine A for both the jobs together.

Time taken by machine B to make 40 widgets = x – 4

Rate of work done by machine B = Work done/Time taken = 40/(x – 4).

We know the combined rate of the machines is 25/3

So here is the equation:

40/x + 40/(x – 4) = 25/3

The steps till here are not complicated. Getting the value of x poses a bit of a problem.

Notice here that that the right hand side is not an integer. This will make the question a little harder for us, right? Wrong! Everything has its pros and cons. The 3 of the denominator gives us ideas for the values of x (as do the options).  To get a 3 in the denominator, we need a 3 in the denominator on the left hand side too.

x cannot be 3 but it can be 6. If x = 6, 40/(6 – 4) = 20 i.e. the sum will certainly not be 20 or more since we have 25/3 = 8.33 on the right hand side.

The only other option that makes sense is x = 12 since it has 3 in it.

40/12 + 40/(12 – 4) = 10/3 + 5 = 25/3

Answer (E)

If we did not have the options, we might have tried x = 9 too before landing on x = 12. Nevertheless, these calculations are not time consuming at all since you can get rid of the incorrect numbers orally. Making a quadratic and solving it is certainly much more time consuming.

Another method could be to bring 3 to the left hand side to get the following equation:

120/x + 120/(x – 4) = 25

This step doesn’t change anything but it helps if you face a mental block while working with fractions. Try to practice such questions using these techniques – they will save you a lot of time.

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!

Quarter Wit, Quarter Wisdom: An Official Assumption Question

Quarter Wit, Quarter WisdomToday we will look at an OG question of critical reasoning (as promised last week). We will use the concept discussed last week – remember what an assumption is. An assumption is a missing necessary premise. It will bring in new information essential to the conclusion.

Now let’s jump on to the OG question.

Question: A recent report determined that although only three percent of drivers on Maryland highways equipped their vehicles with radar detectors, thirty-three percent of all vehicles ticketed for exceeding the speed limit were equipped with them. Clearly, drivers who equip their vehicles with radar detectors are more likely to exceed the speed limit regularly than are drivers who do not.

The conclusion drawn above depends on which of the following assumptions?

(A) Drivers who equip their vehicles with radar detectors are less likely to be ticketed for exceeding the speed limit than are drivers who do not.

(B) Drivers who are ticketed for exceeding the speed limit are more likely to exceed the speed limit regularly than are drivers who are not ticketed.

(C) The number of vehicles that were ticketed for exceeding the speed limit was greater than the number of vehicles that were equipped with radar detectors.

(D) Many of the vehicles that were ticketed for exceeding the speed limit were ticketed more than once in the time period covered by the report.

(E) Drivers on Maryland highways exceeded the speed limit more often than did drivers on other state highways not covered in the report.

Solution:

Let’s look at the question stem first. We need to find an assumption. An assumption is a missing necessary premise. Something that will not only strengthen the conclusion but also be essential to the argument.
An assumption is a statement that needs to be added to the premises for the conclusion to be true. Let’s first find the premises and the conclusion of this argument.

Premises:

– Only 3% of drivers on Maryland highways had radar detectors.

– 33% of vehicles that got speeding tickets had radar detectors.

Conclusion:

Drivers with radar detectors are more likely to exceed the speed limit regularly than other drivers.

There must be some disconnect between the premises and conclusion since there is an assumption in the argument. Look carefully. Premises give you the connection between ‘vehicles that have radar detectors’ and ‘vehicles that get speeding tickets’. The conclusion, on the other hand, concludes a relation between ‘vehicles that have radar detectors’ and ‘vehicles that exceed the speed limit’. The assumption must then give a connection between ‘vehicles that get speeding tickets’ and ‘vehicles that exceed speed limit’.

To clarify it further,

A – vehicles that have radar detectors

B – vehicles that get speeding tickets/vehicles that were ticketed for speeding

C – vehicles that exceed the speed limit

Premises:

– Only 3% of all vehicles are A

– 33% of B are A

Conclusion:

– A are  more likely to be C

The assumption needs to be something that links B to C i.e. that links ‘vehicles that get speeding tickets’

to ‘vehicles that exceed the speed limit’. Option (B) gives us that relation. It says ‘B are more likely to be C’.

Lets add it to premises and see if the conclusion makes more sense now:

– Only 3% of drivers on Maryland highways had radar detectors.

– 33% of vehicles that got speeding tickets had radar detectors.

– Drivers who get speeding tickets are more likely to exceed the speed limit regularly than others.

Conclusion: Drivers with radar detectors are more likely to exceed the speed limit regularly than other drivers.

Now it makes sense!

Let’s take a quick look at the other options and see why they don’t work. We will retain the A, B, C structure given above.

Option (A) says ‘A are less likely to be B’ – Cannot be our assumption

Option (C) only tells us that number of B are greater than number of A.

Option (D) tells us that many vehicles were ticketed multiple times.

Option (E) compares drivers on Maryland highways with drivers on other state highways. This is out of scope.

It is clear that option (B) is the outright winner. This question is one of the tougher questions. You can easily handle it using this technique. We hope you will be able to put this technique to good 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!

Quarter Wit, Quarter Wisdom: Making Sense of Assumptions

Quarter Wit, Quarter WisdomToday we would like to discuss a technique which is very useful in solving assumption questions. No, I am not talking about the ‘Assumption Negation Technique’ (ANT), which, by the way, is extremely useful, no doubt. The point is that ANT is explained beautifully and in detail in your book so there is no point of re-doing it here. You already know how to use it.

What we are going to discuss today is not so much a technique as a revelation of how you can identify the assumption in a particular argument by just fully understanding this – ‘what is an assumption?’. Actually, this is also already discussed in your CR book but I would like to draw your attention to it. We usually end up ignoring the finer points here and hence get stuck on something that is supposed to be quite obvious.

Ok so, what is an assumption?

An assumption is a missing necessary premise. (Doesn’t seem like much a revelation, right? You already knew that! Right! Focus on every word now.)

An assumption is a premise – it gives you some new fact/information.

It is also necessary – necessary for the conclusion to be true. The conclusion cannot be true if the assumption doesn’t hold. Our ANT is based on this premise.

To add, it is also missing – it is not something already mentioned in the argument.

Let’s take a very simplistic example to understand the implication of a missing necessary premise.

Argument: A implies B. B implies C. Hence, A implies D.

Premises given in the argument:
– A implies B
– B implies C

Conclusion given in the argument:
– A implies D

Is it apparent that something is missing here? Sure! The premises give us the relations between A, B and C. They do not mention D. But while drawing the conclusion, we are concluding about the relation between A and D. We can’t do that. We must know something about D too to be able to conclude a relation between A and D. Hence, there is a necessary premise that is missing here. What we are looking for is something that says ‘C implies D’.

When we add this to our premises, our argument makes sense.

Argument: A implies B. B implies C. C implies D. Hence, A implies D.

This little point will help us in solving the trickiest of questions. We get so lost in the n number of things mentioned in the argument that we forget to consider this aspect.

We will discuss an LSAT question today because it seems to be created just to exemplify this concept! Many people falter on this question. After going through it with us here, you will wonder why.

Question:

Therapist: The ability to trust other people is essential to happiness, for without trust there can be no meaningful emotional connection to another human being, and without meaningful emotional connections to others we feel isolated.
Which one of the following, if assumed, allows the conclusion of the therapist’s argument to be properly inferred?
(A) No one who is feeling isolated can feel happy.
(B) Anyone who has a meaningful emotional connection to another human being can be happy.
(C) To avoid feeling isolated, it is essential to trust other people.
(D) At least some people who do not feel isolated are happy.
(E) Anyone who is able to trust other people has a meaningful emotional connection to at least one other human being.

Solution:

First, we break down the argument into premises and conclusion.

Premises:

– Without trust there can be no meaningful emotional connection.

– Without meaningful emotional connections, we feel isolated.

Conclusion:

Ability to trust is essential to happiness.

Do you see something missing here? We are concluding about trust and happiness but in the premises, the link between ‘feeling isolated’ and ‘happiness’ is missing. The premises do not talk about happiness at all. So we need a premise which says, ‘feeling isolated’ means ‘not happy’ for the conclusion to make sense.
Look at the premises now:

Premises:

– Without trust there can be no meaningful emotional connection.

– Without meaningful emotional connections, we feel isolated.

– When we feel isolated, we cannot be happy. (The assumption)

Conclusion:

Ability to trust is essential to happiness.

Now it all makes sense, doesn’t it?

Look at the options now.

Option (A) says – ‘No one who is feeling isolated can feel happy.’ – exactly what we needed.

Hence (A) will be your answer.

Hope this makes sense to you. Next week, we will see how you can easily solve OG questions using this concept.

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!

Quarter Wit, Quarter Wisdom: The Efficiency of Using Variation

Quarter Wit, Quarter WisdomToday, we would like to discuss one of our own work questions. The intent is to show you how simple your calculations can get when you use the methods we discussed in the last few weeks. I couldn’t say it enough – develop a love for ratios. You will save a huge amount of time in lots of questions. If you haven’t been following the last few weeks’ posts, take a look at this link before checking out the question. Otherwise the method may not make sense to you.

Question: 16 horses can haul a load of lumber in 24 minutes. 12 horses started hauling a load and after 14 minutes, 12 mules joined the horses. Will it take less than a quarter-hour for all of them together to finish hauling the load?

(1) Mules work more slowly than horses.
(2) 48 mules can haul the same load of lumber in 16 minutes.

Solution: First do this question on your own and see the calculations involved. Thereafter, check out the solution given below to know how we can solve the question using our joint variation method.

We are given that 16 horses can complete the work in 24 mins. Let’s find out how much work is done by 12 horses in 14 mins (before the mules join in)

16 horses ……… 24 mins ………. 1 work
12 horses ……… 14 mins ………. ?? work

Work done = 1*(14/24)*(12/16) = 7/16 work (if you don’t know how we arrived at this, seriously, check out last week’s post first)

So in 14 mins, the 12 horses can complete 7/16 of the work i.e. they do 1/16 of the work every 2 mins.

How much work is leftover for the mules and horses to do together? 1 – 7/16 = 9/16

Leftover work = 9/16

This makes us think that 12 horses alone will take 9*2 = 18 mins to finish the work. When 12 mules join in, depending on the rate of work of mules, time taken to complete this work could be less than or more than 15 mins.

Statement 1: Mules work more slowly than horses.

This statement doesn’t give us enough information. It just tells us that mules work slower than horses. Say if they work very slowly so that, effectively, they are not adding much to the work done, the work will get done in approximately 18 mins. If they work faster, time taken will keep decreasing. If they work as fast as the horses, the rate at which the work will be done will double (because we already have 12 horses and we will add 12 mules which will be equivalent to 12 horses) and time taken will become half i.e. it will be 9 mins. So the time taken will vary in the range 9 mins to 18 mins depending on the rate of work of mules. This statement alone is not sufficient.

Statement 2: 48 mules can haul the same load of lumber in 16 minutes.

We now know the rate of work of mules. The point is that now we can easily calculate the exact time taken by 12 horses and 12 mules to complete 9/16 of the work. Once we calculate the exact time, we will be able to say whether the time taken will be less than or more than 15 mins. Hence this statement alone is sufficient to answer the question. We don’t really need to find out exactly how much is taken by the 24 animals together since it is a DS question. Ideally, we should mark the answer as (B) and move on.

Nevertheless, let’s do the calculations if only to practice application of work concepts.

Let’s try to find the equivalence of mules and horses (the way we did with cars in the previous post)

We know that 16 horses can haul a load of lumber in 24 minutes. Let’s find out the number of mules that are needed to complete the work in 24 mins.

48 mules …….16 mins.

?? mules ……..24 mins

No. of mules required = 48*16/24 = 32 mules

So, 32 mules do the same work in the same time as done by 16 horses. Or we can say that 2 mules are equivalent to 1 horse. Hence, 12 mules are equivalent to 6 horses. When 12 mules join the 12 horses, equivalently we get 12+6 = 18 horses.

16 horses ……… 24 mins ………. 1 work

18 horses ……….. ?? mins ……….. 9/16 work

Time taken by 18 horses (i.e. 12 horses and 12 mules) = 24*(9/16)*(16/18) = 12 mins

Yes, the horses and mules together will take less than a quarter-hour to finish hauling the load.

Answer (B)

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!

Quarter Wit, Quarter Wisdom: Work-Rate Using Joint Variation

Quarter Wit, Quarter WisdomThis week, let’s look at some work-rate questions which use joint variation. Check out the last three posts of QWQW series if you are not comfortable with joint variation.

Question 1: A contractor undertakes to do a job within 100 days and hires 10 people to do it. After 20 days, he realizes that one fourth of the work is done so he fires 2 people. In how many more days will the work get over?

(A) 60

(B) 70

(C) 75

(D) 80

(E) 100

Solution: Can we say that 10 people can finish the work in 100 days? No. If that were the case, after 20 days, only 1/5th of the work would have been over. But actually 1/4th of the work is over. This means that ‘10 people can complete the work in 100 days’ was just the contractor’s estimate (which turned out to be incorrect). Actually 10 people can do 1/4th of the work in 20 days. The contractor fires 2 people. So the question is how many days are needed to complete 3/4th of the work if 8 people are working?

We need to find the number of days. How is ‘no. of days’ related to ‘no. of people’ and ‘work done’?

If we have more ‘no. of days’ available, we need fewer people. So ‘no. of days’ varies inversely with ‘no. of people’.

If we have more ‘no. of days’ available, ‘work done’ will be more too. So ‘no. of days’ varies directly with ‘work done’.

Therefore,

‘no. of days’ * ‘no. of people’/’work done’ = constant

20*10/(1/4) = ‘no. of days’*8/(3/4)

No. of days = 75

So, the work will get done in 75 days if 8 people are working.

We can also do this question using simpler logic. The concept used is joint variation only. Just the thought process is simpler.

10 people can do 1/4th of the work in 20 days.

8 people can do 3/4th of the work in x days.

Start with the no. of days since you want to find the no of days:

x = 20*(10/8)*(3/1) = 75

From where do we get 10/8? No. of people decreases from 10 to 8. If no. of people is lower, the no of days taken to do the work will be more. So 20 (the initial no. of days) is multiplied by 10/8, a number greater than 1, to increase the number of days.

From where do we get (3/1)? Amount of work increases from 1/4 to 3/4. If more work has to be done, no. of days required will be more. So we further multiply by (3/4)/(1/4) i.e. 3/1, a number greater than 1 to further increase the number of days.

This gives us the expression 20*(10/8)*(3/1)

We get that the work will be complete in another 75 days.

Answer (C)

Let’s take another question to ensure we understand the logic.

Question 2: A company’s four cars running 10 hrs a day consume 1200 lts of fuel in 10 days. In the next 6 days, the company will need to run 9 cars for 12 hrs each so it rents 5 more cars which consume 20% less fuel than the company’s four cars. How many lts of fuel will be consumed in the next 6 days?

(A) 1200 lt

(B) 1555 lt

(C) 1664 lt

(D) 1728 lt

(E) 4800 lt

Solution: First let’s try to figure out what is meant by ‘consume 20% less fuel than the company’s cars’. It means that if company’s each car consumes 1 lt per hour, the hired cars consume only 4/5 lt per hour. So renting 5 more cars is equivalent to renting 4 cars which are same as the company’s cars. Hence, the total number of cars that will be run for the next 6 days is 8 company-equivalent cars.

4 cars running 10 hrs for 10 days consume 1200 lt of fuel

8 cars running 12 hrs for 6 days consume x lt of fuel

x = 1200*(8/4)*(12/10)*(6/10) = 1728 lt

We multiply by 8/4 because more cars implies more fuel so we multiply by a number greater than 1.

We multiply by 12/10 because more hours implies more fuel so we multiply by a number greater than 1.

We multiply by 6/10 because fewer days implies less fuel so we multiply by a number smaller than 1.

Answer (D)

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!

Quarter Wit, Quarter Wisdom: Varying Jointly

Quarter Wit, Quarter WisdomNow that we have discussed direct and inverse variation, joint variation will be quite intuitive. We use joint variation when a variable varies with (is proportional to) two or more variables.

Say, x varies directly with y and inversely with z. If y doubles and z becomes half, what happens to x?

“x varies directly with y” implies x/y = k (keeping z constant)

If y doubles, x doubles too.

“x varies inversely with z” implies xz = k (now keeping y constant)

If z becomes half, x doubles.

So the overall effect is that x becomes four times of its initial value.

The joint variation expression in this case will be xz/y = k. Notice that when z is constant, x/y = k and when y is constant, xz = k; hence both conditions are being met. Once you get the expression, it’s very simple to solve for any given conditions.

x1*z1/y1 = x2*z2/y2 = k (In any two instances, xz/y must remain the same)

x1*z1/y1 = x2*(1/2)z1/2*y1

x2 = 4*x1

Let’s look at some more examples. How will you write the joint variation expression in the following cases?

1. x varies directly with y and directly with z.

2. x varies directly with y and y varies inversely with z.

3. x varies inversely with y^2 and inversely with z^3.

4. x varies directly with y^2 and y varies directly with z.

5. x varies directly with y^2, y varies inversely with z and z varies directly with p^3.

Solution: Note that the expression has to satisfy all the conditions.

1. x varies directly with y and directly with z.

x/y = k

x/z = k

Joint variation: x/yz = k

2. x varies directly with y and y varies inversely with z.

x/y = k

yz = k

Joint variation: x/yz = k

3. x varies inversely with y^2 and inversely with z^3.

x*y^2 = k

x*z^3 = k

Joint variation: x*y^2*z^3 = k

4. x varies directly with y^2 and y varies directly with z.

x/y^2 = k

y/z = k which implies that y^2/z^2 = k

Joint variation: x*z^2/y^2 = k

5. x varies directly with y^2, y varies inversely with z and z varies directly with p^3.

x/y^2 = k

yz = k which implies y^2*z^2 = k

z/p^3 = k which implies z^2/p^6 = k

Joint variation: (x*p^6)/(y^2*z^2) = k

Let’s take a GMAT prep question now to see these concepts in action:

Question 1: The rate of a certain chemical reaction is directly proportional to the square of the concentration of chemical M present and inversely proportional to the concentration of chemical N present. If the concentration of chemical N is increased by 100 percent, which of the following is closest to the percent change in the concentration of chemical M required to keep the reaction rate unchanged?

(A) 100% decrease
(B) 50% decrease
(C) 40% decrease
(D) 40% increase
(E) 50% increase

Solution:

Rate/M^2 = k

Rate*N = k

Rate*N/M^2 = k

If Rate has to remain constant, N/M^2 must remain the same too.

If N is doubled, M^2 must be doubled too i.e. M must become ?2 times. Since ?2 = 1.4 (approximately),

M must increase by 40%.

Answer (D)

Simple enough?

Quarter Wit, Quarter Wisdom: Varying Inversely

Quarter Wit, Quarter WisdomAs promised, we will discuss inverse variation today. The concept of inverse variation is very simple – two quantities x and y vary inversely if increasing one decreases the other proportionally.

If x takes values x1, x2, x3… and y takes values y1, y2, y3 … correspondingly, then x1*y1 = x2*y2 = x3*y3 = some constant value

This means that if x doubles, y becomes half; if x becomes 1/3, y becomes 3 times etc. In other words, product of x and y stays the same in different instances.

Notice that x1/x2 = y2/y1; The ratio of x is inverse of the ratio of y.

The concept will become clearer after working on a few examples.

Question 1: The price of a diamond varies inversely with the square of the percentage of impurities. The cost of a diamond with 0.02% impurities is $2500. What is the cost of a diamond with 0.05% impurities (keeping everything else constant)?

(A)   $400

(B)   $500

(C)   $1000

(D)   $4000

(E)    $8000

Solution:

Price1*(% of Impurities1)^2 = Price2*(% of Impurities2)^2

2500*(.02)^2 = Price2*(.05)^2

Price = $400

Answer (A)

The answer is quite intuitive in the sense that if % of impurities in the diamond increases, the price of the diamond decreases.

There is an important question type related to inverse variation. It often uses the formula:

Total Price = Number of units*Price per unit

If, due to budgetary constraints, we need to keep the total money spent on a commodity constant, number of units consumed varies inversely with price per unit. If price per unit increases, we need to reduce the consumption proportionally.

Question 2: The cost of fuel increases by 10%. By what % must the consumption of fuel decrease to keep the overall amount spent on the fuel same?

(A)   5%

(B)   9%

(C)   10%

(D)   11%

(E)    20%

Solution: Do you think the answer is 10%? Think again.

Total Cost = Number of units*Price per unit

If the price per unit increases by 10%, it becomes 11/10 of its original value. To keep the total cost same, you need to multiply number of units by 10/11. i.e. you need to decrease the number of units by 1/11 i.e. 9.09%. In that case,

New Total Cost = (10/11)*Number of units*(11/10)*Price per unit

This new total cost will be the same as the previous total cost.

Answer (B)

Let’s look at one more example of the same concept but this one is a little trickier.

Question 3: Recently, fuel price has seen a hike of 20%. Mr X is planning to buy a new car with better mileage as compared to his current car. By what % should the new mileage be more than the previous mileage to ensure that Mr X’s total fuel cost stays the same for the month? (assuming the distance traveled every month stays the same)

(A)   10%

(B)   17%

(C)   20%

(D)   21%

(E)    25%

Solution:  The problem here is ‘how is mileage related to fuel price?’

Total fuel cost = Fuel price * Quantity of fuel used

Since the ‘total fuel cost’ needs to stay the same, ‘fuel price’ varies inversely with ‘quantity of fuel used’.

Quantity of fuel used = Distance traveled/Mileage

Distance traveled = Quantity of fuel used*Mileage

Since the same distance needs to be traveled, ‘quantity of fuel used’ varies inversely with the ‘mileage’.

We see that ‘fuel price’ varies inversely with ‘quantity of fuel used’ and ‘quantity of fuel used’ varies inversely with ‘mileage’. So, if fuel price increases, quantity of fuel used decreases proportionally and if quantity of fuel used decreases, mileage increases proportionally. Hence, if fuel price increases, mileage increases proportionally or we can say that fuel price varies directly with mileage.

If fuel price becomes 6/5 (20% increase) of previous fuel price, we need the mileage to become 6/5 of the previous mileage too i.e. mileage should increase by 20% too.

Another method is that you can directly plug in the expression for ‘Quantity of fuel used’ in the original equation.

Total fuel cost = Fuel price * Distance traveled/Mileage

Since ‘total fuel cost’ and ‘distance traveled’ need to stay the same, ‘fuel price’ is directly proportional to ‘mileage’.

Answer (C)

We hope you are comfortable with fundamentals of direct and inverse variation now. More next week!

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!

 

Quarter Wit, Quarter Wisdom: Varying Directly

Quarter Wit, Quarter WisdomWe can keep working on ‘pattern recognition’ questions for a long time and not run out of questions of different types on which it can be used. We hope you have understood the basic concepts involved. So let’s move on to another topic now: Variation.

Basically, variation describes the relation between two or more quantities. e.g. workers and work done, children and noise, entrepreneurs and start ups. More workers means more work done; more children means more noise; more entrepreneurs means more start ups and so on… These are examples of direct variation i.e. if one quantity increases, the other increases proportionally. Then there are quantities that have inverse variation between them e.g. workers and time taken. If there are more workers, time taken to complete a work will be less.

Let’s discuss direct variation today.

Formally, let’s say x varies directly with y. If x takes values x1, x2, x3… and y takes values y1, y2, y3 … correspondingly, then x1/y1 = x2/y2 = x3/y3 = Some constant value

In other words, ratio of x and y stays the same in different instances.

(Notice that this is the same as x1/x2 = y1/y2)

It might seem a little cumbersome when put this way but the truth is that direct variation is quite intuitive. A couple of questions will make it clear.

Question 1: 20 workmen can make 35 widgets in 5 days. How many workmen are needed to make 105 widgets in 5 days?

(A)   7

(B)   20

(C)   25

(D)   30

(E)    60

Solution: Notice that the number of days stays the same so we can ignore it. Now think, how are workmen and widgets related? If the number of workmen increases, the number of widget made also increases proportionally. You need to find the new number of workmen required. The number of widgets has become thrice (105/35 = 3) so number of workmen needed will become thrice as well (remember, the number of workmen will increase in the same proportion).

We need 20*3 = 60 workmen

Answer (E)

The concept of variation is very intuitive. If the number of widgets required doubles, the number of workmen required to make them in the same amount of time will double too. If the number of widgets required becomes one fourth, the number of workmen required to make them in the same amount of time will become one fourth too.

A quantity can directly vary with some power of another quantity. Let’s take an example of this scenario too.

Question 2: If the ratio of the volumes of two right circular cylinders is given by 64/9, what is the ratio of their radii? (Both the cylinders have the same height)

(A)   4/3

(B)   8/3

(C)   16/9

(D)   4/1

(E)    16/3

Solution: This question involves a little bit of geometry too. The volume of a right circular cylinder is given by Area of base * height i.e.

Volume of a right circular cylinder = pi*radius^2 * height

So volume varies directly with the square of radius.

Va/Vb = 64/9 = Ra^2/Rb^2

Ra/Rb = 8/3

Answer (B)

We hope this little concept is not hard to understand. We will work on inverse proportion next week and then work on problems involving both (that’s where the good questions are!).

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!

Quarter Wit, Quarter Wisdom: Pattern Recognition or Number Properties?

Quarter Wit, Quarter WisdomContinuing our quest to master ‘pattern recognition’, let’s discuss a tricky little question today. It is best done using divisibility and remainders logic we discussed in some previous posts. We suggest you check out these divisibility posts if you haven’t yet.

We are first going to see how to solve the question conceptually. The interesting thing is – what do you do if you are under immense pressure during the test and are unable to remember anything you ever read on divisibility? It is a fairly common phenomenon – students have reported that they had blanked out during the test and couldn’t think of an appropriate approach. Our suggestion is that in that case, you should lean on trying to figure out the pattern. Try out a couple of values and see what you get. You may not understand why you are getting what you are getting but that doesn’t stop you from getting the correct answer. Let’s jump on to the question – we will first discuss the ideal approach and then go on to what happens if you don’t have a clue of what to do in the question.

Question: If p, m, and n are positive integers, n is odd, and p = m^2 + n^2, is m divisible by 4?

Statement 1: When p is divided by 8, the remainder is 5.
Statement 2: m – n = 3

Solution:  Given that p, m and n are positive integers. If n is odd, n^2 must also be odd. How do you represent an odd integer? As (2k + 1)

n^2 = (2k + 1)^2 = 4k^2 + 4k + 1 = 4k(k + 1) + 1
Since one of k and (k+1) will definitely be even (out of any two consecutive integers, one is always even, the other is always odd – discussed in detail here), 4k(k+1) will be divisible by 8. Therefore, when n^2 is divided by 8, it will leave a remainder of 1.

Statement 1: When p is divided by 8, the remainder is 5.

When p ( i.e. m^2 + n^2) is divided by 8, we get a remainder of 5. When n^2 is divided by 8, the remainder will be 1. To get a remainder of 5, when m^2 is divided by 8, we should get a remainder of 4.
m^2 = 8a + 4 (i.e. we can make ‘a’ groups of 8 and 4 will be leftover)
m^2 = 4(2a + 1)

This implies m = 2*?(Odd Number) because (2a+1) is an odd number. Square root of an odd number will also be odd.
Therefore, we can say that m is not divisible by 4.

This statement alone is sufficient.

Statement 2: m – n = 3

The difference between m and n is 3 i.e. an odd number. Since n is odd, we can say that m will be even (Even – Odd = Odd). But whether m is divisible by 2 only or by 4 as well, we cannot say since here we have no constraints on p.

This statement alone is not sufficient.

Answer (A)

In this question, analyzing the question stem and statement 1 is a little complicated. Say we don’t analyze the question stem and jump to statement 1. Let’s see how we can use pattern recognition to make it easier.

Question: If p, m, and n are positive integers, n is odd, and p = m^2 + n^2, is m divisible by 4?

Statement 1: When p is divided by 8, the remainder is 5.
Statement 2: m – n = 3

Solution:

Statement 1: When p is divided by 8, the remainder is 5.
When p ( i.e. m^2 + n^2) is divided by 8, we get a remainder of 5.  We need a remainder of 5 when m^2 + n^2 is divided by 8. Let’s try to find the remainders when m^2 and n^2 are divided by 8.

We are given that n is odd. Let’s try to figure out what this implies.

n = 1; When n^2 ( = 1) is divided by 8, the remainder is 1.
n = 3; When n^2 ( = 9) is divided by 8, the remainder is 1.
n = 5; When n^2 ( = 25) is divided by 8, the remainder is 1.
n = 7; When n^2 (= 49) is divided by 8, the remainder is 1.

There is a pattern here! Whenever you divide the square of an odd number by 8, you get the remainder 1. (We have discussed ‘why’ above in the logical explanation of statement 1)

This implies that when we divide m^2 by 8, we get 4 as remainder. If m^2 gives 4 as remainder, it means it is of the form m^2 = 8a + 4 = 4(2a + 1). So m must be of the form 2?(Odd Number). Hence m is not divisible by 4.

This statement is sufficient alone.

Through this example, you can see that pattern recognition is a very important tool (in easy as well as difficult questios)

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!

Quarter Wit, Quarter Wisdom: Pattern Recognition – Part II

Quarter Wit, Quarter Wisdom
Today’s post comes from Karishma, a Veritas Prep GMAT instructor. Before reading, be sure to check out Part I from last week!

Last week we saw how to use pattern recognition. Today, let’s take up another question in which this concept will help us. Mind you, there are various ways of solving a question. Most questions we solve using pattern recognition can be solved using another method. But pattern recognition is a method we can use in various cases. It is something that comes to our aid when we forget everything else. If you don’t know from where to start on a question, try to give some values to the variables. You might see a pattern. You may not ‘know’ something. Even then, you can ‘figure out’ the answer because GMAT is not a test of your knowledge; it is a test of your wits. It is a test of whether you can keep your cool when faced with the unknown and use whatever you know to solve the question.

Let’s look at a question now.
Continue reading “Quarter Wit, Quarter Wisdom: Pattern Recognition – Part II”

Quarter Wit, Quarter Wisdom: Pattern Recognition

Quarter Wit, Quarter WisdomIf you are hoping for a 700+ in GMAT, you need to develop the ability to recognize patterns. GMAT does not test advanced concepts but you can certainly get advanced questions on simple concepts. For such questions, the ability to quickly observe patterns can come in quite handy. We will discuss a complicated question today which can be easily solved by observing the pattern.

Question: If a and b are distinct integers and a^b = b^a, how many solutions does the ordered pair (a, b) have? Continue reading “Quarter Wit, Quarter Wisdom: Pattern Recognition”

Quarter Wit Quarter Wisdom: GCF and LCM of Fractions

Quarter Wit, Quarter WisdomLast week we discussed some concepts of GCF. Today we will talk about GCF and LCM of fractions.

LCM of two or more fractions is given by: LCM of numerators/GCF of denominators

GCF of two or more fractions is given by: GCF of numerators/LCM of denominators

Why do we calculate LCM and GCF of fractions in this way? Let’s look at the algebraic explanation first. Then we will look at a more intuitive reason.

Algebraic Approach:

Consider 2 fractions a/b and c/d in their lowest form, their LCM is Ln/Ld and GCF is Gn/Gd, also in their lowest forms.
Let’s work on figuring out the LCM first.

LCM is a multiple of both the numbers so Ln/Ld must be divisible by a/b. This implies (Ln/Ld)/(a/b) is an integer. We can re-write this as:

Ln*b/Ld*a is an integer.

Since a/b and Ln/Ld are in their lowest forms, Ln must be divisible by a; also, b must be divisible by Ld (because a and b have no common factors and Ln and Ld have no common factors).

Using the same logic, Ln must be divisible by c; also d must be divisible by Ld.

Ln, the numerator of LCM, must be divisible by both a and c and hence should be the LCM of a and c, the numerators. Ln cannot be just any multiple of a and c; it must be the lowest common multiple so that Ln/Ld is the lowest multiple of the two fractions.

b and d both must be divisible by Ld, the denominator of LCM, and hence Ld must be their highest common factor. Mind you, it cannot be just any common factor; it needs to be the highest common factor so that Ln/Ld is the lowest multiple possible.

This is why LCM of two or more fractions is given by: LCM of numerators/GCF of denominators.

Using similar reasoning, you can figure out why we find GCF of fractions the way we do.

Now let me give you some feelers. They are more important than the algebraic explanation above. They build intuition.

Intuitive Approach:

Let me remind you first that LCM is the lowest common multiple. It is that smallest number which is a multiple of both the given numbers.

Say, we have two fractions: 1/4 and 1/2. What is their LCM? It’s 1/2 because 1/2 is the smallest fraction which is a multiple of both 1/2 and 1/4. It will be easier to understand in this way:

1/2 = 2/4. (Fractions with the same denominator are comparable.)

LCM of 2/4 and 1/4 will obviously be 2/4.

If this is still tricky to see, think about their equivalents in decimal form:

1/2 = 0.50 and 1/4 = 0.25. You can see that 0.50 is the lowest common multiple they have.

Let’s look at GCF now.

What is GCF of two fractions? It is that greatest factor which is common between the two fractions. Again, let’s take 1/2 and 1/4. What is the greatest common factor between them?

Think of the numbers as 2/4 and 1/4. The greatest common factor between them is 1/4.

(Note that 1/2 and 1/4 are both divisible by other factors too e.g. 1/8, 1/24 etc but 1/4 is the greatest such common factor)

Now think, what will be the LCM of 2/3 and 1/8?

We know that 2/3 = 16/24 and 1/8 = 3/24.
LCM = 16*3/24 = 48/24 = 2

LCM is a fraction greater than both the fractions or equal to one or both of them (when both fractions are equal). When you take the LCM of the numerator and GCF of the denominator, you are making a fraction greater than (or equal to) the numbers.

Also, what will be the GCF of 2/3 and 1/8?

We know that 2/3 = 16/24 and 1/8 = 3/24.

GCF = 1/24

GCF is a fraction smaller than both the fractions or equal to one or both of them (when both fractions are equal). When you take the GCF of the numerator and LCM of the denominator, you are making a fraction smaller than (or equal to) the numbers.

We hope the concept of GCF and LCM of fractions makes sense to you now.

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!

Quarter Wit, Quarter Wisdom: Some GCF Concepts

Quarter Wit, Quarter WisdomSometimes students come up looking for explanations of concepts they come across in books. Actually, in Quant, you can establish innumerable inferences from the theory of any topic. The point is that you should be comfortable with the theory. You should be able to deduce your own inferences from your understanding of the topic. If you come across some so-called rules, you should be able to say why they hold. Let’s discuss a couple of such rules from number properties regarding GCF (greatest common factor). Many of you might read them for the first time. Stop and think why they must hold.

Rule 1: Consecutive multiples of ‘x’ have a GCF of ‘x’

Explanation: What do we mean by consecutive multiples of x? They are the consecutive terms in the multiplication table of x. For example, 4x and 5x are consecutive multiples of x. So are 18x and 19x…

What will be the greatest common factor of 18x and 19x? We know that x is their common factor. Do 18 and 19 have any common factors (except 1)? No. So greatest common factor will be x. Take any two consecutive numbers. They will have no common factors except 1. Hence, if we have two consecutive factors of x, their GCF will always be x.

For more on common factors of consecutive numbers, check:
http://www.veritasprep.com/blog/2011/09 … c-or-math/
http://www.veritasprep.com/blog/2011/09 … h-part-ii/

Can you derive some of your own ‘rules’ based on this now? Let’s give you some ideas:

Two consecutive integers have GCF of 1.

Two consecutive odd multiples of x have GCF of x.

Rule 2: The G.C.F of two distinct numbers cannot be larger than the difference between the two numbers.

Explanation: GCF is a factor of both the numbers. Say, the GCF of two distinct numbers is x. This means the two numbers are mx and nx where m and n have no common factor. What can be the smallest difference between m and n? m and n cannot be equal since the numbers are distinct. The smallest difference between them can be 1 i.e. they can be consecutive numbers. In that case, the difference between mx and nx will be x which is equal to the GCF. If m and n are not consecutive integers, the difference between them will be much larger than x. The difference between mx and nx cannot be less than x.
Say, GCF of two numbers is 6. The numbers can be 12 and 18 (GCF = 6) or 12 and 30 (GCF = 6) etc but they cannot be 12 and 16 since both numbers must have 6 as a factor. So after a multiple of 6, the other multiple of 6 must be at least 6 away.

Let’s look at a third party question based on these concepts now.

Question 1: What is the greatest common factor of x and y?

Statement 1: Both x and y are divisible by 4.
Statement 2: x – y = 4

Solution:

Statement 1: Both x and y are divisible by 4

We know that 4 is a factor of both x and y. But is it the highest common factor? We do not know. There could be another factor common between x and y and hence highest common factor could be greater than 4. e.g. 4 and 16 have 4 as the highest common factor but 12 and 36 have 12 as the highest common factor though both pairs have 4 as a common factor.

Statement 2: x – y = 4

We know that x and y differ by 4. So their GCF cannot be greater than 4 (as discussed above). The GCF could be any of 1/2/4 e.g. 7 and 11 have GCF of 1 while 2 and 6 have GCF of 2.

Taking both statements together: From statement 1, we know that x and y have 4 as a common factor. From statement 2, we know that x and y have one of 1/2/4 as highest common factor. Hence 4 is the highest common factor.

Answer (C)

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!

Quarter Wit, Quarter Wisdom: Have a Game Plan

Quarter Wit, Quarter WisdomFor the past few weeks, we have been discussing conditional statements. Let’s switch back to Quant today. I have been meaning to discuss a question for a while. We can easily solve it by plugging in the right values. The only issue is in figuring out the right values quickly. The point we are going to discuss is that there has to be a plan.

Question 1: Six countries in a certain region sent a total of 75 representatives to an international congress. No two countries sent the same number of representatives. Of the six countries, if Country A sent the second greatest number of representatives, did Country A send at least 10 representatives?

Statement 1: One of the six countries sent 41 representatives to the congress.
Statement 2: Country A sent fewer than 12 representatives to the congress. Continue reading “Quarter Wit, Quarter Wisdom: Have a Game Plan”

Quarter Wit, Quarter Wisdom: Necessary Conditions

Quarter Wit, Quarter WisdomLast week we discussed a very tricky CR question based on conditional statements. This week, we would like to discuss another CR question based on necessary conditions. Note that you don’t need to be given ‘only if’ or ‘only when’ to mark a necessary condition. The wording of the statement could imply it. You need to keep a keen eye to figure out necessary and sufficient conditions.

Question: Successfully launching a new product for supermarket sale requires either that supermarkets give prominent shelf space to the product or that plenty of consumers who have not tried it seek it out. Consumers will seek out a new product only if it is extensively advertised, either on television or in the press. One way for a manufacturer to obtain prominent shelf space for a new product is to promote it in trade journals.
Continue reading “Quarter Wit, Quarter Wisdom: Necessary Conditions”

Quarter Wit, Quarter Wisdom: A Question on Conditional Statements

Quarter Wit, Quarter WisdomWe hope that you have understood conditional sentences we discussed in the last post. The concept is very important and you will come across questions using this concept often. Now, let’s discuss the GMAT question we gave you last week.

Question: A newborn kangaroo, or joey, is born after a short gestation period of only 39 days. At this stage, the joey’s hind limbs are not well developed, but its forelimbs are well developed, so that it can climb from the cloaca into its mother’s pouch for further development. The recent discovery that ancient marsupial lions were also born with only their forelimbs developed supports the hypothesis that newborn marsupial lions must also have needed to climb into their mothers’ pouches.
Continue reading “Quarter Wit, Quarter Wisdom: A Question on Conditional Statements”

Quarter Wit, Quarter Wisdom: Conditional Statements

Quarter Wit, Quarter WisdomLast week we discussed a Critical Reasoning question in detail. Today, I want to discuss a very important concept of CR — analyzing a conditional statement. You will often encounter these even though they may not be in the exact same format as the one we will discuss below. We will discuss the basic framework and then we will look at questions where this concept will be very helpful. Mind you, without this framework, it can get a little tricky to wrap your head around these questions.

Statement 1:

If you trouble your teacher, you will be punished.

What does this imply? It implies that ‘troubling your teacher’ is a sufficient condition to get punished. If you trouble, you will get punished.
Continue reading “Quarter Wit, Quarter Wisdom: Conditional Statements”

Quarter Wit, Quarter Wisdom: Evaluate the Conclusion

Quarter Wit, Quarter WisdomAt the risk of generalizing from a relatively small sample, let me say that people who are good at Quant, tend to be good at Critical Reasoning in Verbal. I certainly cannot comment about their SC and RC prowess but they are either good at CR right from the start or improve dramatically after just a couple of our sessions. The reason for this is very simple – CR is more like Quant than like Verbal. CR is very mathematical. You need to keep in mind some basic rules and based on those, you can easily crack the most difficult of problems. There is only one catch – don’t get distracted by options put there to distract you!

Today I would like to discuss a great CR question from our book. It upsets a lot of students even though it is simple – just like a GMAT question is supposed to be. Here is the question:
Continue reading “Quarter Wit, Quarter Wisdom: Evaluate the Conclusion”

Quarter Wit, Quarter Wisdom: Rates Revisited

Quarter Wit, Quarter WisdomPeople often complain about getting stuck in work-rate problems. Hence, I would like to take some 700+ level questions on rate today. I have discussed the basic concepts of work-rate (using ratios) in a previous post:

Cracking the Work Rate Problems

You  might want to go through that post before you set out to work on these problems. Ensure that you are very comfortable with the relation: Work = Rate*Time and its implications: If rate doubles, work done doubles too if the time remains constant; if one work is done, rate = 1/time etc. Thorough understanding of these implications is fundamental to ‘reasoning out’ the answer.
Continue reading “Quarter Wit, Quarter Wisdom: Rates Revisited”

Quarter Wit, Quarter Wisdom: Questions on Factorials

Quarter Wit, Quarter WisdomLast week we discussed factorials – how we can take something common when we have factorials in some equations. Today let’s discuss a couple of questions based on factorials. They look intimidating but they are pretty simple. Factorial is all about multiplication and hence there is a high probability that you will be able to take something common and cancel something. These techniques reduce our work significantly. Hence, seeing a factorial in a question should bring a smile to your face!

Question 1: Given that x, y and z are positive integers, is y!/x! an integer?

Statement 1: (x + y)(x – y) = z! + 1

Statement 2: x + y = 121
Continue reading “Quarter Wit, Quarter Wisdom: Questions on Factorials”

Quarter Wit, Quarter Wisdom: Managing Factorials in Equations

Quarter Wit, Quarter WisdomA concept we have not yet covered in this series is factorials (though we used some factorials in the post Power in Factorials). Let’s first discuss the basics of factorials. Once we do, we will see that most factorial expressions can be easily solved using a single method: taking common!

First of all, what is (n!)?

n! = 1*2*3*4*5*6*…*(n – 2)*(n -1)*n

Let’s take some examples:
Continue reading “Quarter Wit, Quarter Wisdom: Managing Factorials in Equations”

Quarter Wit, Quarter Wisdom: Successive Division

Quarter Wit, Quarter WisdomWe discussed divisibility and remainders many weeks ago. Today, we will use those concepts and discuss another type of question – successive division. But before we do, you need to go through the previous related posts on division if you haven’t read them already:

Divisibility Unraveled

Divisibility Applied on the GMAT

Divisibility Applied to Remainders
Continue reading “Quarter Wit, Quarter Wisdom: Successive Division”

Quarter Wit, Quarter Wisdom: Working on Getting the Full Picture Again!

Quarter Wit, Quarter WisdomToday, we will continue our discussion on why it is important to understand the workings behind seemingly miraculous shortcuts. We will use another example from probability.

Question 1: A bag contains 4 white balls, 2 black balls & 3 red balls. One by one three balls are drawn out with replacement (i.e. a ball is drawn and then put back. Thereafter, another ball is drawn). What is the probability that the third ball is red?

Solution:

The question is simple, isn’t it? When you draw balls with replacement, the probability stays the same at the beginning of every cycle. On first draw, the probability of drawing a red ball is 3/9 (since there are 3 red balls and 9 balls in all). When you draw the ball and put in back, the probability of drawing a red ball again stays the same i.e. 3/9 (since again there are 3 red balls and 9 balls in all). The situation at the beginning of every draw is the same.
Continue reading “Quarter Wit, Quarter Wisdom: Working on Getting the Full Picture Again!”

Quarter Wit, Quarter Wisdom: Get the Full Picture – Part II

Quarter Wit, Quarter WisdomLet’s refer back to last week’s post. We discussed why it is important to fully understand what you are doing and why you are doing it, especially while using an innovative method. We talked about it using a combinatorics example.

Let’s revisit it here:

Question 1: What is the probability that you will get a sum of 8 when you throw three dice simultaneously?

We discussed the ways of obtaining various sums. The regular way of obtaining a sum of 8 is enumerating all the possibilities. An innovative way was using 7C2 (as discussed last week).
Continue reading “Quarter Wit, Quarter Wisdom: Get the Full Picture – Part II”

Quarter Wit, Quarter Wisdom: Get the Full Picture

Quarter Wit, Quarter WisdomToday we will discuss why ‘understanding’ rather than just ‘learning’ a concept is important. Most questions can be solved using different methods. Sometimes, a particular method seems really easy and quick and we tend to ‘learn’ it without actually knowing why we are doing what we are doing. We need to understand the strengths and the weaknesses of the method before we use it. Let me elaborate with an example.

Question: What is the probability that you will get a sum of 8 when you throw three dice simultaneously?

When you throw three dice simultaneously, you can obtain a total sum ranging from 3 (1 on each die) to 18 (6 on each die). Let’s consider all the possible sums that we can obtain:

Sum of 3: This can happen in only 1 way. Each die shows 1 in this case

Sum of 4: This can happen in 3 ways. One die shows 2 and the other two show 1 each. Any of the three dice could show 2 so there are a total of 3 ways of getting a sum of 4.

Sum of 5: There are different ways of obtaining a sum of 5:

1, 1, 3 – Any of the three dice could show 3 so there are a total of 3 ways of obtaining 5 in this way.

1, 2, 2 – Any of the three dice could show 1 so there are a total of 3 ways of obtaining 5 in this way.

Total number of ways of obtaining a sum of 5 = 3 + 3 = 6

Sum of 6: There are different ways of obtaining a sum of 6:

1, 1, 4 – Any of the three dice could show 4 so there are a total of 3 ways of obtaining 6 in this way.

1, 2, 3 – These 3 numbers could be arranged among the 3 dice in 3! ways (using basic counting principle). There are a total of 3! = 6 ways of obtaining 6 in this way.

2, 2, 2 – This can happen in only one way. All dice show 2.

Total number of ways of obtaining a sum of 6 = 3 + 6 + 1 = 10

Similarly, we can find the number of ways in which all other sums can be obtained. Below I will list the number of ways in each case.

Sum of 3: 1 way                          Sum of 18: 1 way

Sum of 4: 3 ways                        Sum of 17: 3 ways

Sum of 5: 6 ways                        Sum of 16: 6 ways

Sum of 6: 10 ways                     Sum of 15: 10 ways

Sum of 7: 15 ways                    Sum of 14: 15 ways

Sum of 8: 21 ways                    Sum of 13: 21 ways

Sum of 9: 25 ways                    Sum of 12: 25 ways

Sum of 10: 27 ways                 Sum of 11: 27 ways

Notice the symmetry here. There is only 1 way of obtaining 3 and only one way of obtaining 18. Similarly, there are 3 ways in which you can obtain 4 and 3 ways in which you can obtain 17. Is it a co-incidence? No. The second half of the column can be obtained by replacing 1 by 6, 2 by 5, 3 by 4, 4 by 3, 5 by 2 and 6 by 1. The number of ways of obtaining a sum is symmetrical about the center.

We have simplified one aspect of this problem. If we need to find the number of ways of obtaining 15, we can instead find the number of ways of obtaining a sum of 6 (which is psychologically easier to handle). Now the problem is whether there is an easier way of obtaining the number of ways in which you can get a sum of 6 or do we have to enumerate it in every case.

You might have come across something like this:

Sum of 3: 2C2 = 1 way

Sum of 4: 3C2 = 3 ways

Sum of 5: 4C2 = 6 ways

Sum of 6: 5C2 = 10 ways

Sum of 7: 6C2 = 15 ways

Sum of 8: 7C2 = 21 ways

Perfect till now! Matches the numbers we obtained above. It seems like a good method to use instead of writing down all the cases, doesn’t it? But what happens further on?

Sum of 9: 8C2 = 28 ways

Sum of 10: 9C2 = 36 ways

These don’t match! The key to understanding this is to understand why it works in the above given cases. First review this post.

Focus on method II of question 2. Notice how you divide n identical objects among m distinct groups. Let’s take the example of a sum of 7. You have to divide 7 among 3 dice such that each die must have at least 1 (no die face can show 0). First step is to take 3 out of the 7 and give one each to the three dice. Now you have 4 left to distribute among 3 distinct groups such that it is possible that some groups may get none of the four. Think of partitioning 4 in 3 groups. This can be done in (4+2)!/4!*2! = 6C2 ways (check out the given link if you do not understand this)

This is how you obtain 6C2 for the sum of 7.

The concept works perfectly till the sum of 8. Thereafter it fails. Think why. I will explain it next week.

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!