# Quarter Wit, Quarter Wisdom: The Power of Deduction on GMAT Data Sufficiency Questions

In a previous post, we have discussed how to find the total number of factors of a number. What does the total number of factors a number has tell us about that number? One might guess, “Not a lot,” but it actually does tell us quite a bit! If the total number of factors is odd, you know the number must be a perfect square. If the total number of factors is even, you know the number is not a perfect square.

We know that the total number of factors of a number A (prime factorised as X^p * Y^q *…) is given by (p+1)*(q+1)… etc.

So, if we know that a number has, say, 6 total factors, what can we say about the number?

6 = (p+1)*(q+1) = 2*3, so p = 1 and q = 2 or vice versa.

A = X^1 * Y^2 where X and Y are distinct prime numbers.

Today, we will look at a data sufficiency question in which we can use factors to deduce much more information than what we might first guess:

When the digits of a two-digit, positive integer M are reversed, the result is the two-digit, positive integer N. If M > N, what is the value of M?

Statement 1: The integer (M – N) has 12 unique factors.
Statement 2: The integer (M – N) is a multiple of 9.

With this question, we are told that M is a two-digit integer and N is obtained by reversing it. So if M = 21, then N = 12; if M = 83, then N = 38 (keeping in mind that M must be greater than N). In the generic form:

M = 10a + b and N =10b + a (where a and b are single-digit numbers from 1 to 9. Neither can be 0 or greater than 9 since both M and N are two-digit numbers.)

We also know that no matter what M and N are, M > N. Therefore:

10a + b > 10b + a
9a > 9b
a > b

Let’s examine both of our given statements:

Statement 1: The integer (M – N) has 12 unique factors.

First, let’s figure out what M – N is:

M – N = (10a + b) – (10b + a) = 9a – 9b

Say M – N = A. This would mean A = 9(a-b) = 3^2 * (a-b)

The total number of factors of A where A = X^p * Y^q *… can be calculated using the formula (p+1)*(q+1)* …

We know that A has 3^2 as a factor, so X = 3 and p = 2. Therefore, the total number of factors would be (2+1)*(q+1)*… = 3*(q+1)*… = 12, so (q+1)*… must be 4.

Case 1:
This means q may be 3 so that (q+1) is 4. Since a-b must be less than or equal to 9 and must also be the cube of a number, (a-b) must be 8. (Note that a-b cannot be 1 because then the total number of factors of A would only be 3.)

So, a must be 9 and b must be 1 in this case (since a > b). The integers will be 91 and 19, and since M > N, M = 91.

Case 2:
Another possibility is that (a-b) is a product of two prime factors (other than 3), both with the power of 1. In that case, the total number of factors = (2+1)*(1+1)*(1+1) = 12

Note, however, that the two prime factors (other than 3) with the smallest product is 2*5 = 10, but the difference of two single-digit positive integers cannot be 10. This means that only Case 1 can be true, therefore, Statement 1 alone is sufficient. This is certainly not what we expected to find from just the total number of factors!

Statement 2: The integer (M – N) is a multiple of 9.

M – N = (10a + b) – (10b + a) = 9a – 9b, so M – N = 9 (a-b) . This is already a multiple of 9.

We get no new information with this statement; (a-b) can be any integer, such as 2 (a = 5, b = 3 or a = 7, b = 5), etc. This statement alone is insufficient, therefore our answer is A.

Don’t take the given data of a GMAT question at face value, especially if you are expecting questions from the 700+ range. Ensure that you have deduced everything that you can from it before coming to a conclusion.

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!

# GMAT Tip of the Week: Making Your GMAT Score SupeRIOr to Ryan Lochte’s

What’s the worst thing that can happen on your GMAT exam? Is it running out of time well before you’re done? Or blanking on nearly every math formula you’ve studied?

Whatever it is, it can’t be nearly as bad as being pulled over by fake cops – no lights or nothing, just a badge – then being told to get on the ground and having a gun placed on your forehead and being like, “whatever.” So your big event of 2016 will already go a lot better than Ryan Lochte’s did; you have that going for you.

What else do you have going for you on the GMAT? The ability to learn from the most recent few days of Lochte’s life. Lochte’s biggest mistake wasn’t vandalizing a gas station bathroom at 4am, but rather making up his own story and creating an even larger mess. And that’s a huge lesson that you need to keep in mind for the GMAT:

Don’t make up your own story.

Here’s what that means, on three major question types:

DATA SUFFICIENCY
People make up their own story on Data Sufficiency all the time. And like a prevailing theory about Lochte (he didn’t connect the vandalism of the bathroom to the men coming after him for restitution; he really did think that he had been robbed for no reason), it’s not that they’re intentionally lying. They’re just “conveniently” misremembering what they’ve read or connecting dots that weren’t actually connected in real life. Consider the question:

The product of consecutive integers a, b, c, and d is 5040. What is the value of integer d?

(1) d is prime
(2) d < c < b < a

Once people have factored 5040 into 7*8*9*10, they can then quickly recognize that Statement 1 is sufficient: the only prime number in that bunch is 7, so d must be 7. But then when it comes to Statement 2, they’ve often made up their own story. By saying “d is the smallest, and, yep, that’s 7!” they’re making up the fact that these consecutive integers are positive. That was not specifically stated! So it could be 7, 8, 9, and 10 or it could be -7, -8, -9, and -10, making d either -10 or 7. And the GMAT (maybe like an NBC interviewer?) makes it easy for you to make up your own story.

With Statement 1, prime numbers must be positive, so if you weren’t already thinking only about positives, the question format nudges you further in that direction. The answer is A when people often mistakenly choose D, and the reason is that the question makes it easy for you to make up your own story when looking at Statement 2. So before you submit an answer, always ask yourself, “Am I only using the facts explicitly provided to me, or am I somehow making up my own story?”

CRITICAL REASONING
Think of your friends who are good storytellers. We hate to break it to you, but they’re probably making at least 10-20% of those stories up. Which makes sense. “It was a pretty big fish,” is a lot less compelling than, “It was the biggest fish any of us had ever seen!” Case in point, the Olympics themselves.

No commentator this week has said that Michael Phelps, Lochte’s teammate, is “a really good swimmer.” They’re posing, “Is he the greatest athlete of all time?” because words that end in -st capture attention (and pageviews). Even Lochte was guilty of going overly-specific for dramatic effect: there was, indeed, a gun pointed at his taxi, but not resting on his forehead. His version just makes the story more exciting and dramatic…and you may very well be guilty of such a mistake on the GMAT. Consider:

About two million years ago, lava dammed up a river in western Asia and caused a small lake to form. The lake existed for about half a million years. Bones of an early human ancestor were recently found in the ancient lake bottom sediments on top of the layer of lava. Therefore, ancestors of modern humans lived in Western Asia between 2 million and 1.5 million years ago.

Which one of the following is an assumption required by the argument?

(A) There were not other lakes in the immediate area before the lava dammed up the river.
(B) The lake contained fish that the human ancestors could have used for food.
(C) The lava under the lake-bottom sediments did not contain any human fossil remains.
(D) The lake was deep enough that a person could drown in it.
(E) The bones were already in the sediments by the time the lake disappeared.

The correct answer here is E (if the bones were not already there, then they’re not good evidence that people were there during that time), but the popular trap answer is C. Consider what would happen if C were untrue: that means that there were human fossil remains that pre-date the time period in question.

But here’s where Lochte Logic is dangerous: you’re not trying to prove that the FIRST humans lived in this period at this time; you’re just trying to prove that humans lived here during that time. And whether or not there were fossils from 2.5 million or 4 million years ago doesn’t change that you still have this evidence of people in that 2 million-1.5 million years ago timeframe.

When people choose C, it’s almost always because they made up their own story about the argument – they read it as, “The earliest human ancestors lived in this place and time,” and that’s just not what’s given. Why do they do that? For Lochte’s very own reasons: it makes the story a little more interesting and a little more favorable.

After all, the average pre-MBA doesn’t spend much time reading about archaeology, but if some discovery is that level of exciting (We’ve discovered the first human! We’ve discovered evidence of aliens!) then it crosses your Facebook/Twitter feeds. You’re used to reading stories about the first/fastest/greatest/last, and so when you get dry subject matter your mind has a tendency to put those words in there subconsciously. Be careful – do not make up your own story about the conclusion!

A similar phenomenon occurs with Reading Comprehension. When you read a long passage, your mind tends to connect dots that aren’t there as it fills in the rest of the story for you. Just like Lochte, who had to fill in the gap of, “Hey what would I have said if someone pointed a gun at me and told me to get on the ground? Oh right…’whatever’ is my default answer for most things,” your mind will start to fill in details that make logical sense.

The problem then comes when you’re asked an Inference question, for which the correct answer must be true based on the passage. For example, if two details in a passage are:

1. Michael swam the fastest race of his life.
2. Ryan’s race was one of the slowest he’s ever swam.

You might answer the question, “Which of the following is a conclusion that can be drawn from the passage?” with:

(A) Michael swam faster than Ryan.

Your mind – particularly amidst a lot of other text between those two facts – wants to logically arrange those two swims together, and with “fastest” for Michael and “slowest” for Ryan, it kind of seems logical that Michael was faster. But those two races are never compared directly to each other. Consider that if Michael and Ryan aren’t Phelps and Lochte, but rather filmmaker Michael Moore and Olympic champion Ryan Lochte, then of course Lochte’s slowest swim would still be way, way faster than Moore’s fastest.

Importantly, Reading Comprehension questions love to bait unwitting test-takers with comparisons as answer choices, knowing that your mind is primed to create your own story and draw comparisons that are probably true, but just not proven. So again, any time you’re faced with an answer that seems obvious, go back and ask yourself if the details you’re using were provided to you, or if instead, you’re making up your own story.

So learn a valuable lesson from Ryan Lochte and avoid making up your own story, sticking only to the clean facts of the matter. Stay true to the truth, and you’ll walk out of the test center saying “Jeah!”

By Brian Galvin.

# GMAT Math Cheat Sheet: Formulas and Tips for Success

An individual who is creating a study plan for the GMAT knows that math must be a part of the equation. Though many people love all sorts of math, there are some who become worried about the Quantitative portion of the exam.

If you’re concerned about the math questions on the GMAT, it can be useful to become more familiar with the specific content in this section. Find out about the types of problems in the Quantitative section and consider some GMAT geometry formulas. Also, check out a gathering of tips on how to prep in an effective way:

What is in the Quantitative Section?
Data Sufficiency and Problem-Solving are the two types of questions in the Quantitative section. The Problem-Solving questions are multiple-choice and test your skills in algebra, basic arithmetic, and geometry. The basic arithmetic questions involve decimals, positive and negative integers, fractions, percentages, and averages. The problems you find in this section are on par with the level of material taught in high school math classes. Though many of the questions on the exam involve basic arithmetic, it’s helpful to have a GMAT formula sheet to refer to when preparing for algebra and geometry problems.

GMAT Formulas for the Math Section
Your GMAT math formulas cheat sheet should include the Pythagorean Theorem. This formula helps you to find the measurement of the third side of a right triangle when given the measurements of the other two sides. Another item on your GMAT math cheat sheet should be A = 1/2 bh, which is the formula for finding the area of a triangle. Distance = rate*time is a very helpful formula to know, too. Find the area of a rectangle in fast fashion by using the formula A = lw. The formula A = s2 will help you discover the area of a square.

Moving Beyond Memorization

A GMAT math formulas cheat sheet is an effective study tool, but it’s equally important to know which formula to apply to a problem, so you should spend time practicing problems that employ each of those formulas. This way, on test day, you’ll be familiar with the formulas and feel comfortable using them. The easiest way to do this, of course, is to let us help you.

The expert instructors at Veritas Prep partner with students to help them learn and to practice these formulas for the Quantitative section. We hire tutors who have excellent teaching skills as well as GMAT scores in the 99th percentile. When you study with us, you know you’re learning from the best! Our instructors work through practice math problems with you to ensure that you understand how to solve them in the most efficient way.

Get the Timing Right
Test-takers are given 75 minutes to tackle the 37 questions in the Quantitative section. This sounds like a long time, but if you get hung up on one question for several minutes, you could end up running out of time for this section. In order to avoid this, you should take timed practice tests. Taking timed tests allows you to establish a rhythm for solving problems and answering questions. Once you establish a rhythm, you don’t have to be so concerned about running out of time before you finish all of the problems.

More Tips for Mastering the Quantitative Section
Studying with a GMAT math cheat sheet is one way to prepare for the test. Another way to save test time and make questions more manageable is to eliminate answer options that are clearly wrong – this allows your mind to focus only on the legitimate choices. Estimating the answer to a problem as you read through it is another way to save test time and arrive at answers more quickly.

Our GMAT curriculum teaches you how to approach questions on the separate math topics within the Quantitative section. Our strategies give you the tools you need to problem-solve like a business professional! We are proud to provide both online and in-person courses that prepare you for the GMAT. Veritas Prep instructors offer solid instruction as well as encouragement to individuals with the goal of acing the GMAT and getting into a preferred business school. Let us partner with you on the road to GMAT success! Contact us to talk with one of our course advisers today.

# When to Pick Your Own Numbers on GMAT Quant Questions

The other day, while working with a tutoring student, I was enumerating the virtues of various test-taking strategies when the student sheepishly interrupted my eloquent paean to picking numbers. She’d read somewhere that these strategies were fine for easy to moderate questions, but that for the toughest questions, you just had to bear down and solve the problem formally. Clearly, she is not a regular reader of our fine blog.

As luck would have it, on her previous practice exam she’d received the following problem, which both illustrates the value of picking numbers and demonstrates why this approach works so well.

A total of 30 percent of the geese included in a certain migration study were male. If some of the geese migrated during the study and 20 percent of the migrating geese were male, what was the ratio of the migration rate for the male geese to the migration rate for the female geese?

[Migration rate for geese of a certain sex = (number of geese of that sex migrating) / (total number of geese of that sex)]

A) 1/4
B) 7/12
C) 2/3
D) 7/8
E) 8/7

This is a perfect opportunity to break out two of my favorite GMAT tools: picking numbers and making charts. So, let’s say there are 100 geese in our population. That means that if 30% are male, we’ll have 30 male geese and 70 females geese, giving us the following chart:

 Male Female Total Migrating Not-Migrating Total 30 70 100

Now, let’s say 10 geese were migrating. That means that 90 were not migrating. Moreover, if 20 percent of the migrating geese were male, we know that we’ll have 2 migrating males and 8 migrating females, giving us the following:

 Male Female Total Migrating 2 8 10 Not-Migrating Total 30 70 100

(Note that if we wanted to, we could fill out the rest of the chart, but there’s no reason to, especially when we’re trying to save as much time as possible.)

Our migration rate for the male geese is 2/30 or 1/15. Our migration rate for the female geese is 8/70 or 4/35. Ultimately, we want the ratio of the male migration rate (1/15) to the female migration rate (4/35), so we need to simplify (1/15)/(4/35), or (1*35)/(15*4) = 35/60 = 7/12. And we’re done – B is our answer.

My student was skeptical. How did we know that 10 geese were migrating? What if 20 geese were migrating? Or 50? Shouldn’t that change the result? This is the beauty of picking numbers – it doesn’t matter what number we pick (so long as we don’t end up with an illogical scenario in which, say, the number of migrating male geese is greater than the number of total male geese). To see why, watch what happens when we do this algebraically:

Say that we have a total of “t” geese. If 30% are male, we’ll have 0.30t male geese and 0.70t females geese.  Now, let’s call the migrating geese “m.” If 20% are male, we’ll have 0.20m migrating males and 0.80m migrating females. Now our chart will look like this:

 Male Female Total Migrating 0.20m 0.80m m Not-Migrating Total 0.30t 0.70t t

The migration rate for the male geese is 0.20m/0.30t or 2m/3t. The migration rate for the female geese is 0.80m/0.70t or 8m/7t. We want the ratio of the male migration rate (2m/3t) to the female migration rate (8m/7t), so we need to simplify (2m/3t)/(8m/7t) = (2m*7t)/(3t * 8m) = 14mt/24mt = 7mt/12mt = 7/12. It’s clear now why the numbers we picked for m and t don’t matter – they cancel out in the end.

Takeaway: We cannot say this enough: the GMAT is not testing your ability to do formal algebra. It’s testing your ability to make good decisions in a stressful environment. So your goal, when preparing for this test, isn’t to become a virtuoso mathematician, even for the toughest questions. It’s to practice the kind of simple creative thinking that will get you to your answer with the smallest investment of your time.

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

# Quarter Wit, Quarter Wisdom: Know Your Subtraction for the GMAT!

Your first reaction to the title of this post is probably, “I already know my subtraction!” No surprise there. But what is surprising is that our statistics tell us that the following GMAT question – which is nothing extraordinary, but does involve some tricky subtraction – is a 700-level question. That made us decide to write this post. We will discuss this concept along with the question:

The last digit of 12^12 + 13^13 – 14^14 × 15^15 =

(A) 0
(B) 1
(C) 5
(D) 8
(E) 9

This is a simple question based on the cyclicity of units digits. There are 3 terms here: 12^12, 13^13 and (14^14)*(15^15). Let’s find the last digit of each of these terms:

12^12
The units digit of 12 is 2.
2 has a cyclicity of 2 – 4 – 8 – 6.
The cycles end at the powers 4, 8, 12 … etc. So, twelve 2’s will end in a units digit of 6.

13^13
The units digit of 13 is 3.
3 has a cyclicity of 3 – 9 – 7 – 1.
A new cycle starts at the powers 1, 5, 9, 13 … etc. So, thirteen 3’s will end in a units digit of 3.

(14^14)*(15^15)
This term is actually the most simple to manage in the case of its units digit – an even number multiplied by a multiple of 5 will end in 0. Also, note that this will be a huge term compared to the other two terms.

This is what our expression looks like when we consider just the units digits of these terms:

(A number ending in 6) + (A number ending in 3) – (A much greater number ending in 0)

Looking at our most basic options, a number ending in 6 added to a number ending in 3 will give us a number ending in 9 (as 3 + 6 = 9). So, the expression now looks like this:

(A number ending in 9) – (A much greater number ending in 0)

It is at this point that many people mess up. They deduce that 9-0 will end in a 9, and hence, the answer will be E. All their effort goes to waste when they do this. Let’s see why:

How do you subtract one number out of another? Take, for example, 10-7 = 3

This can also be written as 7-10 = -3. (Here, you are still subtracting the number with a lower absolute value from the number with a greater absolute value, but giving it a negative sign.)

Let’s try to look at this in tabular form. The number with the greater absolute value goes on the top and the number with the smaller absolute value goes under it. You then subtract and the result gets the sign of the number with the greater absolute value.

(i) 100-29
100
-29
071

(ii) 29-100
100
-29
071
(But since the sign of 100 is negative, your answer is actually -71.)

So, the number with greater absolute value is always on top. Going back to our original question now, (A number ending in 9) – (A much greater number ending in 0) will look like:

abcd0
–  pq9
ghjk1

Ignoring the letter variables (these are simply placeholders), note that the greater number ending in 0 will be on the top and the smaller one ending in 9 will be below it. This means the answer will be a negative number ending in a units digit of 1. Therefore, our answer is B.

As we learn more advanced concepts, make sure you are not taking your basic principles for granted!

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!

# GMAT Tip of the Week: What Simone Biles and the Final Five Can Teach You About GMAT Math

On this Friday, ending the first week of the Rio Olympics, your office has undoubtedly said the name “Simone” exponentially more than ever before. Michael Phelps’ blowout win – his 4th straight – in the 200 IM was incredible, but last night belonged to two Texans named Simone.

Swimmer Simone Manuel and gymnast Simone Biles each won historic gold medals, and if you’re at all inspired to pursue your own “go for the gold” success in business school (maybe Stanford like Manuel, or UCLA like Biles), you can learn a lot from the Olympic experience. Two lessons, in particular, stand out from the performance of Biles and her “Final Five” teammates:

There’s no way to watch Olympic gymnastics and not be overwhelmingly impressed by the skills that each gymnast brings to competition. So at times it’s frustrating and saddening to hear the TV announcers discuss deduction after deduction; shouldn’t everyone at all times just be yelling, “Wow!!!!” at the otherworldly talents of each athlete?

Much like the GMAT, though, Olympic gymnastics is not about the sheer possession of these skills – at that level, everyone has them. It’s more about the ability to execute them and, as becomes evident from the expert commentary of Tim Dagget and Nastia Liukin, to connect them. It’s not the uneven bars handstand or release itself that wins the gold, it’s the ability to connect skill after skill as part of a routine. The line, “She was supposed to connect that skill to another…” is always followed by, “That will be a deduction” – both in Olympic gymnastics and on the GMAT.

How does that affect you?

By test day, you had better have all of the necessary skills to compete on the GMAT Quant Section. Area of a triangle, Pythagorean Theorem, Difference of Squares…if you don’t know these rules, you’re absolutely sunk. But to do really well, you need to quickly connect skill to skill, and connect items in the problems to the skills necessary to work with them. For example:

If a problem includes a term x^4 – 1, you should immediately be thinking, “That connects really well to the Difference of Squares rule: a^2 – b^2 = (a + b)(a – b), and since x^4 is a square [it’s (x^2)^2] and 1 is a square (it’s 1^2), I can write that as (x^2 + 1)(x^2 – 1), and for good measure I could apply Difference of Squares to the (x^2 – 1) term too.” The GMAT won’t ever specifically tell you, “Use the Difference of Squares,” so it’s your job to immediately connect the symptoms of Difference of Squares (an even exponent, a subtraction sign, a square of some kind, even if it’s 1) to the opportunity to use it.

If you see a right triangle, you should recognize that Area and Pythagorean Theorem easily connect. In a^2 + b^2 = c^2, sides a and b are perpendicular and allow you to use them as the base and height in the area formula. And the Pythagorean Theorem includes three squares with the opportunity to create subtraction [you could write it as a^2 = c^2 – b^2, allowing you to say that a^2 = (c + b)(c – b)…], so you could connect yet another skill to it to help solve for variables.

Similarly, if you see a square or rectangle, its diagonal is the hypotenuse of a right triangle, allowing you to use the sides as a and b in the Pythagorean setup, which could also connect to Difference of Squares…etc.

When you initially learned most of these skills in high school (much like when Biles, Aly Raisman, Gabby Douglas, etc. learned handstands and cartwheels in Gymboree), you learned them as individual, isolated skills. “Here’s the formula, and here are 10 questions that test it.” On the GMAT – as in the Olympics – you’re being tested more on your ability to connect them, to see opportunities to use a skill that’s not obvious at first (“Well, I’m not sure what to do but I do have multiple squared terms so let me try to apply Difference of Squares…or maybe I can use a and b in the Area calculation.”), but that helps you build more knowledge of the problem.

So as you study, don’t just learn individual skills. Look for opportunities to connect them, and look for signals that will tell you that a connection is possible. A rectangle problem with a square root of 3 in the answer choices should tell you “the diagonal of this rectangle may very well be connected to a 30-60-90 triangle, since those have the 1, √3, 2 side ratio…” The GMAT is about connections more so than just skills, so study accordingly.

Stick the Landing
If you’re like most in the “every four years I love gymnastics for exactly one week” camp, the single most important thing you look for on any apparatus is, “Did he/she stick the landing?” A hop or a step on the landing is the most noticeable deduction on a gymnastics routine…and the same holds true for the GMAT.

Again, the GMAT is testing you on how well you connect a variety of skills, so naturally there are places for you to finish the problem a step short. A problem that requires you to leverage the Pythagorean Theorem and the Area of a Triangle may ask for the sum of sides A and B, for example, but if you’ve solved for the sides individually first, you might see a particular value (A = 6) on your noteboard and in the answer choices and choose it without double checking that you answered the proper question.

That is a horrible and unnecessary “deduction” on your GMAT score: you did all the work right, all the hard part right (akin to the flip-and-two-twists in the air on your vault or the dazzling array of jumps and handstands on the tiny beam) and then botched the landing.

On problems that include more than one variable, circle the variable that the test is looking for and then make sure that you submit the proper answer for that variable. If a problem asks for a combination of variables (a + b, for example), write that down at the top of your scratchwork and go back to it after you’ve calculated. Take active steps to ensure that you stick the landing, because nothing is worse than doing all the work right and then still getting the problem wrong.

In summary, recognize that there are plenty of similarities between the GMAT and GyMnAsTics [the scoring system is too complex for the layman to worry about, the “Final Five” are more important than you think (hint: the test can’t really use the last five questions of a section for research purposes since so many people are rushing and guessing), etc.]. So take a lesson from Simone Biles and her gold-medal-winning teammates: connect your skills, stick the landing, and you’ll see your score vault to Olympian heights.

By Brian Galvin.

# Quarter Wit, Quarter Wisdom: Linear Relations in GMAT Questions

We have covered the concepts of direct, inverse and variation in previous posts. Today, we will discuss what we mean by “linearly related”. A linear relation is one which, when plotted on a graph, is a straight line. In linear relationships, any given change in an independent variable will produce a corresponding change in the dependent variable, just like a change in the x-coordinate produces a corresponding change in the y-coordinate on a line.

We know the equation of a line: it is y = mx + c, where m is the slope and c is a constant.

Let’s illustrate this concept with a GMAT question. This question may not seem like a geometry question, but using the concept of linear relations can make it easy to find the answer:

A certain quantity is measured on two different scales, the R-scale and the S-scale, that are related linearly. Measurements on the R-scale of 6 and 24 correspond to measurements on the S-scale of 30 and 60, respectively. What measurement on the R-scale corresponds to a measurement of 100 on the S-scale?

(A) 20
(B) 36
(C) 48
(D) 60
(E) 84

Let’s think of the two scales R and S as x- and y-coordinates. We can get two equations for the line that depicts their relationship:

30 = 6m + c ……. (I)
60 = 24m + c ……(II)

(II) – (I)
30 = 18m
m = 30/18 = 5/3

Plugging m = 5/3 in (I), we get:

30 = 6*(5/3) + c
c = 20

Therefore, the equation is S = (5/3)R + 20. Let’s plug in S = 100 to get the value of R:

100 = (5/3)R + 20
R = 48

Alternatively, we have discussed the concept of slope and how to deal with it without any equations in this post. Think of each corresponding pair of R and S as points lying on a line – (6, 30) and (24, 60) are points on a line, so what will (r, 100) be on the same line?

We see that an increase of 18 in the x-coordinate (from 6 to 24) causes an increase of 30 in the y-coordinate (from 30 to 60).

So, the y-coordinate increases by 30/18 = 5/3 for every 1 point increase in the x-coordinate (this is the concept of slope).

From 60 to 100, the increase in the y-coordinate is 40, so the x-coordinate will also increase from 24 to 24 + 40*(3/5) = 48. Again, C is our 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!

# GMAT Math Help: Understanding and Solving Combinatorics Problems

Students who are taking the GMAT are going to encounter combinatorics problems. If you are a little rusty on your math topics, you may be asking, “What is combinatorics?” Combinatorics has to do with counting and evaluating the possibilities within a scenario that involve various amounts of people or things. Learn more about GMAT combinatorics questions and how to arrive at the right answers to be better prepared for the test.

Permutations
Picture a certain number of people or objects. Permutations are the possible arrangements that those people or objects can be in. One of the things you have to decide when looking at combinatorics problems is whether order is an important factor. If order is important in a problem, then the answer has to do with permutations. If order is not important in a problem, then the answer deals with combinations.

For example, say you line up five postcards from different cities on a tabletop. You may wonder how many different orders you can put these postcards in. Another way to say that would be, “How many different permutations can I make with these five postcards?” To figure out this problem, you would need the help of an equation: 5! = (5) (4) (3) (2) (1) = 120. The exclamation point in the formula is a symbol that means “factorial.”

Combinations
When working on combinatorics questions that deal with combinations, the order/arrangement of items is not important. For example, say that you have eight books and you want to know how many ways you can group three of those books on a library shelf. You could plug numbers into the three places in this formula to figure out the answer: (8) (7) (6) = 336 ways. This is the slot method of solving a combination problem.

Combinations With a Large Amount of Numbers
You will quickly find yourself needing combinatorics help if you try to count up a lot of numbers in one combination problem on the GMAT. Furthermore, you’ll use a lot of valuable test time with this counting method. Knowing the formula for combinations can help you to find the solution to a problem in a much shorter amount of time. The formula is nCr = n!/r!(n-r)! Here, n is the total number of options, r is the number of options chosen, and ! is the symbol for factorial.

Preparing for Applied Combinatorics Questions on the GMAT
One of the most effective ways of preparing for applied combinatorics questions is to take practice tests and review the various steps of problems. You want to get into the habit of approaching a problem by asking yourself whether order is a factor in a problem. This will help you determine whether a problem deals with permutations or combinations. Then, you can start to attack a problem from the right angle.

In addition, it’s important to time yourself when taking a practice Quantitative test. Though there are not many of these problems on the test, you have to get into the habit of spending only a certain amount of minutes on each problem so you don’t run out of test time before finishing.

We have a program of study at Veritas Prep that prepares you for questions on combinatorics as well as all of the other problems in the Quantitative section. We instruct you on how to approach test questions instead of just coaching you on how to memorize facts. Pair up with one of our skilled instructors at Veritas Prep and you will be studying with someone who scored in the 99th percentile on the GMAT. We believe that in order to perform at your best on the GMAT, you have to learn from a first-rate instructor! Our instructors can work through a combinatorics tutorial with you to determine what your strengths and weaknesses are in this branch of math. Then, we give you strategies that help you to improve.

For your convenience, we offer both in-person and online GMAT prep courses. We recognize that professionals in the business world have busy schedules, so we provide several study options to fit your life. When it comes to the topic of combinatorics, GMAT tips, instruction, and encouragement, we are your test prep experts. Contact us today and let us know how we can help you achieve your top GMAT score!

# Probability and Combinations: What You’ll Need to Know for the GMAT

If you’ve been paying attention to the exciting world of GMAT prep, you know thatfairly recently. I’d mentioned in a previous post that I was going to write about any conspicuous trends I noted, and one unmistakable pattern I’ve seen with my students is that probability questions seem to be cropping up with greater and greater frequency.

While these questions don’t seem fundamentally different from what we’ve seen in the past, there does seem to be a greater emphasis on probability questions for which a strong command of combinations and permutations will prove indispensable.

First, recall that the probability of x is the number ways x can occur/number of total possible outcomes (or p(x) = # desired/ # total). Another way to think about this equation is to see it as a ratio of two combinations or permutations. The number of ways x can occur is one combination (or permutation), and the total number of possible outcomes is another.

Keeping this in mind, let’s tackle this new official prompt:

From a group of 8 volunteers, including Andrew and Karen, 4 people are to be selected at random to organize a charity event. What is the probability that Andrew will be among the 4 volunteers selected and Karen will not?

(A) 3/7
(B) 5/12
(C) 27/70
(D) 2/7
(E) 9/35

Typically, I’ll start by calculating the total number of possible outcomes, as this calculation tends to be the more straightforward one. We’ve got 8 volunteers, and we want to know the number of total ways we can select 4 people from these 8 volunteers. Note, also, that the order does not matter – group of Tiffany, Mike, Louis, and Amy is the same as a group of Louis, Amy, Mike, and Tiffany. We’re not assigning titles or putting anyone in seats, so this is a combination.

If we use our combination formula N!/[(K!*(N-K)!] then N, our total pool of candidates, is 8, and K, the number we’re selecting, is 4. We get 8!/(4!*4!), which comes out to 70. At this point, we know that the denominator must be a factor of 70, so anything that doesn’t meet this criterion is out. In this case, this only allows us to eliminate B.

Now we want our desired outcomes, in which Andrew is selected and Karen is not. Imagine that you’re responsible for assembling this group of four from a total pool of eight people. You plan on putting your group of four in a conference room. Your supervisor tells you that Andrew must be in and Karen must not be, so you take Andrew and put him in the conference room. Now you’ve got three more spots to fill and seven people remaining. But remember that Karen cannot be part of this group. That means you only have 6 people to choose from to fill those other 3 spots in the room.

Put another way, think of the combination as the number of choices you have. Andrew and Karen are not choices – you’ve been ordered to include one of them and not the other. Of the 4 spots in the conference room, you only get to choose 3. And you’re only selecting from the other 6 people for those spots. Now N = 6 and K = 3. Plugging these into our trusty combination formula, we get 6!/(3!*3!), which comes out to 20.

Summarizing, we know that there are 20 ways to create our desired group of 4, and 70 total ways to select 4 people from a pool of 8, giving us a probability of 20/70, or 2/7, so the correct answer is D.

Takeaway: Probability questions can be viewed as ratios of combinations or permutations, so when you brush up on combinatorics, you’re also bolstering your probability fundamentals. Anytime you’re stuck on a complex probability question, break your calculation down into its component parts – find the total number of possible outcomes first, then find the total number of desired outcomes. Like virtually every hard question on the GMAT, probability questions are never as hard as they first seem.

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

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

# How to Use Units Digits to Avoid Doing Painful Calculations on the GMAT

During the first session of each new class I teach, we do a quick primer on the utility of units digits. Imagine I want to solve 130,467 * 367,569. Without a calculator, we are surely entering a world of hurt. But we can see almost instantaneously what the units digit of this product would be.

The units digit of 130,467 * 367,569 would be the same as the units digit of 7*9, as only the units digits of the larger numbers are relevant in such a calculation. 7*9 = 63, so the units digit of 130,467 * 367,569 is 3. This is one of those concepts that is so simple and elegant that it seems too good to be true.

And yet, this simple, elegant rule comes into play on the GMAT with surprising frequency.

Take this question for example:

If n is a positive integer, how many of the ten digits from 0 through 9 could be the units digit of n^3?

A) three
B) four
C) six
D) nine
E) ten

Surely, you think, the solution to this question can’t be as simple as cubing the easiest possible numbers to see how many different units digits result. And yet that’s exactly what we’d do here.

1^3 = 1

2^3 = 8

3^3 = 27 à units 7

4^3 = 64 à units 4

5^3 = ends in 5 (Fun fact: 5 raised to any positive integer will end in 5.)

6^3 = ends in 6 (Fun fact: 6 raised to any positive integer will end in 6.)

7^3 = ends in 3 (Well 7*7 = 49. 49*7 isn’t that hard to calculate, but only the units digit matters, and 9*7 is 63, so 7^3 will end in 3.)

8^3 = ends in 2 (Well, 8*8 = 64, and 4*8 = 32, so 8^3 will end in 2.)

9^3 = ends in 9 (9*9 = 81 and 1 * 9 = 9, so 9^3 will end in 9.)

10^3 = ends in 0

Amazingly, when I cube all the integers from 1 to 10 inclusive, I get 10 different units digits. Pretty neat. The answer is E.

Of course, this question specifically invoked the term “units digit.” What are the odds of that happening? Maybe not terribly high, but any time there’s a painful calculation, you’d want to consider thinking about the units digits.

Take this question, for example:

A certain stock exchange designates each stock with a one, two or three letter code, where each letter is selected from the 26 letters of the alphabet. If the letters may be replaced and if the same letters used in a different order constitute a different code, how many different stocks is it possible to uniquely designate with these codes?

A) 2,951
B) 8,125
C) 15,600
D) 16,302
E) 18,278

Conceptually, this one doesn’t seem that bad.

If I wanted to make a one-letter code, there’d be 26 ways I could do so.

If I wanted to make a two-letter code, there’d be 26*26 or 26^2 ways I could do so.

If I wanted to make a three-letter code, there’d be 26*26*26, or 26^3 ways I could so.

So the total number of codes I could make, given the conditions of the problem, would be 26 + 26^2 + 26^3. Hopefully, at this point, you notice two things. First, this arithmetic will be deeply unpleasant to do.  Second, all of the answer choices have different units digits!

Now remember that 6 raised to any positive integer will always end in 6. So the units digit of 26 is 6, and the units digit of 26^2 is 6 and the units digit of 26^3 is also 6. Therefore, the units digit of 26 + 26^2 + 26^3 will be the same as the units digit of 6 + 6 + 6. Because 6 + 6 + 6 = 18, our answer will end in an 8. The only possibility here is E. Pretty nifty.

Takeaway: Painful arithmetic can always be avoided on the GMAT. When calculating large numbers, note that we can quickly find the units digit with minimal effort. If all the answer choices have different units digits, the question writer is blatantly telegraphing how to approach this problem.

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

# Quarter Wit Quarter Wisdom: What is Your Favorite Number?

Fans of The Big Bang Theory will remember Sheldon Cooper’s quote from an old episode on his favorite number:

The best number is 73. Why? 73 is the 21st prime number. Its mirror, 37, is the 12th and its mirror, 21, is the product of multiplying 7 and 3… and in binary 73 is a palindrome, 1001001, which backwards is 1001001.”

Though Sheldon’s logic is infallible, my favorite number is 1001 because it has a special role in standardized tests.

1001 is 1 more than 1000 and hence, is sometimes split as (1000 + 1). It sometimes appears in the a^2 – b^2 format such as 1001^2 – 1, and its factors are 7, 11 and 13 (not the factors we usually work with).

Due to its unusual factors and its convenient location (right next to 1000), it could be a part of some tough-looking GMAT questions and should be remembered as a “special” number. Let’s look at a question to understand how to work with this  number.

Which of the following is a factor of 1001^(32) – 1 ?

(A) 768
(B) 819
(C) 826
(D) 858
(E) 924

Note that 1001 is raised to the power 32. This is not an exponent we can easily handle. If  we try to use a binomial here and split 1001 into (1000 + 1), all we will achieve is that upon expanding the given expression, 1 will be cancelled out by -1 and all other terms will have 1000 in common. None of the answer choices are factors of 1000, however, so we must look for some other factor of 1001^(32) – 1.

Without a calculator, it is not possible for us to find the factors of 1001^(32) – 1, but we do know the prime factors of 1001 and hence, the prime factors of 1001^32. We may not be able to say which numbers are factors of 1001^(32) – 1, but we will be able to say which numbers are certainly not factors of this!

Let me explain:

1001 = 7 * 11 * 13 (Try dividing 1001 by 7 and you’ll get 143. 143 is divisible by 11, giving you 13.)

1001^32 = 7^32 * 11^32 * 13^32

Now, what can we say about the prime factors of 1001^(32) – 1? Whatever they are, they are certainly not 7, 11 or 13 – two consecutive integers cannot have any common prime factor (discussed here and continued here).

Now look at the answer choices and try dividing each by 7:

(A) 768 – Not divisible by 7

(B) 819 – Divisible by 7

(C) 826 – Divisible by 7

(D) 858 – Not divisible by 7

(E) 924 – Divisible by 7

Options B, C and E are eliminated. They certainly cannot be factors of 1001^(32) – 1 since they have 7 as a prime factor, and we know 1001^(32) – 1 cannot have 7 as a prime factor.

Now try dividing the remaining options by 11:

(A) 768 – Not divisible by 11

(D) 858 – Divisible by 11

D can also be eliminated now because it has 11 as a factor. By process of elimination, the answer is A; it must be a factor of 1001^(32) – 1.

I hope you see how easily we used the factors of 1001 to help us solve this difficult-looking question. And yes, another attractive feature of 1001 – it is a palindrome in the decimal representation itself!

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!

# GMAT Tip of the Week: How to Avoid GMAT (and Pokemon Go) Traps

In seemingly the most important development in world history since humans learned to create fire, Pokemon Go has arrived and is taking the world by storm. Rivaling Twitter and Facebook for mobile phone attention and battling the omnipresent selfie as a means of death-by-mobile-phone, Pokemon Go is everywhere you want to be…and often in places you don’t.

And that is why Pokemon Go is responsible for an ever-important GMAT lesson.

Perhaps most newsworthy about Pokemon Go these days is the dangerous and improper places that it has led its avid users. On the improper side,  such solemn and dignified places as the national Holocaust Museum and Arlington National Cemetery have had to actively prohibit gamers from descending upon mourners/commemorators while playing the game. And as for danger, there have been several instances of thieves luring gamers into traps and therefore robbing them of valuable (if you’re playing the game, you definitely have a smartphone) items.

And the GMAT can and will do the same thing.

How?

If you’re reading this on our GMAT blog, you’ve undoubtedly already learned that, on Data Sufficiency problems, you cannot assume that a variable is positive, or that it is an integer. But think about what makes Pokemon Go users so vulnerable to being lured into a robbery or to losing track of basic human decency. They’re so invested in the game that they lose track of the situations they’re being lured into.

Similarly, the most dangerous GMAT traps are those for which you should absolutely know better, but the testmaker has gotten your mind so invested in another “game” that you lose track of something basic. Consider the example:

If y is an odd integer and the product of x and y equals 222, what is the value of x?

(1) x is a prime number
(2) y is a 3 digit number

Statement 1 is clearly sufficient. Since y is odd, and an integer, and the product of integers x and y is an even integer, that means that x must be even. And since x also has to be prime (which is how you know it’s an integer, too), the only even prime is 2, making x = 2.

From there your mind is fixated on the game. You can quickly see that in that case y = 111 and x = 2. Which you then have to forget about as you attack Statement 2. But here’s the reason that less than 25% of users in the Veritas Prep Question Bank get this right, while nearly half incorrectly choose D. Statement 1 has gotten your mind fixated on the even/odd/prime game, meaning that you may only be thinking about integers (and positive integers at that) at this point.

That y is a 3-digit number DOES NOT mean that it has to be 111. It could be -111 (making x = -2) or 333 (making x = 2/3). So only Statement 1 alone is sufficient, but the larger lesson is more important. Just like Pokemon Go has the potential to pollute your mind and have you see the real world through its “enhanced reality” lens, so does a statement that satisfies your intellect (“Ah, 2 is the only even prime number!”) give you just enough tunnel vision that you make poor decisions and fall for traps.

The secret here is that almost no one scoring above a 500 carries over all of Statement 1 (“Oh, well I already know that x = 2!”) – a total rookie mistake. It’s that Statement 1 got you fixated on definitions of types of integers (prime, even, odd) and therefore got your mind looking through the “enhanced reality” of integers-only.

The lesson? Much like Pokemon Go, the GMAT has tools to get you so invested in a particular facet of a game that you lose your universal awareness of your surroundings. Know that going in – that you have to consciously step back from that enhanced reality you’ve gained after Statement 1 and look at the whole picture. So take a lesson from Pokemon Go and know when to stop and step back.

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, YouTubeand Twitter!

By Brian Galvin.

# How to Approach Difficult GMAT Problems

My students have a hard time understanding what makes a difficult GMAT question difficult. They assume that the tougher questions are either testing something they don’t know, or that these problems involve a dizzying level of complexity that requires an algebraic proficiency that’s simply beyond them.

One of my main goals in teaching a class is to persuade everyone that this is not, in fact, how hard questions work on this test. Hard questions don’t ask you do to something you don’t know how to do. Rather, they’re cleverly designed to provoke an anxiety response that makes it difficult to realize that you do know exactly how to solve the problem.

Take this official question, for example:

Let a, b, c and d be nonzero real numbers. If the quadratic equation ax(cx + d) = -b(cx +d) is solved for x, which of the following is a possible ratio of the 2 solutions?

A) –ab/cd
B) –ac/bd
D) ab/cd

Most students see this and panic. Often, they’ll start by multiplying out the left side of the equation, see that the expression is horrible (acx^2 + adx), and take this as evidence that this question is beyond their skill level. And, of course, the question was designed to elicit precisely this response. So when I do this problem in class, I always start by telling my students, much to their surprise, that every one of them already knows how to do this. They’ve just succumbed to the question writer’s attempt to convince them otherwise.

So let’s start simple. I’ll write the following on the board: xy = 0. Then I’ll ask what we know about x or y. And my students shrug and say x or y (or both) is equal to 0. They’ll also wonder what on earth such a simple identity has to do with the algebraic mess of the question they’d been struggling with.

I’ll then write this: zx + zy = 0. Again, I’ll ask what we know about the variables. Most will quickly see that we can factor out a “z” and get z(x+y) = 0. And again, applying the same logic, we see that one of the two components of the product must equal zero – either z = 0 or x + y = 0.

Next, I’ll ask if they would approach the problem any differently if I’d given them zx = -zy – they wouldn’t.

Now it clicks. We can take our initial equation in the aforementioned problem: ax(cx +d) = -b(cx+d), and see that we have a ‘cx + d’ on both sides of the equation, just as we’d had a “z” on both sides of the previous example. If I’m able to get everything on one side of the equation, I can factor out the common term.

Now ax(cx +d) = -b(cx+d) becomes ax(cx +d) + b(cx+d) = 0.

Just as we factored out a “z” in the previous example, we can factor out “cx + d” in this one.

Now we have (cx + d)(ax + b) = 0.

Again, if we multiply two expressions to get a product of zero, we know that at least one of those expressions must equal 0. Either cx + d = 0 or ax + b = 0.

If cx + d = 0, then x = -d/c.

If ax + b = 0, then x = -b/a.

Therefore, our two possible solutions for x are –d/c and –b/a. So, the ratio of the two would simply be (-d/c)/(-b/a). Recall that dividing by a fraction is the equivalent of multiplying by the reciprocal, so we’re ultimately solving for (-d/c)(-a/b). Multiplying two negatives gives us a positive, and we end up with da/cb, which is equivalent to answer choice E.

Takeaway: Anytime you see something on the GMAT that you think you don’t know how to do, remind yourself that the question was designed to create this false impression. You know how to do it – don’t hesitate to dive in and search for how to apply this knowledge.

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

# How to Go From a 48 to 51 in GMAT Quant – Part VII

Both a test-taker at the 48 level and one at the 51 level in the GMAT Quant section, are conceptually strong – given an unlimited time frame, both will be able to solve most GMAT questions correctly. The difference lies in the two things a test-taker at the 51 level does skillfully:

1. Uses holistic, big-picture methods to solve Quant questions.
2. Handles questions he or she finds difficult in a timely manner.

We have been discussing holistic methods on this blog for a long time now and will continue discussing them. (Before you continue reading, be sure to check out parts I, II, III, IVV and VI of this series.)

Today we will focus on “handling the hard questions in a timely manner.” Note that we do not say “solving the hard questions in a timely manner.” Occasionally, one might be required to make a quick call and choose to guess and move on – but again, that is not the focus of this post. We are actually going to talk about the “lightbulb” moment that helps us save on time. There are many such moments for the 51 level test-taker – in fact, the 51 scorers often have time left over after attempting all these questions.

Test takers at the 48 level will also eventually reach the same conclusions but might need much more time. That will put pressure on them the next time they look at the ticking clock, and once their cool is lost, “silly errors” will start creeping in. So it isn’t about just that one question – one can end up botching many other questions too.

There are many steps that can be easily avoided by a lightbulb moment early on. This is especially true for Data Sufficiency questions.

Let’s take an official example:

Pam owns an inventory of unopened packages of corn and rice, which she has purchased for \$17 and \$13 per package, respectively. How many packages of corn does she have ?

Statement 1: She has \$282 worth of packages.

Statement 2: She has twice as many packages of corn as of rice.

A high scorer will easily recognize that this question is based on the concept of “integral solutions to an equation in two variables.” Since, in such real world examples, x and y cannot be negative or fractional, these equations usually have a finite number of solutions.

After we find one solution, we will quickly know how many solutions the equation has, but getting the first set of values that satisfy the equation requires a little bit of brute force.

The good thing here is that this is a Data Sufficiency question – you don’t need to find the actual solution. The only thing we need is to establish that there is a single solution only. (Obviously, there has to be a solution since Pam does own \$282 worth of packages.)

So, the test-taker will start working on finding the first solution (using the method discussed in this post). We are told:

Price of a packet of corn = \$17
Price of a packet of rice = \$13

Say Pam has “x” packets of corn and “y” packets of rice.

Statement 1: She has \$282 worth of packages

Using Statement 1, we know that 17x + 13y = 282.

We are looking for the integer values of x and y.

If x = 0, y will be 21.something (not an integer)
If x = 1, y = 20.something
If x = 2, y = 19.something
If x = 3, y = 17.something

This is where the 51 level scorer stops because they never lose sight of the big picture. The “lightbulb” switches on, and now he or she knows that there will be only one set of values that can satisfy this equation. Why? Because y will be less than 17 in the first set of values that satisfies this equation. So if we want to get the next set that satisfies, we will need to subtract y by 17 (and add 13 to x), which will make y negative.

So in any case, there will be a unique solution to this equation. We don’t actually need to find the solution and hence, nothing will be gained by continuing these calculations. Statement 1 is sufficient.

Statement 2: She has twice as many packages of corn as of rice.

Statement 2 gives us no information on the total number of packages or the total amount spent. Hence, we cannot find the total number of packages of corn using this information alone. Therefore, our answer is A.

I hope you see how you can be alert to what you want to handle these Quant questions in a timely manner.

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!

# GMAT Tip of the Week: The Overly Specific Question Stem

For most of our lives, we ask and answer relatively generic questions: “How’s it going?” “What are you up to this weekend?” “What time do the Cubs play tonight?”

And think about it, what if those questions were more specific: “Are you in a melancholy mood today?” “Are you and Josh going to dinner at Don Antonio’s tonight and ordering table-side guacamole?” “Do the Cubs play at 7:05 tonight on WGN?” If someone is asking those questions instead, you’re probably a bit suspicious. Why so specific? What’s your angle?

The same is true on the GMAT. Most of the question stems you see are relatively generic: “What is the value of x?” “Which of the following would most weaken the author’s argument?” So when the question stem get a little too specific, you should become a bit suspicious. What’s the test going for there? Why so specific?

The overly-specific Critical Reasoning question stem is a great example. Consider the problem:

Raisins are made by drying grapes in the sun. Although some of the sugar in the grapes is caramelized in the process, nothing is added.
Moreover, the only thing removed from the grapes is the water that evaporates during the drying, and water contains no calories or nutrients.
The fact that raisins contain more iron per food calorie than grapes do is thus puzzling.

Which one of the following, if true, most helps to explain why raisins contain more iron per calorie than do grapes?

(A) Since grapes are bigger than raisins, it takes several bunches of grapes to provide the same amount of iron as a handful of raisins does.
(B) Caramelized sugar cannot be digested, so its calories do not count toward the food calorie content of raisins.
(C) The body can absorb iron and other nutrients more quickly from grapes than from raisins because of the relatively high water content of grapes.
(D) Raisins, but not grapes, are available year-round, so many people get a greater share of their yearly iron intake from raisins than from grapes.
(E) Raisins are often eaten in combination with other iron-containing foods, while grapes are usually eaten by themselves.

Look at that question stem: a quick scan naturally shows you that you need to explain/resolve a paradox, but the question goes into even more detail for you. It reaffirms the exact nature of the paradox – it’s not about “iron,” but instead that that raisins contain more iron per calorie than grapes do. By adding that extra description into the question stem, the testmaker is practically yelling at you, “Make sure you consider calories…don’t just focus on iron!” And therefore, you should be prepared for the correct answer B, the only one that addresses calories, and deftly avoid answers A, C, D, and E, which all focus only on iron (and do so tangentially to the paradox).

Strategically speaking, if a Critical Reasoning question stem gets overly specific, you should pay particular attention to the specificity there…it’s most likely directing you to the operative portion of the argument.

Overly specific questions are most helpful in Data Sufficiency questions (and that same logic will help on Problem Solving too, as you’ll see). The testmaker knows that you’ve trained your entire algebraic life to solve for individual variables. So how can a question author use that lifetime of repetition against you? By asking you to solve for a specific combination that doesn’t require you to find the individual values. Consider this example, which appears courtesy the Official Guide for GMAT Quantitative Review:

If x^2 + y^2 = 29, what is the value of (x – y)^2?

(1) xy = 10
(2) x = 5

Two major clues should stand out to you that you need to Leverage Assets on this problem. For one, using both statements together (answer choice C) is dead easy. If xy = 10 and x = 5 then y = 2 and you can solve for any combination of x and y that anyone could ever ask for. But secondly and more subtly, the question stem should jump out as a classic way-too-specific, Leverage Assets question stem. They asked for a really, really specific value: (x – y)^2.

Now, immediately upon seeing that specificity you should be thinking, “That’s too specific…there’s probably a way to solve for that exact value without getting x and y individually.” That thought process alone tells you where to spend your time – you want to really leverage Statement 1 to try to make it work alone.

And if you’re still unconvinced, consider what the specificity does: the “squared” portion removes the question of negative vs. positive from the debate, removing one of the most common reasons that a seemingly-sufficient statement just won’t work. And, furthermore, the common quadratic (x – y)^2 shares an awful lot in common with the x^2 and y^2 elsewhere in the question stem. If you expand the parentheses, you have “What is x^2 – 2xy + y^2?” meaning that you’re already 2/3 of the way there (so to speak), since they’ve spotted you the sum x^2 + y^2.

The important strategy here is that the overly-specific question stem should scream “LEVERAGE ASSETS” and “You don’t need to solve for x and y…there’s probably a way to solve directly for that exact combination.” Since you know that you’re solving for the expanded x^2 – 2xy + y^2, and you already know that x^2 + y^2 = 29, you’re really solving for 29 – 2xy. Since you know from Statement 1 that xy = 20, then 29 – 2xy will be 29 – 2(10), which is 9.

Statement 1 alone is sufficient, even though you don’t know what x and y are individually. And one of the major signals that you should recognize to help you get there is the presence of an overly specific question stem.

So remember, in a world of generic questions, the oddly specific question should arouse a bit of suspicion: the interrogator is up to something! On the GMAT, you can use that to your advantage – an overly specific Critical Reasoning question usually tells you exactly which keywords are the most important, and an overly specific Data Sufficiency question stem begs for you to leverage assets and find a way to get the most out of each statement.

By Brian Galvin.

# Don’t Swim Against the Arithmetic Currents on the GMAT Quant Section

When I was a child, I was terrified of riptides. Partially, this was a function of having been raised by unusually neurotic parents who painstakingly instilled this fear in me, and partially this was a function of having inherited a set of genes that seems to have predisposed me towards neuroticism. (The point, of course, is that my parents are to blame for everything. Perhaps there is a better venue for discussing these issues.)

If there’s a benefit to fears, it’s that they serve as potent motivators to find solutions to the troubling predicaments that prompt them. The solution to dealing with riptides is to avoid struggling against the current. The water is more powerful than you are, so a fight is a losing proposition – rather, you want to wait for an opportunity to swim with the current and allow the surf to bring you back to shore. There’s a profound wisdom here that translates to many domains, including the GMAT.

In class, whenever we review a strategy, my students are usually comfortable applying it almost immediately. Their deeper concern is about when to apply the strategy, as they’ll invariably find that different approaches work with different levels of efficacy on different problems. Moreover, even if one has a good strategy in mind, the way the strategy is best applied is often context-dependent. When we’re picking numbers, we can say that x = 2 or x = 100 or x = 10,000; the key is not to go in with a single approach in mind. Put another way, don’t swim against the arithmetic currents.

Let’s look at some questions to see this approach in action:

At a picnic there were 3 times as many adults as children and twice as many women as men. If there was a total of x men, women, and children at the picnic, how many men were there, in terms of x?

A) x/2
B) x/3
C) x/4
D) x/5
E) x/6

The moment we see “x,” we can consider picking numbers. The key here is contemplating how complicated the number should be. Swim with the current – let the question tell you. A quick look at the answer choices reveals that x could be something simple. Ultimately, we’re just dividing this value by 2, 3, 4, 5, or 6.

Keeping this in mind, let’s think about the first line of the question. If there are 3 times as many adults as children, and we’re keeping things simple, we can say that there are 3 adults and 1 child, for a total of 4 people. So, x = 4.

Now, we know that among our 3 adults, there are twice as many women as men. So let’s say there are 2 women and 1 man. Easy enough. In sum, we have 2 women, 1 man, and 1 child at this picnic, and a total of 4 people. The question is how many men are there? There’s just 1! So now we plug x = 4 into the answers and keep going until we find x = 1. Clearly x/4 will work, so C is our answer. The key was to let the question dictate our approach rather than trying to impose an approach on the question.

Let’s try another one:

Last year, sales at Company X were 10% greater in February than in January, 15% less in March than in February, 20% greater in April than in March, 10% less in May than in April, and 5% greater in June than in May. On which month were sales closes to the sales in January?

A) February
B) March
C) April
D) May
E) June

Great, you say. It’s a percent question. So you know that picking 100 is often a good idea. So, let’s say sales in January were 100. If we want the month when sales were closest to January’s level, we want the month when sales were closest to 100, Sales in February were 10% greater, so February sales were 110. (Remember that if sales increase by 10%, we can multiply the original number by 1.1. If they decrease by 10% we could multiply by 0.9, and so forth.)

So far so good. Sales in March were 15% less than in February. Well, if sales in Feb were 110, then the sales in March must be 110*(0.85). Hmm… A little tougher, but not insurmountable. Now, sales in April were 20% greater than they were in March, meaning that April sales would be 110*(0.85)*1.2. Uh oh.  Once you see that sales are 10% less in May than they were in April, we know that sales will be 110*(0.85)*1.2*0.9.

Now you need to stop. Don’t swim against the current. The arithmetic is getting hard and is going to become time-consuming. The question asks which month is closest to 100, so we don’t have to calculate precise values. We can estimate a bit. Let’s double back and try to simplify month by month, keeping things as simple as possible.

Our February sales were simple: 110. March sales were 110*0.85 – an unpleasant number. So, let’s try thinking about this a little differently. 100*0.85 = 85.  10*0.85 = 8.5. Add them together and we get 85 + 8.5 = 93.5.  Let’s make life easier on ourselves – we’ll round up, and call this number 94.

April sales are 20% more than March sales. Well, 20% of 100 is clearly 20, so 20% of 94 will be a little less than that. Say it’s 18. Now sales are up to 94 + 18 = 112. Still not close to 100, so we’ll keep going.

May sales are 10% less than April sales. 10% of 112 is about 11. Subtract 11 from 112, and you get 101. We’re looking for the number closest to 100, so we’ve got our answer – it’s D, May.

Takeaway: Don’t try to impose your will on GMAT questions. Use the structural clues of the problems to dictate how you implement your strategy, and be prepared to adjust midstream. The goal is never to conquer the ocean, but rather, to ride the waves to calmer waters.

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

# How to Simplify Percent Questions on the GMAT

One of the most confounding aspects of the GMAT is its tendency to make simple concepts seem far more complex than they are in reality. Percent questions are an excellent example of this.

When I introduce this topic, I’ll typically start by asking my class the following question: If you’ve completed 10% of a project how much is left to do?  I have never, in all my years of teaching, had a class that was unable to tell me that 90% of the project remains. It’s more likely that they’ll react as though I’m insulting their collective intelligence. And yet, when test-takers see this concept under pressure, they’ll often fail to recognize it.

Take the following question, for example:

Dara ran on a treadmill that had a readout indicating the time remaining in her exercise session. When the readout indicated 24 min 18 sec, she had completed 10% of her exercise session. The readout indicated which of the following when she had completed 40% of her exercise session.

(A) 10 min. 48 sec.
(B) 14 min. 52 sec.
(C) 14 min. 58 sec.
(D) 16 min. 6 sec.
(E) 16 min. 12 sec.

Hopefully, you’ve noticed that this question is testing the same simple concept that I use when introducing percent problems to my class. And yet, in my experience, a solid majority of students are stumped by this problem. The reason, I suspect, is twofold. First, that figure – 24 min. 18 sec. – is decidedly unfriendly. Painful math often lends itself to careless mistakes and can easily trigger a panic response. Second, anxiety causes us to work faster, and when we work faster, we’re often unable to recognize patterns that would be clearer to us if we were calm.

There’s interesting research on this. Psychologists, knowing that the color red prompts an anxiety response and that the color blue has a calming effect, conducted a study in which test-takers had to answer math questions – the questions were given to some subjects on paper with a red background and to other subjects on paper with a blue background. (The control group had questions on standard white paper.) The red anxiety-producing background noticeably lowered scores and the calming blue background boosted scores.

Now, the GMAT doesn’t give you a red background, but it does give you unfriendly-seeming numbers that likely have the same effect. So, this question is as much about psychology as it is about mathematical proficiency. Our job is to take a deep breath or two and rein in our anxiety before we proceed.

If Dara has completed 10% of her workout, we know she has 90% of her workout remaining. So, that 24 min. 18 sec. presents 90% of her total workout. If we designate her total workout time as “t,” we end up with the following equation:

24 min. 18 sec. = 0.90t

Let’s work with fractions to solve. 18 seconds is 18/60 minutes, which simplifies to 3/10 minutes. 0.9 is 9/10, so we can rewrite our equation as:

24 + 3/10 = (9/10)t
(243/10) = (9/10)t
(243/10)*(10/9) = t
27 = t

Not so bad. Dara’s full workout is 27 minutes long.

We want to know how much time is remaining when Dara has completed 40% of her workout. Well, if she’s completed 40% of her workout, we know she has 60% of her workout remaining. If her full workout is 27 minutes, then 60% of this value is 0.60*27 = (3/5)*27 = 81/5 = 16 + 1/5, or 16 minutes 12 seconds. And we’ve got our answer: E.

Now, let’s say you get this problem with 20 seconds remaining on the clock and you simply don’t have time to solve it properly. Let’s estimate.

Say, instead of 24 min 18 seconds remaining, Dara had 24 minutes remaining (so we know we’re going to underestimate the answer). If that’s 90% of her workout time, 24 = (9/10)t, or 240/9 = t.

We want 60% of this, so we want (240/9)*(3/5).

Because 240/5 = 48 and 9/3 = 3, (240/9)*(3/5) = 48/3 = 16.

We know that the correct answer is over 16 minutes and that we’ve significantly underestimated – makes sense to go with E.

Takeaway: Don’t let the question-writer trip you up with figures concocted to make you nervous. Take a breath, and remember that the concepts being tested are the same ones that, when boiled down to their essence, are a breeze when we’re calm.

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

# GMAT Tip of the Week: The Least Helpful Waze To Study

If you drive in a large city, chances are you’re at least familiar with Waze, a navigation app that leverages user data to suggest time-saving routes that avoid traffic and construction and that shave off seconds and minutes with shortcuts on lesser-used streets.

And chances are that you’ve also, at some point or another, been inconvenienced by Waze, whether by a devout user cutting blindly across several lanes to make a suggested turn, by the app requiring you to cut through smaller streets and alleys to save a minute, or by Waze users turning your once-quiet side street into the Talladega Superspeedway.

To its credit, Waze is correcting one of its most common user  that it often leads users into harrowing and time-consuming left turns. But another major concern still looms, and it’s one that could damage both your fender and your chances on the GMAT:

Beware the shortcuts and “crutches” that save you a few seconds, but in doing so completely remove all reasoning and awareness.

With Waze, we’ve all seen it happen: someone so beholden to, “I must turn left on 9th Street because the app told me to!” will often barrel through two lanes of traffic – with no turn signal – to make that turn…not realizing that the trip would have taken the exact same amount of time, with much less risk to the driver and everyone else on the road, had he waited a block or two to safely merge left and turn on 10th or 11th. By focusing so intently on the app’s “don’t worry about paying attention…we’ll tell you when to turn” features, the driver was unaware of other cars and of earlier opportunities to safely make the merge in the desired direction.

The GMAT offers similar pitfalls when examinees rely too heavily on “turn your brain off” tricks and techniques. As you learn and practice them, strategies like the “plumber butt” for rates and averages may seem quick, easy, and “turn your brain off” painless. But the last thing you want to do on a higher-order thinking test like the GMAT is completely turn your brain off. For example, a “turn your brain off” rate problem might say:

John drives at an average rate of 45 miles per hour. How many miles will he drive in 2.5 hours?

And using a Waze-style crutch, you could remember that to get distance you multiply time by rate so you’d get 112.5 miles. That may be a few seconds faster than performing the algebra by thinking “Rate = Distance over Time”; 45 = D/2.5; 45(2.5) = D; D = 112.5.

But where a shortcut crutch saves you time on easier problems, it can leave you helpless on longer problems that are designed to make you think. Consider this Data Sufficiency example:

A factory has three types of machines – A, B, and C – each of which works at its own constant rate. How many widgets could one machine A, one Machine B, and one Machine C produce in one 8-hour day?

(1) 7 Machine As and 11 Machine Bs can produce 250 widgets per hour

(2) 8 Machine As and 22 Machine Cs can produce 600 widgets per hour

Here, simply trying to plug the information into a simple diagram will lead you directly to choice E. You simply cannot separate the rate of A from the rate of B, or the rate of B from the rate of C. It will not fit into the classic “rate pie / plumber’s butt” diagram that many test-takers use as their “I hate rates so I’ll just do this trick instead” crutch.

However, those who have their critical thinking mind turned on will notice two things: that choice E is kind of obvious (the algebra doesn’t get you very close to solving for any one machine’s rate) so it’s worth pressing the issue for the “reward” answer of C, and that if you simply arrange the algebra there are similarities between the number of B and of C:

7(Rate A) + 11(Rate B) = 250
8(Rate A) + 22(Rate C) = 600

Since 11 is half of 22, one way to play with this is to double the first equation so that you at least have the same number of Bs as Cs (and remember…those are the only two machines that you don’t have “together” in either statement, so relating one to the other may help). If you do, then you have:

14(A) + 22(B) = 500
8(A) + 22(C) = 600

Then if you sum the questions (Where does the third 22 come from? Oh, 14 + 8, the coefficients for A.), you have:

22A + 22B + 22C = 1100

So, A + B + C = 50, and now you know the rate for one of each machine. The two statements together are sufficient, but the road to get there comes from awareness and algebra, not from reliance on a trick designed to make easy problems even easier.

The lesson? Much like Waze, which can lead to lack-of-awareness accidents and to shortcuts that dramatically up the degree of difficulty for a minimal time savings, you should take caution when deciding to memorize and rely upon a knee-jerk trick in your GMAT preparation.

Many are willing (or just unaware that this is the decision) to sacrifice mindfulness and awareness to save 10 seconds here or there, but then fall for trap answers because they weren’t paying attention or become lost when problems are more involved because they weren’t prepared.

So, be choosy in the tricks and shortcuts you decide to adopt! If a shortcut saves you a or two of calculations, it’s worth the time it takes to learn and master it (but probably never worth completely avoiding the “long way” or knowing the general concept). But if its time savings are minimal and its grand reward is that, “Hey, you don’t have to understand math to do this!” you should be wary of how well it will serve your aspirations of scores above around 600.

Don’t let these slick shortcut waze of avoiding math drive you straight into an accident. Unless the time savings are game-changing, you shouldn’t make a trade that gains you a few seconds of efficiency on select, easier problems in exchange for your awareness and understanding.

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, YouTubeGoogle+ and Twitter!

By Brian Galvin.

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

Happy 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.

By Brian Galvin.

# Quarter Wit, Quarter Wisdom: Squares and Square Roots on the GMAT

In today’s post, we will try to clear up your doubts regarding positive and negative solutions in the case of squares and square roots. We will explain the reasons behind each case, which will help you recall the fundamentals when you need to use them. While preparing for the GMAT, you have probably come across a discussion that says x^2 = 4 has two roots, 2 and -2, while √4 has only one value, 2.

Now, let’s try to understand why this is so:

1) x^2 = 4
Basic algebra tells us that quadratics have two roots. Here, x can be either 2 or -2; each, when squared, will give you 4.

x^2 – 4 = 0 and (x + 2)*(x – 2) = 0 when x equals -2 or 2.

2) √x is positive, only
Now this is odd, right? √4 must be 2. Why is that? Shouldn’t it be 2 or -2. After all, when we square both 2 and -2, we get 4 (as discussed above). So, √4 should be 2 or -2.

Here is the concept: √x denotes only the principal square root. x has two square roots – the positive square root (or principal square root) written as √x and the negative square root written as -√x. Therefore, when you take the square root of 4, you get two roots: √4 and -√4, which  is 2 and -2 respectively.

On a GMAT question, when you see √x, this is specifically referring to the positive square root of the number. So √4 is 2, only.

3) (√x)^2 = x
This is fairly straightforward – since x has a square root, it must be non-negative. When you square it, just the square root sign vanishes and you are left with x.

4) √(x^2) = |x|
Now this isn’t intuitive either. √(x^2) should simply be x – why do we have absolute value of x, then? Again, this has to do with the principal square root concept. First you will square x, and then when you write √, it is by default just the principal square root. The negative square root will be written as -√(x^2). So, irrespective of whether x was positive or negative initially, √(x^2) will definitely be positive x. Therefore, we will need to take the absolute value of x.

Here’s a quick recap with some examples:

• √9 = 3
• x^2 = 16 means x is either 4 or -4
• √(5^2) = 5
• √(-5^2) = 5
• (√16)^2 = 16
• √100 = 10

To see this concept in action, let’s take a look at a very simple official problem:

If x is not 0, then √(x^2)/x =

(A) -1
(B) 0
(C) 1
(D) x
(E) |x|/x

We know that √(x^2) is not simply x, but rather |x|. So, √(x^2)/x = |x|/x.

Depending on whether x is positive or negative, |x|/x will be 1 or -1 – we can’t say which one. Hence, there is no further simplification that we can do, and our answer must be E.

Now that you are all warmed up, let’s examine a higher-level question:

Is √[(x – 3)^2] = (3 – x)?

Statement 1: x is not 3
Statement 2: -x * |x| > 0

We know that √(x^2) = |x|, so √[(x – 3)^2] = |x – 3|.

This means that our question is basically:

Is |x – 3| = 3 – x?

Note that 3 – x can also be written as -(x – 3).

Is |x – 3| = -(x – 3)?

Recall the definition of absolute values: |a| = a if a is greater than or equal to 0, and -a if a < 0.

So, “Is |x – 3| = -(x – 3)?” depends on whether (x – 3) is positive or negative. If (x – 3) is negative (or 0), then |x – 3| is equal to -(x – 3).

So our question now boils down to:

Is (x – 3) negative (or 0)?

Statement 1: x is not 3

This means we know that (x – 3) is not 0, but we still don’t know whether it is negative or positive. This statement is not sufficient.

Statement 2: -x * |x| > 0

|x| is always non-negative, so for the product to be positive, “-x” must also be positive. This means x must be negative. If x is negative, x – 3 must be negative, too.

If (x – 3) is negative, |x – 3| is equal to -(x – 3). Hence, this statement alone is sufficient, and our answer is 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!

# GMAT Rate Questions: Tackling Problems with Multiple Components

A few posts ago, I tackled rate/work questions, which are invariably a source of consternation for GMAT test-takers. On the latest official practice tests that GMAC has released, these questions showed up with surprising frequency, so I thought it might be worthwhile to tackle a challenging incarnation of this question type: one in which a single machine begins a project and then multiple machines complete the partially-finished work.

To review, the key for dealing with this type of question is to apply the following rules:

1. Rate * Time = Work
2. Rates are additive in work questions.
3. Rate and time have a reciprocal relationship.

For the questions involving partially completed jobs, we’ll throw in the addendum that a completed job can be designated as “1”’

And that’s it!

Here’s a question I saw on my recent practice test:

Working alone at its constant rate, pump X pumped out ¼ of the water in a tank in 2 hours. Then pumps Y and Z started working and the three pumps, working simultaneously at their respective constant rates, pumped out the rest of the water in 3 hours. If pump Y, working alone at its constant rate, would have taken 18 hours to pump out the rest of the water, how many hours would it have taken pump Z, working alone at its constant rate, to pump out all of the water that was pumped out of the tank?

A) 6
B) 12
C) 15
D) 18
E) 24

Okay, deep breath. Recall our three aforementioned rules. Next, let’s designate the rates for the pumps as x, y, and z, respectively.

If pump x can pump out ¼ of the water in 2 hours, then it would take 4*2 = 8 hours to pump out all the water alone. If pump x can complete 1 tank in 8 hours, then x = 1/8.

If x removes ¼ of the water on its own, then all three pumps working together have to remove the ¾ of the water left in the tank. We’re told that together they can do this in 3 hours. If x, y, and z together can do ¾ of the work in 3 hours, then x + y + z = (¾)/3 = 3/12 = ¼.

We’re told that y, alone, could have pumped out the rest of the water in 18 hours – again, there was ¾ of a tank left, so y = (¾)/18 = 1/24.

To summarize, we know that x = 1/8, y = 1/24, and x + y + z = ¼;  Not so hard to solve for z, right?

1/8 + 1/24 + z = ¼

Multiply everything by 24, and we get:

3 + 1 + 24z = 6

24z = 2

z = 1/12.

That’s z’s rate. If rate and time have a reciprocal relationship, we know that it would take z 12 hours to pump out all the water of one tank alone. The answer is, therefore, B.

Takeaway: The joy of seeing new material from GMAC (Is joy the right word?) is the realization that no matter how many additional layers of complexity the question-writers throw at us, the old verities hold true. So when you see tough questions, slow down. Remind yourself that the strategies you’ve cultivated will unlock even the toughest problems. Then, dive in and discover, yet again, that these questions are never quite as hard as they appear at first glance.

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

# Avoid Obtaining the Wrong Values in Percent Increase Questions

Many test takers make mistakes in percent increase quantitative GMAT questions, not because they do not understand the principle of percent increase, but rather, because they don’t evaluate the correct values.

A quick recap: percent increase questions can be identified (often literally) by the words “percent increase,” and tend to be word problems that don’t read in the most straightforward manner. The first step to take when working towards answering these questions is to be cautious and evaluate them carefully.

The second step is to, of course, use the percent increase formula – (new value – initial value) / (initial value) x 100%.

Let’s start by going through a sample GMAT practice problem:

In 2005, 25 percent of the math department’s 40 students were female, and in 2007, 40 percent of the math department’s 65 students were female. What was the percent increase from 2005 to 2007 in the number of female students in the department?

A) 15%
B) 50%
C) 62.5%
D) 115%
E) 160%

At first can be difficult to determine what the answer is for this question, but keep in mind that the best place to start looking is in the last sentence and/or the actual question that is posed. In this case, the new value is the number of female students in 2007, “the number of female students in the department?”

By working backwards through this problem, we would take  40% of 65 (our final value), which we can easily calculate as 0.4*65 (or 2/5*65), giving us a total of 26 students in 2007.

Our initial value must then be the number of female students in 2005, which we can get by calculating 25% of 40. 0.25*40 (or 1/4*40) leaves us with a total of 10 female students in 2005.

Breaking up the question up into smaller, more manageable chunks gives us the ability to plug 26 and 10 into the percent increase formula – (26‐10)/10 = 16/10 = 1.6 = 160%. Therefore, the correct answer is E.

This strategy of not trying to figure out the conclusion without evaluating all the separate parts of the question is important to tackle percent change GMAT problems, but can be applied across a variety of quantitative questions. Understanding that these questions can be much more manageable, and are more about strategy versus understanding complex math concepts, is the key to success on the Quantitative Section.

By Ashley Triscuit, a Veritas Prep GMAT instructor based in Boston.

# How to Simplify Sequences on the GMAT

The GMAT loves sequence questions. Test-takers, not surprisingly, do not feel the same level of affection for this topic. In some ways, it’s a peculiar reaction. A sequence is really just a set of numbers. It may be infinite, it may be finite, but it’s this very open-endedness, this dizzying level of fuzzy abstraction, that can make sequences so difficult to mentally corral.

If you are one of the many people who fear and dislike sequences, your main consolation should come from the fact that the main weapon in the question writer’s arsenal is the very fear these questions might elicit. And if you have been a reader of this blog for any length of time, you know that the best way to combat this anxiety is to dive in and convert abstractions into something concrete, either by listing out some portion of the sequence, or by using the answer choices and working backwards.

Take this question for example:

For a certain set of numbers, if x is in the set, then x – 3 is also in the set. If the number 1 is in the set, which of the following must also be in the set?

I. 4
II. -1
III. -5

A) I only
B) II only
C) III only
D) I and II
E) II and III

Okay, so let’s list out the elements in this set. We know that 1 is in the set. If x= 1, then x – 3 = -2. So -2 is in the set. If x = -2 is in the set, then x – 3 = -5. So -5 is in the set.

By this point, the pattern should be clear: each term is three less than the previous term, giving us a sequence that looks like this: 1, -2, -5, -8, -11….

So we look at our options, and see we that only III is true. And we’re done. That’s it. The answer is C.

Sure, Dave, you may say. That is much easier than any question I’m going to see on the GMAT. First, this is an official question, so I’m not sure where you’re getting the idea that you’d never see a question like this. Second, you’d be surprised by how many test-takers get this wrong.

There is the temptation to assume that if 1 is in the set, then 4 must also be in the set. And note that this is, in fact, a possibility. If x = 4, then x – 3 = 1. But the question asks us what “must be” in the set. So it’s possible that 4 is in our set. But it’s also possible our set begins with 1, in which case 4 would not be included. This little wrinkle is enough to generate a substantial number of incorrect responses.

Still, surely the questions get harder than this. Well, yes. They do. So what are you waiting for? I’m not sure where this testy impatience is coming from, but if you insist:

The sequence a1, a2, a3, . . , an of n integers is such that ak = k if k is odd and ak = -ak-1 if k is even. Is the sum of the terms in the sequence positive?

1) n is odd

2) an is positive

Yikes! Hey, you asked for a harder one. This question looks far more complicated than the previous one, but we can attack it the same way. Let’s establish our sequence:

a1 is the first term in the sequence. We’re told that ak = k if k is odd. Well, 1 is odd, so now we know that a1 = 1. So far so good.

a2 is the second term in the sequence. We’re told that ak = -ak-1 if k is even. 2 is even, so a2 = -a2-1 , meaning that a2 = -a1. Well, we know that a1 = 1, so if a2 = -a1 then a2 = -1.

So, here’s our sequence so far: 1, -1…

Let’s keep going.

a3 is the third term in the sequence. Remember that ak = k if k is odd. 3 is odd, so now we know that a3 = 3.

a4 is the fourth term in the sequence. Remember that ak = -ak-1 if k is even. 4 is even, so a4 = -a4-1 , meaning that a4 = -a3We know that a3 = 3, so if a4 = -a3 then a4 = -3.

Now our sequence looks like this: 1, -1, 3, -3…

By this point we should see the pattern. Every odd term is a positive number that is dictated by its place in the sequence (the first term = 1, the third term = 3, etc.) and every even term is simply the previous term multiplied by -1.

After one term, we have 1.

After two terms, we have 1 + (-1) = 0.

After three terms, we have 1 + (-1) + 3 = 3.

After four terms, we have 1 + (-1) + 3 + (-3) = 0.

Notice the trend: after every odd term, the sum is positive. After every even term, the sum is 0.

So the initial question, “Is the sum of the terms in the sequence positive?” can be rephrased as, “Are there an ODD number of terms in the sequence?”

Now to the statements. Statement 1 tells us that there are an odd number of terms in the sequence. That clearly answers our rephrased question, because if there are an odd number of terms, the sum will be positive. This is sufficient.

Statement 2 tells us that an is positive. an is the last term in the sequence. If that term is positive, then, according to the pattern we’ve established, that term must be odd, meaning that the sum of the sequence is positive. This is also sufficient. And the answer is D, either statement alone is sufficient to answer the question.

Takeaway: sequence questions are nothing to fear. Like everything else on the GMAT, the main obstacle we need to overcome is the self-fulfilling prophesy that we don’t know how to proceed, when, in fact, all we need to do is simplify things a bit.

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

# Solving GMAT Standard Deviation Problems By Using as Little Math as Possible

The other night I taught our Statistics lesson, and when we got to the section of class that deals with standard deviation, there was a familiar collective groan – not unlike the groan one encounters when doing compound interest, or any mathematical concept that, when we learned it in school, involved an intimidating-looking formula.

So, I think it’s time for me to coin an axiom: the more painful the traditional formula associated with a given topic, the simpler the actual calculations will be on the GMAT. (Please note, though the axiom is awaiting official mathematical verification by Veritas’ hard-working team of data scientists, the anecdotal evidence in support of the axiom is overwhelming.)

So, let’s talk standard deviation. If you’re like my students, your first thought is to start assembling a list of increasingly frantic questions: Do we need to know that horrible formula I learned in Stats class? (No.) Do we need to know the relationship between variance and Standard deviation? (You just need to know that there is a relationship, and that if you can solve for one, you can solve for the other.) Etc.

So, rather than droning on about what we don’t need to know, let’s boil down what we do need to know about standard deviation. The good news – it isn’t much. Just make sure you’ve internalized the following:

• The standard deviation is a measure of the dispersion the elements of the set around mean. The farther away the terms are from the mean, the larger the standard deviation.
• If we were to increase or decrease each element of the set by “x,” the standard deviation would remain unchanged.
• If we were to multiply each element of the set by “x,” the standard deviation would also be multiplied by “x.”
• If the mean of a set is “m” and the standard deviation is “d,” then to say that something is within 3 standard deviations of a set is to say that it falls within the interval of (m – 3d) to (m + 3d.) And to say that something is within 2 standard deviations of the mean is to say that it falls within the interval of (m – 2d) to (m + 2d.)

That’s basically it. Not anything to get too worked up about. So, let’s see some of these principles in action to substantiate the claim that we won’t have to do too much arithmetical grinding on these types of questions:

If d is the standard deviation of x, y, z, what is the standard deviation of x+5, y+5, z+5 ?

A) d
B) 3d
C) 15d
D) d+5
E) d+15

If our initial set is x, y, z, and our new set is x+5, y+5, and z+5, then we’re adding the same value to each element of the set. We already know that adding the same value to each element of the set does not change the standard deviation. Therefore, if the initial standard deviation was d, the new standard deviation is also d. We’re done – the answer is A. (You can see this with a simple example. If your initial set is {1, 2, 3} and your new set is {6, 7, 8} the dispersion of the set clearly hasn’t changed.)

Surely the questions get harder than this, you say. They do, but if you know the aforementioned core concepts, they’re all quite manageable. Here’s another one:

Some water was removed from each of 6 tanks. If standard deviation of the volumes of water at the beginning was 10 gallons, what was the standard deviation of the volumes at the end?

1) For each tank, 30% of water at the beginning was removed
2) The average volume of water in the tanks at the end was 63 gallons

We know the initial standard deviation. We want to know if it’s possible to determine the new standard deviation after water is removed. To the statements we go!

Statement 1: If 30% of the water is removed from each tank, we know that each term in the set is multiplied by the same value: 0.7. Well, if each term in a set is multiplied by 0.7, then the standard deviation of the set is also multiplied by 0.7. If the initial standard deviation was 10 gallons, then the new standard deviation would be 10*(0.7) = 7 gallons. And we don’t even need to do the math – it’s enough to see that it’s possible to calculate this number. Therefore, Statement 1 alone is sufficient.

Statement 2: Knowing the average of a set is not going to tell us very much about the dispersion of the set. To see why, imagine a simple case in which we have two tanks, and the average volume of water in the tanks is 63 gallons. It’s possible that each tank has exactly 63 gallons and, if so, the standard deviation would be 0, as everything would equal the mean. It’s also possible to have one tank that had 126 gallons and another tank that was empty, creating a standard deviation that would, of course, be significantly greater than 0. So, simply knowing the average cannot possibly give us our standard deviation. Statement 2 alone is not sufficient to answer the question.

Maybe at this point you’re itching for more of a challenge. Let’s look at a slightly tougher one:

7.51; 8.22; 7.86; 8.36
8.09; 7.83; 8.30; 8.01
7.73; 8.25; 7.96; 8.53

A vending machine is designed to dispense 8 ounces of coffee into a cup. After a test that recorded the number of ounces of coffee in each of 1000 cups dispensed by the vending machine, the 12 listed amounts, in ounces, were selected from the data above. If the 1000 recorded amounts have a mean of 8.1 ounces and a standard deviation of 0.3 ounces, how many of the 12 listed amounts are within 1.5 standard deviations of the mean?

A)Four
B) Six
C) Nine
D) Ten
E) Eleven

Okay, so the standard deviation is 0.3 ounces. We want the values that are within 1.5 standard deviations of the mean. 1.5 standard deviations would be (1.5)(0.3) = 0.45 ounces, so we want all of the values that are within 0.45 ounces of the mean. If the mean is 8.1 ounces, this means that we want everything that falls between a lower bound of (8.1 – 0.45) and an upper bound of (8.1 + 4.5). Put another way, we want the number of values that fall between 8.1 – 0.45 = 7.65 and 8.1 + 0.45 = 8.55.

Looking at our 12 values, we can see that only one value, 7.51, falls outside of this range. If we have 12 total values and only 1 falls outside the range, then the other 11 are clearly within the range, so the answer is E.

As you can see, there’s very little math involved, even on the more difficult questions.

Takeaway: remember the axiom that the more complex-looking the formula is for a concept, the simpler the calculations are likely to be on the GMAT. An intuitive understanding of a topic will always go a lot further on this test than any amount of arithmetical virtuosity.

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

# How to Avoid Trap Answers On GMAT Data Sufficiency Questions

When I’m not teaching GMAT classes or writing posts for our fine blog, I am, unfortunately, writing fiction. Anyone who has taken a stab at writing fiction knows that it’s hard, and because it’s hard, it is awfully tempting to steer away from pain and follow the path of least resistance.

This tendency can manifest itself in any number of ways. Sometimes it means producing a cliché rather than straining for a more precise and original way to render a scene. More often, it means procrastinating – cleaning my desk or refreshing espn.com for the 700th time – rather than doing any writing at all. The point is that my brain is often groping for an easy way out. This is how we’re all wired; it’s a dangerous instinct, both in writing and on the GMAT.

This problem is most acute on Data Sufficiency questions. Most test-takers like to go on auto-pilot when they can, relying on simple rules and heuristics rather than proving things to themselves – if I have the slope of a line and one point on that line, I know every point on that line; if I have two linear equations and two variables I can solve for both variables, etc.

This is not in and of itself a problem, but if you find your brain shifting into path-of-least-resistance mode and thinking that you’ve identified an answer to a question within a few seconds, be very suspicious about your mode of reasoning. This is not to say that you should simply assume that you’re wrong, but rather to encourage you to try to prove that you’re right.

Here’s a classic example of a GMAT Data Sufficiency question that appears to be easier than it is:

Joanna bought only \$.15 stamps and \$.29 stamps. How many \$.15 stamps did she buy?

1) She bought \$4.40 worth of stamps
2) She bought an equal number of \$.15 stamps and \$.29 stamps

Here’s how the path-of-least-resistance part of my brain wants to evaluate this question. Okay, for Statement 1, there could obviously be lots of scenarios. If I call “F” the number of 15 cent stamps and “T” the number of 29 cent stamps, all I know is that .15F + .29T = 4.40. So that statement is not sufficient. Statement 2 is just telling me that F = T. Clearly no good – any number could work. And together, I have two unique linear equations and two unknowns, so I have sufficiency and the answer is C.

This line of thinking only takes a few seconds, and just as I need to fight the urge to take a break from writing to watch YouTube clips of Last Week Tonight with John Oliver because it’s part of my novel “research,” I need to fight the urge to assume that such a simple line of reasoning will definitely lead me to the correct answer to this question.

So let’s rethink this. I know for sure that the answer cannot be E – if I can solve for the unknowns when I’m testing the statements together, I clearly have sufficiency there. And I know for sure that the answer cannot be that Statement 2 alone is sufficient. If F = T, there are an infinite number of values that will work.

So, let’s go back to Statement 1. I know that I cannot purchase a fraction of a stamp, so both F and T must be integer values. That’s interesting. I also know that the total amount spent on stamps is \$4.40, or 440 cents, which has a units digit of 0. When I’m buying 15-cent stamps, I can spend 15 cents if I buy 1 stamp, 30 cents if I buy two, etc.

Notice that however many I buy, the units digit must either be 5 or 0. This means that the units digit for the amount I spend on 29 cent stamps must also be 5 or 0, otherwise, there’d be no way to get the 0 units digit I get in 440. The only way to get a units digit of 5 or 0 when I’m multiplying by 29 is if the other number ends in 5 or 0 . In other words, the number of 29-cent stamps I buy will have to be a multiple of 5 so that the amount I spend on 29-cent stamps will end in 5 or 0.

Here’s the sample space of how much I could have spent on 29-cent stamps:

Five stamps: 5*29 = 145 cents
Ten stamps: 10*29 = 290 cents
Fifteen stamps: 15* 29 = 435 cents

Any more than fifteen 29-cent stamps and I ‘m over 440, so these are the only possible options when testing the first statement.

Let’s evaluate: say I buy five 29-cent stamps and spend 145 cents. That will leave me with 440 – 145 = 295 cents left for the 15-cent stamps to cover. But I can’t spend exactly 295 cents by purchasing 15-cent stamps, because 295 is not a multiple of 15.

Say I buy ten 29-cent stamps, spending 290 cents. That leaves 440 – 290 = 150. Ten 15-cent stamps will get me there, so this is a possibility.

Say I buy fifteen 29-cent stamps, spending 435 cents. That leaves 440 – 435 = 5. Clearly that’s not possible to cover with 15-cent stamps.

Only one option works: ten 29-cent stamps and ten 15-cent stamps. Because there’s only one possibility, Statement 1 alone is sufficient, and the answer here is actually A.

Takeaway: Don’t take the GMAT the way I write fiction. Following the path of least-resistance will often lead you right into the trap the question writer has set for unsuspecting test-takers. If something feels too easy on a Data Sufficiency, it probably is.

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

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

# Quarter Wit, Quarter Wisdom: Using the Standard Deviation Formula on the GMAT

We have discussed standard deviation (SD). We know what the formula is for finding the standard deviation of a set of numbers, but we also know that GMAT will not ask us to actually calculate the standard deviation because the calculations involved would be way too cumbersome. It is still a good idea to know this formula, though, as it will help us compare standard deviations across various sets – a concept we should know well.

Today, we will look at some GMAT questions that involve sets with similar standard deviations such that it is hard to tell which will have a higher SD without properly understanding the way it is calculated. Take a look at the following question:

Which of the following distribution of numbers has the greatest standard deviation?

(A) {-3, 1, 2}
(B) {-2, -1, 1, 2}
(C) {3, 5, 7}
(D) {-1, 2, 3, 4}
(E) {0, 2, 4}

At first glance, these sets all look very similar. If we try to plot them on a number line, we will see that they also have similar distributions, so it is hard to say which will have a higher SD than the others. Let’s quickly review their deviations from the arithmetic means:

For answer choice A, the mean = 0 and the deviations are 3, 1, 2
For answer choice B, the mean = 0 and the deviations are 2, 1, 1, 2
For answer choice C, the mean = 5 and the deviations are 2, 0, 2
For answer choice D, the mean = 2 and the deviations are 3, 0, 1, 2
For answer choice E, the mean = 2 and the deviations are 2, 0, 2

We don’t need to worry about the arithmetic means (they just help us calculate the deviation of each element from the mean); our focus should be on the deviations. The SD formula squares the individual deviations and then adds them, then the sum is divided by the number of elements and finally, we find the square root of the whole term. So if a deviation is greater, its square will be even greater and that will increase the SD.

If the deviation increases and the number of elements increases, too, then we cannot be sure what the final effect will be – an increased deviation increases the SD but an increase in the number of elements increases the denominator and hence, actually decreases the SD. The overall effect as to whether the SD increases or decreases will vary from case to case.

First, we should note that answers C and E have identical deviations and numbers of elements, hence, their SDs will be identical. This means the answer is certainly not C or E, since Problem Solving questions have a single correct answer.

Let’s move on to the other three options:

For answer choice A, the mean = 0 and the deviations are 3, 1, 2
For answer choice B, the mean = 0 and the deviations are 2, 1, 1, 2
For answer choice D, the mean = 2 and the deviations are 3, 0, 1, 2

Comparing answer choices A and D, we see that they both have the same deviations, but D has more elements. This means its denominator will be greater, and therefore, the SD of answer D is smaller than the SD of answer A. This leaves us with options A and B:

For answer choice A, the mean = 0 and the deviations are 3, 1, 2
For answer choice B, the mean = 0 and the deviations are 2, 1, 1, 2

Now notice that although two deviations of answers A and B are the same, answer choice A has a higher deviation of 3 but fewer elements than answer choice B. This means the SD of A will be higher than the SD of B, so the SD of A will be the highest. Hence, our answer must be A.

Let’s try another one:

Which of the following data sets has the third largest standard deviation?

(A) {1, 2, 3, 4, 5}
(B) {2, 3, 3, 3, 4}
(C) {2, 2, 2, 4, 5}
(D) {0, 2, 3, 4, 6}
(E) {-1, 1, 3, 5, 7}

How would you answer this question without calculating the SDs? We need to arrange the sets in increasing SD order. Upon careful examination, you will see that the number of elements in each set is the same, and the mean of each set is 3.

Deviations of answer choice A: 2, 1, 0, 1, 2
Deviations of answer choice B: 1, 0, 0, 0, 1 (lowest SD)
Deviations of answer choice C: 1, 1, 1, 1, 2
Deviations of answer choice D: 3, 1, 0, 1, 3
Deviations of answer choice E: 4, 2, 0, 2, 4 (highest SD)

Obviously, option B has the lowest SD (the deviations are the smallest) and option E has the highest SD (the deviations are the greatest). This means we can automatically rule these answers out, as they cannot have the third largest SD.

Deviations of answer choice A: 2, 1, 0, 1, 2
Deviations of answer choice C: 1, 1, 1, 1, 2
Deviations of answer choice D: 3, 1, 0, 1, 3

Out of these options, answer choice D has a higher SD than answer choice A, since it has higher deviations of two 3s (whereas A has deviations of two 2s). Also, C is more tightly packed than A, with four deviations of 1. If you are not sure why, consider this:

The square of deviations for C will be 1 + 1+ 1 + 1  + 4 = 8
The square of deviations for A will be 4 + 1 + 0 + 1 + 4 = 10

So, A will have a higher SD than C but a lower SD than D. Arranging from lowest to highest SD’s, we get: B, C, A, D, E. Answer choice A has the third highest SD, and therefore, A is our answer

Although we didn’t need to calculate the actual SD, we used the concepts of the standard deviation formula to answer these questions.

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!

# Breaking Down Changes in the New Official GMAT Practice Tests: Unit Conversions in Shapes

Recently, GMAC released two more official practice tests. Though the GMAT is not going to test completely new concepts – if the test changed from year to year, it wouldn’t really be standardized – we can get a sense of what types of questions are more likely to be emphasized by noting how official materials change over time. I thought it might be interesting to take these practice tests and break down down any conspicuous trends I detected.

In the Quant section of the first new test, there was one type of question that I’d rarely encountered in the past, but saw multiple times within a span of 20 problems. It involves unit conversions in two or three-dimensional shapes.

Like many GMAT topics, this concept isn’t difficult so much as it is tricky, lending itself to careless mistakes if we work too fast. If I were to draw a line that was one foot long, and I asked you how many inches it was, you wouldn’t have to think very hard to recognize that it would be 12 inches.

But what if I drew a box that had an area of 1 square foot, and I asked you how many square inches it was? If you’re on autopilot, you might think that’s easy. It’s 12 square inches. And you better believe that on the GMAT, that would be a trap answer. To see why it’s wrong, consider a picture of our square:

We see that each side is 1 foot in length. If each side is 1 foot in length, we can convert each side to 12 inches in length. Now we have the following:

Clearly, the area of this shape isn’t 12 square inches, it’s 144 square inches: 12 inches * 12 inches = 144 inches^2.

Another way to think about it is to put the unit conversion into equation form. We know that 1 foot = 12 inches, so if we wanted the unit conversion from feet^2 to inches^2, we’d have to square both sides of the equation in order to have the appropriate units. Now (1 foot)^2 = (12 inches)^2, or 1 foot^2 = 144 inches^2.  So converting from square feet to square inches requires multiplying by a factor of 144, not 12.

Let’s see this concept in action. (I’m using an older official question to illustrate – I don’t want to rob anyone of the joy of encountering the recently released questions with a fresh pair of eyes.)

If a rectangular room measures 10 meters by 6 meters by 4 meters, what is the volume of the room in cubic centimeters? (1 meter = 100 centimeters)

A) 24,000
B) 240,000
C) 2,400,000
D) 24,000,000
E) 240,000,000

First, we can find the volume of the room by multiplying the dimensions together: 10*6*4 = 240 cubic meters. Now we want to avoid the trap of thinking, “Okay, 100 centimeters is 1 meter, so 240 cubic meters is 240*100 = 24,000 cubic centimeters.”  Remember, the conversion ratio we’re given is for converting meters to centimeters – if we’re dealing with 240 cubic meters, or 240 meters^3, and we want to find the volume in cubic centimeters, we’ll need to adjust our conversion ratio accordingly.

If 1 meter = 100 centimeters, then (1 meter)^3 = (100 centimeters)^3, and 1 meter^3 = 1,000,000 centimeters^3. [100 = 10^2 and (10^2)^3 = 10^6, or 1,000,000.] So if 1 cubic meter = 1,000,000 cubic centimeters, then 240 cubic meters = 240*1,000,000 cubic centimeters, or 240,000,000 cubic centimeters, and our answer is E.

Alternatively, we can do all of our conversions when we’re given the initial dimensions. 10 meters = 1000 centimeters. 6 meters = 600 centimeters. 4 meters  = 400 centimeters. 1000 cm * 600 cm * 400 cm = 240,000,000 cm^3. (Notice that when we multiply 1000*600*400, we can simply count the zeroes. There are 7 total, so we know there will be 7 zeroes in the correct answer, E.)

Takeaway: Make sure you’re able to do unit conversions fluently, and that if you’re dealing with two or three-dimensional space, that you adjust your conversion ratios accordingly. If you’re dealing with a two-dimensional shape, you’ll need to square your initial ratio. If you’re dealing with a three-dimensional shape, you’ll need to cube your initial ratio. The GMAT is just as much about learning what traps to avoid as it is about relearning the elementary math that we’ve long forgotten.

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

# Quarter Wit, Quarter Wisdom: An Innovative Use of the Slope of a Line on the GMAT

Let’s continue our discussion on coordinate geometry today.

The concept of slope is extremely important on the GMAT – it is not sufficient to just know how to calculate it using (y2 – y1)/(x2 – x1).

In simple terms, the slope of a line specifies the units by which the y-coordinate changes and the direction in which it changes with each 1 unit increase in the x-coordinate. If the slope (m) is positive, the y-coordinate changes in the same direction as the x-coordinate. If m is negative, however, the y-coordinate changes in the opposite direction.

For example, if the slope of a line is 2, it means that every time the x-coordinate increases by 1 unit, the y-coordinate increases by 2 units. So if the point (3, 5) lies on a line with a slope of 2, the point (4, 7) will also lie on it. Here, when the x-coordinate increases from 3 to 4, the y-coordinate increases from 5 to 7 (by an increase of 2 units). Similarly,  the point (2, 3) will also lie on this same line – if the x-coordinate decreases by 1 unit (from 3 to 2), the y-coordinate will decrease by 2 units (from 5 to 3). Since the slope is positive, the direction of change of the x-coordinate will be the same as the direction of change of the y-coordinate.

Now, if we have a line where the slope is -2 and the point (3, 5) lies on it, when the x-coordinate increases by 1 unit, the y-coordinate DECREASES by 2 units – the point (4, 3) will also lie on this line. Similarly, if the x-coordinate decreases by 1 unit, the y-coordinate will increase by 2 units. So, for example, the point (2, 7) will also lie on this line.

This understanding of the concept of slope can be very helpful, as we will see in this GMAT question:

Line L and line K have slopes -2 and 1/2 respectively. If line L and line K intersect at (6,8), what is the distance between the x-intercept of line L and the y-intercept of line K?

(A) 5
(B) 10
(C) 5√(5)
(D) 15
(E) 10√(5)

Traditionally, one would solve this question like this:

The equation of a line with slope m and constant c is given as y = mx + c. Therefore, the equations of lines L and K would be:

Line L: y = (-2)x + a
and
Line K: y = (1/2)x + b

As both these lines pass through (6,8), we would substitute x=6 and y=8 to get the values of a and b.

Line L: 8 = (-2)*6 + a
a = 20

Line K: 8 = (1/2)*6 + b
b = 5

Thus, the equations of the 2 lines become:

Line L: y = (-2)x + 20
and
Line K: y = (1/2)x + 5

The x-intercept of a line is given by the point where y = 0. So, the x-intercept of line L is given by:

0 = (-2)x + 20
x = 10

This means line L intersects the x-axis at the point (10, 0).

Similarly, the y-intercept of a line is given by the point where x = 0. So, y-intercept of line K is given by:

y = (1/2)*0 + 5
y = 5

This means that line K intersects the y-axis at the point (0, 5).

Looking back at our original question, the distance between these two points is given by √((10 – 0)^2 + (0 – 5)^2) = 5√(5). Therefore, our answer is C.

Method 2: Using the Slope Concept
Although the using the traditional method is effective, we can answer this question much quicker using the concept we discussed above.

Line L has a slope of -2, which means that for every 1 unit the x-coordinate increases, the y-coordinate decreases by 2. Line L also passes through the point (6, 8). We know the line must intersect the x-axis at y = 0, which is a decrease of 8 y-coordinates from the given point (6,8). If y increases by 8, according to our slope concept, x will increase by 4 to give 6 + 4 = 10. So the x-intercept of line L is at (10, 0).

Line K has slope of 1/2 and also passes through (6, 8). We know the this line must intersect the y-axis at x = 0, which is a decrease of 6 x-coordinates from the given point (6,8). This means y will decrease by 1/2 of that (6*1/2 = 3) and will become 8 – 3 = 5. So the y-intercept of line K is at (0, 5).

The distance between the two points can now be found using the Pythagorean Theorem – √(10^2 + 5^2) = 5√(5), therefore our answer is, again, C.

Using the slope concept makes solving this question much less tedious and saves us a lot of precious time. That is the advantage of using holistic approaches over the more traditional approaches in tackling GMAT questions.

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!

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

Today 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.

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:

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.

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!

# GMAT Tip of the Week: Death, Taxes, and the GMAT Items You Know For Certain

Here on April 15, it’s a good occasion to remember the Benjamin Franklin quote: “In this world nothing can be said to be certain, except death and taxes.” Franklin, of course, never took the GMAT (which didn’t become a thing until a little ways after his own death, which he accurately predicted above). But if he did, he’d have plenty to add to that quote.

On the GMAT, several things are certain. Here’s a list of items you will certainly see on the GMAT, as you attempt to raise your score and therefore your potential income, thereby raising your future tax bills in Franklin’s honor:

Integrated Reasoning
You will struggle with pacing on the Integrated Reasoning section. 12 prompts in 30 minutes (with multiple problems per prompt) is an extremely aggressive pace and very few people finish comfortably. Be willing to guess on a problem that you know could sap your time: not only will that help you finish the section and protect your score, it will also help save your stamina and energy for the all-important Quant Section to follow.

Word Problems
On the Quantitative Section, you will certainly see at least one Work/Rate problem, one Weighted Average problem, and one Min/Max problem. This is good news! Word problems reward repetition and preparation – if you’ve put in the work, there should be no surprises.

Level of Difficulty
If you’re scoring above average on either the Quant or Verbal sections, you will see at least one problem markedly below your ability level. Because each section contains several unscored, experimental problems, and those problems are delivered randomly, probability dictates that every 700+ scorer will see at least one problem designed for the 200-500 crowd (and probably more than that). Do not try to read in to your performance based on the difficulty level of any one problem! It’s easy to fear that such a problem was delivered to you because you’re struggling, but the much more logical explanation is that it was either random or difficult-but-sneakily-so, so stay confident and move on.

Data Sufficiency
You will see at least one Data Sufficiency problem that seems way too easy to be true. And it’s probably not true: make sure that you think critically any time the testmaker is directly baiting you into a particular answer.

Sentence Correction
You will have to pick an answer that you don’t like, that doesn’t catch the ear the way you’d write or say it. Make sure that you prioritize the major errors that you know you can routinely catch and correct, and not let the GMAT bait you into a decision you’re just not qualified to make.

You will see a passage that takes you a few re-reads to even get your mind to process it. Remember to be– get enough out of the passage to know where to look when they ask you a specific question, but don’t worry about becoming a subject-matter expert on the topic. GMAT passages are designed to be difficult to read (particularly toward the end of a long test), so know that your competitive advantage is that you’ll be more efficient than your competition.

Critical Reasoning
You will have the opportunity to make quick work of several Critical Reasoning problems if you notice the tiny that each argument provides, and if you’re able to notice the subtle-but-significant words that make conclusions extra specific (and therefore harder to prove).

Few things are certain in life, but as you approach the GMAT there are plenty of certainties that you can prepare for so that you eliminate surprises and proceed throughout your test day confidently. On this Tax Day, take inventory of the things you know to be certain about the GMAT so that your test day isn’t so taxing.

By Brian Galvin.

# What Makes GMAT Quant Questions So Hard?

We know that the essentials of the GMAT Quant section are pretty simple: advanced topics such as derivatives, complex numbers, matrices and trigonometry are not included, while fundamentals we all learned from our high school math books are included. So it would be natural to think that the GMAT Quant section should not pose much of a problem for most test-takers (especially for engineering students, who have actually covered far more advanced math during their past studies).

Hence, it often comes as a shock when many test-takers, including engineering students, receive a dismal Quant score on the first practice test they take. Of course, with practice, they usually wise up to the treachery of the GMAT, but until then, the Quant section is responsible for many a nightmare!

Today, let’s see what kind of treachery we are talking about – problems like this make some people laugh out loud and others pull at their hair!

Is the product pqr divisible by 12?
Statement 1: p is a multiple of 3
Statement 2: q is a multiple of 4

This seems like an easy C (Statements 1 and 2 together are sufficient, but alone are not sufficient), doesn’t it? P is a multiple of 3 and q is a multiple of 4, so together, p*q would be a multiple of 3*4 = 12. If p * q is already a multiple of 12, then obviously it would seem that p*q*r would be a multiple of 12, too.

But here is the catch – where is it mentioned that r must be an integer? Just because p and q are integers (multiples of 3 and 4 respectively), it does not imply that r must also be an integer.

If r is an integer, then sure, p*q*r will be divisible by 12. Imagine, however, that p = 3, q = 4 and r = 1/12. Now the product p*q*r = 3*4*(1/12) = 1. 1 is not divisible by 12, so in this case, pqr is not divisible by 12. Hence, both statements together are not sufficient to answer the question, and our answer is in fact E!

This question is very basic, but it still tricks us because we want to assume that p, q and r are clean integer values.

Along these same lines, let’s try the another one:

If 10^a * 3^b * 5^c = 450^n, what is the value of c?
Statement 1: a is 1.
Statement 2:  b is 2.

The first thing most of us will do here is split 450 into its prime factors:

450 = 2 * 3^2 * 5^2

450^n = 2^n * 3^2n * 5^2n

And do the same thing with the left side of the equation:

10^a * 3^b * 5^c = 2^a * 3^b * 5^(a+c)

Bringing the given equation back, we get:

2^a * 3^b * 5^(a+c) = 2^n * 3^2n * 5^2n

Statement 1: a is 1.

Equating the power of 2 on both sides, we see that a = n = 1.

a + c = 2n (equating the power of 5 on both sides)

1 + c = 2

c = 1

Statement 2:  b is 2.

Equating the power of 3 on both sides, we see that b = 2n = 2, so n = 1.

If n = 1, a = 1 by equating the powers of 2 on both sides.

a + c = 2n (equating the power of 5 on both sides)

1 + c = 2

c = 1

So it seems that both statements are separately sufficient. But hold on – again, the variables here don’t need to be cleanly fitting integers. The variables could pan out the way discussed in our first problem, or very differently.

Say, n = 1. When Statement 1 gives you that a = 1, you get 10^1 * 3^b * 5^c = 450^1.

3^b * 5^c = 45

Now note that value of c depends on the value of b, which needn’t be 2.

If b  = 3, then 3^3 * 5^c = 45.

5^c = 45/27

C will take a non-integer value here.

c = .3174

The question does not mention that all variables are integers, therefore there are infinite values that c can take depending on the values of b. Similarly, we can see that Statement 2 alone is also not sufficient. Using both statements together, you will get:

2^a * 3^b * 5^(a+c) = 450^n

2^1 * 3^2 * 5^(1 + c) = 450^n

5^(1 + c) = 450^n/18

By now, you’ve probably realized that depending on the value of n, c can take infinite different values. If n = 1, c = 1. If n = 2, c = 4.8. And so on… We don’t need to actually find these values – it is enough to know that different values of n will give different values of c.

With this in mind, we can see that both statements together are not sufficient, and therefore our answer must be E.

Hopefully, in future, this sneaky trick will not get you!

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: Dealing with Tangents on the GMAT

Considering a two dimensional figure, a tangent is a line that touches a curve at a single point.  Here are some examples of tangents:

In each of these cases, the line touches the curve at a single point. In the case of a circle, when you draw the radius of the circle from the center to the point of contact with the tangent, the radius is perpendicular to the tangent (as demonstrated in the figure on the right, above). A question discussing this concept is given in our post here.

Today, we will look at a question involving a tangent to a parabola:

If f(x) = 3x^2 – tx + 5 is tangent to the x-axis, what is the value of the positive number t?

(A) 2√15
(B) 4√15
(C) 3√13
(D) 4√13
(E) 6√15

Let’s first try to understand what the question is saying.

f(x) is a tangent to the x-axis. We know that the x-axis is a straight line, so f(x) must be a curve. A quadratic equation, such as our given equation of f(x) = 3x^2 -tx +5, gives a parabola. Since the x^2 term in the equation is positive, the parabola would be facing upwards and touching the x-axis at a single point, such as:

Since the parabola touches the x-axis in only one point, it means the quadratic has only one root, or in other words, the quadratic must be a perfect square.

Therefore, f(x) = 3x^2 – tx + 5 = √3(x)^2 – tx + (√5)^2

To get f(x) in the form a^2 – 2ab + b^2 = (a – b)^2,

tx = 2ab = (2√3)x * √5

t = 2√15

Note that if t takes this value, the quadratic will have only one root.

Plugging this value of t back into our equation, we will get: f(x) = √3(x)^2 – 2(√15)(x) + (√5)^2

f(x) = (√3)x – (√5)^2

We know that the root of f(x) is the point where the value of the y coordinate is 0. Therefore:

(√3)x – (√5)^2  = 0

x = (√5)/(√3)

At this x co-ordinate, the parabola will touch the x axis.

[This calculation was shown only to help you completely understand the question. We could have easily stopped at t = 2(√15).]

The question can be solved in various other ways – think of how, and write your thoughts in the comments below!

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!

# 2 Tips to Make GMAT Remainder Questions Easy

Several months ago, I wrote an article aboutBecause this concept shows up so often on the GMAT, I thought it would be useful to revisit the topic. At times, it will be helpful to know the kind of terminology we’re taught in grade school, while at other times, we’ll simply want to select simple numbers that satisfy the parameters of a Data Sufficiency statement.

So let’s explore each of these scenarios in a little more detail. A simple example can illustrate the terminology: if we divide 7 by 4, we’ll have 7/4 = 1 + ¾.

7, the term we’re dividing by something else, is called the dividend. 4, which is doing the dividing, is called the divisor. 1, the whole number component of the mixed fraction, is the quotient. And 3 is the remainder. This probably feels familiar.

In the abstract, the equation is: Dividend/Divisor = Quotient + Remainder/Divisor. If we multiply through by the Divisor, we get: Dividend = Quotient*Divisor + Remainder.

Simply knowing this terminology will be sufficient to answer the following official question:

When N is divided by T, the quotient is S and the remainder is V. Which of the following expressions is equal to N?

A) ST
B) S + V
C) ST + V
D) T(S+V)
E) T(S – V)

In this problem, N – which is getting divided by something else – is our dividend, T is the divisor, S is the quotient, and V is the remainder. Plugging the variables into our equation of Dividend = Quotient*Divisor + Remainder, we get N = ST + V… and we’re done! The answer is C.

(Note that if you forgot the equation, you could also pick simple numbers to solve this problem. Say N = 7 and T = 3. 7/3 = 2 + 1/3.  The Quotient is 2, and the remainder is 1, so V = 1. Now, if we plug in 3 for T, 2 for S, and 1 for V, we’ll want an N of 7. Answer choice C will give us an N of 7, 2*3 + 1 = 7, so this is correct.)

When we need to generate a list of potential values to test in a data sufficiency question, often a statement will give us information about the dividend in terms of the divisor and the remainder.

Take the following example: when x is divided by 5, the remainder is 4. Here, the dividend is x, the divisor is 5, and the remainder is 4. We don’t know the quotient, so we’ll just call it q. In equation form, it will look like this: x = 5q + 4. Now we can generate values for x by picking values for q, bearing in mind that the quotient must be a non-negative integer.

If q = 0, x = 4. If q = 1, x = 9. If q=2, x = 14. Notice the pattern in our x values: x = 4 or 9 or 14… In essence, the first allowable value of x is the remainder. Afterwards, we’re simply adding the divisor, 5, over and over. This is a handy shortcut to use in complicated data sufficiency problems, such as the following:

If x and y are integers, what is the remainder when x^2 + y^2 is divided by 5?

1) When x – y is divided by 5, the remainder is 1
2) When x + y is divided by 5, the remainder is 2

In this problem, Statement 1 gives us potential values for x – y. If we begin with the remainder (1) and continually add the divisor (5), we know that x – y = 1 or 6 or 11, etc. If x – y = 1, we can say that x = 1 and y = 0. In this case, x^2 + y^2 = 1 + 0 = 1, and the remainder when 1 is divided by 5 is 1. If x – y = 6, then we can say that x = 7 and y = 1. Now x^2 + y^2 = 49 + 1 = 50, and the remainder when 50 is divided by 5 is 0. Because the remainder changes from one scenario to another, Statement 1 is not sufficient alone.

Statement 2 gives us potential values for x + y. If we begin with the remainder (2) and continually add the divisor (5), we know that x + y = 2 or 7 or 12, etc. If x + y = 2, we can say that x = 1 and y = 1. In this case, x^2 + y^2 = 1 + 1 = 2, and the remainder when 2 is divided by 5 is 2. If x + y = 7, then we can say that x = 7 and y = 0. Now x^2 + y^2 = 49 + 0 = 49, and the remainder when 49 is divided by 5 is 4. Because the remainder changes from one scenario to another, Statement 2 is also not sufficient alone.

Now test them together – simply select one scenario from Statement 1 and one scenario from Statement 2 and see what happens. Say x – y = 1 and x + y = 7. Adding these equations, we get 2x = 8, or x = 4. If x = 4, y = 3. Now x^2 + y^2 = 16 + 9 = 25, and the remainder when 25 is divided by 5 is 0.

We need to see if this will ever change, so try another scenario. Say x – y = 6 and x + y = 12. Adding the equations, we get 2x = 18, or x = 9. If x =  9, y = 3, and x^2 + y^2 = 81 + 9 = 90. The remainder when 90 is divided by 5 is, again, 0. No matter what we select, this will be the case – we know definitively that the remainder is 0. Together the statements are sufficient, so the answer is C.

Takeaway: You’re virtually guaranteed to see remainder questions on the GMAT, so you want to make sure you have this concept mastered. First, make sure you feel comfortable with the following equation: Dividend = Divisor*Quotient + Remainder. Second, if you need to select values, you can simply start with the remainder and then add the divisor over and over again. If you internalize these two ideas, remainder questions will become considerably less daunting.

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

# All You Need to Know About Using Interest Equations on the GMAT

As an undergraduate, I concentrated in Finance. When I tell people this, they make two unwarranted assumptions: the first is that I work in Finance (I don’t), and the second is that I am a glutton for mathematical punishment (debatable).

The reason people are intimidated by the kinds of compound interest equations we encounter in finance classes is that they look complicated. GMAT test-takers get anxious whenever I introduce this topic in class. But, as with most seemingly abstruse topics, these concepts are far less difficult than they appear at first glance.

Here’s all we really need to know about interest equations: if we’re talking about simple interest, the interest will be the same in every time period, and the equation you assemble will end up being straightforward linear algebra (if you choose to do algebra, that is). If we’re talking about compound interest, we’re really talking about an exponent question. The rest involves a bit of logic and algebraic manipulation.

Look at this official question that many of my students have initially struggled with:

An investment of \$1000 was made in a certain account and earned interest that was compounded annually. The annual interest rate was fixed for the duration of the investment, and after 12 years the \$1000 increased to 4000 by earning interest. In how many years after the initial investment was made would the 1000 have increased to 8000 by earning interest at that rate?

(A) 16
(B) 18
(C) 20
(D) 24
(E) 30

Looking at this question, the first instinct of most test-takers is to start frantically rummaging through their memory banks for that compound interest formula – there’s no need. Take a deep breath and remind yourself that these questions are just exponent questions involving a bit of algebra. With this in mind, let’s call the factor that the principal is multiplied by in each time period “x”. (If you’re accustomed to working with the formula, “x” is basically standing in for your standard (1 + r/100.) If you’re not accustomed to this formula, feel free to retroactively erase this parenthetical from your memory banks.)

If the principal is getting multiplied by “x” each year, then after one year, the investment will be 1000x. After two years the investment will be 1000x^2. After three years, it will be 1000x^3… and so on. In our problem, we’re talking about an investment after 12 years, which would be 1000x^12. If this value is 4000, we get the following equation: 1000x^12 = 4000 (and file away for now that the exponent represents the number of years elapsed).

Ultimately, we want to know what the exponent should be when the investment is at \$8000. If you’re looking at the answer choices now and think that 24 seems just a little too easy, your instincts are sound.

We need to work with 1000x^12 = 4000. Let’s simplify:

Divide both sides by 1000 to get x^12 = 4.  Solving for x seems unnecessarily complicated, so let’s consider our options. x^12 = 4 is the same as x^12 = 2^2, so if we take the square root of both sides, we will get x^6 = 2.

Essentially, this means that every 6 years (the exponent) the investment is doubling, or multiplied by 2. But we want to know how long it will take for that initial \$1000 to become \$8000, or to be multiplied by a factor of 8.

What can we do to x^6 = 2 so that we have an 8 on the right side? We can cube both sides!

(x^6)^3 = 2^3

x^18 = 8

This means that it will take 18 years to increase the investment by a factor of 8. Therefore, our answer is B.

Alternatively, once we see that the investment doubles every 6 years, we can ask ourselves how many times we need to double an investment to go from \$1000 to \$8000. Doubling once gets us to \$2000. Doubling twice gets us to \$4000. Doubling a third time gets us to \$8000. So if we double the investment every 6 years, and we need the investment to double 3 times, it will take a total of 6*3 = 18 years.

Takeaway: There are plenty of formulas that could come in handy on the GMAT – just know that a little logic and conceptual understanding will allow you to solve many of the questions that seem to require a particular formula. Memorization has limits that logic and mental agility don’t.

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

# Quarter Wit, Quarter Wisdom: Ratios in GMAT Data Sufficiency

We know that ratios are the building blocks for a lot of other concepts such as time/speed, work/rate and mixtures. As such, we spend a lot of time getting comfortable with understanding and manipulating ratios, so the GMAT questions that test ratios seem simple enough, but not always! Just like questions from all other test areas, questions on ratios can be tricky too, especially when they are formatted as Data Sufficiency questions.

Let’s look at two cases today: when a little bit of data is sufficient, and when a lot of data is insufficient.

When a little bit of data is sufficient!
Three brothers shared all the proceeds from the sale of their inherited property. If the eldest brother received exactly 5/8 of the total proceeds, how much money did the youngest brother (who received the smallest share) receive from the sale?

Statement 1: The youngest brother received exactly 1/5 the amount received by the middle brother.

Statement 2: The middle brother received exactly half of the two million dollars received by the eldest brother.

First impressions on reading this question? The question stem gives the fraction of money received by one brother. Statement 1 gives the fraction of money received by the youngest brother relative to the amount received by the middle brother. Statement 2 gives the fraction of money received by the middle brother relative to the eldest brother and an actual amount. It seems like the three of these together give us all the information we need. Let’s dig deeper now.

From the Question stem:

Eldest brother’s share = (5/8) of Total

Statement 1: Youngest Brother’s share = (1/5) * Middle brother’s share

We don’t have any actual number – all the information is in fraction/ratio form. Without an actual value, we cannot find the amount of money received by the youngest brother, therefore, Statement 1 alone is not sufficient.

Statement 2: Middle brother’s share = (1/2) * Eldest brother’s share, and the eldest brother’s share = 2 million dollars

Middle brother’s share = (1/2) * 2 million dollars = 1 million dollars

Now, we might be tempted to jump to Statement 1 where the relation between youngest brother’s share and middle brother’s share is given, but hold on: we don’t need that information. We know from the question stem that the eldest brother’s share is (5/8) of the total share.

So 2 million = (5/8) of the total share, therefore the total share = 3.2 million dollars.

We already know the share of the eldest and middle brothers, so we can subtract their shares out of the total and get the share of the youngest brother.

Youngest brother’s share = 3.2 million – 2 million – 1 million = 0.2 million dollars

Statement 2 alone is sufficient, therefore, the answer is B.

When a lot of data is insufficient!
A department manager distributed a number of books, calendars, and diaries among the staff in the department, with each staff member receiving x books, y calendars, and z diaries. How many staff members were in the department?

Statement 1: The numbers of books, calendars, and diaries that each staff member received were in the ratio 2:3:4, respectively.

Statement 2: The manager distributed a total of 18 books, 27 calendars, and 36 diaries.

First impressions on reading this question? The question stem tells us that each staff member received the same number of books, calendars, and diaries. Statement 1 gives us the ratio of books, calendars and diaries. Statement 2 gives us the actual numbers. It certainly seems that we should be able to obtain the answer. Let’s find out:

Looking at the question stem, Staff Member 1 recieved x books, y calendars, and z diaries, Staff Member 2 recieved x books, y calendars, and z diaries… and so on until Staff Member n (who also recieves x books, y calendars, and z diaries).

With this in mind, the total number of books = nx, the total number of calendars = ny, and the total number of diaries = nz.

Question: What is n?

Statement 1 tells us that x:y:z = 2:3:4. This means the values of x, y and z can be:

2, 3, and 4,

or 4, 6, and 8,

or 6, 9, and 12,

or any other values in the ratio 2:3:4.

They needn’t necessarily be 2, 3 and 4, they just need the required ratio of 2:3:4.

Obviously, n can be anything here, therefore, Statement 1 alone is not sufficient.

Statement 2 tell us that nx = 18, ny = 27, and nz = 36.

Now we know the actual values of nx, ny and nz, but we still don’t know the values of x, y, z and n.

They could be

2, 3, 4 and 9

or 6, 9, 12 and 3

Therefore, Statement 2 alone is also not sufficient.

Considering both statements together, note that Statement 2 tells us that nx:ny:nz = 18:27:36 = 2:3:4 (they had 9 as a common factor).

Since n is a common factor on left side, x:y:z = 2:3:4 (ratios are best expressed in the lowest form).

This is a case of what we call “we already knew that” – information given in Statement 1 is already a part of Statement 2, so it is not possible that Statement 2 alone is not sufficient but that together Statement 1 and 2 are. Hence, both statements together are not sufficient, and our answer must be E.

A question that arises often here is, “Why can’t we say that the number of staff members must be 9?”

This is because the ratio of 2:3:4 is same as the ratio of 6:9:12, which is same as 18:27:36 (when you multiply each number of a ratio by the same number, the ratio remains unchanged).

If 18 books, 27 calendars, and 36 diaries are distributed in the ratio 2:3:4, we could give them all to one person, or to 3 people (giving them each 6 books, 9 calendars and 12 diaries), or to 9 people (giving them each 2 books, 3 calendars and 4 diaries).

When we see 18, 27 and 36, what comes to mind is that the number of people could have been 9, which would mean that the department manager distributed 2 books, 3 calendars and 4 diaries to each person. But we know that 9 is divisible by 3, which should remind us that the number of people could also be 3, which would mean that the manager distributed 6 books, 9 calendars and 12 diaries to each person. As such, we still don’t know how many staff members there are, and our answer remians 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!

# It’s All Greek to Me: How to Use Greek Concepts to Beat the GMAT

The ancient Greeks were, to put it mildly, really neat. They created or helped to create the foundations of philosophy, theater, science, democracy, and mathematics – no small accomplishment for a small war-torn civilization from over two millennia ago. Many of our contemporary ideas, beliefs, and traditions are rooted in contributions made by Greek thinkers, and the GMAT is no exception.

When I first encountered this problem I couldn’t help but wonder what kind of mad scientist question-writer engineered it. Where would such an idea even come from? It turns out, it wasn’t a GMAC employee at all, but Archimedes, the famous Greek geometer and coiner of the phrase “Eureka!”

The question is based on his attempt to trisect an angle with only a straight edge and a compass. (Alas, Archimedes’ work, though ingenious, was not technically a correct solution to the problem, as it provides only an approximation.) The reader is hereby invited to contemplate the kind of person who encounters a proof by Archimedes and instinctively thinks, “This would make an excellent Data Sufficiency question on the GMAT!” We’d like to believe that the good folks at GMAC are just like you and me, but perhaps not.

So this got me thinking: what other interesting Greek contributions to mathematics might be helpful in analyzing GMAT questions? In Euclid’s work Elements, he offers a simple and elegant proof for why there is no largest prime number. The proof proceeds by positing a hypothetical largest prime number “p.” We can then construct a product that consists of every prime number 2*3*5*7….*p. We’ll call this product “q.”

The next consecutive number will be q + 1. Now, we know that “q” contains 2 as a factor, as “q,” supposedly, contains every prime as a factor. Therefore q +1 will not contain 2 as a factor. (The next number to contain 2 as a factor will be q + 2.) We know that “q” contains 3 as a factor. Therefore q + 1 will not contain 3 as a factor. (The next number to contain 3 as a factor will be q + 3.)

Uh oh. If “p” really is the largest prime number, we’ve got a problem, because q + 1 will not contain any of the primes between 2 and p as factors. So either q + 1 is itself prime, or there is some prime greater than p and less than q + 1 that we’ve failed to consider. Either way, we’ve proven that p can’t be the largest prime number – I told you the Greeks were neat.

One axiom that’s worth internalizing from Euclid’s proof is the notion that two consecutive numbers cannot have any factors in common aside from 1.  When q contains every prime from 2 to p as a factor, q + 1 contains none of those primes. How would this be helpful on the GMAT? Glad you asked. Check out this question:

x is the product of all even numbers from 2 to 50, inclusive. The smallest prime factor of x + 1 must be:

(A) Between 1 and 10

(B) Between 11 and 15

(C) Between 15 and 20

(D) Between 20 and 25

(E) Greater than 25

We’re given information about x, and we’re asked about x + 1. If x is the product of all even numbers from 2 to 50, we can write x = 2 * 4 * 6 …* 50. This is the same as (1*2) * (2*2) * (3*2)… (25*2), which means the product consists of all the integers from 1 to 25, inclusive, and a bunch of 2’s.

So now we know that every prime number between 2 and 25 will be a factor of x. What about x + 1? (Paging Euclid!) We know that 2 is not a factor of x + 1, as 2 is a factor of x, and so the next multiple of 2 would be x + 2. We know that 3 is also not a factor of x + 1, as 3 is a factor of x, and so the next multiple of 3 would be x + 3. And once we’ve internalized that two consecutive numbers cannot have any factors in common aside from 1, we know that if all the primes between 2 and 25 are factors of x, none of those primes can be factors of x + 1, meaning that the smallest prime of x, whatever is, will be greater than 25. The answer, therefore, is E.

Takeaway: One of the beautiful things about mathematics is that fundamental truths do not change over time. What worked for the Greeks will work for us. The same axioms that allowed ancient mathematicians to grapple with problems two millennia ago will allow us to unravel the toughest GMAT questions. Learning a few of these axioms is not only interesting – though I’d caution against bringing up Archimedes’ trisection proof at a dinner party – but also helpful on the GMAT.

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

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

# GMAT Tip of the Week: Kanye, Wiz Khalifa, Twitter Beef…and GMAT Variables

This week, the internet exploded with a massive Twitter feud between rappers Kanye West and Wiz Khalifa, with help from their significant others and exes. For days now, hashtags unpublishable for an education blog have topped the trending lists, all as a result of the epic social media confrontation. And all of THAT originated from a classic GMAT mistake from the Louis Vuitton Don – a man who so loves his hometown Kellogg School of Management that he essentially named his daughter Northwestern – himself:

Kanye didn’t consider all the possibilities when he saw variables.
A brief history of the beef: there was musical origin, as Wiz wanted a bit of credit for his young/wild/free friends for the term “Wave,” as Kanye changed his upcoming album title from Swish to Waves. But where things escalated quickly all stemmed from Wiz’s use of variables in the following tweet:

Hit this kk and become yourself.

Kanye, whose wife bears those exact initials, K.K., immediately interpreted those variables as a reference to Kim and lost his mind. But Wiz had intended those variables kk to mean something entirely different, a reference to his favorite drug of choice. And then…well let’s just say that things got out of hand.

So back to the GMAT: Kanye’s main mistake was that he didn’t consider alternate possibilities for the variables he saw in the tweet, and quickly built in some incorrect assumptions that led to disastrous results. Do not let this happen to you on the GMAT! Here’s how it could happen:

If a problem, for example, defines k as 10 < k < 12, you can’t just think “k = 11” because you don’t know that k has to be an integer. 11.9 or 10.1 are also possibilities. Similarly if k^2 = 121, you have to consider that k could be -11 as well as it could be 11.

Ultimately, that was Yeezy’s mistake: he saw KK and with tunnel vision saw the most obvious possibility. But why couldn’t “KK” have been Krispy Kreme or Kyle Korver or Kato Kaelin? Before you leap to conclusions on a GMAT variable, see if there’s anything else it could be.

2) Assuming that each variable must represent a different number.
This one is a bit more nuanced. Suppose you were asked:

For positive integers a and b, is the product ab > 1?

(1) a = 1

With that statement, you might start thinking, “Well if a is 1, b has to be something else…” but all the variable b really means is “a number we don’t know.” Just because a problem assigns two different variables does not mean that they represent two different numbers! B could also be 1…we just don’t know yet.

Where this manifests itself as a problem most often is on function problems. When people see the setup, for example:

The function f is defined for all values x as f(x) = x^2 – x – 1

They’ll often be confused when that’s paired with a question like, “Is f(a) > 1?” and a statement like:

(1) -2 < a < 2

“I know about f(x) but I don’t know anything about f(a),” they might say, but the way these variables work, f(x) means “the function of any number…we just don’t know which number” so when you then see f(a), a becomes that number you don’t know. You’ll do the same thing for a: f(a) = a^2 – a – 1. What goes in the parentheses is just “the number you perform the function on” – the function doesn’t just apply to the variable in the definition, but to any number, variable, or combination that is then put in the parentheses.

The real lesson here is this: variables on the GMAT are a lot like variables in Wiz Khalifa’s Twitter feed. You might think you know what they mean, but before you stake your reputation (or score) on your response to those variables, consider all the options. Hit this GMAT and become yourself.

By Brian Galvin.

# GMAT Tip of the Week: Make 2016 The Year Of Number Fluency

Whether you were watching the College Football Playoffs or Ryan Seacrest; whether you were at a house party, in a nightclub, or home studying for the GMAT; however you rang in 2016, if 2016 is the year that you make your business school goals come true, hopefully you had one of the following thoughts immediately after seeing the number 2016 itself:

• Oh, that’s divisible by 9
• Well, obviously that’s divisible by 4
• Huh, 20 and 16 are consecutive multiples of 4
• 2, 0, 1, 6 – that’s three evens and an odd
• I wonder what the prime factors of 2016 are…

Why? Because the GMAT – and its no-calculator-permitted format for the Quant Section – is a test that highly values and rewards mathematical fluency. The GMAT tests patterns in, and properties of, numbers quite a bit. Whenever you see a number flash before your eyes, you should be thinking about even vs. odd, prime vs. composite, positive vs. negative, “Is that number a square or not?” etc. And, mathematically speaking, the GMAT is a multiplication/division test more than a test of anything else, so as you process numbers you should be ready to factor and divide them at a moment’s notice.

Those who quickly see relationships between numbers are at a huge advantage: they’re not just ready to operate on them when they have to, they’re also anticipating what that operation might be so that they don’t have to start from scratch wondering how and where to get started.

With 2016, for example:

The last two digits are divisible by 4, so you know it’s divisible by 4.

The sum of the digits (2 + 0 + 1 + 6) is 9, a multiple of 9, so you know it’s divisible by 9 (and also by 3).

So without much thinking or prompting, you should already have that number broken down in your head. 16 divided by 4 is 4 and 2000 divided by 4 is 500, so you should be hoping that the number 504 (also divisible by 9) shows up somewhere in a denominator or division operation (or that 4 or 9 does).

So, for example, if you were given a problem:

In honor of the year 2016, a donor has purchased 2016 books to be distributed evenly among the elementary schools in a certain school district. If each school must receive the same number of books, and there are to be no books remaining, which of the following is NOT a number of books that each school could receive?

(A) 18

(B) 36

(C) 42

(D) 54

(E) 56

You shouldn’t have to spend any time thinking about choices A and B, because you know that 2016 is divisible by 4 and by 9, so it’s definitely divisible by 36 which means it’s also divisible by every factor of 36 (including 18). You don’t need to do long division on each answer choice – your number fluency has taken care of that for you.

From there, you should look at the other numbers and get a quick sense of their prime factors:

42 = 2 * 3 * 7 – You know that 2016 is divisible by 2 and 3, but what about 7?

54 = 2 * 3 * 3 * 3 – You know that 2016 is divisible by that 2 and that it’s divisible by 9, so you can cover two of the 3s. But is 2016 divisible by three 3s?

56 = 2 * 2 * 2 * 7 – You know that two of the 2s are covered, and it’s quick math to divide 2016 by 4 (as you saw above, it’s 504). Since 504 is still even, you know that you can cover all three 2s, but what about 7?

Here’s where good test-taking strategy can give you a quick leg up: to this point, a savvy 700-scorer shouldn’t have had to do any real “work,” but testing all three remaining answer choices could now get a bit labor intensive. Unless you recognize this: for C and E, the only real question to be asked is “Is 2016 divisible by 7?” After all, you’re already accounted for the 2 and 3 out of 42, and you’ve already accounted for the three 2s out of 56.

7 is the only one you haven’t checked for. And since there can only be one correct answer, 2016 must be divisible by 7…otherwise you’d have to say that C and E are both correct.

But even if you’re not willing to take that leap, you may still have the hunch that 7 is probably a factor of 2016, so you can start with choice D. Once you’ve divided 2016 by 9 (here you may have to go long division, or you can factor it out), you’re left with 224. And that’s not divisible by 3. Therefore, you know that 2016 cannot be divided evenly into sets of 54, so answer choice D must be correct. And more importantly, good number fluency should have allowed you to do that relatively quickly without the need for much (if any) long division.

So if you didn’t immediately think “divisible by 4 and 9!” when you saw the year 2016 pop up, make it your New Year’s resolution to start thinking that way. When you see numbers this year, start seeing them like a GMAT expert, taking note of clear factors and properties and being ready to quickly operate on that number.

By Brian Galvin.

# How to Choose the Right Number for a GMAT Variable Problem

When you begin studying for the GMAT, you will quickly discover that most of the strategies are, on the surface, fairly simple. It will not come as a terribly big surprise that selecting numbers and doing arithmetic is often an easier way of attacking a problem than attempting to perform complex algebra. There is, however, a big difference between understanding a strategy in the abstract and having honed that strategy to the point that it can be implemented effectively under pressure.

Now, you may be thinking, “How hard can it possibly be to pick numbers? I see an “x” and I decide x = 5. Not so complicated.” The art is in learning how to pick workable numbers for each question type. Different questions will require different types of numbers to create a scenario that truly is simpler than the algebra. The harder the problem, the more finesse that will be required when selecting numbers. Let’s start with a problem that doesn’t require much strategy:

If n=4p, where p is prime number greater than 2, how many different positive even divisors does n have, including n?

(A) 2

(B) 3

(C) 4

(D) 6

(E) 8

Okay in this problem, “p” is a prime number greater than 2. So let’s say p = 3. If n = 4p, and 4p = 4*3 = 12. Let’s list out the factors of 12: 1, 2, 3, 4, 6, 12. The even factors here are 2, 4, 6, 12. There are 4 of them. So the answer is C. Not so bad, right? Just pick the first simple number that pops into your head and you’re off to the races. Bring on the test!

If only it were that simple for all questions. So let’s try a much harder question to illustrate the pitfalls of adhering to an approach that’s overly mechanistic:

The volume of water in a certain tank is x percent greater than it was one week ago. If r percent of the current volume of water in the tank is removed, the resulting volume will be 90 percent of the volume it was one week ago. What is the value of r in terms of x?

(A) x + 10

(B) 10x + 1

(C) 100(x + 10)

(D) 100 * (x+10)/(x+100)

(E) 100 * (10x + 1)/(10x+10)

You’ll notice quickly that if you simply declare that x = 10 and r =20, you may run into trouble. Say, for example, that the starting value from one week ago was 100 liters. If x = 10, a 10% increase will lead to a volume of 110 liters. If we remove 20% of that 110, we’ll be removing .20*110 = 22 liters, giving us 110-22 = 88 liters. But we’re also told that the resulting volume is 90% of the original volume! 88 is not 90% of 100, therefore our numbers aren’t valid. In instances like this, we need to pick some simple starting numbers and then calculate the numbers that will be required to fit the parameters of the question.

So again, say the volume one week ago was 100 liters. Let’s say that x = 20%, so the volume, after water is added, will be 100 + 20 = 120 liters.

We know that once water is removed, the resulting volume will be 90% of the original. If the original was 100, the volume, once water is removed, will be 100*.90 = 90 liters.

Now, rather than arbitrarily picking an “r”, we’ll calculate it based on the numbers we have. To summarize:

Start: 100 liters

After removing water: 90 liters

We now need to calculate what percent of those 120 liters need to be removed to get down to 90. Using our trusty percent change formula [(Change/Original) * 100] we’ll get (30/120) * 100 = 25%.

Thus, when x = 20, r =25. Now all we have to do is substitute “x” with “20” in the answer choices until we hit our target of 25.

Remember that in these types of problems, we want to start at the bottom of the answer choice options and work our way up:

(E) 100 * (10x + 1)/(10x+10)

100 * (10*20 + 1)/(10*20+10) = 201/210. No need to simplify. There’s no way this equals 25.

(D) 100 * (x+10)/(x+100)

100 * (20+10)/(20+100) = 100 * (30/120) = 25. That’s it! We’re done. The correct answer is D.

Takeaways: Internalizing strategies is the first step in your process of preparing for the GMAT. Once you’ve learned these strategies, you need to practice them in a variety of contexts until you’ve fully absorbed how each strategy needs to be tweaked to fit the contours of the question. In some cases, you can pick a single random number. Other times, there will be multiple variables, so you’ll have to pick one or two numbers to start and then solve for the remaining numbers so that you don’t violate the conditions of the problem. Accept that you may have to make adjustments mid-stream. Your first selection may produce hairy arithmetic. There are no style point on the GMAT, so stay flexible, cultivate back-up plans, and remember that mental agility trumps rote memorization every time.

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

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

# How to Use Difference of Squares to Beat the GMAT

In Michael Lewis’ Flashboys, a book about the hazards of high-speed trading algorithms, Lewis relates an amusing anecdote about a candidate interviewing for a position at a hedge fund. During this interview, the candidate receives the following question: Is 3599 a prime number? Hopefully, your testing Spidey Senses are tingling and telling you that the answer to the question is going to incorporate some techniques that will come in handy on the GMAT. So let’s break this question down.

First, this is an interview question in which the interviewee is put on the spot, so whatever the solution entails, it can’t involve too much hairy arithmetic. Moreover, it is far easier to prove that a large number is NOT prime than to prove that it is prime, so we should be thinking about how we can demonstrate that this number possesses factors other than 1 and itself.

Whenever we’re given unpleasant numbers on the GMAT, it’s worthwhile to think about the characteristics of round numbers in the vicinity. In this case, 3599 is the same as 3600 – 1. 3600, the beautiful round number that it is, is a perfect square: 602. And 1 is also a perfect square: 12. Therefore 3600 – 1 can be written as the following difference of squares:

3600 – 1 = 602 – 12

We know that x– y= (x + y)(x – y), so if we were to designate “x” as “60” and “y” as “1”, we’ll arrive at the following:

60– 1= (60 + 1)(60 – 1) = 61 * 59

Now we know that 61 and 59 are both factors of 3599. Because 3599 has factors other than 1 and itself, we’ve proven that it is not prime, and earned ourselves a plumb job at a hedge fund. Not a bad day’s work.

But let’s not get ahead of ourselves. Let’s analyze some actual GMAT questions that incorporate this concept.

First:

999,9992 – 1 =

A) 1010 – 2

B) (106 – 2) 2

C) 105 (106 -2)

D) 106 (105 -2)

E) 106 (106 -2)

Notice the pattern. Anytime we have something raised to a power of 2 (or an even power) and we subtract 1, we have the difference of squares, because 1 is itself a perfect square. So we can rewrite the initial expression as 999,9992 – 12.

Using our equation for difference of squares, we get:

999,9992 – 12  = (999,999 +1)(999,999 – 1)

(999,999 + 1)(999,999 – 1) = 1,000,000* 999,998.

Take a quick glance back at the answer choices: they’re all in terms of base 10, so there’s a little work left for us to do. We know that 1,000,000 = 106  (Remember that the exponent for base 10 is determined by the number of 0’s in the figure.) And we know that 999,998 = 1,000,000 – 2 = 106 – 2, so 1,000,000* 999,998 = 106 (106 -2), and our answer is E.

Let’s try one more:

Which of the following is NOT a factor of 38 – 28?

A) 97

B) 65

C) 35

D) 13

E) 5

Okay, you’ll see quickly that 38 – 28 will involve same painful arithmetic. But thankfully, we’ve got the difference of two numbers, each of which has been raised to an even exponent, meaning that we have our trusty difference of squares! So we can rewrite 38 – 28 as (34)2 – (24)2. We know that 34 = 81 and 24 = 16, so (34)2 – (24)2 = 812 – 162. Now we’re in business.

812 – 162 = (81 + 16)(81 – 16) = 97 * 65.

Right off the bat, we can see that 97 and 65 are factors of our starting numbers, and because we’re looking for what is not a factor, A and B are immediately out. Now let’s take the prime factorization of 65. 65 = 13 * 5. So our full prime factorization is 97 * 13 * 5. Now we see that 13 and 5 are factors as well, thus eliminating D and E from contention. That leaves us with our answer C. Not so bad.

Takeaways:

• The GMAT is not interested in your ability to do tedious arithmetic, so anytime you’re asked to find the difference of two large numbers, there is a decent chance that the number can be depicted as a difference of squares.
• If you have the setup (Huge Number)2 – 1, you’re definitely looking at a difference of squares, because 1 is a perfect square.
• If you’re given the difference of two numbers, both of which are raised to even exponents, this can also be depicted as a difference of squares, as all integers raised to even exponents are, by definition, perfect squares.