Learn How to Begin a GMAT Problem by Focusing on Keywords in the Question Stem

Quarter Wit, Quarter WisdomToday, we will not begin our post as we usually do by introducing the topic we intend to discuss. Instead, we will directly ask you to think about a question. The reason is this – when faced with similar questions on the GMAT without any preface, we often struggle to identify the concept being tested, which is the starting point of our efforts. Our post today focuses on how to observe the keywords in the question stem, and how to know where to go from there.

Take a look at this example Quant question:

The length and width of a rectangle are integer values. What is the area of the smallest such rectangle that can be inscribed in a circle whose radius is also an integer?

(A) 12 
(B) 24 
(C) 36 
(D) 48 
(E) 60

Now here is the problem – the question stem does not give us any numbers! We don’t know any dimensions of the rectangle or the circle, yet the answer choice options are very specific numbers! So how do we begin? The smallest positive integer is 1, so should we start by testing the radius of the circle as 1, and then try to go on from there? And if 1 doesn’t work, then move on to 2, 3, 4… etc?

No – we are not a computer algorithm and on top of that, the GMAT only gives us around 2 minutes to figure out the answer. With this in mind, the question should enough clues to make all all of that trial and error testing unnecessary. So if plugging in numbers isn’t the way to go, how should we start solving this problem?

Now, the moment we read “rectangle inscribed in a circle”, what comes to mind is that a rectangle has 90 degree angles, and hence, the diagonal of the rectangle is the diameter of the circle (an arc that subtends a 90 degree angle at the circumference is a semicircle). The rectangle inside of the circle will look something like this:

 

 

 

 

Now we can see that we have a circle with a diameter (AB) and 90 degree angles subtended in each semicircle (angle AMB and angle ANB).

Essentially then, we have two right triangles (triangle AMB and triangle ANB) that share the hypotenuse AB. Also, it’s important to note that each side of these triangles is an integer – since we know the radius of the circle is an integer, the diameter has to be an integer too. This should make us think of Pythagorean triples!

Whenever all three sides of a right triangle are integers, they will form a Pythagorean triple. Can you have a right triangle with all integer sides such that the length of one side is 1? No. There are no Pythagorean triples with 1 as a side. The smallest Pythagorean triple we know of is 3, 4, 5 (so there can be no right triangle with all integer sides such that the length of one side is 2, either).

We already know Pythagorean triples are the lengths of the sides of right triangles where all sides are integers. What we need to internalize is that ONLY Pythagorean triples are the lengths of sides of right triangles where all sides are integers. You cannot have a right triangle with all integer sides but whose sides are not a Pythagorean triple.

This means that the smallest right triangle with all integer sides is a 3, 4, 5 triangle.

Now note that in the given question, the hypotenuse is the diameter of the circle. We are given that the radius of the circle is an integer, so the diameter will be twice an integer, i.e. an even integer.

So we know the hypotenuse is an even integer, but as we discussed last week, the hypotenuse of a primitive Pythagorean right triangle must be odd. So this triangle must be a non-primitive Pythagorean triple. The smallest such triple will be twice of 3, 4, 5, i.e. the triangle will have sides with lengths 6, 8, 10.

This means the sides of the rectangle must be 6 and 8, while its diagonal must have a length of length 10. The area of the rectangle, then, must be 6*8 = 48. The answer is D.

Finally, at the end of the post we have figured out that this post is a continuation of last week’s post on properties of Pythagorean triples! We hope you enjoyed it!

Getting ready to take the GMAT? We have free online GMAT seminars running all the time. And, be sure to follow us on FacebookYouTubeGoogle+, and Twitter!

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!

The Pythagorean Triples Properties You’ll See on the GMAT

Quarter Wit, Quarter WisdomToday, let’s discuss a few useful properties of primitive Pythagorean triples. A primitive Pythagorean triple is one in which a, b and c (the length of the two legs and the hypotenuse, respectively) are co-prime. So, for example, (3, 4, 5) is a primitive Pythagorean triple while its multiple, (6, 8, 10), is not.

Now, without further ado, here are the properties of primitive Pythagorean triples that you’ll probably encounter on the GMAT:

I. One of a and b is odd and the other is even.
II. From property I, we can then say that c is odd.
III. Exactly one of a, b is divisible by 3.
IV. Exactly one of a, b is divisible by 4.
V. Exactly one of a, b, c is divisible by 5.

If you keep in mind the first primitive Pythagorean triple that we used as an example (3, 4, 5), it is very easy to remember all these properties.

If we look at some other examples:

(3, 4, 5), (5, 12, 13), (8, 15, 17) (7, 24, 25) (20, 21, 29) (12, 35, 37) (9, 40, 41) (28, 45, 53) (11, 60, 61) (16, 63, 65) (33, 56, 65) (48, 55, 73), etc.

we will see that these properties hold for all primitive Pythagorean triples.

Now, let’s take a look at an example GMAT question which can be easily solved if we know these properties:

The three sides of a triangle have lengths p, q and r, each an integer. Is this triangle a right triangle?

Statement 1: The perimeter of the triangle is an odd integer.
Statement 2: If the triangle’s area is doubled, the result is not an integer.

We know that the three sides of the triangle are all integers. So if the triangle is a right triangle, the three sides will represent a Pythagorean triple. Given that p, q and r are all integers, let’s use the properties of primitive Pythagorean triples to break down each of the statements.

Statement 1: The perimeter of the triangle is an odd integer.

Looking at the properties above, we know that a primitive Pythagorean triple can be represented as:

(Odd, Even, Odd) (The first two are interchangeable.)

Non-primitive triples are made by multiplying each member of the primitive triple by an integer n greater than 1. Depending on whether n is odd or even, the three sides can be represented as:

(Odd*Odd, Even*Odd, Odd*Odd) = (Odd, Even, Odd)
or
(Odd*Even, Even*Even, Odd*Even) = (Even, Even, Even)

However, the perimeter of a right triangle can never be odd because:

Odd + Even + Odd = Even
Even + Even + Even = Even

Hence, the perimeter will be even in all cases. (If the perimeter of the given triangle is odd, we can say for sure that it is not a right triangle.) This statement alone is sufficient.

Statement 2: If the triangle’s area is doubled, the result is not an integer.

If p, q and r are the sides of a right triangle such that r is the hypotenuse (the hypotenuse could actually be either p, q, or r but for the sake of this example, let’s say it’s r), we can say that:

The area of this triangle = (1/2)*p*q
and
Double of area of this triangle = p*q

Double the area of the triangle has to be an integer because we are given that both p and q are integers, but this statement tells us that this is not an integer. In that case, this triangle cannot be a right triangle. If the triangle is not a right triangle, double the area would be the base * the altitude, and the altitude would not be an integer in this case.

This statement alone is sufficient, too. Therefore, our answer is D.

As you can see, understanding the special properties of primitive Pythagorean triples can come in handy on the GMAT – especially in tackling complicated geometry questions.

Getting ready to take the GMAT? We have free online GMAT seminars running all the time. And, be sure to follow us on FacebookYouTubeGoogle+, and Twitter!

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!

Solving GMAT Geometry Problems That Involve Infinite Figures

Quarter Wit, Quarter WisdomSometimes, we come across GMAT geometry questions that involve figures inscribed inside other figures. One shape inside of another shape may not be difficult to work with, but how do we handle problems that involve infinite figures inscribed inside one another? Such questions can unsettle even the most seasoned test takers. Let’s take a look at one of them today:

A square is drawn by joining the midpoints of the sides of a given square. A third square is drawn inside the second square in this way and this process is continued indefinitely. If a side of the first square is 4 cm. Determine the sum of areas of all squares?

(A) 18
(B) 32
(C) 36
(D) 64
(E) None

Now the first thing that might come to our mind is this – how do we mathematically, in the time limit of approximately 2 minutes, calculate areas of infinite squares?

There has to be a formula for this. Recall that we do, in fact, have a formula that calculates the sum of infinite terms – the geometric progression formula! Let’s see if we can use that to find the areas of the squares mentioned in this problem.

ConcentricSquares

First, we’ll see if we can find a pattern in the areas of the squares:

Say the side of the outermost square is “s“. The area of the outermost square will be s^2 and half of the side will be s/2. The side of the next square inside this outermost square (the second square) forms the hypotenuse of a right triangle with legs of length s/2 each. Using the Pythagorean Theorem:

Hypotenuse^2 = (s/2)^2 + (s/2)^2 = s^2/2
Hypotenuse = s/√(2)

So now we know the sides of the second square will each equal s/√(2), and the area of the second square will be s^2/2.

Our calculations will be far easier if we note that the diagonal of the second square will be the same length as the side of the outer square. We know that area of a square given diagonal d is d^2/2, so that would directly bring us to s^2/2 as the area of the second square.

The second square and the square inscribed further inside it (the third square) will have the same relation. The area of the third square will be (s^2/2)*(1/2) = s^2/4.

Now we know the area of every subsequent square will be half the area of the outside square. So the total area of all squares = s^2 + s^2/2 + s^2/4 + s^2/8 + …Each term is half the previous term.

Therefore, the sum of an infinite Geometric Progression where the common ratio is less than 1 is:

Total Sum = a/(1-r)
a: First Term
r: Common Ratio

Sum of areas of all squares = s^2 + s^2/2 + s^2/4 + s^2/8 + …
Sum of areas of all squares = s^2/(1 – 1/2)
Sum of areas of all squares = 2s^2

Since s is the length of the side of the outermost square, and s = 4 (this fact is given to us by the questions stem), the sum of the areas of all the squares = 2*4^2 = 32 cm^2. Therefore, our answer is B.

We hope you understand how we have used the geometric progression formula to get to our answer. To recap, the sum of an infinite geometric progression is a/(1 – r).

Getting ready to take the GMAT? We have free online GMAT seminars running all the time. And, be sure to follow us on FacebookYouTubeGoogle+, and Twitter!

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!

How to Find the Maximum Distance Between Points on a 3D Object

Quarter Wit, Quarter WisdomHow do we find the the two farthest points on a 3D object? For example, we know that on a circle, any two points that are diametrically opposite will be the farthest from each other than from any other points on the circle. Which two points will be the farthest from each other on a square? The diagonally opposite vertices. Now here is a trickier question – which two points are farthest from each other on a rectangular solid? Again, they will be diagonally opposite, but the question is, which diagonal?

A rectangular box is 10 inches wide, 10 inches long, and 5 inches high. What is the greatest possible (straight-line) distance, in inches, between any two points on the box?

(A) 15
(B) 20
(C) 25
(D) 10 * √(2)
(E) 10 * √(3)

There are various different diagonals in a rectangular solid. Look at the given figure:
MaxDistRectangularSolid

 

 

 

 

 

BE is a diagonal, BG is a diagonal, GE is a diagonal, and BH is a diagonal. So which two points are farthest from each other? B and E, G and E, B and G, or B and H?

The inside diagonal BH can be seen as the hypotenuse of the right triangle BEH. So both BE and EH will be shorter in length than BH.

The inside diagonal BH can also be seen as the hypotenuse of the right triangle BHG. So both HG and BG will also be shorter in length than BH.

The inside diagonal BH can also be seen as the hypotenuse of the right triangle BDH. So both BD and DH will also be shorter in length than BH.

Thus, we see that BH will be longer than all other diagonals, meaning B and H are the points that are the farthest from each other. Solving for the exact value of BH then should not be difficult.

In our question we know that:

l = 10 inches
w = 10 inches
h = 5 inches

Let’s consider the right triangle DHB. DH is the length, so it is 10 inches.

DB is the diagonal of the right triangle DBC. If DC = w = 10 and BC = h = 5, then we can solve for DB^2 using the Pythagorian Theorem:

DB^2 = DC^2 + BC^2
DB^2 = 10^2 + 5^2 = 125

Going back to triangle DHB, we can now say that:

BH^2 = HD^2 + DB^2
BH^2 = 10^2 + 125
BH = √(225) = 15

Thus, our answer to this question is A.

Similarly, which two points on a cylinder will be the farthest from each other? Let’s examine the following practice GMAT question to find out:

The radius of cylinder C is 5 inches, and the height of cylinder C is 5 inches. What is the greatest possible straight line distance, in inches, between any two points on a cylinder C?

(A) 5 * √2
(B) 5 * √3
(C) 5 * √5
(D) 10
(E) 15

Look at where the farthest points will lie – diametrically opposite from each other and also at the opposite sides of the length of the cylinder:
MaxDistanceCylinder

 

 

 

 

 

 

The diameter, the height and the distance between the points forms a right triangle. Using the given measurements, we can now solve for the distance between the two points:

Diameter^2 + Height^2 = Distance^2
10^2 + 5^2 = Distance^2
Distance = 5 * √5

Thus, our answer is C.

In both cases, if we start from one extreme point and traverse every length once, we reach the farthest point. For example, in case of the rectangular solid, say we start from H, cover length l and reach D – from D, we cover length w and reach C, and from C, we cover length h and reach B. These two are the farthest points.

Getting ready to take the GMAT? We have free online GMAT seminars running all the time. And, be sure to follow us on FacebookYouTubeGoogle+, and Twitter!

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!

How to Use the Pythagorean Theorem With a Circle

Pie ChartIt does not surprise anyone when they learn that the properties of circles are tested on the GMAT. Most test-takers will nod and rattle off the relevant equations by rote: Area = Π*radius ^2; Circumference  = 2Π* radius; etc. However, many of my students are caught off guard to learn that the equation for a circle on the coordinate plane is our good friend the Pythagorean theorem. Why on earth would an equation for a right triangle describe a circle?

Take a look at the following diagram in which a circle is centered on the origin (0,0) in the coordinate plane:

DG Circle 1

 

 

 

 

 

 

 

Designate a random point on the circle (x,y.) If we draw a line from the center of the circle to x,y, that line is a radius of the circle. Call it r. If we drop a line down from (x,y) to the x-axis, we’ll have a right triangle:

DG Blog 2

 

 

 

 

 

 

 

 

Note that the base of the triangle is x, and the height of the triangle is y. So now we have our Pythagorean theorem: x^2 + y^2 = r^2. This is also the equation for a circle centered on the origin on the coordinate plane. [The more general equation for a circle with a center (a,b) is (x-a)^2 + (y-b)^2 = r^2. When a circle is centered on the origin, (a,b) is simply (0,0.)]

This ends up being an immensely useful tool to use on the GMAT. Take the following question, for example:

A certain circle in the xy-plane has its center at the origin. If P is a point on the circle, what is the sum of the squares of the coordinates of P?

(1) The radius of the circle is 4
(2) The sum of the coordinates of P is 0

So let’s draw this, designating P as (x,y):

DG Blog 3

 

 

 

 

 

 

Now we draw our trust right triangle by dropping a line down from P to the x-axis, which will give us this:

DG Blog 4

 

 

 

 

 

 

We’re looking for x^2 + y^2. Hopefully, at this point, you notice what the question is going for – because we have a right triangle, x^2 + y^2 = r^2, meaning that all we need is the radius!

Statement 1 is pretty straightforward – if r = 4, we can insert this into our equation of x^2 + y^2 = r^2 to get x^2 + y^2 = 4^2. So x^2 + y^2 = 16. Clearly, this is sufficient.

Now look at Statement 2. If the sum of x and y is 0, we can say x = 1 and y = -1 or x = 2 and y = -2 or x = 100 and y = -100, etc. Each of these will yield a different value for x^2 + y^2, so this statement alone is clearly not sufficient. Our answer is A.

Takeaway: any shape can appear on the coordinate plane. If the shape in question is a circle, remember to use the Pythagorean theorem as your equation for the circle, and what would have been a challenging question becomes a tasty piece of baklava. (We are talking about principles elucidated by the ancient Greeks, after all.)

Plan on taking the GMAT soon? We have GMAT prep courses starting all the time. And be sure to follow us on FacebookYouTubeGoogle+ and Twitter!

By David Goldstein, a Veritas Prep GMAT instructor based in Boston. You can find more articles written by him here.