GMAT Tip of the Week: The Song Remains the Same

Welcome back to hip hop month in the GMAT Tip of the Week space, where we’re constantly asking ourselves, “Wait, where have I heard that before?” If you listen to enough hip hop, you’ll recognize that just about every beat or lyric you hear either samples from or derives from another track that came before it (unless, of course, the artist is Ol’ Dirty Bastard, for whom, as his nickname derives, there ain’t no father to his style).

Biggie’s “Hypnotize” samples directly from “La Di Da Di” (originally by Doug E. Fresh – yep, he’s the one who inspired “The Dougie” that Cali Swag District wants to teach you – and Slick Rick). “Biggie Biggie Biggie, can’t you see, sometimes your words just hypnotize me…” was originally “Ricky, Ricky, Ricky…” And right around the same time, Snoop Dogg and 2Pac just redid the entire song just about verbatim, save for a few brand names.

The “East Coast edit” of Chris Brown’s “Loyal”? French Montana starts his verse straight quoting Jay-Z’s “I Just Wanna Love U” (“I’m a pimp by blood, not relation, I don’t chase ’em, I replace ’em…”), which (probably) borrowed the line “I don’t chase ’em I replace ’em” from a Biggie track, which probably got it from something else. And these are just songs we heard on the radio this morning driving to work…

The point? Hip hop is a constant variation on the same themes, one of the greatest recycling centers the world has ever known.

And so is the GMAT.

Good test-takers – like veteran hip hop heads – train themselves to see the familiar within what looks (or sounds) unique. A hip hop fan often says, “Wait, where I have heard that before?” and similarly, a good test-taker sees a unique, challenging problem and says, “Wait, where have I seen that before?”

And just like you might recite a lyric back and forth in your mind trying to determine where you’ve heard it before, on test day you should recite the operative parts of the problem or the rule to jog your memory and to remind yourself that you’ve seen this concept before.

Is it a remainder problem? Flip through the concepts that you’ve seen during your GMAT prep about working with remainders (“the remainder divided by the divisor gives you the decimals; when the numerator is smaller then the denominator the whole numerator is the remainder…”).

Is it a geometry problem? Think of the rules and relationships that showed up on tricky geometry problems you have studied (“I can always draw a diagonal of a rectangle and create a right triangle; I can calculate arc length from an inscribed angle on a circle by doubling the measure of that angle and treating it like a central angle…”).

Is it a problem that asks for a seemingly-incalculable number? Run through the strategies you’ve used to perform estimates or determine strange number properties on similar practice problems in the past.

The GMAT is a lot like hip hop – just when you think they’ve created something incredibly unique and innovative, you dig back into your memory bank (or click to a jazz or funk station) and realize that they’ve basically re-released the same thing a few times a decade, just under a slightly different name or with a slightly different rhythm.

The lesson?

You won’t see anything truly unique on the GMAT. So when you find yourself stumped, act like the old guy at work when you tell him to listen to a new hip hop song: “Oh I’ve heard this before…and actually when I heard it before in the ’90s, my neighbor told me that she had heard it before in the ’80s…” As you study, train yourself to see the similarities in seemingly-unique problems and see though the GMAT’s rampant plagiarism of itself.

The repetitive nature of the GMAT and of hip hop will likely mean that you’re no longer so impressed by Tyga, but you can use that recognition to be much more impressive to Fuqua.

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

By Brian Galvin.

3.14 Reasons to Love Pi

Pie ChartEvery March 14, numerically expressed as 3/14, math nerds and test prep instructors celebrate the time-honored tradition of “Pi Day,” deriving plenty of happiness from the fact that the date looks like the number 3.14, the approximation of π. Pi (π) is, of course, the lynchpin value in all circle calculations. The area of a circle is π(r^2), and the circumference of a circle is 2πr or πd.

As you study for a major standardized test, you know that you’ll be working with circles at some point, so here are 3.14 reasons that you should learn to love the number π:

1) Pi should make you salivate.
On any standardized test question, if you see the value π, whether in the question itself of in the answer choices, that π tells you that you’re dealing with a circle. Some test questions disguise what they want you to do – you may have to draw in a triangle to find the diagonal of a square, for example – but circle problems cannot hide from you! π is a dead giveaway that you’re dealing with a circle, so like Pavlov’s Dog, when you see that signal, π, you should respond with a biological response and conjure up all your knowledge of circles immediately.

2) Pi can be easily cut into slices.
Whether you’re dealing with a section of the area of a circle or a section of the circumference (arc length), the fact that a circle is perfectly symmetrical makes the job of cutting that circle into slices an easy one. With arc length, all you end up doing is using the central angle to determine the proportion of that section (angle/360 = proportion of what you want), making it very easy to slice up a circle using π. With the area of a section, as long as the arms of that section are equal to the radius of the circle, you can do the exact same thing. Just like an apple pie or pizza pie, if you’re cutting into slices from the center of the circle, cutting that pie into slices is a relatively simple task.

3) You can take your pi to go.
You will almost never have to calculate the value of pi on a standardized test: almost always, the symbol π will appear in the answer choices (e.g. 5π, 7π, etc.), meaning that you can just carry π through your calculations and bring it with you to the answer choices. If, for example, you need to calculate the area of a circle with radius 3, you’ll plug the radius into your formula [π(3^2)] and just end up with 9π, which you’ll find in the answer choices. With most other symbols (x, y, r, etc.) you’ll need to do some work to turn them into numbers. Pi is great because you can take it to go.

3.14) The decimals in pi are just a sliver.
If you ever are asked to “calculate” pi (which typically means that the question is asking you to approximate a value, not to directly calculate it), you can use the fact that the .14 in 3.14 is a tiny sliver of a decimal. For example, if you had to estimate a value for 5π, 5 times 3 is clearly 15, but 5 times .14 is so small that it won’t require you to go all the way to 16. So if your answer choices were 15.7, 16.1, 16.4, etc., you could rely on the fact that the decimal .14 is so small that you can eliminate all the 16s.

Other irrational numbers like the square root of 2 and square root of 3 have decimal places more in the neighborhood of .5, so you will probably need to work a little harder to estimate how they’ll react when you multiply them even by relatively small numbers. But π’s decimals come in small slivers, allowing you to manage your calculations in bite size pieces.

So remember – there are 3.14 (and counting) reasons to love pi, and learning to love pi can help turn your test day into a piece of cake.

Are you getting ready to take the SAT, ACT, GMAT or GRE? Check out our website for a variety of helpful test prep resources. And as always, be sure to follow us on Facebook, YouTubeGoogle+ and Twitter!

By Brian Galvin.

GMAT Tip of the Week: Evolving Your GMAT Quant Score with Help from The Evolution Of Rap

GMAT Tip of the WeekIf it’s March, it must be Hip Hop Month at the GMAT Tip of the Week space, where this year we’ve been transfixed by Vox’s video on the evolution of rhyme schemes in the rap world.

The video below (which is absolutely worth a watch during a designated study break) explores the way that rap has evolved from simple rhyme schemes (yada yada yada Bat, yada yada yada Hat, yada yada yada Rat, yada yada yada Cat…) to the more complex “wait did he just say what I thought he said?” inside-out rhyme schemes that make you rewind an Eminem or Kendrick Lamar track because your ears must be playing tricks on you.

And if you don’t have the study break time right now, we’ll summarize. While a standard rhyme might have a one-syllable rhyme at the end of each bar (do you like green eggs and HAM, yes I like them Sam I AM), rappers have continued to evolve to the point where nowadays each bar can contain multiple rhyme schemes. Consider Eminem’s “Lose Yourself”:

Snap back to reality, oh there goes gravity
Oh there goes Rabbit he choked, he’s so mad but he won’t
Give up that easy, nope, he won’t have it he knows
His whole back’s to these ropes, it don’t matter he’s dope
He knows that but he’s broke, he’s so stagnant he knows…

Where “gravity,” “Rabbit, he,” “mad but he,” “that easy,” “have it he,” “back’s to these,” “matter he’s,” “that but he’s,” and “stagnant, he” all rhyme with one another, the list of goes/goes/choked/so/won’t/knows/whole/ropes/don’t/dope… keeps that hard “O” sound rhyming consistently throughout, too. And that was 15 years ago…since them, Eminem, Kendrick, and others have continued to build elaborate rhyme schemes that reward those listeners who don’t just listen for the simple rhyme at the end of each bar, but pick up the subtle rhyme flows that sometimes don’t come back until a few lines later.

So what does this have to do with your GMAT score?

One of the most common study mistakes that test-takers make is that they study skills as individual, standalone entities, and don’t look for the subtle ways that the GMAT testmaker can layer in those sophisticated Andre-3000-style combinations. Consider an example of an important GMAT skill, the “Difference of Squares” rule that (x + y)(x – y) = x^2 – y^2. A standard (think early 1980s Sugarhill Gang or Grandmaster Flash) GMAT question might test it in a relatively “obvious” way:

What is the value of (x + y)?

(1) x^2 – y^2 = 0
(2) x does not equal y

Here if you factor Statement 1 you’ll get (x + y)(x – y) = 0, and then Statement 2 tells you that it’s not (x – y) that equals zero, so it must be x + y. This Data Sufficiency answer is C, and the test is essentially just rewarding you for knowing the Difference of Squares.

The GMAT it cares
’bout the Difference of Squares
When there’s squares and subtraction
Put this rule into action

A slightly more sophisticated question (think late 1980s/early 1990s Rob Bass or Run DMC) won’t so obviously show you the Difference of Squares. It might “hide” that behind a square that few people tend to see as a square, the number 1:

If y = 2^(16) – 1, the greatest prime factor of y is:

(A) Less than 6
(B) Between 6 and 10
(C) Between 10 and 14
(D) Between 14 and 18
(E) Greater than 18

Here, many people don’t recognize 1 as a perfect square, so they don’t see that the setup is 2^(16) – 1^(2), which can be factored as:

(2^8 + 1)(2^8 – 1)

And that 2^8 – 1 can be factored again, since 1 remains 1^2:

(2^8 + 1)(2^4 + 1)(2^4 – 1)

And that ultimately you could do it again with 2^4 – 1 if you wanted, but you should know that 2^4 is 16 so you can now get to work on smaller numbers. 2^8 is 256 and 2^4 is 16, so you have:

257 * 17 * 15

And what really happens now is that you have to factor out 257 to see if you can break it into anything smaller than 17 as a factor (since, if not, you can select “greater than 18”). Since you can’t, you know that 257 must have a prime factor greater than 18 (it turns out that it’s prime) and correctly select E.

The lesson here? This problem directly tests the Difference of Squares (you don’t want to try to calculate 2^16, then subtract 1, then try to factor out that massive number) but it does so more subtly, layering it inside the obvious “prime factor” problem like a rapper might embed a secondary rhyme scheme in the middle of each bar.

But in really hard problems, the testmaker goes full-on Greatest of All Time rapper, testing several things at the same time and rewarding only the really astute for recognizing the game being played. Consider:

The size of a television screen is given as the length of the screen’s diagonal. If the screens were flat, then the area of a square 21-inch screen would be how many square inches greater than the area of a square 19-inch screen?

(A) 2
(B) 4
(C) 16
(D) 38
(E) 40

Now here you KNOW you’re dealing with a geometry problem, and it also looks like a word problem given the television backstory. As you start calculating, you’ll know that you have to take the diagonal of each square TV and use that to determine the length of each side, using the 45-45-90 triangle ratio, where the diagonal = x√2. So the length of a side of the smaller TV is 19/√2 and the length of a side of the larger TV is 21/√2.

Then you have to calculate the area, which is the side squared, so the area of the smaller TV is (19/√2)^2 and the area of the larger TV is (21/√2)^2. This is starting to look messy (Who knows the squares for 21 and 19 offhand? And radicals in denominators never look fun…) UNTIL you realize that you have to subtract the two areas. Which means that your calculation is:

(21/√2)^2 – (19/√2)^2

This fits perfectly in the Difference of Squares formula, meaning that you can express x^2 – y^2 as (x + y)(x – y). Doing that, you have:

[(21 + 19)/√2][(21-19)/√2]

Which is really convenient because the math in the numerators is easy and leaves you with:

(40/√2) * (2/√2)

And when you multiply them, the √2 terms in the denominators square out to 2, which factors with the 2 in the numerator of the right-side fraction, and everything simplifies to 40. And then, in classic “oh this guy’s effing GOOD” hip-hop style (like in the Eminem lyric “you’re witnessing a massacre like you’re watching a church gathering take place” and you realize that he’s using “massacre” and “mass occur” – the church gathering taking place – simultaneously), you realize that you should have seen it coming all along. Because when you subtract the area of one square minus the area of another square you’re LITERALLY taking the DIFFERENCE of two SQUARES.

So what’s the point?

Too often people study for the GMAT like they’d listen to 1980s rap. They expect the Difference of Squares to pair nicely at the end of an Algebra-with-Exponents bar, and the Isosceles Right Triangle formula to pair nicely with a Triangle question. They learn skills in distinct silos, memorize their flashcards in nice, tidy sets, and then go into the test and realize that they’re up against an exam that looks a lot more like a 2017 mixtape with layers of rhyme schemes and motives.

You need to be prepared to use skills where they don’t seem to obviously belong, to jot down and rearrange your scratchwork, label your unknowns, etc., looking for how you might reposition the math you’re given to help you bring in a skill or concept that you’ve used countless times, just in totally different contexts. The GMAT testmaker has a much more sophisticated flow than the one you’re likely studying for, so pay attention to that nuance when you study and you’ll have a much better chance of keeping your score 800.

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

By Brian Galvin.

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!

Quarter Wit, Quarter Wisdom: Beware of Sneaky Answer Choices on the GMAT!

Quarter Wit, Quarter WisdomTest-takers often ask for tips and short cuts to cut down the amount of work necessary to solve a GMAT problem. As such, the Testmaker might want to award the test-taker who pays attention to detail and puts in the required effort.

Today, we will look at an example of this concept – if it seems to be too easy, it is a trap!

QWQW_11_21

 

 

 

 

 

In the figure given above, the area of the equilateral triangle is 48. If the other three figures are squares, what is the perimeter, approximately, of the nine-sided shape they form?

(A) 8√(2)
(B) 24√(3)
(C) 72√(2)
(D) 144√(2)
(E) 384

The first thing I notice about this question is that we have an equilateral triangle. So I am thinking, the area = s^2 * √(3)/4 and/or the altitude = s*√(3)/2.

The irrational number in play is √(3). There is only one answer choice with √(3) in it, so will this be the answer?

Now, it actually makes me uncomfortable that  there is only one option with √(3). At first glance, it seems that the answer has been served to us on a silver plate. But the question format doesn’t seem very easy – it links two geometrical figures together. So I doubt very much that the correct answer would be that obvious.

The next step will be to think a bit harder:

The area of the triangle has √(3) in it, so the side would be a further square root of √(3). This means the actual irrational number would be the fourth root of 3, but we don’t have any answer choice that has the fourth root of 3 in it.

Let’s go deeper now and actually solve the question.

The area of the equilateral triangle = Side^2 * (√(3)/4) = 48

Side^2 = 48*4/√(3)
Side^2 = 4*4*4*3/√(3)
Side = 8*FourthRoot(3)

Now note that the side of the equilateral triangle is the same length as the sides of the squares, too. Hence, all sides of the three squares will be of length 8*FourthRoot(3).

All nine sides of the figure are the sides of squares. Hence:

The perimeter of the nine sided figure = 9*8*FourthRoot(3)
The perimeter of the nine sided figure =72*FourthRoot(3)

Now look at the answer choices. We have an option that is 72√(2). The other answer choices are either much smaller or much greater than that.

Think about it – the fourth root of 3 = √(√(3)) = √(1.732), which is actually very similar to √(2). Number properties will help you figure this out. Squares of smaller numbers (that are still greater than 1) are only a bit larger than the numbers themselves. For example:

(1.1)^2 = 1.21
(1.2)^2 = 1.44
(1.3)^2 = 1.69
(1.414)^2 = 2

Since 1.732 is close to 1.69, the √(1.732) will be close to the √(1.69), i.e. 1.3. Also, √(2) = 1.414. The two values are quite close, therefore, the perimeter is approximately 72√(2). This is the reason the question specifically requests the “approximate” perimeter.

We hope you see how the Testmaker could sneak in a tempting answer choice – beware the “easiest” option!

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.

GMAT Geometry Practice Questions and Problems

SAT/ACTWould you call yourself a math person? If so, you’ll be glad to know that there are plenty of algebra, geometry, arithmetic, and other types of math problems on the GMAT. Perhaps you like math but need a little review when it comes to the topic of geometry. If so, learn some valuable tips on how to prep for GMAT geometry problems before you get started studying for the exam.

Learn and Practice the Basic Geometry Formulas
Knowing some basic formulas in geometry is an essential step to mastering these questions on the GMAT. One formula you should know is the Pythagorean Theorem, which is a^2 + b^2 = c^2, where c stands for the longest side of a right triangle, while a and b represent the other two sides.

Another formula to remember is the area of a triangle, which is A = 1/2bh, where A is the area, b is the length of the base, and h is the height. The formula for finding the area of a rectangle is l*w = A (length times width equals the area). Once you learn these and other basic geometry formulas for the GMAT, the next step is to put them into practice so you know how to use them when they’re called for on the exam.

Complete Practice Quizzes and Questions
Reviewing problems and their answers and completing GMAT geometry practice questions are two ways to sharpen your skills for this section of the test. This sort of practice also helps you become accustomed to the timing when it comes to GMAT geometry questions. These questions are found within the Quantitative section of the GMAT.

You are given just 75 minutes to finish 37 questions in this section. Of course, not all 37 questions involve geometry – GMAT questions in the Quantitative section also include algebra, arithmetic, and word problems – but working on completing each geometry problem as quickly as possible will help you finish the section within the time limit. In fact, you should work on establishing a rhythm for each section of the GMAT so you don’t have to worry about watching the time.

Use Simple Study Tools to Review Problems
Another way to prepare for GMAT geometry questions is to use study tools such as flashcards to strengthen your skills. Some flashcards are virtual and can be accessed as easily as taking your smartphone out of your pocket. If you prefer traditional paper flashcards, they can also be carried around easily so you can review them during any free moments throughout the day. Not surprisingly, a tremendous amount of review can be accomplished at odd moments during a single day.

In addition, playing geometry games online can help you hone your skills and add some fun to the process at the same time. You could try to beat your previous score on an online geometry game or even compete against others who have played the same game. Challenging another person to a geometry game can sometimes make your performance even better.

Study With a Capable Tutor
Preparing with a tutor can help you to master geometry for GMAT questions. A tutor can offer you encouragement and guide you in your studies. All of our instructors at Veritas Prep have taken the GMAT and earned scores that have put them in the 99th percentile of test-takers. When you study with one of our tutors, you are learning from an experienced instructor as well as someone who has been where you are in the GMAT preparation process.

Our prep courses instruct you on how to approach geometry questions along with every other topic on the GMAT. We know that memorizing facts is not enough: You must apply higher-order thinking to every question, including those that involve geometry. GMAT creators have designed the questions to test some of the skills you will need in the business world.

Taking a practice GMAT gives you an idea of what skills you’ve mastered and which you need to improve. Our staff invites you to take a practice GMAT for free. We’ll give you a score report and a performance analysis so you have a clear picture of what you need to focus on. Then, whether you want help with geometry or another subject on the GMAT, our team of professional instructors is here for you.

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!

GMAT Tip of the Week: The Curry Twos Remind You To Keep The GMAT Simple

GMAT Tip of the WeekHappy Friday from Veritas Prep headquarters, where we’re actively monitoring the way that Twitter is reacting to UnderArmour’s release of the new Steph Curry shoes. What’s the problem with the Curry Twos? Essentially they’re too plain and buttoned up – much more Mickelson than Michael, son.

OK, so what? The Curry 2s are more like the Curry 401(k)s. Why should that matter for your GMAT score?

Because on the GMAT, you want to be as simple and predictable as a Steph Curry sneaker.

What does that mean? One of the biggest study mistakes that people make is that once they’ve mastered a core topic like “factoring” or “verb tenses,” they move on to more obscure topics and spend their valuable study time on those.

There are two major problems with this: 1) the core topics appear much more often and are much more repeatable, and 2) in chasing the obscure topics later in their study regimen, people spend the most valuable study time – that coming right before the test – feverishly memorizing things they probably won’t see or use at the expense of practicing the skills and strategies that they’ll need to use several times on test day.

Consider an example: much like Twitter is clowning the Curry Twos, a handful of Veritas Prep GMAT instructors were laughing this time last week about an explanation in a practice test (by a company that shall remain nameless…) for a problem similar to:

Two interconnected, circular gears travel at the same circumferential rate. If Gear A has a diameter of 30 centimeters and Gear B has a diameter of 50 centimeters, what is the ratio of the number of revolutions that Gear A makes per minute to the number of revolutions that Gear B makes per minute?

(A) 3:5
(B) 9:25
(C) 5:3
(D) 25:9
(E) Cannot be determined from the information provided

Now, the “Curry Two” approach – the tried and true, “don’t-overcomplicate-this-for-the-sake-of-overcomplicating-it” method – is to recognize that the distance around any circle (a wheel, a gear, etc.) is its circumference. And circumference is pi * diameter. So, if each gear travels the same circumferential distance, that distance for any given period of time is “circumference * number of revolutions.” That then means that the circumference of A times the number of revolutions of A is equal to the circumference of B times the number of revolutions for B, and you know that’s:

30π * A = 50π * B (where A = # of revolutions for A, and B = # of revolutions for B). Since you want the ratio of A:B, divide both sides by B and by 30, and you have A/B = 50/30, or A:B = 5:3 (answer choice C).

Why were our instructors laughing? The explanation began, “There is a simple rule for interconnected gears…” Which is great to know if you see a gear-based question on the test or become CEO of a pulley factory, but since the GMAT officially tests “geometry,” you’re much better off recognizing the relationship between circles, circumferences, and revolutions (for questions that might deal with gears, wheels, windmills, or any other type of spinning circles) than you are memorizing a single-use rule about gears.

Problems like this offer the “Curry Two” students a fantastic opportunity to reinforce their knowledge of circles, their ability to think spatially about shapes, etc. But, naturally, there are students who will add “gear formula” to their deck of flashcards and study that single-use rule (which 99.9% of GMAT examinees will never have the opportunity to use) with the same amount of time/effort/intensity as they revisit the Pythagorean Theorem (which almost everyone will use at least twice).

Hey, the Curry Twos are plain, boring, and predictable, as are the core rules and skills that you’ll use on the GMAT. But simple, predictable, and repeatable are what win on this test, so heed this lesson. As 73 regular season opponents learned this basketball season, Curry Twos lead to countless Curry 3s, and on the GMAT, “Curry Two” strategies will help you curry favor with admissions committees by leading to Curry 700+ scores.

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

By Brian Galvin.

Coordinate Geometry: Solving GMAT Problems With Lines Crossing Either the X-Axis or the Y-Axis

Quarter Wit, Quarter WisdomToday let’s learn about the cases in which lines on the XY plane cross, or do not cross, the x- or y-axis. Students often struggle with questions such as this:

Does the line with equation ax+by = c, where a,b and c are real constants, cross the x-axis?

What concepts will you use here? How will you find whether or not a line crosses the x-axis? What conditions should it meet? Think about this a little before you move ahead.

We know that most lines on the XY plane cross the x-axis as well as the y-axis. Even if it looks like a given line doesn’t cross either of these axes, eventually, it will if it has a slope other than 0 or infinity.

QWQW pic 1

 

 

 

 

 

Note that by definition, a line extends infinitely in both directions – it has no end points (otherwise it would be a line “segment”). We cannot depict a line extending infinitely, which is why we will only show a small section of it. Ideally, a line on the XY plane should be shown with arrowheads to depict that it extends infinitely on both sides, but we often omit them for our convenience. For instance, if we try to extend the example line above, we see that it does, in fact, cross the x-axis:

QWQW pic 2

 

 

 

 

 

So what kind of lines do not cross either the x-axis or the y-xis? We know that the equation of a line on the XY plane is given by ax + by  + c = 0. We also know that if we want to find the slope of a line, we can use the equation y = (-a/b)x – c/b, where the slope of the line is -a/b.

A line with a slope of 0 is parallel to the x-axis. For the slope (i.e. -a/b) to be 0, a must equal 0. So if a = 0, the line will not cross the x-axis – it is parallel to the x-axis. The equation of the line, in this case, will become y = k. In all other cases, a line will cross the x-axis at some point.

Similarly, it might appear that a line doesn’t cross the y-axis but it does at some point if its slope is anything other than infinity. A line with a slope of infinity is parallel to the y-axis. For -a/b to be infinity, b must equal 0. So if b = 0, the line will not cross the y-axis. The equation of the line in this case will become x = k. In all other cases, a line will cross the y-axis at some point.

Now, we can easily solve this official question:

Does the line with equation ax+by = c, where a, b and c are real constants, cross the x-axis?

Statement 1: b not equal to 0

Statement 2: ab > 0

As we discussed earlier, all lines cross the x-axis except lines which have a slope of 0, i.e. a = 0.

Statement 1: b not equal to 0

This tells statement us that b is not 0 – which means the line is not parallel to y-axis – but it doesn’t tell us whether or not a is 0, so we don’t know whether the line is parallel to the x-axis or crosses it. Therefore, this statement alone is not sufficient.

Statement 2: ab>0

If ab > 0, it means that neither a nor b is 0 (since any number times 0 will equal 0). This means the line is parallel to neither x-axis nor the y-xis, and therefore must cross the x-axis. This statement alone is sufficient and our answer is B.

Hopefully this has helped clear up some coordinate geometry concepts today.

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!

Improve Your Speed on the ACT Math Section Using Math Fluidity

stopwatch-620Speed is key on the Math Section of the ACT – you have only 60 minutes to complete 60 questions. However, this doesn’t mean you should spend one minute on each question, as not every question on in this section is created equal. Many questions (particularly Questions 1-30) are problems that you can solve in under one minute. In fact, you should aim to solve Questions 1-30 in less than 30 minutes – around 25 minutes is the goal.

That’s because some of the later questions, particularly the questions from Questions 40-60, will require more than a minute. Basically, you want to put aside extra time for the tricky questions at end of the section by completing the easier, earlier questions as quickly as possible. If you do Questions 1-30 in 25 minutes, then you have 35 minutes to do Questions 31-60.

One way to improve your speed on the Math Section is to develop what I call “math fluidity.” That means recognizing how common patterns, formulas and special rules can help you solve any particular problem. To illustrate, take a look at the following triangle problem:

Triangle ABC (below) is an equilateral triangle with side of length 4. What is the area of triangle ABC?

ACT Triangle 1

 

 

 

The first step to any geometry problem is writing down what relevant common formula you’ll need to solve the problem; i.e. whenever I’m asked the area of a triangle, at the top of my work space I’ll write:

A = (b*h)/2

Having the formula in front of you will be helpful because right away, it’s clear that although we have some information, we don’t have all the information we need to solve this problem – we have the base of the triangle (4), but not the height. Since the height of an equilateral triangle always goes from one angle to the opposite side, where it forms two 90-degree angles, drawing the height of an equilateral triangle creates two identical triangles, as shown below:

ACT Triangle 2

 

 

 

Many students would now conclude that they need the Pythagorean theorem to solve for the height (that line bisecting the equilateral triangle). This is where math fluidity comes in. Although you could use the Pythagorean theorem, it’s much faster to instead recognize what type of triangle you are dealing with.

Whenever you split an equilateral triangle in half, you create two 30-60-90 triangles. These are also called “special right triangles” because they always follow the rule that the shortest side is always “x,” the side opposite the 60-degree angle is always x√3, and the hypotenuse is always 2x. See the triangle below:

ACT Triangle 3

 

 

 

So, rather than spend any time solving for the height of the our triangle by using the Pythagorean Theorem, recognize that because the hypotenuse is 4 and the base is 2 (of either of the smaller triangles), and because the triangle is a right triangle, the height must be 2√3. Therefore, the area of the larger triangle is  (2√3)(4)(1/2), which equals 4√3.

Instantly recognizing that the two smaller triangles are 30-60-90 triangles only saves a little bit of time – if you can regularly shave off 20 seconds on question after question by recognizing special rules or how best to apply formulas, you’ll accrue saved time that can later be spent on harder math questions. Speaking of which, math fluidity also applies to tricky questions – similar to what we previously saw, recognition will break down hard questions into easier, faster steps.

So, let’s take a look at a more difficult question. Note, this next example is especially relevant for students shooting for 99th percentile or perfect scores. Although many students can solve the following question if given enough time, few students can solve it quickly enough to get it correct on the ACT. Here’s the problem:

In triangle ABC below, angle BAE measures 30 degrees. What is the value of angle AED minus angle ABE?

A) 30ACT Triangle 4
B) 60
C) 90
D) 120
E) 150

 

Although there are several ways to solve this problem, math fluidity will help with whatever approach you choose. As I mentioned earlier, it is always best to start by writing down a relevant formula, as it will include what information you have and what information you need. In this case, I’m looking for AED-ABE. Because I’ve also been given the measure of angle BAE, I’ll write down:

BAE = 30 and BAE + ABE = AED

Here’s where math fluidity comes in; the second formula is based off a theorem that you probably learned (and then forgot!) in your geometry class. I do recommend (re)memorizing it for the ACT as follows: a measure of an exterior angle of a triangle is equal to the sum of the measures of the two non-adjacent interior angles.

ACT Triangle 5Are you drawing a blank? If so, take a moment to think about why that statement is true. If the smaller two angles of a right-angle triangle, as shown at left, are 40 and 50, then if we extend a line as shown to form the adjacent exterior angle x, then x + 50 = 180, so x = 130.

 

Also, 40 + 50 + 90 = 180, since the sum of interior angles of a triangle always add up to 180. So, if x + 50 = 180, and  40 + 50 + 90 = 180, then x+ 50 = 40 + 50 + 90.

Removing the 50 from both sides, we can conclude that x = 40 + 90, or x (the adjacent  exterior angle of one interior angle) is equal to the sum of the other two interior angles.

Now, returning to our original problem:

If BAE = 30 and BAE + ABE = AED, then:

30 + ABE = AED

AED – ABE = 30

Therefore, our answer is A, 30.

Still need to take the ACT? We run a free online ACT prep seminar every few weeks. And be sure to find us on Facebook, YouTube, Google+ and Twitter!

By Rita Pearson

A 750+ Level GMAT Geometry Question

Quarter Wit, Quarter WisdomToday we will discuss a pretty advanced GMAT question, because we can still use our basic GMAT concepts to find the answer. It may seem like we will need trigonometry to handle this question, but that is not so. In fact, the question will look familiar at first, but will present unforeseen problems later on.

While going through this exercise, we will learn a few tips and tricks which will be useful in our mainstream GMAT questions, hence, it will add value to our GMAT repertoire (especially in elimination techniques). Let’s go on to the question now:

In triangle ABC, if angle ABC is 30 degrees, AC = 2*sqrt(2) and AB = BC = X, what is the value of X?

(A) Sqrt(3) – 1

(B) Sqrt(3) + 2

(C) (Sqrt(3) – 1)/2

(D) (Sqrt(3) + 1)/2

(E) 2*(Sqrt(3) + 1)

What we see here is an isosceles triangle with one angle as 30 degrees and other two angles as (180 – 30)/2 = 75 degrees each.

The side opposite the 30 degrees angle is 2*sqrt(2). One simple observation is that X must be greater than 2*sqrt(2) because these sides are opposite the greater angles (75 degrees).

2*sqrt(2) is a bit less than 2*1.5 because Sqrt(2) = 1.414. So 2*sqrt(2) is a bit less than 3. Note that options (A), (C), and (D) are much smaller than 3, so these cannot be the value of X. We have already improved our chances of getting the correct answer by eliminating three options! Now we have to choose out of (B) and (E).
2Triangles

 

 

 

 

 

 

Here is what is given: Angle ABC = 30 degrees, and AC = 2*sqrt(2). We need to find the value of X. Now, our 30 degree angle reminds us of a 30-60-90 triangle in which we know the ratio of the sides – given one side, we can find the other two.

The problem is this: if we drop an altitude from angle B to AC, the angle 30 degrees will be split in half and we will actually get a 15-75-90 triangle, instead. We won’t have a 30-60-90 triangle anymore, so what do we do now? Let’s try to maintain the 30 degree angle as it is to get the 30-60-90 triangle, and drop an altitude from angle C to AB instead, calling it CE. Now we have a 30-60-90 triangle! Since BCE is a 30-60-90 triangle, its sides are in the ratio 1:sqrt(3):2. Side X corresponds to 2 on the ratio, so CE = x/2.

Area of triangle ABC = (1/2)*BD*AC = (1/2)*CE*AB

(1/2)*BD*2*sqrt(2) = (1/2)*(X/2)*X

BD = X^2/4*Sqrt(2)

Now DC = (1/2)AC = 2*sqrt(2)/2 = sqrt(2)

Let’s use the pythagorean theorem on triangle BDC:

BD^2 + DC^2 = BC^2

(X^2/4*Sqrt(2))^2 + (Sqrt(2))^2 = X^2

X^4/32 + 2 = X^2

X^4 – 32*X^2 + 64 = 0

X^4 – 16X^2 + 8^2 – 16X^2 = 0

(X^2 – 8)^2 – (4X)^2 = 0

(X^2 -8 + 4X) * (X^2 – 8 – 4X) = 0

Normally, this would require us to use the quadratic roots formula, but let’s not get that complicated. We can just plug in the the two shortlisted options and see if either of the factors is 0. If one of the factors becomes 0, the equation will be satisfied and we will have the root of the equation.

Since both options have both terms positive, it means the co-efficient corresponding to B in Ax^2 + Bx + C = 0 must be negative.

x = [-B +- Sqrt(B^2 – 4AC)]/2A

-B will give us a positive term if B is negative, so we will get the answer by plugging into (X^2 – 4X – 8):

Put X = Sqrt(3) + 2 in X^2 – 4X – 8 and you do not get 0.

Put X = 2*(Sqrt(3) + 1) in X^2 – 4X – 8 and you do get 0.

This means that X is 2*(Sqrt(3) + 1), so our answer must be (E).

To recap:

Tip 1: A greater side of a triangle is opposite a greater angle.

Tip 2: We can get the relation between sides and altitudes of a triangle by using the area of the triangle formula.

Tip 3: The quadratic formula can help identify the sign of the irrational roots.

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!