# GMAT Tip of the Week: Big Sean Says Your GMAT Score Will Bounce Back

Welcome back to Hip Hop Month in the GMAT Tip of the Week space, where naturally, we woke up in beast mode (with your author legitimately wishing he was bouncing back to D-town from LAX this weekend, but blog duty calls!).

If you have a car stereo or Pandora account, you’ve undoubtedly heard Big Sean talking about bouncing back this month. “Bounce Back” is a great anthem for anyone hitting a rough patch – at work, in a relationship, after a rough day for your brackets during next week’s NCAA tournament – but this isn’t a self-help, “it’s always darkest before dawn,” feel-good article. Big Sean has some direct insight into the GMAT scoring algorithm with Bounce Back, and if you pay attention, you can leverage Bounce Back (off the album “I Decided” – that’ll be important, too) to game-plan your test day strategy and increase your score.

So, what’s Big Sean’s big insight?

The GMAT scoring (and question delivery) algorithm is designed specifically so that you can “take an L” and bounce back. And if you understand that, you can budget your time and focus appropriately. The test is designed so that just about everybody misses multiple questions – the adaptive system serves you problems that should test your upper threshold of ability, and can also test your lower limit if you’re not careful.

What does that mean? Say you, as Big Sean would say, “take an L” (or a loss) on a question. That’s perfectly fine…everyone does it. The next question should be a bit easier, providing you with a chance to bounce back. The delivery system is designed to use the test’s current estimate of your ability to deliver you questions that will help it refine that estimate, meaning that it’s serving you questions that lie in a difficulty range within a few percentile points of where it thinks you’re scoring.

If you “take an L” on a problem that’s even a bit below your true ability, missing a question or two there is fine as long as it’s an outlier. No one question is a perfect predictor of ability, so any single missed question isn’t that big of a deal…if you bounce back and get another few questions right in and around that range, the system will continue to test your upper threshold of ability and give you chances to prove that the outlier was a fluke.

The problem comes when you don’t bounce back. This doesn’t mean that you have to get the next question right, but it does mean that you can’t afford big rough patches – a run of 3 out of 4 wrong or 4 out of 5 wrong, for example. At that point, the system’s estimate of you has to change (your occasional miss isn’t an outlier anymore) and while you can still bounce back, you now run the risk of running out of problems to prove yourself. As the test serves you questions closer to its new estimate of you, you’re not using the problems to “prove how good you are,” but instead having to spend a few problems proving you’re “not that bad, I promise!”

So, okay. Great advice – “don’t get a lot of problems wrong.” Where’s the real insight? It can be found in the lyrics to “Bounce Back”:

Everything I do is righteous
Betting on me is the right risk
Even in a ***** crisis…

During the test you have to manage your time and effort wisely, and that means looking at hard questions and determining whether betting on that question is the right risk. You will get questions wrong, but you also control how much you let any one question affect your ability to answer the others correctly. A single question can hurt your chances at the others if you:

• Spend too much time on a problem that you weren’t going to get right, anyway
• Let a problem get in your head and distract you from giving the next one your full attention and confidence

Most test-takers would be comfortable on section pacing if they had something like 3-5 fewer questions to answer, but when they’re faced with the full 37 Quant and 41 Verbal problems they feel the need to rush, and rushing leads to silly mistakes (or just blindly guessing on the last few problems). And when those silly mistakes pile up and become closer to the norm than to the outlier, that’s when your score is in trouble.

You can avoid that spiral by determining when a question is not the right risk! If you recognize in 30-40 seconds (or less) that you’re probably going to take an L, then take that L quickly (put in a guess and move on) and bank the time so that you can guarantee you’ll bounce back. You know you’re taking at least 5 Ls on each section (for most test-takers, even in the 700s that number is probably closer to 10) so let yourself be comfortable with choosing to take 3-4 Ls consciously, and strategically bank the time to ensure that you can thoroughly get right the problems that you know you should get right.

Guessing on the GMAT doesn’t have to be a panic move – when you know that the name of the game is giving yourself the time and patience to bounce back, a guess can summon Big Sean’s album title, “I Decided,” as opposed to “I screwed up.” (And if you need proof that even statistics PhDs who wrote the GMAT scoring algorithm need some coaching with regard to taking the L and bouncing back, watch the last ~90 seconds of )

So, what action items can you take to maximize your opportunity to bounce back?

Right now: pay attention to the concepts, question types, and common problem setups that you tend to waste time on and get wrong. Have a plan in mind for test day that “if it’s this type of problem and I don’t see a path to the finish line quickly, I’m better off taking the L and making sure I bounce back on the next one.”

Also, as you review those types of problems in your homework and practice tests, look for techniques you can use to guess intelligently. For many, combinatorics with restrictions is one of those categories for which they often cannot see a path to a correct answer. Those problems are easy to guess on, however! Often you can eliminate a choice or two by looking at the number of possibilities that would exist without the restriction (e.g. if Remy and Nicki would just patch up their beef and stand next to each other, there would be 120 ways to arrange the photo, but since they won’t the number has to be less than 120…). And you can also use that total to ask yourself, “Does the restriction take away a lot of possibilities or just a few?” and get a better estimate of the remaining choices.

On test day: Give yourself 3-4 “I Decided” guesses and don’t feel bad about them. If your experience tells you that betting your time and energy on a question is not the right risk, take the L and use the extra time to make sure you bounce back.

The GMAT, like life, guarantees that you’ll get knocked down a few times, but what you can control is how you respond. Accept the fact that you’re going to take your fair share of Ls, but if you’re a real one you know how to bounce back.

By Brian Galvin.

# Quarter Wit, Quarter Wisdom: When Can You Divide by a Variable?

We have often come across test takers confused about division by a variable. When is it allowed, when is it not allowed? Why is it allowed in some cases and not in others? What are the constraints we need to look out for?

For example:

Is division by x allowed here: x^2 = 10x?
Is division by x allowed here: y = 4x?
Is division by x allowed here: x^2 < 4x?

Let’s take a detailed look at all these questions today.

The basic guidelines:

1. Division by 0 is not allowed, hence you cannot divide by a variable until and unless we know that it cannot be 0.
2. In the case of an inequality, when you divide by a negative number, the sign of the inequality flips. So we cannot divide by a variable until and unless we know that it cannot be 0 AND whether it is positive or negative.

Let’s look at the three questions given above and try to solve them using these guidelines:

Is division by x allowed here: x^2 = 10x?

The first thing to find out here is whether or not x can equal 0.

Case 1: If no other information has been given, then x can be 0 and we cannot divide by it. This is how we proceed in that case:

x^2 – 10x = 0
x(x – 10) = 0
x = 0 or 10

Case 2: If the question stem tells us that x is not 0, then we can divide by x.

x^2/x = 10x/x
x = 10

Obviously, we don’t get the second solution (x = 0) in this case, as we already know that x cannot be 0. Now let’s look at the second problem:

Is division by x allowed here: y = 4x?

Again, this is an equation and we need to know whether or not x can equal 0.

Case 1: If x can be 0, you cannot divide by it. In this case, x = 0 and y = 0 is one of the infinite possible solutions.

Case 2: If the question stem states that x cannot be 0, then we can do the following:

y/x = 4

Now let’s look at the final question:

Is division by x allowed here: x^2 > -4x?

Here, we have an inequality. Before deciding whether we can divide by x or not, we need to know not only whether x can be equal to 0, but also whether x is positive or negative.

Case 1: If we know nothing about the possible values that x can take, then this is how we proceed:

x^2 + 4x > 0
x(x + 4) > 0

Now we can use the method discussed in the first problem to arrive at the range of x.

x > 0 or x < -4

Case 2: If we know that x is positive, then we can proceed like this:

x^2/x > -4x/x
x > -4

Since we are given that x is positive, we know that that x > 0 (looking at the two options above).

Case 3: If we know that x is negative, then this is how we will proceed:

x^2/x < -4x/x (we flip the sign of the inequality because we divide by x, which is negative)
x < -4

The results obtained are logical, right? When x can be anywhere on the number line, we get the range as x > 0 or x < -4.

If x has to be positive, the range is x > 0.
If x has to be negative, the range is x < -4.

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: Evolving Your GMAT Quant Score with Help from The Evolution Of Rap

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

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

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

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

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

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

What is the value of (x + y)?

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

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

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

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

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

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

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

(2^8 + 1)(2^8 – 1)

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

(2^8 + 1)(2^4 + 1)(2^4 – 1)

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

257 * 17 * 15

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

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

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

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

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

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

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

(21/√2)^2 – (19/√2)^2

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

[(21 + 19)/√2][(21-19)/√2]

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

(40/√2) * (2/√2)

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

So what’s the point?

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

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

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

By Brian Galvin.

# Take the 2017 MBA Applicant Survey and Win \$500!

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Take the survey here.

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This online survey should take just a few minutes to complete. We would love to receive as many responses as possible before the survey closes in early April – and we would like to see one of our readers win the \$500 cash prize!

More about the Association of International Graduate Admissions Consultants: AIGAC promotes high ethical standards and professional development among graduate admissions consultants, increases public understanding of graduate admissions consulting, and enhances channels of communication with complementary organizations. The annual MBA Applicant Survey is just one way in which AIGAC serves the admissions and admissions consulting communities.

Thanks in advance for your participation, and good luck with the drawing!

# Quarter Wit, Quarter Wisdom: When a Little Information is Enough to Solve a GMAT Problem

We have reviewed what standard deviation is in a past post. We know what data is necessary to calculate the standard deviation of a set, but in some cases, we could actually do with a lot less information than the average test-taker may think they need.

Let’s explore this idea through an example GMAT data sufficiency question:

What is the standard deviation of a set of numbers whose mean is 20?

Statement 1: The absolute value of the difference of each number in the set from the mean is equal.
Statement 2: The sum of the squares of the differences from the mean is greater than 100.

We need to determine whether the information we have been given is sufficient to get us the exact value of the standard deviation of a particular set of numbers. To find the standard deviation of a set, we need to know the deviation of each term from the mean so that we can square those deviations, sum the squares, divide them by the number of terms, and then find the square root.

Essentially, to find the standard deviation we either need to know each element of the set, or we need to know the deviation of each element from the mean (which will also give us the number of terms), or we need to know the sum of the square of deviations and the number of terms in the set.

The question stem here tells us that the mean of the set is 20. We have no other information about any of the actual elements of the set or the number of elements. With this in mind, let’s examine each of the statements:

Statement 1: The absolute value of the difference of each number in the set from the mean is equal.

With this statement, we don’t actually know what the absolute value of the difference is. We also don’t know how many elements there are. The set could be something like:

19, 21 (each term is exactly 1 away from the mean 20)
or
18, 18, 22, 22 (each term is exactly 2 away from the mean 20)
etc.

The standard deviation in each case will be different. We don’t know the elements of the set and we don’t know the number of elements in the set. Because of this, there is no way for us to know the value of the standard deviation – this statement alone is not sufficient.

Statement 2: The sum of the squares of the differences from the mean is greater than 100.

“Greater than 100” encompasses a large range of numbers – it could be any value larger than 100. Again, we cannot find the exact standard deviation of the set, so this statement is also not sufficient alone.

Using both statements together, we still do not have any idea of what the elements of the set are or what the sum of the squares of the differences from the mean is. We also still don’t know the number of elements. Hence, both statements together are not sufficient, so the answer is E.

Now, let us add just one more piece of information to the problem in this similar question:

What is the standard deviation of a set of 7 numbers whose mean is 20?

Statement 1: The absolute value of the difference of each number in the set from the mean is equal.
Statement 2: The sum of the squares of the differences from the mean is greater than 100.

What would you expect the answer to be? Still E, right? The sum of the deviations are still unknown and the exact elements of the set are still unknown – all we know is the number of elements. Actually, this information is already too much. All we need to know is that the number of elements is odd and suddenly we can find the standard deviation.

Here is why:

Statement 1 is quite tricky.

If we have an odd number of elements, in which case can the absolute values of the differences of each number in the set from the mean be equal?

Think about it – the mean of the set is 20. What could a possible set look like such that the mean is 20 and the absolute values of the differences of each number in the set from the mean are equal. Try to think of such a set with just 3 elements. Can you come up with one?

19, 19, 21? No, the mean is not 20

19, 20, 21? No, the absolute value of the difference of each number in the set from the mean is not equal. 19 is 1 away from mean but 20 is 0 away from mean.

Note that in this case, the only possible set that could fit the given criteria is one consisting of just an odd number of 20s (all elements in this set must be 20). Only then can each number be equidistant from the mean, i.e. each number would be 0 away from mean. If the numbers of the set all have equal elements, then obviously the standard deviation of the set is 0. It doesn’t matter how many elements it has; it doesn’t matter what the mean is! In this case, Statement 1 alone is sufficient so the answer would be A.

Takeaway:
If a set has an even number of distinct terms, the absolute values of the distances of each term from the mean could be equal. But if a set has an odd number of terms and the absolute values of the distances of each term from the mean are equal, all the terms in the set must be the same and will be equal to the mean.

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: How to Read GMAT Questions Carefully

We all know that we need to be very careful while reading GMAT questions – that every word is important. Even small oversights can completely change an answer for you. This is what happens with many test takers who try to tackle this official question. Even though the question looks very simple, the way it is worded causes test-takers to often ignore one word, which changes the solution entirely. Let’s look at this question now:

Alice’s take-home pay last year was the same each month, and she saved the same fraction of her take-home pay each month. The total amount of money that she had saved at the end of the year was 3 times the amount of that portion of her monthly take-home pay that she did NOT save. If all the money that she saved last year was from her take-home pay, what fraction of her take-home pay did she save each month?

(A) 1/2
(B) 1/3
(C) 1/4
(D) 1/5
(E) 1/6

Let’s consider the question stem sentence by sentence:

“Alice’s take-home pay last year was the same each month, and she saved the same fraction of her take-home pay each month.”

Say Alice’s take-home pay last year was \$100 each month. She saves a fraction of this every month – let the amount saved be x.

“The total amount of money that she had saved at the end of the year was 3 times the amount of that portion of her monthly take-home pay that she did NOT save.”

What would be “the total amount of money that she had saved at the end of the year”? Since Alice saves x every month, she would have saved 12x by the end of the year.

What would be “the amount of that portion of her monthly take-home pay that she did NOT save”? Note that this is going to be (100 – x). Many test takers end up using (100 – x)*12, however this equation is not correct. The key word here is “monthly” – we are looking for how much Alice does not save each month, not how much she does not save during the whole year.

The total amount of money that Alice saved at the end of the year is 3 times the amount of that portion of her MONTHLY take-home pay that she did not save. Now we know we are looking for:

12x = 3*(100 – x)
x = 20

“If all the money that she saved last year was from her take-home pay, what fraction of her take-home pay did she save each month?”

From our equation, we have determined that Alice saved \$20 out of every \$100 she earned every month, so she saved 20/100 = 1/5 of her take-home pay.

Often, test-takers make the mistake of writing the equation as:

12x = 3*(100 – x)*12
x = 300/4

However, this will give them the fraction (300/4)/100 = 3/4, and that’s when they will wonder what went wrong.

Be extra careful when reading GMAT questions so that precious minutes are not wasted on such avoidable errors.

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: Solving the Fuel-Up Puzzles

We hope you are enjoying the puzzles we have been putting up in the last few weeks. Though all of them may not be directly convertible to GMAT questions, they are great mathematical brain teasers!

(Before we tackle today’s puzzle, first take a look at our posts on how to solve pouring water puzzles, weighing puzzles, and hourglass puzzles.)

Another variety of puzzle involves distributing fuel among vehicles to reach a destination. Let’s look at this type of question today:

A military car carrying an important letter must cross a desert. There is no petrol station in the desert, and the car’s fuel tank is just enough to take it halfway across. There are other cars with the same fuel capacity that can transfer their petrol to one another. There are no canisters to carry extra fuel or rope to tow the cars.

How can the letter be delivered?

Here, we are given that a single car can only reach the midpoint of the desert on its own tank of gas. Since there are no canisters, the car cannot carry extra fuel, so it will need to be fueled up by other cars traveling along with it.

Let’s fill up 4 cars and get them to start crossing the desert together. By the time they cover a quarter of the desert, half of their fuel tanks will be empty. Hence, we will have 4 cars with half tanks, and the status of their fuel tanks will be:

(0.5, 0.5, 0.5, 0.5)

If we transfer the fuel from two of the cars into two other cars, we will have:

(1, 1, 0, 0)

The two cars with fuel in their tanks will continue to cross the desert and cover another quarter of it. Now both of the cars will have half tanks again, and they will have reached the middle of the desert:

(0.5, 0.5, 0, 0)

Now one car will transfer all of its fuel to the other car, allowing that car to have one full tank:

(1, 0, 0, 0)

That car can then carry the letter through the remaining half of the desert.

For this problem, we didn’t really care about the stalled cars in the middle of the desert since we are not required to bring them back. The only important thing is to get the letter completely across the desert. Now, how do we handle a puzzle that asks us to get all of the vehicles back, too? Let’s look at an example question with those constraints:

A distant planet “X” has only one airport located at the planet’s North Pole. There are only 3 airplanes and lots of fuel at the airport. Each airplane has just enough fuel capacity to get to the South Pole (which is diametrically opposite the North Pole). The airplanes can land anywhere on the planet and transfer their fuel to one another.

The mission is for at least one airplane to fly completely around the globe and stay above the South Pole; in the end, all of the airplanes must return to the airport at the North Pole.

For this problem, we are given that a plane with a full tank of fuel can only reach the South Pole, i.e. cover half the distance it needs to travel for the mission. We need it to take a full trip around the planet – from the North Pole, to the South pole, and back again to North Pole. Obviously, we will need more than one plane to fuel the plane which will fly above the South pole.

Let’s divide the distance from pole to pole into thirds (from the North Pole to the South Pole we have three thirds, and from the South Pole to the North Pole we have another three thirds).

Step #1: 2 airplanes will fly to the first third. A third of their fuel will be used, so the status of their fuel tanks will be:

(2/3, 2/3)

One airplane will then fuel up the other plane and go back to the airport. Now the status of their tanks is:

(3/3, 1/3)

Step #2: 2 airplanes will again fly from the airport to the first third – one airplane will fuel up the other plane and go back to the airport. So the status of these two airplanes is this:

(3/3, 1/3)

Step #3: Now there are two airplanes at the first third mark with their tanks full. They will now fly to the second third point, giving us:

(2/3, 2/3)

One of the airplanes will fuel up the second one (until its tank is full) and go back to the first third, where it will meet the third airplane (which has just come back from the airport to support it with fuel) so that they both can return to the airport.

In the meantime, the airplane at the second third, with a full tank of fuel, will fly as far as it can – over the South Pole and towards the North pole, to the last third before the airport.

Step #4: One of the two airplanes from the airport can now go to the first third (on the opposite side of the North pole as before), and share its 1/3 fuel so that both airplanes safely land back at the airport.

And that is how we can have one plane travel completely around the planet and still have all airplanes arrive back safely!

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: Solving the Weighing Puzzle (Part 2)

A couple of weeks back, we discussed how to handle puzzles involving a two pan balance. In those problems, we learned how to tackle problems that ask you to measure items against one another.

Today, we will look at some puzzles that require the use of a traditional weighing scale. When we put an object on this scale, it shows us the weight of the object.

This is what such a scale looks like:

Puzzles involving a weighing scale can be quite tricky! Let’s take a look at a couple of examples:

You have 10 bags with 1000 coins in each. In one of the bags, all of the coins are forgeries. A true coin weighs 1 gram; each counterfeit coin weighs 1.1 grams.

If you have an accurate weighing scale, which you can use only once, how can you identify the bag with the forgeries?

We are allowed only a single weighing, so we cannot weigh all 10 bags on the scale individually to measure which one has counterfeit coins. We need to find the bag in only one weighing, so we need to somehow make the coins in the bags distinctive.

How do we do that? We can take out one coin from the first bag, two coins from the second bag, three coins from the third bag and so on. Finally, we will have 1 + 2 + 3 + … + 10 = 10*11/2 = 55 coins.

Let’s weigh these 55 coins now.

If all coins were true, the total weight would have been 55 grams. But since some coins are counterfeit, the total weight will be more. Say, the total weight comes out to be 55.2 grams. What can we deduce from this? We can deduce that there must be two counterfeit coins (because each counterfeit coin weighs 0.1 gram extra). So the second bag must be the bag of counterfeit coins.

Let’s try one more:

A genuine gummy bear has a mass of 10 grams, while an imitation gummy bear has a mass of 9 grams. You have 7 cartons of gummy bears, 4 of which contain real gummy bears while the others contain imitation bears.

Using a scale only once and the minimum number of gummy bears, how can you determine which cartons contain real gummy bears?

Now this has become a little complicated! There are three bags with imitation gummy bears. Taking a cue from the previous question, we know that we should take out a fixed number of gummy bears from each bag, but now we have to ensure that the sum of any three numbers is unique. Also, we have to keep in mind that we need to use the minimum number of gummy bears.

So from the first bag, take out no gummy bears.

From the second bag, take out 1 gummy bear.

From the third bag, take out 2 gummy bears (if we take out 1 gummy bear, the sum will be the same in case the second bag has imitation gummy bears or in case third bag has imitation gummy bears.

From the fourth bag, take out 4 gummy bears. We will not take out 3 because otherwise 0 + 3 and 1 + 2 will give us the same sum. So we won’t know whether the first and fourth bags have imitation gummy bears or whether second and third bags have imitation gummy bears.

From the fifth bag, take out 7 gummy bears. We have obtained this number by adding the highest triplet: 1 + 2 + 4 = 7. Note that anything less than 7 will give us a sum that can be made in multiple ways, such as:

0 + 1 + 6 = 7 and 1 + 2 + 4 = 7
or
0 + 1 + 5 = 6 and 0 + 2 + 4 = 6

But we need the sum to be obtainable in only one way so that we can find out which three bags contain the imitation gummy bears.

At this point, we have taken out 0, 1, 2, 4, and 7 gummy bears.

From the sixth bag, take out 13 gummy bears. We have obtained this number by adding the highest triplet: 2 + 4 + 7 = 13. Note that anything less than 13 will, again, give us a sum that can be made in multiple ways, such as:

12 + 1 + 0 = 13 and 2 + 4 + 7 = 13
or
0 + 1 + 9 = 10 and 1 + 2 + 7 = 10
…etc.

Note that this way, we are also ensuring that we measure only the minimum number of gummy bears, which is what the question asks us to do.

From the seventh bag, take out 24 gummy bears. We have obtained this number by adding the highest triplet again: 4 + 7 + 13 = 24. Again, anything less than 24 will give us a sum that can be made in multiple ways, such as:

0 + 1 + 15 = 16 and 1 + 2 + 13 = 16
or
0 + 1 + 19 = 20 and 0 + 7 + 13 = 20
or
0 + 1 + 23 = 24 and 4 + 7 + 13 = 24
…etc.

Thus, this is the way we will pick the gummy bears from the 7 bags: 0, 1, 2, 4, 7, 13, 24.

In all, 51 gummy bears will be weighed. Their total weight should be 510 grams (51*10 = 510) but because three bags have imitation gummy bears, the weight obtained will be less.

Say the weight is less by 8 grams. This means that the first bag (which we pulled 0 gummy bears from), the second bag (which we pulled 1 gummy bear from) and the fifth bag (which we pulled 7 gummy bears from) contain the imitation gummy bears. This is because 0 + 1 + 7 = 8 – note that we will not be able to make 8 with any other combination.

We hope this tricky little problem got you thinking. Work those grey cells and the GMAT will not seem hard at all!

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: Keep Your GMAT Score Safe from the Bowling Green Massacre

The hashtag of the day is #bowlinggreenmassacre, inspired by an event that never happened. Whether intentionally or accidentally (we’ll let you and your news agency of choice decide which), White House staffer Kellyanne Conway referenced the “event” in an interview, inspiring an array of memes and references along the way.

Whatever Ms. Conway’s intentions (or lack thereof; again we’ll let you decide) with the quote, she is certainly guilty of inadvertently doing one thing: she didn’t likely intend to help you avoid a disaster on the GMAT, but if you’re paying attention she did.

Your GMAT test day does not have to be a Bowling Green Massacre!

Here’s the thing about the Bowling Green Massacre: it never happened. But by now, it’s lodged deeply enough in the psyche of millions of Americans that, to them, it did. And the same thing happens to GMAT test-takers all the time. They think they’ve seen something on the test that isn’t there, and then they act on something that never happened in the first place. And then, sadly, their GMAT hopes and dreams suffer the same fate as those poor souls at Bowling Green (#thoughtsandprayers).

Here’s how it works:

The Quant Section’s Bowling Green Massacre
On the Quant section, particularly with Data Sufficiency, your mind will quickly leap to conclusions or jump to use a rule that seems relevant. Consider the example:

What is the perimeter of isosceles triangle LMN?

(1) Side LM = 4
(2) Side LN = 4√2

A. Statement (1) ALONE is sufficient, but statement (2) alone is insufficient
B. Statement (2) ALONE is sufficient, but statement (1) alone is insufficient
C. BOTH statements TOGETHER are sufficient, but NEITHER statement ALONE is sufficient
D. EACH statement ALONE is sufficient
E. Statements (1) and (2) TOGETHER are NOT sufficient

When people see that square root of 2, their minds quickly drift back to all those flash cards they studied – flash cards that include the side ratio for an isosceles right triangle: x, x, x√2. And so they then leap to use that rule, inferring that if one side is 4 and the other is 4√2, the other side must also be 4 to fit the ratio and they can then calculate the perimeter. With both statements together, they figure, they can derive that perimeter and select choice C.

But think about where that side ratio comes from: an isosceles right triangle. You’re told in the given information that this triangle is, indeed, isosceles. But you’re never told that it’s a right triangle. Much like the Bowling Green Massacre, “right” never happened. But the mere suggestion of it – the appearance of the √2 term that is directly associated with an isosceles, right triangle – baits approximately half of all test-takers to choose C here instead of the correct E (explanation: “isosceles” means only that two sides match, so the third side could be either 4, matching side LM, or 4√2, matching side LN).

Your mind does this to you often on Data Sufficiency problems: you’ll limit the realm of possible numbers to integers, when that wasn’t defined, or to positive numbers, when that wasn’t defined either. You’ll see symptoms of a rule or concept (like √2 leads to the isosceles right triangle side ratio) and assume that the entire rule is in play. The GMAT preys on your mind’s propensity for creating its own story when in reality, only part of that story really exists.

The Verbal Section’s Bowling Green Massacre
This same phenomenon appears on the Verbal section, too – most notably in Critical Reasoning. Much like what many allege that Kellyanne Conway did, your mind wants to ascribe particular significance to events or declarations, and it will often exaggerate on you. Consider the example:

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 that lie on top of the layer of lava. Therefore, ancestors of modern humans lived in Western Asia between two million and one-and-a-half 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 that lay 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 key to most Critical Reasoning problems is finding the conclusion and knowing EXACTLY what the conclusion says – nothing more and nothing less. Here the conclusion is the last sentence, that “ancestors of modern humans lived” in this region at this time. When people answer this problem incorrectly, however, it’s almost always for the same reason. They read the conclusion as “the FIRST/EARLIEST ancestors of modern humans lived…” And in doing so, they choose choice C, which protects against humans having come before the ones related to the bones we have.

“First/earliest” is a classic Bowling Green Massacre – it’s a much more noteworthy event (“scientists have discovered human ancestors” is pretty tame, but “scientists have discovered the FIRST human ancestors” is a big deal) that your brain wants to see. But it’s not actually there! It’s just that, in day to day life, you’d rarely ever read about a run-of-the-mill archaeological discovery; it would only pop up in your social media stream if it were particularly noteworthy, so your mind may very well assume that that notoriety is present even when it’s not.

In order to succeed on the GMAT, you need to become aware of those leaps that your mind likes to take. We’re all susceptible to:

• Assuming that variables represent integers, and that they represent positive numbers
• Seeing the symptoms of a rule and then jumping to apply it
• Applying our own extra superlatives or limits to conclusions

So when you make these mistakes, commit them to memory – they’re not one-off, silly mistakes. Our minds are vulnerable to Bowling Green Massacres, so on test day #staywoke so that your score isn’t among those that are, sadly, massacred.

By Brian Galvin.

# Quarter Wit, Quarter Wisdom: Solving the Hourglass Puzzle

Let’s continue our puzzles discussion today with another puzzle type – time measurement using an hourglass. (Before you continue reading this article, check out our posts on how to solve pouring water puzzles and weighing and balancing puzzles)

First, understand what an hourglass is – it is a mechanical device used to measure the passage of time. It is comprised of two glass bulbs connected vertically by a narrow neck that allows a regulated trickle of sand from the upper bulb to fall into the lower one. The sand also takes a fixed amount of time to fall from the upper bulb to the lower bulb. Hourglasses may be reused indefinitely by inverting the bulbs once the upper bulb is empty.

This is what they look like:

Say a 10-minute hourglass will let us measure time in intervals of 10 minutes. This means all of the sand will flow from the upper bulb to the lower bulb in exactly 10 minutes. We can then flip the hourglass over – now sand will start flowing again for the next 10 minutes, and so on. We cannot measure, say, 12 minutes using just a 10-minute hourglass, but we can measure more time intervals when we have two hourglasses of different times. Let’s look at this practice problem to see how this can be done:

A teacher of mathematics used an unconventional method to measure a 15-minute time limit for a test. He used a 7-minute and an 11-minute hourglass. During the whole time, he turned the hourglasses only 3 times (turning both hourglasses at once counts as one flip). Explain how the teacher measured out 15 minutes.

Here, we have a 7-minute hourglass and an 11-minute hourglass. This means we can measure time in intervals of 7 minutes as well as in intervals of 11 minutes. But consider this: if both hourglasses start together, at the end of 7 minutes, we will have 4 minutes of sand leftover in the top bulb of the 11-minute hourglass. So we can also measure out 4 minutes of time.

Furthermore, if we flip the 7-minute hourglass over at this time and let it flow for that 4 minutes (until the sand runs out of the top bulb of the 11-minute hourglass), we will have 3 minutes’ worth of sand leftover in the 7-minute hourglass. Hence, we can measure a 3 minute time interval, too, and so on…

Now, let’s see how we can measure out 15 minutes of time using our 7-minute and 11-minute hourglasses.

First, start both hourglasses at the same time. After the top bulb of the 7-minute hourglass is empty, flip it over again. At this time, we have 4 minutes’ worth of sand still in the top bulb of the 11-minute hourglass. When the top bulb of the 11-minute hourglass is empty, the bottom bulb of 7-minute hourglass will have 4 minutes’ worth of sand in it. At this point, 11 minutes have passed

Now simply flip the 7-minute hourglass over again and wait until the sand runs to the bottom bulb, which will be in 4 minutes.

This is how we measure out 11 + 4 = 15 minutes of time using a 7-minute hourglass and an 11-minute hourglass.

Let’s look at another problem:

Having two hourglasses, a 7-minute one and a 4-minute one, how can you correctly time out 9 minutes?

Now we need to measure out 9 minutes using a 7-minute hourglass and a 4-minute hourglass. Like we did for the last problem, begin by starting both hourglasses at the same time. After 4 minutes pass, all of the sand in the 4-minute hourglass will be in the lower bulb. Now flip this 4-minute hourglass back over again. In the 7-minute hourglass, there will be 3 minutes’ worth of sand still in the upper bulb.

After 3 minutes, all of the sand from the 7-minute hourglass will be in the lower bulb and 1 minute’s worth of sand will be in the upper bulb of the 4-minute hourglass.

This is when we will start our 9-minute interval.

The 1 minute’s worth of sand will flow to the bottom bulb of the 4-minute hourglass. Then we just need to flip the 4-minute hourglass over and let all of the sand flow out (which will take 4 minutes), and then flip the hourglass over to let all of the sand flow out again (which will take another 4 minutes).

In all, we have measured out a 1 + 4 + 4 = 9-minute interval, which is what the problem has asked us to find.

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: Solving the Weighing and Balancing Puzzle

Let’s continue the discussion on puzzles that we began last week. Today we look at another kind of puzzle – weighing multiple objects using a two-pan balance while we are given a limited number of times to weight the objects against each other.

First of all, do we understand what a two-pan balance looks like?

Here is a picture.

As you can see, it has two pans that will be even if the weights in them are equal. If one pan has heavier objects in it, that pan will go down due to the weight. With this in mind, let’s try our first puzzle:

One of twelve coins is a bit lighter than the other 11 (which have the same weight). How would you identify this lighter coin if you could use a two-pan balance scale only 3 times? (You can only balance one set of coins against another, i.e. you have no weight measurements.)

There are various ways in which we can solve this.

We are given 12 coins, all of same weight, except one which is a bit lighter.

Let’s split the coins into two groups of 6 coins each and put them in the two pans. Since there is one lighter coin, one pan will be lighter than the other and will rise higher. So now we know that one of these 6 coins is the lighter coin.

Now split these 6 coins into another two groups of 3 coins each. Again, one pan will rise higher since it will have the lighter coin on it. Now we know that one of these three coins is the lighter coin.

Now what do we do? We have 3 coins and we cannot split them equally. What we can do is put one coin in each pan. What happens if the pans are not balanced? Then we know the pan that rises higher has the lighter coin on it (and thus, we have identified our coin). But what if both pans are balanced? The catch is that then the leftover coin is the lighter one! In any case, we would be able to identify the lighter coin using this strategy.

We hope you understand the logic here. Now let’s try another puzzle:

One of 9 coins is a bit lighter than the other 8. How would you identify this lighter coin if you could use a two-pan balance scale only 2 times?

Now we can use the balance only twice, and we are given an odd number of coins so we cannot split them evenly. Recall what we did in the first puzzle when we had an odd number of coins – we put one coin aside. What should we do here? Can we try putting 1 coin aside and splitting the rest of the 8 coins into two groups of 4 each? We can but once we have a set of 4 coins that contain the lighter coin, we will still need 2 more weighings to isolate the light coin, and we only have a total of 2 weighings to use.

Instead, we should split the 9 coins into 3 groups of 3 coins each. If we put one group aside and put the other two groups into the two pans of the scale, we will be able to identify the group which has the lighter coin. If one pan rises up, then that pan is holding the lighter coin; if the pans weight the same, then the group put aside has the lighter coin in it.

Now the question circles back to the strategy we used in the first puzzle. We have 3 coins, out of which one is lighter than the others, and we have only one chance left to weigh the coins. Just like in the first puzzle, we can put one coin aside and weigh the other two against each other – if one pan rises, it is holding the lighter coin, otherwise the coin put aside is the lighter coin! Thus, we were able to identify the lighter coin in just two weighings. Can you use the same method to answer the first puzzle now?

We will leave you with a final puzzle:

On a Christmas tree there were two blue, two red, and two white balls. All seemed the same, however in each color pair, one ball was heavier. All three lighter balls weighed the same, just like all three heavier balls weighed the same. Using a 2-pan balance scale only twice, identify the lighter balls.

Can you solve this problem using the strategies above? Let us know in the comments!

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: Solving the Pouring Water Puzzle

Some time back, we came across a GMAT Data Sufficiency word problem question based on the “pouring water puzzle”. That made us think that it is probably a good idea to be comfortable with the various standard puzzle types. From this week on, we will look at some fundamental puzzles to acquaint ourselves with these mind benders in case we encounter them on test day.

Today, we will look at the popular “pouring water puzzle”. You may remember a similar puzzle from the movie Die Hard with a Vengeance, where Bruce Willis and Samuel L. Jackson had to diffuse a bomb by placing a 4 gallon jug of water on a set of scales.

Here is the puzzle:

You have a 3- and a 5-liter water container – each container has no markings except for that which gives us its total volume. We also have a running tap. We must use the containers and the tap in such a way that we measure out exactly 4 liters of water. How can this be done?

Don’t worry that this question is not written in a traditional GMAT format! We need to worry only about the logic behind the puzzle – we can then answer any question about it that is given in any GMAT format.

Let’s break down what we are given. We have only two containers – one of 3-liter and the other of 5-liter capacity. The containers have absolutely no markings on them other than those which give us the total volumes, i.e. the markings for 3 liters and 5 liters respectively. There is no other container. We also have a tap/faucet of running water, so basically, we have an unlimited supply of water. Environmentalists may not like my saying this, but this fact means we can throw out water when we need to and just refill again.

STEP 1: Let’s fill up the 5-liter container with water from the tap. Now we are at (5, 0), with 5 being the liters of water in the 5-liter container, and 0 being the liters of water in the 3-liter container.

STEP 2: Now, there is nothing we can do with this water except transfer it to the 3-liter container (there is no other container and throwing out the water will bring us back to where we started). After we fill up the 3-liter container, we are left with 2 liters of water in the 5-liter container. This brings us to (2, 3).

STEP 3: We gain nothing from transferring the 3 liters of water back to 5-liter container, so let’s throw out the 3 liters that are in the 3-liter container. Because we just threw out the water from the 3-liter container, we will gain nothing by simply refilling it with 3 liters of water again. So now we are at (2, 0).

STEP 4: The next logical step is to transfer the 2 liters of water we have from the 5-liter container to the 3-liter container. This means the 3-liter container has space for 1 liter more until it reaches its maximum volume mark. This brings us to (0, 2).

STEP 5: Now fill up the 5-liter container with water from the tap and transfer 1 liter to the 3-liter container (which previously had 2 liters of water in it). This means we are left with 4 liters of water in the 5-liter container. Now we are at (4, 3).

This is how we are able to separate out exactly 4 liters of water without having any markings on the two containers. We hope you understand the logic behind solving this puzzle. Let’s take a look at another question to help us practice:

We are given three bowls of 7-, 4- and 3-liter capacity. Only the 7-liter bowl is full of water. Pouring the water the fewest number of times, separate out the 7 liters into 2, 2, and 3 liters (in the three bowls).

This question is a little different in that we are not given an unlimited supply of water. We have only 7 liters of water and we need to split it into 2, 2 and 3 liters. This means we can neither throw away any water, nor can we add any water. We just need to work with what we have.

We start off with (7, 0, 0) – with 7 being the liters of water in the 7-liter bowl, the first 0 being the liters of water in the 4-liter bowl, and the second 0 being the liters of water in the 3-liter bowl – and we need to go to (2, 2, 3). Let’s break this down:

STEP 1: The first step would obviously be to pour water from the 7-liter bowl into the 4-liter bowl. Now you will have 3 liters of water left in the 7-liter bowl. We are now at (3, 4, 0).

STEP 2: From the 4-liter bowl, we can now pour water into the 3-liter bowl. Now we have 1 liter in the 4-liter bowl, bringing us to (3, 1, 3).

STEP 3: Empty out the 3-liter bowl, which is full, into the 7-liter bowl for a total of 6 liters – no other transfer makes sense [if we transfer 1 liter of water to the 7-liter bowl, we will be back at the (4, 0, 3) split, which gives us nothing new]. This brings us to (6, 1, 0).

STEP 4: Shift the 1 liter of water from the 4-liter bowl to the 3-liter bowl. We are now at (6, 0, 1).

STEP 5: From the 7-liter bowl, we can now shift 4 liters of water into the 4-liter bowl. This leaves us with with 2 liters of water in the 7-liter bowl. Again, no other transfer makes sense – pouring 1 liter of water into some other bowl takes us back to a previous step. This gives us (2, 4, 1).

STEP 6: Finally, pour water from the 4-liter bowl into the 3-liter bowl to fill it up. 2 liters will be shifted, bringing us to (2, 2, 3). This is what we wanted.

We took a total of 6 steps to solve this problem. At each step, the point is to look for what helps us advance forward. If our next step takes us back to a place at which we have already been, then we shouldn’t take it.

Keeping these tips in mind, we should be able to solve most of these pouring water puzzles in the future!

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!

# 4 Predictions for Test Prep and Admissions in 2017

There goes another year. Seemingly no sooner than it started, 2016 has packed up and stormed off, leaving many dizzy in its wake. Now that 2017 is underway, it’s time to dust off the old Veritas Prep crystal ball and see what may be in store for 2017 in the worlds of test preparation and admissions. Odds are that we won’t be right on all of these — and we may even manage to get all four wrong — but let’s dig in and predict a few things that we expect to see in 2017:

One-year MBA programs will reach a tipping point in the United States.
For decades, one-year programs have been more popular in Europe than in the United States, although some prominent American programs, such as Kellogg, have moved to expand their one-year programs in recent years. With more and more articles appearing in the media about students and their families questioning the costs of higher education, accelerated programs will keep looking more and more appealing to applicants who don’t want to spend six figures on an MBA. The globalization of management graduate education will continue, and more American business schools will start to embrace what’s traditionally been a more Euro-flavored program type.

Video prompts will become much more common in business school applications.
Yes, we predicted this last year, and it didn’t quite come to fruition. But, schools are becoming more and more comfortable with video as a medium for learning about applicants, and — probably more importantly — applicants themselves mostly seem to be comfortable with video. In AIGAC’s 2016 MBA Applicant Survey, only 16% of applicants surveyed said that video responses were the most challenging part of the application. That’s far smaller than the percentage of applicants who said that standardized tests (61%) and written essays (46%) were the most challenging! Rotman, Yale, Kellogg, and McCombs have helped blaze a video trail that we expect others will soon follow.

An Asia-scale cheating scandal will hit the SAT or ACT in the United States.
News articles about standardized test cheating scandals like this one and this one seem to come out nearly every month. Much of the blame lies with the pressure that students — and especially their families — put on themselves to do well on these exams.

It’s also greed. For every student that will do anything to do well on an exam, there’s a person or company who’s happy to take their money and do whatever it takes to give that student a leg up. Sometimes that means legally and ethically training that student to perform to the best of their ability, but many other times it means falsifying documents or providing students with live test questions for large sums of money. This kind of greed exists everywhere in the world, and it’s only a matter of time until a similar large-scale scandal happens in the U.S.

Community colleges will gain a lot more recognition.
Did you know that more than half of students who enroll in college first do so at a community college? Most Americans don’t know that, even though community colleges have been the engine that educates millions of Americans each year. We’ll see the federal government putting more emphasis on jobs and job training in the coming year, and community colleges are perfectly positioned to serve that role. While it remains to be seen whether community colleges get all of the funding they need to keep serving their mission, we expect that, at a minimum, they’ll start to get more recognition for the job they do to train and retrain America’s workforce.

Happy New Year, everyone. We can’t wait to check back in 2018 and see how this year turned out!

By Scott Shrum

# How to Answer GMAT Questions That are About an Unfamiliar Topic

Usually, GMAT questions that are based on your field of work or interests are simple to comprehend and relatively easy to answer correctly. But what happens when, say, an engineer gets a question based on psychiatry? Is he or she bound to fail? No.

Remember that the GMAT offers a level playing field for test takers from different backgrounds – it doesn’t matter whether your major was literature or physics. If you feel lost on a question about renaissance painters, remember that the guy next to you is lost on the problem involving planetary systems.

So how can you successfully handle GMAT questions on any topic? By sticking to the basics. The logic and reasoning required to answer these questions will stay the same no matter which field the information in the question stem comes from.

To give an example of this, let’s today take a look at a GMAT question involving psychoanalysis:

Studies in restaurants show that the tips left by customers who pay their bill in cash tend to be larger when the bill is presented on a tray that bears a credit-card logo. Consumer psychologists hypothesize that simply seeing a credit-card logo makes many credit-card holders willing to spend more because it reminds them that their spending power exceeds the cash they have immediately available.

Which of the following, if true, most strongly supports the psychologists’ interpretation of the studies?

(A) The effect noted in the studies is not limited to patrons who have credit cards.
(B) Patrons who are under financial pressure from their credit-card obligations tend to tip less when presented with a restaurant bill on a tray with credit-card logo than when the tray has no logo.
(C) In virtually all of the cases in the studies, the patrons who paid bills in cash did not possess credit cards.
(D) In general, restaurant patrons who pay their bills in cash leave larger tips than do those who pay by credit card.
(E) The percentage of restaurant bills paid with given brand of credit card increases when that credit card’s logo is displayed on the tray with which the bill is prepared.

Let’s break down the argument:

Argument: Studies show that cash tips left by customers are larger when the bill is presented on a tray that bears a credit-card logo.

Why would that be? Why would there be a difference in customer behavior when the tray has no logo from when the tray has a credit card logo? Psychologists’ hypothesize that seeing a credit-card logo reminds people of the spending power given by the credit card they carry (and that their spending power exceeds the actual cash they have right now).

The question asks us to support the psychologists’ interpretation. And what is the psychologists’ interpretation of the studies? It is that seeing a logo reminds people of their own credit card status. Say we change the argument a little by adding a line:

Argument: Studies show that cash tips left by customers are larger when the bill is presented on a tray that bears a credit-card logo. Patrons under financial pressure from credit-card obligations tend to tip less when presented with a restaurant bill on a tray with credit-card logo than when the tray has no logo.

Now, does the psychologists’ interpretation make even more sense? The psychologists’ interpretation is only that “seeing a logo reminds people of their own credit card status.” The fact “that their spending power exceeds the cash they have right now” explains the higher tips. If we are given that some customers tip more upon seeing that card logo and some tip less upon seeing it, it makes sense, right? Different people have different credit card obligation status, hence, people are reminded of their own card obligation status and they tip accordingly.

Answer choice B increases the probability that the psychologists’ interpretation is true because it tells you that in the cases of very high credit card obligations, customers tip less. This is what you would expect if the psychologists’ interpretation were correct.

In simpler terms, the logic here is similar to the following situation:

A: After 12 hours of night time sleep, I can’t study.
B: Yeah, because your sleep pattern is linked to your level of concentration. After a long sleep, your mind is still muddled and lazy so you can’t study.
A: After only 4 hrs of night time sleep, I can’t study either.

Does B’s theory make sense? Sure! B’s theory is that “sleep pattern is linked to level of concentration.” If A sleeps too much, her concentration is affected. If she sleeps too little, again her concentration is affected. So B’s theory certainly makes more sense.

(E) The percentage of restaurant bills paid with given brand of credit card increases when that credit card’s logo is displayed on the tray with which the bill is prepared.

This option supports the hypothesis that credit card logos remind people of their own card – not of their card obligations. The psychologists’ interpretation talks about the logo reminding people of their card status (high spending power or high obligations). Hence, this option is not correct.

Now let’s examine the rest of the answer choices to see why they are also incorrect:

(A) The effect noted in the studies is not limited to patrons who have credit cards.

This argument is focused only on credit cards, not on credit cards and their logos, so this is irrelevant.

(C) In virtually all of the cases in the studies, the patrons who paid bills in cash did not possess credit cards.

This option questions the validity of the psychologists’ interpretation. Hence, this is also not correct.

(D) In general, restaurant patrons who pay their bills in cash leave larger tips than do those who pay by credit card.

This argument deals with people who have credit cards but are tipping by cash, hence this is also irrelevant.

We hope you see that if you approach GMAT questions logically and stick to the basics, it is not hard to interpret and solve them, even if they include information from an unfamiliar field.

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: Taking the Least Amount of Time to Solve “At Least” Probability Problems

In its efforts to keep everyone from getting perfect 800s, the GMAT has two powerful tools to stop you from perfection. For one, it can bait you into wrong answers (with challenging content, tempting trap answers, or a combination thereof). And secondly, it can waste your time, making it look like you need to do a lot of work when there’s a much simpler way.

Fortunately, and contrary to popular belief, the GMAT isn’t “pure evil.” Wherever it provides opportunities for less-savvy examinees to waste their time, it also provides a shortcut for those who have put in the study time to learn it or who have the patience to look for the elevator, so to speak, before slogging up the stairs. And one classic example of that comes with the “at least one” type of probability question.

To illustrate, let’s consider an example:

In a bowl of marbles, 8 are yellow, 6 are blue, and 4 are black. If Michelle picks 2 marbles out of the bowl at random and at the same time, what is the probability that at least one of the marbles will be yellow?

(A) 5/17
(B) 12/17
(C) 25/81
(D) 56/81
(E) 4/9

Here, you can first streamline the process along the lines of one of those “There are two types of people in the world: those who _______ and those who don’t _______” memes. Your goal is to determine whether you get a yellow marble, so you don’t care as much about “blue” and “black”…those can be grouped into “not yellow,” thereby giving you only two groups: 8 yellow marbles and 10 not-yellow marbles. Fewer groups means less ugly math!

But even so, trying to calculate the probability of every sequence that gives you one or two yellow marbles is labor intensive. You could accomplish that “not yellow” goal several ways:

First marble: Yellow; Second: Not Yellow
First: Not Yellow; Second: Yellow
First: Yellow; Second: Yellow

That’s three different math problems each involving fractions and requiring attention to detail. There ought to be an easier way…and there is. When a probability problem asks you for the probability of “at least one,” consider the only situation in which you WOULDN’T get at least one: if you got none. That’s a single calculation, and helpful because if the probability of drawing two marbles is 100% (that’s what the problem says you’re doing), then 100% minus the probability of the unfavorable outcome (no yellow) has to equal the probability of the favorable outcome. So if you determine “the probability of no yellow” and subtract from 1, you’re finished. That means that your problem should actually look like:

PROBABILITY OF NO YELLOW, FIRST DRAW: 10 non-yellow / 18 total
PROBABILITY OF NO YELLOW, SECOND DRAW: 9 remaining non-yellow / 17 remaining total

10/18 * 9/17 reduces to 10/2 * 1/17 = 5/17. Now here’s the only tricky part of using this technique: 5/17 is the probability of what you DON’T want, so you need to subtract that from 1 to get the probability you do want. So the answer then is 12/17, or B.

More important than this problem is the lesson: when you see an “at least one” probability problem, recognize that the probability of “at least one” equals 100% minus the probability of “none.” Since “none” is always a single calculation, you’ll always be able to save time with this technique. Had the question asked about three marbles, the number of favorable sequences for “at least one yellow” would be:

Yellow Yellow Yellow
Yellow Not-Yellow Not-Yellow
Yellow Not-Yellow Yellow
Yellow Yellow Not-Yellow
Not-Yellow Yellow Yellow

(And note here – this list is not yet exhaustive, so under time pressure you may very well forget one sequence entirely and then still get the problem wrong even if you’ve done the math right.)

Whereas the probability of No Yellow is much more straightforward: Not-Yellow, Not-Yellow, Not-Yellow would be 10/18 * 9/17 * 8/16 (and look how nicely that last fraction slots in, reducing quickly to 1/2). What would otherwise be a terrifying slog, the “long way” becomes quite quick the shorter way.

So, remember, when you see “at least one” probability on the GMAT, employ the “100% minus probability of none” strategy and you’ll save valuable time on at least one Quant problem on test day.

By Brian Galvin.

# Solving GMAT Geometry Problems That Involve Infinite Figures

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

Karishma, a Computer Engineer with a keen interest in alternative Mathematical approaches, has mentored students in the continents of Asia, Europe and North America. She teaches the GMAT for Veritas Prep and regularly participates in content development projects such as this blog!

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

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

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

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

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

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

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

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

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

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

In our question we know that:

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

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

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

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

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

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

Thus, our answer to this question is A.

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

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

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

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

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

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

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

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 Sentence Correction: How to Tackle Inverted Sentence Structures

One of the challenges test-takers encounter on Sentence Correction questions is the tendency of question writers to structure sentences in a way that departs from the way we typically write or speak. Take a simple example: “My books are on the table,” could also be written as “On the table are my books.” If you’re like me, you cringe a little bit with the second option – it sounds starchy and pretentious, but it’s a perfectly legitimate sentence, and an example of what’s called “inverted structure.”

In a standard structure, the subject will precede the verb. In an inverted structure, the subject comes after the verb. The tipoff for such a construction is typically a prepositional phrase – in this case, “on the table,” followed by a verb. It is important to recognize that the object of the prepositional phrase, “table,” cannot be the subject of the verb, “are,” so we know that the subject will come after the verb.

Let’s look at an example from an official GMAT question:

The Achaemenid empire of Persia reached the Indus Valley in the fifth century B.C., bringing the Aramaic script with it, from which was derived both northern and southern Indian alphabets.

(A) the Aramaic script with it, from which was derived both northern and
(B) the Aramaic script with it, and from which deriving both the northern and the
(C) with it the Aramaic script, from which derive both the northern and the
(D) with it the Aramaic script, from which derives both northern and
(E) with it the Aramaic script, and deriving from it both the northern and

The first thing you might notice is the use of the relative pronoun “which.” We’d like for “which” to be as close to as possible to its referent. So what do we think the alphabets were derived from? From the Aramaic script.

Notice that in options A and B, the closes referent to “which” is “it.” There are two problems here. One, it would be confusing for one pronoun, “which,” to have another pronoun, “it,” as its antecedent. Moreover, “it” here seems to refer to the Achaemenid Empire. Do we think that the alphabets derived from the empire? Nope. Eliminate A and B. Though E eliminates the “which,” this option also seems to indicate that the alphabets derived from the empire, so E is out as well.

We’re now down to C and D. Notice that our first decision point is to choose between “from which derive” and “from which derives.” This is an instance of inverted sentence structure. We have the prepositional phrase “from which,” followed immediately by a verb “derive or “derives.” Thus, we know that the subject for this verb is going to come later in the sentence, in this case, the northern and southern alphabets.  If we were to rearrange the sentences so that they had a more conventional structure, our choice would be between the following options:

C) Both the northern and the southern Indian alphabets derive from [the empire.]

or

D) Both northern and southern Indian alphabets derives from [the empire.]

Because “alphabets” is plural, we want to pair this subject with the plural verb, “derive.” Therefore, the correct answer is C.

Takeaway: anytime we see the construction “prepositional phrase + verb,” we are very likely looking at a sentence with an inverted sentence structure. In these cases, make sure to look for the subject of the sentence after the verb, rather than before.

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

# Advanced Number Properties on the GMAT – Part VII

We have seen a number of posts on divisibility, odd-even concepts and perfect squares. Individually, each topic has very simple concepts but when they all come together in one GMAT question, it can be difficult to wrap one’s head around so many ideas. The GMAT excels at giving questions where multiple concepts are tested. Let’s take a look at one such Data Sufficiency question today:

If p, x, and y are positive integers, y is odd, and p = x^2 + y^2, is x divisible by 4?

1) When p is divided by 8, the remainder is 5.
2) x – y = 3

This Data Sufficiency question has a lot of information in the question stem.  First, we need to sort through this information before we move on to the statements.

We know that p, x and y are positive integers. y is an unknown odd number, so it can be written in the form 2n + 1. We also know that p = x^2 + y^2.

Because y is written in the form 2n + 1, y^2 can be written as:

y^2 =(2n + 1)^2

y^2 = 4n^2 + 4n + 1

y^2 = 4n(n + 1) + 1

An interesting thing to note here is that one case of n and (n+1) will be odd and the other will be even. In every case, n(n + 1) is even. Therefore, y^2 is 1 more than a multiple of 8. In other words, we can write it as y^2 = 8m + 1.

Now we can say p = x^2 + 8m + 1.

With this in mind, is x divisible by 4? Let’s examine the statements to find out:

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

Because y^2 = 8m + 1, we can see that when y^2 is divided by 8, the remainder will be 1. Therefore, to get a remainder of 5 when p is divided by 8, when x^2 is divided by 8, we should get a remainder of 4.

Now we know that x^2 can be written in the form 8a + 4 (i.e. we can make “a’” groups of 8 each and have 4 leftover).

x^2 = 4*(2a + 1)

So x = 2 * √(an odd number)

Note that square root of an odd number will be an odd number only. If there is no 2 in the perfect square, obviously there was no 2 in the number, too.

So, x = 2 * some other odd number, which means x will be a multiple of 2, but not of 4 definitely. This statement alone is sufficient.

Now let’s look at the next statement:

Statement 2: x – y = 3

Since y is odd, we can say that x will be even (an even – an odd = an odd). But whether x is divisible by 2 only or by 4 as well, we cannot say since we have no constraints on p.

This statement alone is not sufficient to answer the question. Therefore, our answer is A.

Test takers might feel that not every step in this solution is instinctive. For example, how do we know that we should put y^2 in the form 4n(n+1) + 1? Keep the target in mind – we know that we need to find whether x is divisible by 4. Hence, try to get everything in terms of multiples of 4 + a remainder.

See you next week!

(For more advanced number properties on the GMAT, check out Parts IIIIIIIV, V and VI of this series.)

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: 3 Guiding Principles for Exponent Problems

If you’re like many GMAT examinees, you’ve found yourself in this familiar situation. You KNOW the rules for exponents. You know them cold. When you’re multiplying the same base and different exponents, you add the exponents. When you’re taking one exponent to another power, you multiply those exponents. A negative exponent? Flip that term into the denominator. A number to the zero power? You’ve got yourself a 1.

But as thoroughly and quickly as you know those rules, this exponent-based problem in front of you has you stumped. You know what you need to KNOW, but you’re not quite sure what you need to DO. And that’s an ever-important part about taking the GMAT – it’s necessary to know the core rules, facts, and formulas, but it’s also every bit as important to have action items for how you’ll apply that knowledge to tricky problems.

For exponents, there are three “guiding principles” that you should keep in mind as your action items. Any time you’re stuck on an exponent-based problem, look to do one (or more) of these things:

1) Find Common Bases
Most of the exponent rules you know only apply when you’re dealing with two exponents of the same base. When you multiply same-base exponents, you add the exponents; when you divide two same-base exponents, you subtract. And if two exponents of the same base are set equal, then you know that the exponents are equal. But keep in mind – these major rules all require you to be using exponents with the same base! If the GMAT gives you a problem with different bases, you have to find ways to make them common, usually by factoring them into their prime bases.

So for example, you might see a problem that says that:

2^x * 4^2x = 8^y. Which of the following must be true?

(A) 3x = y
(B) x = 3y
(C) y = (3/5)x
(D) x = (3/5)y
(E) 2x^2 = y

In order to apply any rules that you know, you must get the bases in a position where they’ll talk to each other. Since 2, 4, and 8 are all powers of 2, you should factor them all in to base 2, rewriting as:

2^x * (2^2)^2x = (2^3)^y

Which simplifies to:

2^x * 2^4x = 2^3y

Now you can add together the exponents on the left:

2^5x = 2^3y

And since you have the same base set equal with two different exponents, you know that the exponents are equal:

5x = 3y

This means that you can divide both sides by 5 to get x = (3/5)y, making answer choice D correct. But more importantly in a larger context, heed this lesson – when you see an exponent problem with different bases for multiple exponents, try to find ways to get the bases the same, usually by prime-factoring the bases.

2) Factor to Create Multiplication
Another important thing about exponents is that they represent recurring multiplication. x^5, for example, is x * x * x * x * x…it’s a lot of x’s multiplied together. Naturally, then, pretty much all exponent rules apply in cases of multiplication, division, or more exponents – you don’t have rules that directly apply to addition or subtraction. For that reason, when you see addition or subtraction in an exponent problem, one of your core instincts should be to factor common terms to create multiplication or division so that you’re in a better position to leverage the rules you know. So, for example, if you’re given the problem:

2^x + 2^(x + 3) = (6^2)(2^18). What is the value of x?

(A) 18
(B) 20
(C) 21
(D) 22
(E) 24

You should see that in order to do anything with the left-hand side of the equation, you’ll need to factor the common 2^x in order to create multiplication and be in a position to divide and cancel terms from the right. Doing so leaves you with:

2^x(1 + 2^3) = (6^2)(2^18)

Here, you can simplify the 1 + 2^3 parenthetical: 2^3 = 8, so that term becomes 9, leaving you with:

9(2^x) = (6^2)(2^18)

And here, you should heed the wisdom from above and find common bases. The 9 on the left is 3^2, and the 6^2 on the right can be broken into 3^2 * 2^2. This gives you:

(3^2)(2^x) = (3^2)(2^2)(2^18)

Now the 3^2 terms will cancel, and you can add the exponents of the base-2 exponents on the right. That means that 2^x = 2^20, so you know that x = 20. And a huge key to solving this one was factoring the addition into multiplication, a crucial exponent-based action item on test day.

3) Test Small Numbers and Look For Patterns
Remember: exponents are a way to denote repetitive, recurring multiplication. And when you do the same thing over and over again, you tend to get similar results. So exponents lend themselves well to finding and extrapolating patterns. When in doubt – when a problem involves too much abstraction or too large of numbers for you to get your head around – see what would happen if you replaced the large or abstract terms with smaller ones, and if you find a pattern, then look to extrapolate it. With this in mind, consider the problem:

What is the tens digit of 11^13?

(A) 1
(B) 2
(C) 3
(D) 4
(E) 5

Naturally, calculating 11^13 without a calculator is a fool’s errand, but you can start by taking the first few steps and seeing if you establish a pattern:

11^1 = 11 –> tens digit of 1

11^2 = 121 –> tens digit of 2

11^3 = 1331 –> tens digit of 3

And depending on how much time you have you could continue:

11^4 = 14641 –> tens digit of 4

But generally feel pretty good that you’ve established a recurring pattern: the tens digit increases by 1 each time, so by 11^13 it will be back at 3. So even though you’ll never know exactly what 11^13 is, you can be confident in your answer.

Remember: the GMAT is a test of how well you apply knowledge, not just of how well you can memorize it. So for any concept, don’t just know the rules, but also give yourself action items for what you’ll do when problems get tricky. For exponent problems, you have three guiding principles:

1) Find Common Bases
2) Factor to Create Multiplication
3) Test Small Numbers to Find a Pattern

By Brian Galvin.

# How NOT to Write the Equation of a Line on the GMAT

A question brought an interesting situation to our notice. Let’s start by asking a question: How do we write the equation of a line? There are two formulas:

y = mx + c (where m is the slope and c is the y-intercept)
and
yy1 = m * (xx1) [where m is the slope and (x1,y1) is a point on the line]

We also know that m = (y2y1)/(x2x1) – this is how we find the slope given two points that lie on a line. The variables are x1, y1 and x2, y2, and they represent specific values.

But think about it, is m = (y2y)/(xx1) really the equation of a line? Let’s further clarify this idea using a GMAT practice question:

In the coordinate plane, line k passes through the origin and has slope 2. If points (3,y) and (x,4) are on line k, then x + y =

(A) 3.5
(B) 7
(C) 8
(D) 10
(E) 14

We have been given that the line passes through (0, 0) and has a slope of 2. We can find the equation of the line from this information.

y = mx + c
y = 2x + 0 (Since the line passes through (0, 0), its y-intercept is 0 – when x is 0, y is also 0.)
y = 2x

Since we are given two other points, (3, y) and (x, 4), on the line and we have a slope of 2, many test-takers will be tempted to make another equation for the line using this information.

(4 – y)/(x – 3) = 2
(4 – y) = 2*(x – 3)
Thus, 2+ y = 10

Here, test-takers will use the two equations to solve for x and y and get x = 5/2 and y = 5.

After adding x and y together, they then wonder why 7.5 is not one of the answer choices. If this were an actual GMAT question, it is quite likely that 7.5 would have been one of the options. So all in all, the test-taker would not even have realized that he or she made a mistake, and would choose 7.5 as the (incorrect) answer.

The error is conceptual here. Note that the equation of the line, 2x + y = 10, is not the same as the equation we obtained above, y = 2x. They represent two different lines, but we have only a single line in the question. So which is the actual equation of that line?

To get the second equation, we have used m = (y2y)/(xx1). But is this really the equation of a line? No. This formula doesn’t have y and x, the generic variables for the x– and y-coordinates in the equation of a line.

To further clarify, instead of x and y, try using the variables a and b in the question stem and see if it makes sense:

“In the coordinate plane, line k passes through the origin and has slope 2. If points (3, a) and (b, 4) are on line k, then a + b =”

You can write (4 – a)/(b – 3) = 2 and this would be correct. But can we solve for both a and b here? No – we can write one of them in terms of the other, but we can’t get their exact values.

We know a and b must have specific values. (3, a) is a point on the line y = 2x. For x = 3, the value of of the y-coordinate, a, will be y = 2*3 = 6. Therefore, a = 6.

(b, 4) is also on the line y = 2x. So if the y-coordinate is 4, the x-coordinate, b, will be 4 = 2b, i.e. b = 2. Thus, a + b = 6 + 2 = 8, and our answer is C.

This logic remains the same even if the variables used are x and y, although test-takers often get confused because of it. Let’s solve the question in another way using the variables as given in the original question.

Recall what we have learned about slope in the past. If the slope of the line is 2 and the point (0, 0) lies on the line, the value of y – if point (3, y) also lies on the line – will be 6 (a slope of 2 means a 1-unit increase in x will lead to a 2-unit increase in y).

Again, if point (x, 4) lies on the line too, an increase of 4 in the y-coordinate implies an increase of 2 in the x-coordinate. So x will be 2, and again, x + y = 2 + 6 = 8.

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!

# 3 Formats for GMAT Inequalities Questions You Need to Know

As if solving inequalities wasn’t already hard enough, sometimes the way a GMAT question is framed will make us wonder which answer option to choose, even after we have already solved solved the problem.

Let’s look at three different question formats today to understand the difference between them:

1. Must Be True
2. Could Be True
3. Complete Range

Case 1: Must Be True
If |-x/3 + 1| < 2, which of the following must be true?
(A) x > 0
(B) x < 8
(C) x > -4
(D) 0 < x < 3
(E) None of the above

We have two linked inequalities here. One is |-x/3 + 1| < 2 and the other is the correct answer choice. We need to think about how the two are related.

We are given that |-x/3 + 1| < 2. So we know that x satisfies this inequality. That will give us the universe which is relevant to us. x will take one of those values only. So let’s solve this inequality. (We will not focus on how to solve the inequality in this post – it has already been discussed here. We will just quickly show the steps.)

|x/3 – 1| < 2
(1/3) * |x – 3| < 2
|x – 3| < 6

The distance of x from 3 is less than 6, so -3 < x < 9. Now we know that every value that x can take will lie within this range.

The question now becomes: what must be true for each of these values of x? Let’s assess each of our answer options with this question:

(A) x > 0
Will each of the values of x be positive? No – x could be a negative number greater than -3, such as -2.

(B) x < 8
Will each of the values of x be less than 8? No – x could be a number between 8 and 9, such as 8.5

(C) x > -4
Will each of the values of x be more than -4? Yes! x will take values ranging from -3 to 9, and each of the values within that range will be greater than -4. So this must be true.

(D) 0 < x < 3
Will each of these values be between 0 and 3. No – since x can take any of the values between -3 and 9, not all of these will be just between 0 and 3.

Therefore, the answer is C (we don’t even need to evaluate answer choice E since C is true).

Case 2: Could Be True
If −1 < x < 5, which is the following could be true?
(A) 2x > 10
(B) x > 17/3
(C) x^2 > 27
(D) 3x + x^2 < −2
(E) 2x – x^2 < 0

Again, we have two linked inequalities, but here the relation between them will be a bit different. One of the inequalities is  −1 < x < 5 and the other will be the correct answer choice.

We are given that -1 < x < 5, so x lies between -1 and 5. We need an answer choice that “could be true”. This means only some of the values between -1 and 5 should satisfy the condition set by the correct answer choice – all of the values need not satisfy. Let’s evaluate our answer options:

(A) 2x > 10
x > 5
No values between -1 and 5 will be greater than 5, so this cannot be true.

(B) x > 17/3
x > 5.67
No values between -1 and 5 will be greater than 5.67, so this cannot be true.

(C) x^2 > 27
x^2 – 27 > 0
x > 3*√(3) or x < -3*√(3)
√(3) is about 1.73 so 3*1.73 = 5.19. No value of x will be greater than 5.19. Also, -3*1.73 will be -5.19 and no value of x will be less than that. So this cannot be true.

(Details on how to solve such inequalities are discussed here.)

(D) 3x + x^2 < −2
x^2 + 3x + 2 < 0
(x + 1)(x + 2) < 0
-2 < x < -1
No values of x will lie between -2 and -1, so this also cannot be true.

(E) 2x – x^2 < 0
x * (x – 2) > 0
x > 2 or x < 0
If -1 < x < 5, then x could lie between -1 and 0 (x < 0 is possible) or between 2 and 5 (x > 2 is possible). Therefore, the correct answer is E.

Case 3: Complete Range
Which of the following represents the complete range of x over which x^3 – 4x^5 < 0?
(A) 0 < |x| < ½
(B) |x| > ½
(C) -½ < x < 0 or ½ < x
(D) x < -½ or 0 < x < ½
(E) x < -½ or x > 0

We have two linked inequalities, but the relation between them will be a bit different again. One of the inequalities is  x^3 – 4x^5 < 0 and the other will be the correct answer choice.

We are given that x^3 – 4x^5 < 0. This inequality can be solved to:

x^3 ( 1 – 4x^2) < 0
x^3*(2x + 1)*(2x – 1) > 0
> 1/2 or -1/2 < x < 0

This is our universe of the values of x. It is given that all values of x lie in this range.

Here, the question asks us the complete range of x. So we need to look for exactly this range. This is given in answer choice C, and therefore C is our answer.

We hope these practice problems will help you become able to distinguish between the three cases now.

Karishma, a Computer Engineer with a keen interest in alternative Mathematical approaches, has mentored students in the continents of Asia, Europe and North America. She teaches the GMAT for Veritas Prep and regularly participates in content development projects such as this blog!

# Investing in Success: The Best In-Person or Online GMAT Tutors Can Make a Difference

Making sure that you’re ready to take the GMAT requires study, time, and effort. Earning a high score on the GMAT can help to impress admissions officials at preferred business schools. One way to make the studying process easier is to work with a private GMAT tutor. A tutor can help you prep for the test in a variety of ways. Naturally, you want to find the tutor who can be the most help to you. Discover some of the qualities to look for when there’s a GMAT tutor needed to complete your study plan.

Knowledge of All Aspects of the GMAT
The best private GMAT tutor has more than just general advice regarding the GMAT. The person has thorough knowledge of the exam and its contents. There are several parts to the GMAT, including the Verbal, Quantitative, Integrated Reasoning, and Analytical Writing sections. A qualified tutor will have plenty of tips to share that can help you to navigate all of the sections on the GMAT.
Plus, an experienced tutor will be able to evaluate the results of your practice GMAT to determine where you need to focus most of your study efforts. This puts the element of efficiency into your test prep.

The GMAT instructors at Veritas Prep achieved scores on the exam that placed them in the 99th percentile, so if you work with a Veritas Prep tutor, you know you’re studying with someone who has practical experience with the exam. Our tutors are experts at describing the subtle points of the GMAT to their students.

If you want to thoroughly prepare for the GMAT, you must use quality study materials. At Veritas Prep, we have a GMAT curriculum that guides you through each section of the test. Your instructor will show you the types of questions on the test and reveal proven strategies you can use to answer them correctly. Of course, our curriculum teaches you the facts you need to know for the test. But just as importantly, we show you how to apply those facts to the questions on the exam. We do this in an effort to help you think like a business executive as you complete the GMAT. Private tutoring services from Veritas Prep give you the tools you need to perform your best on the exam.

The best GMAT tutors can offer you several options when it comes to preparing for the exam. Perhaps you work full-time as a business professional. You want to prepare for the GMAT but don’t have the time to attend traditional courses. In that case, you should search for an online GMAT tutor. As a result, you can prep for the GMAT without disrupting your busy work schedule. At Veritas Prep, we provide you with the option of online tutoring as well as in-person classes. We recognize that flexibility is important when it comes to preparing for the GMAT, and we want you to get the instruction you need to earn a high score on this important test.

An Encouraging Instructor
Naturally, when you take advantage of GMAT private tutoring services, you will learn information you need to know for the test. But a tutor should also take the time to encourage you as you progress in your studies. It’s likely that you’ll face some stumbling blocks as you prepare for the different sections of the GMAT. A good instructor must be ready with encouraging words when you’re trying to master difficult skills.

Encouraging words from a tutor can give you the push you need to conquer especially puzzling questions on the test. The understanding tutors at Veritas Prep have been through preparation for the GMAT as well as the actual test, so we understand the tremendous effort it takes to master all of its sections.

If you want to partner with the best GMAT tutor as you prep for the test, we have you covered at Veritas Prep! When you sign up to study for the GMAT with Veritas Prep, you are investing in your own success. Give us a call or write us an email today to let us know when you want to start gearing up for excellence on the GMAT!

# GMAT Integrated Reasoning Practice: Sample Questions and Prep Tips

On one section of the GMAT, you’ll encounter Integrated Reasoning questions. These questions test your ability to solve problems using several forms of data. Though you’ve found plenty of advice on studying for the GMAT, you may feel a little concerned about these particular questions. Consider some information about the nature of these questions, then learn how to prep for them with our help.

Take a Timed Practice Test
One way you can get GMAT Integrated Reasoning practice is to take a timed practice test. When you take the entire test or a set of GMAT Integrated Reasoning practice questions, you get an idea of what to expect on test day. More importantly, your results will reveal which skills need improvement.

Timing yourself is an important factor when taking a practice test. You get just 30 minutes to complete the 12 Integrated Reasoning questions on the GMAT. Establishing a reasonable testing pace can lower your stress level and help you to finish all of the questions in the allotted time. At Veritas Prep, we have a free GMAT test that you can take advantage of for this purpose.

Get Into the Mindset of a Business Executive
Taking the GMAT is one of the steps necessary on your path to business school, so it makes perfect sense that the GMAT gauges your skills in business. One of the best prep tips you can follow is to complete all GMAT Integrated Reasoning sample questions with the mindset of a business executive. Think of the questions as real-life scenarios that you will encounter in your business career. Taking this approach allows you to best highlight your skills to GMAT scorers.

Become Familiar With the Question Formats
As you tackle a set of GMAT Integrated Reasoning sample questions, you’ll see that there are a few different question formats – Graphics Interpretation, Two-Part Analysis, Multi-Source Reasoning, and Table Analysis are the different types of questions on the GMAT.

The Graphics Interpretation questions feature a chart, graph, or diagram. For instance, you may see a question that features a bar chart that asks you to answer two questions based on the data in the chart. Other graphics you may see include scatterplots, pie charts, bubble charts, and line charts.

Two-Part Analysis problems involve a chart with three columns of data and accompanying questions. One tip to remember about these questions is that you have to answer the first question presented before you tackle the second one because the answers will work together in some way. Multi-Source Reasoning questions contain a lot of data. These questions test your ability to combine the data contained in different graphs, formulas, and diagrams to arrive at the correct answer choice. Table Analysis questions ask you to look at a table that may contain four or more columns of data. You have to examine this data closely to answer the questions.

Practice Working With Different Types of Graphs and Diagrams
Effective GMAT Integrated Reasoning practice involves learning the details about the different types of graphs, charts, and diagrams featured on the test. Financial magazines and newspapers are great resources for different graphics that you may see on the GMAT. Take some time to make sure you understand the purpose behind various graphs and charts so you feel at ease with them on test day.

Work With a Capable Tutor
When studying for the section on Integrated Reasoning, GMAT practice questions can be very useful. Another way to boost your preparation for this section is to partner with an experienced tutor. The instructors at Veritas Prep follow a thorough GMAT curriculum as they prep you for Integrated Reasoning questions as well as the other questions on the exam. We provide you with proven test-taking strategies and show you how to showcase what you know on the GMAT. With our guidance, you can move through each section of the test with confidence.

The professional tutors at Veritas Prep have the skills and knowledge to prepare you for the section on Integrated Reasoning. GMAT questions in all of the sections are easier to navigate after working through our unique GMAT curriculum. We offer both online and in-person courses, so you can choose the option that best suits your schedule. Contact our offices today and get first-rate prep for the GMAT!

# Using Special Formats on GMAT Variable Problems

In today’s post, we will discuss some special formats when we assume variables on the GMAT. These will allow us to minimize the amount of manipulations and calculations that are required to solve certain Quant problems.

Here are some examples:

An even number: 2a
Logic: It must be a multiple of 2.

An odd number: (2a + 1) or (2a – 1)
Logic: It will not be a multiple of 2. Instead, it will be 1 more (or we can say 1 less) than a multiple of 2.

Two consecutive integers: 2a, (2a + 1) or (2a – 1), 2a
Logic: One number will be even and the other will be the next odd number (or the other way around).

Four consecutive odd numbers: (2a – 3), (2a – 1), (2a + 1), (2a + 3)
In this case, the sum of the numbers comes out to be a clean 8a. This can be very useful in many cases.

Five consecutive even numbers: (2a – 4), (2a – 2), 2a, (2a + 2), (2a + 4)
In this case, the sum of the numbers comes out to be a clean 10a. This can also be very useful in many cases.

A prime number: (6a+1) / (6a – 1)
Every prime number greater than 3 is of the form (6a + 1) or (6a – 1). Note, however, that every number of this form is not prime.

Three consecutive numbers:
If we know one number is even and the other two are odd, we will have: (2a – 1), 2a, (2a + 1).
Logic: They add up to give 6a.
In a more generic case, we will have: 3a, (3a+1), (3a+2).
This gives us some important information. It tells us that one of the numbers will definitely be a multiple of 3 and the other two numbers will not be. Note that the numbers can be in a different order such as (3a + 1), (3a + 2) and (3a + 3). (3a + 3) can be written as 3b, so the three numbers will still have the same properties.

Basically, try to pick numbers in a way that will make it easy for you to manage them. Remember, three numbers do not need to be a, b and c – there could be, and in fact often are, several other hints which will give you the relations among the numbers.

Now, let’s see how picking the right format of these numbers can be helpful using a 700-level GMAT question:

The sum of four consecutive odd numbers is equal to the sum of 3 consecutive even numbers. Given that the middle term of the even numbers is greater than 101 and lesser than 200, how many such sequences can be formed?

(A) 12
(B) 17
(C) 25
(D) 33
(E) 50

Let’s have the four consecutive odd numbers be the following, where “a” is any integer: (2a – 3), (2a – 1), (2a + 1), (2a + 3)

The sum of these numbers is: (2a – 3) + (2a – 1) + (2a + 1) + (2a + 3) = 8a

Now let’s have the three consecutive even numbers be the following, where “b” is any integer: (2b – 2), 2b, (2b + 2)

The sum of these numbers is: (2b – 2) + 2b + (2b + 2) = 6b

Note here that instead of 2a, we used 2b. There is no reason that the even numbers would be right next to the odd numbers, hence we used different variables so that we don’t establish relations that don’t exist between these seven numbers.

We are given that the sum 8a is equal to the sum 6b.

8a = 6b, or a/b = 3/4, where a and b can be any integers. So “a” has to be a multiple of 3 and “b” has to be a multiple of 4.

With this in mind, possible solutions for a and b are:

a = 3, b = 4;
a = 6, b = 8;
a = 9, b = 12
etc.

We are also given that the middle term of the even numbers is greater than 101 and less than 200.

So 101 < 2b < 200, i.e. 50.5 < b < 100.

B must be an integer, hence, 51 ≤ b ≤ 99.

Also, b has to be a multiple of 4, so the values that b can take are 52, 56, 60, 64 … 96

The number of values b can take = (Last term – First term)/Common Difference + 1 = (96 – 52)/4 + 1 = 12

For each of these 12 values of b, there will be a corresponding value of a and, hence, we will get 12 such sequences. Therefore, the answer to our question is A.

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!

# Assumption vs. Strengthen Critical Reasoning Questions: What’s the Difference?

I had a discussion with a tutoring student the other day about the distinction between Assumption and Strengthen questions in the Critical Reasoning section. The two categories feel similar, after all. They are different, however, and the difference, as with most Critical Reasoning questions, lies mainly in the texture of the language that would be most appropriate for a correct answer in either category.

To illustrate, let’s take a simple argument: Dave opens a coffee shop in Veritasville called Dave’s Blends. According to surveys, Dave’s Blends has the best tasting coffee in the city. Therefore, Dave’s Blends will garner at least 50% the local market.

First, imagine that this is a simple Strengthen question. In order to strengthen this somewhat fanciful conclusion, we’re going to want strong language. For example: Virtually all coffee drinkers in Veritasville buy coffee daily from Dave’s. That’s a pretty good strengthener. The statement increases the likelihood that Dave’s Blends will dominate the local market. But an answer choice such as, “Some people buy coffee at Dave’s,” would be a lousy choice, as the fact that Dave’s has at least one customer is hardly a compelling reason to conclude that it will get to at least a 50% market share.

Now imagine that we take the same argument and make it an Assumption question. The first aforementioned answer choice is now much less appealing. Can we really assume that virtually everyone in town will get their coffee at Dave’s? Not really. If Dave’s has 51% of the market share, it doesn’t mean that virtually everyone gets their coffee there. But now consider the second answer choice – if we’re concluding that Dave’s will get at least half of the local market, we are assuming that some people will purchase coffee there, so now this would be a good answer.

The difference is that in a Strengthen question, we’re looking for new information that will make the conclusion more likely. In an Assumption question, we’re looking for what is true based on the conclusion.  Put another way, strong language (“virtually everyone”) is often desirable in a Strengthen question, whereas softer language (“some people”) is usually more desirable in an Assumption question.

Let’s see this in action with a GMAT practice question:

For most people, the left half of the brain controls linguistic capabilities, but some people have their language centers in the right half. When a language center of the brain is damaged, for example by a stroke, linguistic capabilities are impaired in some way. Therefore, people who have suffered a serious stroke on the left side of the brain without suffering any such impairment must have their language centers in the right half.

Which of the following is an assumption on which the reasoning in the argument above depends?

(A) No part of a person’s brain that is damaged by a stroke ever recovers.
(B) Impairment of linguistic capabilities does not occur in people who have not suffered any damage to any language center of the brain.
(C) Strokes tend to impair linguistic capabilities more severely than does any other cause of damage to language centers in the brain.
(D) If there are language centers on the left side of the brain, any serious stroke affecting that side of the brain damages at least one of them.
(E) It is impossible to determine which side of the brain contains a person’s language centers if the person has not suffered damage to either side of the brain.

First, let’s break this argument down:

Conclusion: People who suffer a stroke on the left side of the brain and don’t’ suffer language impairment have language centers in the right half of the brain.

Premises: Most people have language centers on the left side of the brain, while some have them on the right. Damage impairs linguistic capabilities.

This is an Assumption question, so we’re looking for what is be true based on the way the premises lead to the conclusion. Put another way, softer language might be preferable here. Now let’s examine each of the answer choices:

(A) Notice the extreme language, “No part…ever recovers“. Can we really assume that? Of course not – some portion might recover. No good.

(B) We don’t know this. Imagine someone has a part of his or her brain removed and this part of the brain doesn’t contain a language center. Surely we can’t assume that this person will have no language impairment at all. No good.

(C) Again, notice the extreme language, “…more severely than other cause. Can we assume that a stroke is worse than every other kind of brain trauma? Of course not. No good.

(D) Now we’re talking. Here, we are given more generous language: damages at least one of them. “At least one” is a pretty low bar. Remember that the conclusion is that someone who suffers a left-brain stroke and doesn’t have language impairment must have language centers on the right side. Well, that only makes sense if there’s some damage somewhere on the left. This answer choice looks good.

(E) Notice again the extreme language, “…it is impossible“. There may be some other way to assess where the language centers are. No good.

Takeaway: Strengthen questions and Assumption questions are not identical. In a Strengthen question, we want a strong answer choice that will make a conclusion more likely. In an Assumption question we want a soft answer that is indisputable based on how the premises lead to the conclusion. Attention to details in the language (some vs. most vs. all) is the key.

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

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

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

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

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

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

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

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

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

The next step will be to think a bit harder:

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

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

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

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

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

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

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

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

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

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

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

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

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!

# Online GMAT Verbal Practice: Samples and Questions to Guide Your Test Prep

The Verbal section of the GMAT measures your ability to comprehend what you read, evaluate arguments, and change elements of sentences to make them correct. One way to prep for this section is to complete sample GMAT Verbal questions. Sample questions give you an idea of what you can expect when you sit down to take the exam. Learning the different types of problems you might encounter will help you to study for Verbal GMAT questions.

GMAT Verbal practice questions in the Reading Comprehension section require you to read a passage that’s followed by several multiple-choice questions. These questions may ask you to draw an inference or make a conclusion about what you read. Also, there are questions that gauge how well you understood statements made within the passage. A question on a GMAT Verbal practice test might start with, “The primary purpose of the passage is to …” or, “The author is critical of X for the following reasons … .” It’s important to carefully read and evaluate the passage before delving into the questions so you have the information you need to make the right choice.

Taking a GMAT Verbal practice test online is an excellent way to become familiar with the format as well as the content of these questions. Plus, tackling practice questions helps you to get into the habit of reading with the purpose of finding out just what the author is trying to say.

The Critical Reasoning Section
The Critical Reasoning section on the GMAT measures your ability to analyze and evaluate an argument. Practice questions on this topic may include a short argument or one that is several sentences long. There are several multiple-choice options for each question that follows the argument. One example of a typical question might start with, “This argument assumes that … .” Another example of a question you’ll likely encounter starts with, “This argument conveys the following … .” You’ll have to look closely at the points of an argument to determine what the author is trying to convey.

The Sentence Correction Section
To do well on GMAT Verbal practice test questions that deal with Sentence Correction, you must have a grasp of proper grammar and sentence structure. You must also recognize a sentence that conveys meaning in an effective way. Each question starts with a passage that includes an underlined portion. Your job is to consider each of the five options and choose the one that best completes the sentence. This requires you to look at various elements throughout the passage, such as verb tenses and noun usage as well as the use of “like” or “as.” The answer option you select must agree with the elements in the rest of the passage.

Preparing for the Verbal Section With a Professional Tutor
Completing lots of GMAT Verbal practice questions is one way to prepare for this portion of the test. Another way is to study with a tutor who scored in the 99th percentile on the exam. That’s exactly what we offer at Veritas Prep. Our talented instructors prep you for the test using our thorough GMAT curriculum. We teach you how to apply the facts and information you’ve learned so you arrive at the correct answer for each question. We also provide you with strategies, tips, and lessons that strengthen your higher-order thinking skills. These are skills you will need well after you conquer the GMAT. We move way beyond memorization of facts – we teach you to think like a business executive!

Wondering where to begin? You can take one of our GMAT practice tests for free. The results can highlight the skills you’ll need to work on before you sit down to take the actual computer-based test. Our GMAT prep courses are ideal if you want to interact with other students who are as determined as you are to master the exam. Or, if you prefer, you can take advantage of our private online tutoring services. We know you have a busy work schedule as well as family obligations, so we make it easy to study with an expert on the Verbal section as well as all of the other sections on the GMAT. Get in touch with us to begin preparing for the GMAT the right way!

# GMAT Preparation That Works for You: Find Your Best Way to Prepare for the GMAT

So you’ve thought it over and have decided to take the GMAT. That’s great! The next step is to prep for the test.

Of course, not everyone prepares for a test in the same way. The goal is to find what works for you. One way to do that is to look at the different options available to you when it comes to preparing for GMAT questions.

In-Person Prep Courses
You could go with the traditional option and take a GMAT prep course in a classroom with an instructor as well as other students. This is an excellent choice if you enjoy participating in class discussions with other students who are as eager to learn as you. Also, if you benefit from hearing the questions and comments of others, then you may consider this the best way to prepare for the GMAT.

At Veritas Prep, we offer in-person courses taught by instructors who provide you with many GMAT preparation tips. All of our instructors earned a score on the GMAT that landed them in the 99th percentile. So when you learn from a Veritas Prep instructor, you’re learning from one of the best!

Preparing Online with a Tutor
Perhaps you’d prefer to go online to prepare for the GMAT. Test preparation can be completed one-on-one with a Veritas Prep tutor on the Internet. Some people find that they are able to focus better when studying online with a tutor. You’re bound to appreciate the option of choosing your own learning environment when you choose online tutoring. If this is the choice for you, the experienced online tutors at Veritas Prep stand ready to help you prepare for the GMAT.

Choosing the Best Environment for Online Learning
If you think that participating in tutoring sessions online is the best way to prepare for the GMAT, then you should decide on your optimal learning environment. Of course, whatever location you select must have Internet access. You may consider choosing a room in your home where you’ll have very few interruptions. However, if you live in a home that’s always overflowing with activity, you may want to reserve a room at a public library or ask to use a quiet room at your workplace instead. To get the most out of your tutoring sessions, you should choose to study in a place where you’ll be able to focus all of your attention on your online tutor and study resources.

Studying with a Friend or Going it Alone
The question of whether to study alone or with a friend may come up as you begin preparing for the GMAT. Well, having someone else around can end up helping or hurting you. For instance, perhaps you have a coworker who is also planning to take the GMAT and asks to study with you. If the two of you are good friends, you may find that you end up chatting about current events, family and work instead of preparing for GMAT questions. This is a perfect example of how studying with another person can hinder your progress.

Alternatively, studying GMAT vocabulary words can be more effective when done with another person. You can quiz one another on the definitions of words, or you can make up a vocabulary game that puts the element of competition into your study sessions. Along with your tutor, a study partner can give you encouragement as you absorb unfamiliar words and their meanings. You are the best judge of whether it would benefit you to study with a partner or study alone outside of your instructional sessions with Veritas Prep.

Along with online or in-person instruction, Veritas Prep has a variety of other resources available to you as prepare for the GMAT. One of the best places to start your GMAT prep is our free practice test. Your score will help reveal what you need to work on when it comes to mastering skills for the GMAT. We also have a free trial class that gives you a good idea of what to expect from our GMAT study program. Go ahead and check out all of the details regarding our professional GMAT tutoring services and give us a call today!

# The Holistic Approach to Absolute Values – Part V

We will continue our holistic approach to absolute values and add more complications to these types of questions. This article should set you up for any question of this kind. Note that this is a 750+ level concept, so if you are targeting a lower score, it may not be necessary for you to know.

(Before you continue reading, be sure to check out Part I, Part II, Part III and Part IV of this lesson.)

Let’s look at the following GMAT question:

For how many integer values of x, is |x – 6| > |3x + 6|?

(A) 1
(B) 3
(C) 5
(D) 7
(E) Infinite

In this question, we are given the inequality |x – 6| > 3*|x + 2|

Using the same logic as we did in the previous two posts, we will word the inequality like this: the distance from 6 should be more than three times the distance from -2.

At x = -2, the distance from 6 is 8 and the distance from -2 is 0. This means the distance from 6 is more than three times the distance from -2.

At x = -1, the distance from 6 is 7 and the distance from -2 is 1. Three times the distance from -2 is 3. This means the distance from 6 is more than three times the distance from -2.

At some point on the right of -1, the distance from 6 will be equal to three times the distance from -2. The distance between -2 and 6 is 8. If we split this 8 into 4 equal parts to get to x = 0, the distance from 6 will be equal to three times the distance from -2.

Now for every point to the right of 0, the distance from 6 will be less than three times the distance from -2.

Let’s try to go to the left of -2 instead. Will there be a point to the left of -2 where the distance from 6 will be equal to three times the distance from -2? Say that point is “a” units away from -2. -2 must then be 2a units away from 6 to ensure that 6 is a total of 3a units away from that point.

The distance between -2 and 6 is 8 – this 8 needs to be equal to 2a, so “a” must be 4 units.

The point where the distance from 6 will be equal to three times the distance from -2 will be 4 units to the left of -2, i.e. at -6. So at points to the right of -6 (but left of 0), the distance from 6 will be more than three times the distance from -2.

Note that for all values to the left of -6, the distance from 6 will be less than three times the distance from -2.

Hence, our x will lie in the range from -6 to 0.

-6 < x < 0

With these parameters, we will have 5 integer solutions: -5, -4, -3, -2 and -1. Hence, our answer is C.

Let’s look at a second question:

For how many integer values of x, is |x – 8| + |5 – x| > |x + 7|?

(A) 1
(B) 3
(C) 5
(D) 7
(E) Infinite

Now the true value of this method is visible, as we have three or more terms. The arduous algebra involved in this given inequality makes our logical approach much more attractive.

First note that we have the term |5 – x|. This is the same as |x – 5| because |x| = |-x|.

We will word the inequality like this: the distance from 5 + the distance from 8 should be greater than the distance from -7.

Let’s find the point where the sum of the distance from 5 and the distance from 8 is equal to distance from -7. Say that point is “a” units to the left of 5.

a + a + 3 = 12 – a
a = 3

So the point is 3 units to the left of 5, which means it is at 2. For all points to the left of 2, the sum of the distance from 5 and the distance from 8 will be greater than the distance from -7.

How about the points that are to the right of 8? Say there is a point “b” units away from 8 where the sum of the distance from 5 and the distance from 8 is equal to the distance from -7.

3 + b + b = 15 + b
b = 12

So if we go 12 units to the right of 8, i.e. at x = 20, the sum of the distance from 5 and the distance from 8 is equal to the distance from -7.

For all points to the right of 20, the sum of the distance from 5 and the distance from 8 is greater than the distance from -7, so there will be infinite points for which the sum of the distance from 5 and the distance from 8 is greater than the distance from -7. Therefore, our answer is E.

Using this concept, try to answer the following question on your own: For how many integer values of x, is |x – 6| – |3x + 6| > 0?

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 Writing Tips: Analytical Writing for the GMAT

You probably know that the GMAT gauges your skills in reading and math. But did you know that there is also a section called the Analytical Writing Assessment? GMAT creators want to see how well you can analyze an argument, so in this section, you are given an argument and expected to critique it. Is it a valid argument, or is it full of flaws? Discover a few GMAT writing tips that can help you to create a critique that earns you a high score on this portion of the test.

Take a Few Minutes to Plan Your Essay
When it comes to the GMAT writing section, you may think this first tip is a no-brainer. Unfortunately, some students become nervous or anxious about this part of the exam and forget to plan out their essay before diving into the task. This can result in a poorly organized essay or one that is missing important points.

Take the time to carefully read the directions and the argument. Then, create a rough outline of what points you want to include in the essay as well as where you want to include them. If you lose your train of thought while you’re writing, simply look at your outline to regain your focus.

Determine the Flaws in the Argument
Your essay’s plan should include the flaws in the author’s argument. Faulty comparisons and mistaken assumptions as well as vague words are all things to point out when critiquing the argument. Writing a quick note about each flaw you find can be helpful when it comes time to elaborate on them in your essay. Plus, making note of them helps you to remember to include all of them in the final piece.

Use Specific Examples in Your Essay
The use of specific examples is a key element for Analytical Writing. GMAT graders will be looking for specific examples as they score your essay. It’s not enough to state that a piece of the given argument is inaccurate – you have to use the information within the argument to prove your point. Also, using specific examples helps you to demonstrate that you understand the argument.

Read and Evaluate High-Scoring Analytical Essays
When preparing for the GMAT Analytical Writing section, it’s a good idea to read and evaluate essays that received high scores. This can help you see what needs to be adjusted in your own writing to create an essay that earns a high score. In fact, you can break each essay down and highlight the individual elements that earned it a high score.

Study the Scoring System for the GMAT Analytical Writing Section
Studying the scoring rubric for the analytical essay is very helpful in your quest to craft a high-scoring piece. After writing a practice essay, you can compare its contents to the criteria on the rubric. If your essay is missing an element, you can go back and do a rewrite. This sort of practice takes a bit of time, but will prove beneficial on test day.

Study with a GMAT Tutor
A professional tutor can assist you in preparing for the section on Analytical Writing. GMAT tutors at Veritas Prep have taken the exam and earned a score in the 99th percentile. This means that when you prep for the Analytical Writing section with one of our tutors, you’re learning from a teacher with practical experience! Your tutor can help you boost your writing skills by reviewing the outline of your practice essay and giving you tips on how to improve it. Also, your tutor can provide strategies for what you can do to make your analytical essay more convincing.

We have a variety of tutoring options for those who want help preparing for the analytical essay section on the GMAT. At Veritas Prep, we know that you have a busy schedule, and we want to make it convenient to prep for this test. We also offer resources such as the opportunity for you to take a free GMAT test. This is an excellent way to find out how your skills measure up on each section of the exam. Call or contact us online today and let us give you a hand with your essay-writing skills!

# The Holistic Approach to Absolute Values – Part IV

Last week, we looked at some absolute value questions involving inequalities. Today, we’ll continue this discussion by adding some more complications to our questions. Consider the question: What is the minimum value of the expression |x – 3| + |x + 1| + |x|? Technically, |x – 3| + |x + 1| + |x| is the sum of “the distance of x from 3,” “the distance of x from -1” and “the distance of x from 0.” To make solving such questions simpler, we’ll often use a parallel situation:

Imagine that there are 3 friends with houses at points -1, 0 and 3 in a straight line. They decide to meet at the point x.

• |x – 3| will be the distance covered by the friend at 3 to reach x.
• |x + 1| will be the distance covered by the friend at -1 to reach x.
• |x| will be the distance covered by the friend at 0 to reach x.

So, the total distance the friends will cover to meet at x will be |x – 3| + |x + 1| + |x|.

Now we can choose to minimize this total distance, bring it to some particular value or make it more or less than some particular value.

If we want to minimize the total distance, we just make the friends meet at the second guy’s house, i.e. at the point 0. The friend at 3 and the friend at -1 need to travel 4 units total to meet anyway, so there’s no point in making the guy at 0 travel any distance at all. So the minimum total distance would be 4, which would then be the minimum value of |x – 3| + |x + 1| + |x|. This minimum value is given by the expression at x = 0.

With this in mind, when we move to the right or to the left of x = 0, the total distance will increase and, hence, the value of the expression |x – 3| + |x + 1| + |x| will also increase.

Thereafter, it is easy to solve for |x – 3| + |x + 1| + |x| = 10 or |x – 3| + |x + 1| + |x| < 10, etc., as seen in our previous post.

Today, let’s look at how to solve a more advanced GMAT question using the same logic:

For how many integer values of x, is |x – 5| + |x + 1| + |x|  + |x – 7| < 15?

(A) 0
(B) 2
(C) 4
(D) 6
(E) Infinite

In our parallel situation of friends and houses, we now have 4 friends with houses at points -1, 0, 5 and 7.

The friends at -1 and 7 are 8 units apart, so they will need to cover at least this total distance together to meet. It doesn’t matter where they meet between -1 and 7 (inclusive), they will need to cover exactly 8 units.

The friends at 0 and 5 will need to travel a minimum distance of 5 to meet. They can meet anywhere between 0 and 5 (inclusive) and the distance they will cover will still be 5.

So, all four friends can meet anywhere between 0 and 5 (inclusive) and the total distance covered will be 8 + 5 = 13. This would be the minimum total distance, and hence, the minimum value of the expression |x – 5| + |x + 1| + |x|  + |x – 7|.

When we move to the left of 0 or to the right of 5, the total distance covered will be more than 13. At any point between -1 and 7, the total distance covered by the friends at -1 and 7 will be only 8. When we move 1 unit to the left of 0 and reach -1, the total distance covered by the friends at 0 and 5 will be 1 + 6 = 7. So to meet at -1, the total distance traveled by all friends together will be 8 + 7 = 15.

Similarly, when we move 1 unit to the right of 5 and reach 6, the total distance covered by the four friends will be again 8 + 7 = 15. So at points x = -1 and x = 6, the value of the expression will be 15. Between these two points (excluding the points themselves), the value of the expression will be less than 15.

So now we know -1 < x < 6. With these parameters, x can take 6 integer values: 0, 1, 2, 3, 4, 5. Therefore, the answer is D.

Note that when we had 3 points on the number line, the minimum total distance was found at the second point. Now when we have 4 points on the number line, the minimum total distance has been found to be in the range between second and third points.

Let’s look at another question:

For how many integer values of x, is |2x – 5| + |x + 1| + |x| < 10?

(A) 1
(B) 2
(C) 4
(D) 5
(E) Infinite

|2x – 5| + |x + 1| + |x| < 10

2*|x – 5/2| + |x + 1| + |x| < 10

In this sum, now the distance from 5/2 is added twice.

In our parallel situation, this is equivalent to two friends living at 5/2, one living at 0 and one living at -1. Now note that the expression may not take the minimum value of x = 0 because there are 2 people who will need to travel from 5/2.

We have four friends in all, so we can expect to get a range in which we will get the minimum value of the expression. The second and third people are at 0 and 5/2, respectively.

The total distance at x = 0 will be 1 + 2*(5/2) = 6. The total distance at x = 5/2 will be 7/2 + 5/2 = 6.

So if we move to the left of 0 or to the right of 5/2, the total distance will increase. If we move 1 unit to the right of 5/2 and reach 7/2, the total distance covered by the four friends will be 9/2 + 7/2 + 2 = 10. If we move 1 unit to the left of 0 and reach -1, the total distance covered by the four friends will be 0 + 1 + 2*(7/2) = 8. Now all four friends are at -1. To cover a distance of another 2, they should move another 0.5 units to the left of -1 to reach -1.5.

Now the total distance covered by the four friends will be 0.5 + 1.5 + 2*4 = 10, so the total distance when x lies between the points -1.5 and 3.5 (excluding the points themselves) will be less than 10.

Now we know -1.5 < x < 3.5. With these parameters, x can take 5 integer values: -1, 0, 1, 2 and 3. Therefore, the answer is D.

Now use these concepts to solve the following question: For how many integer values of x, is |3x – 3| + |2x + 8| < 15?

Karishma, a Computer Engineer with a keen interest in alternative Mathematical approaches, has mentored students in the continents of Asia, Europe and North America. She teaches the GMAT for Veritas Prep and regularly participates in content development projects such as this blog!

# The Patterns to Solve GMAT Questions with Reversed-Digit Numbers – Part II

, I wrote about the GMAT’s tendency to ask questions regarding the number properties of two two-digit numbers whose tens and units digits have been reversed.

The biggest takeaways from that post were:

1. Anytime we add two two-digit numbers whose tens and units digits have been reversed, we will get a multiple of 11.
2. Anytime we take the difference of two two-digit numbers whose tens and units digits have been reversed, we will get a multiple of 9.

For the hardest GMAT questions, we’re typically mixing and matching different types of number properties and strategies, so it can be instructive to see how the above axioms might be incorporated into such problems.

Take this challenging Data Sufficiency question, for instance:

When the digits of 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?

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

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

The average test-taker looks at Statement 1, sees that it will be very difficult to simply pick numbers that satisfy this condition, and concludes that this can’t possibly be enough information. Well, the average test-taker also scores in the mid-500’s, so that’s not how we want to think.

First, let’s concede that Statement 1 is a challenging one to evaluate and look at Statement 2 first. Notice that Statement 2 tells us something we already know – as we saw above, anytime you have two two-digit numbers whose tens and units digits are reversed, the difference will be a multiple of 9. If Statement 2 is useless, we can immediately prune our decision tree of possible correct answers. Either Statement 1 alone is sufficient, or the statements together are not sufficient, as Statement 2 will contribute nothing. So right off the bat, the only possible correct answers are A and E.

If we had to guess, and we recognize that the average test-taker would likely conclude that Statement 1 couldn’t be sufficient, we’d want to go in the opposite direction – this question is significantly more difficult (and interesting) if it turns out that Statement 1 gives us considerably more information than it initially seems.

In order to evaluate Statement 1, it’s helpful to understand the following shortcut for how to determine the total number of factors for a given number. Say, for example, that we wished to determine how many factors 1000 has. We could, if we were sufficiently masochistic, simply list them out (1 and 1000, 2 and 500, etc.). But you can see that this process would be very difficult and time-consuming.

Alternatively, we could do the following. First, take the prime factorization of 1000. 1000 = 10^3, so the prime factorization is 2^3 * 5^3. Next, we take the exponent of each prime base and add one to it. Last, we multiply the results. (3+1)*(3+1) = 16, so 1000 has 16 total factors. More abstractly, if your number is x^a * y^b, where x and y are prime numbers, you can find the total number of factors by multiplying (a+1)(b+1).

Now let’s apply this process to Statement 1. Imagine that the difference of M and N comes out to some two-digit number that can be expressed as x^a * y^b. If we have a total of 12 factors, then we know that (a+1)(b+1) = 12. So, for example, it would work if a = 3 and b = 2, as a + 1 = 4 and b + 1 = 3, and 4*3 =12. But it would also work if, say, a = 5 and b = 1, as a + 1 = 6 and b + 1 = 2, and 6*2 = 12. So, let’s list out some numbers that have 12 factors:

1. 2^3 * 3^2 (3+1)(2+1) = 12
2. 2^5 * 3^1 (5+1)(1+1) = 12
3. 2^2 * 3^3 (2+1)(3+1) = 12

Now remember that M – N, by definition, is a multiple of 9, which will have at least 3^2 in its prime factorization. So the second option is no longer a candidate, as its prime factorization contains only one 3. Also recall that we’re talking about the difference of two two-digit numbers. 2^2 * 3^3 is 4*27 or 108. But the difference between two positive two-digit numbers can’t possibly be a three-digit number! So the third option is also out.

The only possibility is the first option. If we know that the difference of the two numbers is 2^3 * 3^2, or 8*9 = 72, then only 91 and 19 will work. So Statement 1 alone is sufficient to answer this question, and the answer is A.

Algebraically, if M = 10x + y, then N = 10y + x.

M – N = (10x + y) – (10y + x) = 9x – 9y = 9(x – y).

If 9(x – y) = 72, then x – y = 8. If the difference between the tens and units digits is 8, the numbers must be 91 and 19.

Takeaway: the hardest GMAT questions will require a balance of strategy and knowledge. In this case, we want to remember the following:

• Anytime we take the difference of two two-digit numbers whose tens and units digits have been reversed, we will get a multiple of 9.
• If one statement is easier to evaluate than the other, tackle the easier one first. If it’s the case that one statement gives you absolutely nothing, and the other is complex, there is a general tendency for the complex statement alone to be sufficient.
• For the number x^a * y^b, where x and y are prime numbers, you can find the total number of factors by multiplying (a+1)(b+1).

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

# GMAT Hacks, Tricks, and Tips to Make Studying and Preparing for the GMAT Simpler

The GMAT measures four general types of knowledge: Verbal, Quantitative, Integrated Reasoning, and Analytical Writing. The entire test takes about three hours and 30 minutes to complete.

Preparing for this important exam may seem like a daunting task, but you can simplify the process with the help of some GMAT tips and tricks.

Use Mnemonics to Learn Vocabulary Words
Making a GMAT cheat sheet complete with mnemonics simplifies the process of learning vocabulary words for the Verbal section. Word pictures can help you to retain the words you’re learning. For instance, suppose you’re trying to learn the word “extricate.” “Extricate” means to free something or someone from a constraint or problem. You may pair the word with a mental picture of a group of people being freed from a stuck elevator by a technician. Creating mnemonics that relate to your life, family, or job can make them all the more memorable.

Look for Vocabulary Words in Context
Studying a GMAT cheat sheet full of words and mnemonics shouldn’t be the end of your vocabulary studies. It’s just as important to be able to recognize those words in context. If you’ve signed up to take the GMAT, there’s a good chance that you already read several business publications, so keep an eye out for the words used within those resources. Reading financial newspapers, magazines, and online articles that contain GMAT vocabulary words helps you become more familiar with them. After a while, you’ll know what the words mean without having to think about them.

Learn the Test Instructions Before Test Day
When you read the instructions for each section before test day arrives, you’ll know what to expect on the actual day. This can make you feel more relaxed about tackling each section. Also, you won’t have to use your test time reading instructions because you will already know what you’re doing.

Always Keep Some Study Materials Close By
When it comes to GMAT tips and strategies, the easiest ones can sometimes be the most effective. Even busy working professionals have free moments throughout the day. It’s a smart idea to use those moments for study and review. For instance, you can work on some practice math problems during a lunch or coffee break. If you have a dentist or doctor’s appointment, you can use virtual flashcards to quiz yourself on GMAT vocabulary words while you’re sitting in the waiting room. Taking a few minutes each day to review can add up to a lot of productive study time by the end of a week.

Set a Timer for Practice Tests
If you’re concerned about completing each section of the GMAT within the allotted number of minutes, one of our favorite GMAT hacks is to try setting a timer as you begin each section of a practice test. If the timer goes off before you’re finished with the section, you may be spending too much time on puzzling problems. Or perhaps you’re taking too much time to read the directions for each section rather than familiarizing yourself with them ahead of time.

Timing your practice tests helps you establish a rhythm that allows you to get through each section with a few minutes to spare for review. At Veritas Prep, we provide you with the opportunity to take a free exam. Taking this practice exam allows you to get a clear picture of what you’ll encounter on test day.

Get Into the Habit of Eliminating Wrong Answer Options
Another very effective GMAT strategy is to eliminate answer options that are clearly incorrect. With the exception of the analytical essay, this can be done on every portion of the test. Taking practice tests gives you the chance to establish this habit. By eliminating obviously incorrect answer options, you are making the most efficient use of your test time. Also, you are making the questions more manageable by giving yourself fewer answers to consider.

Here at Veritas Prep, our GMAT instructors follow a unique curriculum that shows you how to approach every problem on the test. We teach you how to strengthen your higher-order thinking skills so you’ll know how to use them to your advantage on the test. Contact our offices today to take advantage of our in-person prep courses or our private tutoring services. Learn GMAT hacks from professional instructors who’ve mastered the test!

# The Holistic Approach to Absolute Values – Part III

A while back, we discussed some holistic approaches to answering absolute value questions. Today, we will enhance our understanding of absolute values with some variations that you might see on the GMAT.

Instead of looking at how to solve equations, like we did in our previous post, we will look at how to solve inequalities using the same concept.

A quick review:

• |x| = The distance of x from 0 on the number line. For example, if |x| = 4, x is 4 away from 0. So x can be 4 or -4.
• |x – 1| = The distance of x from 1 on the number line. For example, if |x – 1| = 4, x is 4 away from 1. so x can be 5 or  -3.
• |x| + |x – 1| = The sum of distance of x from 0 and distance of x from 1 on the number line. for example, if x = 5, the distance of x from 0 is 5 and the distance of x from 1 is 4. The sum of the distances is 5 + 4 = 9. So |x| + |x – 1| = 5 + 4 = 9.

Let’s move ahead now and see how we can use these concepts to solve inequalities:

For how many integer values of x, is |x – 3| + |x + 1| + |x| < 10?

(A) 0
(B) 2
(C) 4
(D) 6
(E) Infinite

In the previous post, we saw the a similar question, except it involved an equation rather than an inequality. For that problem, we found that the two points where the total distance is equal to 10 are -2.667 and 4:

What will be the total distance at any value of x between these two points?

Say, x = 0
|x – 3| + |x + 1| + |x|
= 3 + 1 + 0
= 4

Say, x = 3
|x – 3| + |x + 1| + |x|
0 + 4 + 3
= 7

In both cases, we see that the total distance covered is less than 10. Note that the minimum distance covered will be 4 at x = 0 (discussed in the previous post) so by moving to the right of 0 or to the left of 0 on the number line, we get to the points where the distance increases to 10. So for every point in between, the total distance will be less than 10 (the entire red region).

Hence, at integer points x = -2, -1, 0, 1, 2 and 3 (which are all between -2.667 and 4), the total distance will be less than 10. The total distance will be less than 10 for all non-integer points lying between -2.667 and 4 too, but the question only asks for the integer values, so that is all we need to focus on. (Of course, there are infinite non-integer points between any two distinct points on the number line.) Hence, the answer will be 6 points, or D.

Along the same lines, consider a slight variation of this question:

For how many integer values of x, is |x – 3| + |x + 1| + |x| > 10?

(A) 0
(B) 2
(C) 4
(D) 6
(E) Infinite

What will the answer be here? We hope you immediately jumped to answer choice E – for every integer value of x to the right of 4 or to the left of -2.667, the total distance will be more than 10 (the blue regions). So there will be infinite such integer points (all integers greater than 4 or less than -2.667). Thus, the answer is E.

We hope this logic is clear. We will look at some other variations of this concept next week!

Karishma, a Computer Engineer with a keen interest in alternative Mathematical approaches, has mentored students in the continents of Asia, Europe and North America. She teaches the GMAT for Veritas Prep and regularly participates in content development projects such as this blog!

# Data Sufficiency Questions: How to Know When Both Statements Together Are Not Sufficient

Today we will discuss a problem we sometimes face while attempting to solve Data Sufficiency questions for which the answer is actually E (when both statements together are not sufficient to answer the question). Ideally, we would like to find two possible answers to the question asked so that we know that the data of both statements is not sufficient to give us a unique answer. But what happens when it is not very intuitive or easy to get these two distinct cases?

Let’s try to answer these questions in today’s post using using one of our own Data Sufficiency questions.

A certain car rental agency rented 25 vehicles yesterday, each of which was either a compact car or a luxury car. How many compact cars did the agency rent yesterday?

(1) The daily rental rate for a luxury car was \$15 higher than the rate for a compact car.
(2) The total rental rates for luxury cars was \$105 higher than the total rental rates for compact cars yesterday

We know from the question stem that the total number of cars rented is 25. Now we must find how many compact cars were rented.

There are four variables to consider here:

1. Number of compact cars rented (this is what we need to find)
2. Number of luxury cars rented
3. Daily rental rate of compact cars
4. Daily rental rate of luxury cars

Let’s examine the information given to us by the statements:

Statement 1: The daily rental rate for a luxury car was \$15 higher than the rate for a compact car.

This statement gives us the difference in the daily rental rates of a luxury car vs. a compact car. Other than that, we still only know that a total of 25 cars were rented. We have no data points to calculate the number of compact cars rented, thus, this statement alone is not sufficient. Let’s look at Statement 2:

Statement 2: The total rental rates for luxury cars was \$105 higher than the total rental rates for compact cars yesterday.

This statement gives us the difference in the total rental rates of luxury cars vs. compact cars (we do not know the daily rental rates). Again, we have no data points to calculate the number of compact cars rented, thus, this statement alone is also not sufficient.

Now, let’s try to tackle both statements together:

The daily rate for luxury cars is \$15 higher than it is for compact cars, and the total rental rates for luxury cars is \$105 higher than it is for compact cars. What constitutes this \$105? It is the higher rental cost of each luxury car (the extra \$15) plus adjustments for the rent of extra/fewer luxury cars hired. That is, if n compact cars were rented and n luxury cars were rented, the extra total rental will be 15n. But if more  luxury cars were rented, 105 would account for the \$15 higher rent of each luxury car and also for the rent of the extra luxury cars.

Event with this information, we still should not be able to find the number of compact cars rented. Let’s find 2 cases to ensure that answer to this question is indeed E – the first one is quite easy.

The total extra money collected by renting luxury cars is \$105.

105/15 = 7

Say out of 25 cars, 7 are luxury cars and 18 are compact cars. If the rent of compact cars is \$0 (theoretically), the rent of luxury cars is \$15 and the extra rent charged will be \$105 (7*15 = 105) – this is a valid case.

Now how do we get the second case? Think about it before you read on – it will help you realize why the second case is more of a challenge.

Let’s make a slight change to our current numbers to see if they still fit:

Say out of 25 cars, 8 are luxury cars and 17 are compact cars. If the rent of compact cars is \$0 and the rent of luxury cars is \$15, the extra rent charged should be \$15*8 = \$120, but notice, 9 morecompact cars were rented than luxury cars. In reality, the extra total rent collected is \$105 – the \$15 reduction is because of the 9 additional compact cars. Hence, the daily rental rate of each compact car would be \$15/9 = \$5/3.

This would mean that the daily rental rate of each luxury car is \$5/3 + \$15 = \$50/3

The total rental cost of luxury cars in this case would be 8 * \$50/3 = \$400/3

The total rental cost of compact cars in this case would be 17 * \$5/3 = \$85/3

The difference between the two total rental costs is \$400/3 – \$85/3 = 315/3 = \$105

Everything checks out, so we know that there is no unique answer to this question – for any number of compact cars you use, you will come up with the same answer. Thus, Statements 1 and 2 together are not sufficient.

The strategy we used to find this second case to test is that we tweaked the numbers we were given a little and then looked for a solution. Another strategy is to try plugging in some easy numbers. For example:

Instead of using such difficult numbers, we could have tried an easier split of the cars. Say out of 25 cars, 10 are luxury and 15 are compact. If the rent of compact cars is \$0 and the rent of luxury cars is \$15, the extra rent charged should be 10*\$15 = \$150 extra, but it is actually only \$105 extra, a difference of \$45, due to the 5 additional compact cars. The daily rental rent of 5 extra compact cars would be \$45/5 = \$9. Using these numbers in the calculations above, you will see that the difference between the rental costs is, again, \$105. This is a valid case, too.

Hence, there are two strategies we saw in action today:

• Tweak the numbers slightly to see if you will get the same results
• Go for the easy split when choosing numbers to plug in

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: Taking The GMAT Like It’s Nintendo Switch

The non-election trending story of the day is the announcement of the forthcoming Nintendo Switch gaming system, a system that promises to help you take the utmost advantage of your leisure time…but that may help you maximize the value of your GMAT experience, too.

How?

The main feature of the Switch (and the driving factor behind its name) is its flexibility. It can be an in-home gaming system attached to a fixed TV set, but then immediately Switch to a hand-held portable system that allows you to continue your game on the go. Nintendo’s business plan is primarily based on offering flexibility…and on the GMAT, your plan should be to prove to business schools that you can offer the same.

The GMAT, of course, tests algebra skills and critical thinking skills and grammar skills, but beneath the surface it also has a preference for testing flexibility. Many problems will punish those with pure tunnel vision, but reward those who can identify that their first course of action isn’t working and who can then Switch to another plan. This often manifests itself in:

• Math problems that seem to require algebra…but halfway through beg to be back-solved using answer choices.
• Sentence Correction problems that seem to ask you to make a decision about one major difference…but for which the natural choices leave you with clearer-cut errors elsewhere.
• Critical Reasoning answer choices that seem out of scope at first, but reward those who read farther and then see their relevance.
• Data Sufficiency problems for which you’ve made a clear, confident decision on one statement…but then the other statement shows you something you hadn’t considered before and forces you to reconsider.
• The overall concept that if you’re a one-trick pony – you’re a master of plugging in answer choices, for example – you’ll find questions that just won’t reward that strategy and will force you to do something else.

Flexibility matters on the GMAT! As an example, consider the following Data Sufficiency question:

Is x/y > 3?

1) 3x > 9y
2) y > 3y

If you’re like many, you’ll confidently address the algebra in Statement 1, divide both sides by 3 to get x > 3y, and then see that if you divide both sides by y, you can make it look exactly like the question stem: x/y > 3. And you may very well say, “Statement 1 is sufficient!” and confidently move on to Statement 2.

But when you look at Statement 2 – either conceptually or algebraically – something should stand out. For one, there’s no way that it’s sufficient because it doesn’t help you determine anything about x. And secondly, it brings up the point that “y is negative” (algebraically you’d subtract y from both sides to get 0 > 2y, then divide by 2 to get 0 > y). And here’s where, if it hasn’t already, your mind should Switch to “positive/negative number properties” mode. If you weren’t thinking about positive vs. negative properties when you considered Statement 1, this one gives you a chance to Switch your thinking and reconsider – what if y were negative? Algebraically, you’d then have to flip the sign when you divide both sides by y:

3x > 9y : Divide both sides by 3

x > 3y : Now divide both sides by y, but remember that if y is positive you keep the sign (x/y > 3), and if y is negative you flip the sign (x/y < 3).

With this in mind, Statement 1 doesn’t really tell you anything. x/y can be greater than 3 or less than 3, so all Statement 1 does is eliminate that x/y could be exactly 3. Now you have the evidence to Switch your answer. If you initially thought Statement 1 was sufficient, Statement 2 has given you a chance to reassess (thereby demonstrating flexibility in thinking) and realize that it’s not, until you know whether y is negative or positive.

Statement 2 supplies that missing piece, and the answer is thus C. But more important is the lesson – because the GMAT so values mental flexibility, it will often provide you with clues that can help you change your mind if you’re paying attention. So on the GMAT, take a lesson from Nintendo Switch: flexibility is an incredibly marketable skill, so look for clues and opportunities to Switch your line of thinking and save yourself from trap answers.

By Brian Galvin.

# How to Solve “Unsolvable” Equations on the GMAT

The moment we see an equation involving the variable x, we have a habit of jumping right into attempting to solve it. But what happens when we are not able to solve it? Let’s say, for example, we have an equation such as x^2 + 1 = 0. How would we solve for x here? We can’t because x has no real value. Note that x^2 is non-negative – it would be either 0 or positive. 1, we know, is positive. So together, a positive number and a non-negative number cannot add up to 0.

In this example, it relatively easy to see that the equation has no real solution. In others, it may not be so obvious, so we will need to use other strategies.

We know how to solve third degree equations. The first solution is found by trial and error – we try simple values such as -2, -1, 0, 1, 2, etc. and are usually able to find the first solution. Then the equation of third degree is split into two factors, including a quadratic. We know how to solve a quadratic, and that is how we get all three solutions, if it has any.

But what if we are unable to find the first solution to a third degree equation by trial and error? Then we should force ourselves to wonder if we even need to solve the equation at all. Let’s take a look at a sample question to better understand this idea:

Is x < 0?
(1) x^3 + x^2 + x + 2 = 0
(2) x^2 – x – 2 < 0

In this problem, x can be any real number – we have no constraints on it. Now, is x negative?

Statement 1: x^3 + x^2 + x + 2 = 0

If we try to solve this equation as we are used to doing, look at what happens:

If you plug in x = 2, you get 16 = 0
If you plug in x = 1, you get 5 = 0
If you plug in x = 0, you get 2 = 0
If you plug in x = -1, you get 1 = 0
If you plug in x = -2, you get -4 = 0

We did not find any root for the equation. What should we do now? Note that when x goes from -1 to -2, the value on the left hand side changes from 1 to -4, i.e. from a positive to a negative. So, in between -1 and -2 there will be some value of x for which the left hand side will become 0. That value of x will not be an integer, but some decimal value such as -1.3 or -1.4, etc.

Even after we find the first root, making the quadratic will be very tricky and then solving it will be another uphill task. So we should ask ourselves whether we even need to solve this equation.

Think about it – can x be positive? If x is indeed positive, x^3, x^2 and x all will be positive. Then, if we add four positive numbers (x^3, x^2, x and 2) we will get a positive sum – we cannot get 0. Obviously x cannot be 0 since that will give us 2 = 0.

This means the value of x must be negative, but what it is exactly doesn’t matter. We know that x has to be negative, and that is sufficient to answer the question.

Statement 2: x^2 – x – 2 < 0

This, we can easily solve:

x^2 – 2x + x – 2 < 0
(x – 2)*(x + 1) < 0

We know how to solve this inequality using the method discussed here.

This this will give us -1 < x < 2.

Since x can be a non-integer value too, x can be negative, 0, or positive. This statement alone is not sufficient,and therefore, the answer is A.

To evaluate Statement 1, we didn’t need to solve the equation at all. We figured out everything we wanted to know by simply using some logic.

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: A GMAT Quant Question That Troubles Many!

What determines whether or not a question can be considered a GMAT question? We know that GMAT questions that are based on seemingly basic concepts can be camouflaged such that they may “appear” to be very hard. Is it true that a question requiring a lot of intricate calculations will not be tested in GMAT? Yes, however it is certainly possible that a question may “appear” to involve a lot of calculations, but can actually be solved without any!

In the same way, it is possible that a question may appear to be testing very obscure concepts, while it is really solvable by using only basic ones.

This happens with one of our own practice questions – we have often heard students exclaim that this problem isn’t relevant to the GMAT since it “tests an obscure number property”. It is a question that troubles many people, so we decided to tackle it in today’s post.

We can easily solve this problem with just some algebraic manipulation, without needing to know any obscure properties! Let’s take a look:

† and ¥ represent non-zero digits, and (†¥)² – (¥†)² is a perfect square. What is that perfect square?

(A) 121
(B) 361
(C) 576
(D) 961
(E) 1,089

The symbols † and ¥ are confusing to work with, so the first thing we will do is replace them with the variables A and B.

The question then becomes: A and B represent non-zero digits, and (AB)² – (BA)² is a perfect square. What is that perfect square?

As I mentioned before, we have heard students complain that this question isn’t relevant to the GMAT because it “uses an obscure number property”.  Now here’s the thing – most advanced number property questions CAN be solved in a jiffy using some obscure number property such as, “If you multiply a positive integer by its 22nd multiple, the product will be divisible by …” etc. However, those questions are not actually about recalling these so-called “properties” – they are about figuring out the properties using some generic technique, such as pattern recognition.

For this question, the complaint is often that is that the question tests the property, “(x + y)*(x – y) (where x and y are two digit mirror image positive integers) is a multiple of 11 and 9.” It doesn’t! Here is how we should solve this problem, instead:

Given the term (AB)^2, where A and B are digits, how will you square this while keeping the variables A and B?

Let’s convert (AB)^2 to (10A + B)^2, because A is simply the placeholder for the tens digit of the number. If you are not sure about this, consider the following:

58 = 50 + 8 = 10*5 + 8
27 = 20 + 7 = 10*2 + 7
…etc.

Along those same lines:

AB = 10A + B
BA = 10B + A

Going back to our original question:

(AB)^2 – (BA)^2
= (10A + B)^2 – (10B + A)^2
= (10A)^2 + B^2 + 2*10A*B – (10B)^2 – A^2 – 2*10B*A
= 99A^2 – 99B^2
= 9*11*(A^2 – B^2)

We know now that the expression is a multiple of 9 and 11. We would not have known this beforehand. Now we’ll just use the answer choices to figure out the solution. Only 1,089 is a multiple of both 9 and 11, so the answer must be E.

We hope you see that this question is not as hard as it seems. Don’t get bogged down by unknown symbols – just focus on the next logical step at each stage of the problem.

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: As You Debate Over Answer Choices… Just Answer The Freaking Question!

If you’re like many – to the dismay of the NFL and the advertising industry – you’re planning to watch another presidential debate this coming Sunday. And just like Trump-Clinton I and Pence-Kaine earlier this week, this debate will provide plenty of opportunities to be annoyed, frustrated, and disappointed…but it will also provide an ever-important lesson about the GMAT.

It’s no surprise that candidate approval ratings are low for the same reason that far too many GMAT scores are lower than candidates would hope. Why?

People don’t directly answer the question.

This is incredibly common in the debates, where the poor moderators are helpless against the talking points and stump speeches of the candidates. The public then suffers because people cannot get direct answers to the questions that matter. This is also very common on the GMAT, where students will invest the time in critical thought and calculation, and then levy an answer that just doesn’t hit the mark. Consider the example:

Donald has \$520,000 in campaign money available to spend on advertising for the month of October, and his advisers are telling him that he should spend a minimum of \$360,000 in the battleground states of Ohio, Florida, Virginia, and North Carolina. If he plans to spend the minimum amount in battleground states to appease his advisers, plus impress his friends by a big ad spend specific to New York City (and then he will skip advertising in the rest of the country), how much money will he have remaining if he wants 20% of his ad spend to take place in New York City?

(A) \$45,000
(B) \$52,000
(C) \$70,000
(D) \$90,000
(E) \$104,000

As people begin to calculate, it’s common to try to determine all of the facets of Donald’s ad spend. If he’s spending only the \$360,000 in battleground states plus the 20% he’ll spend in New York City, then \$360,000 will represent 80% of his total ad spend. If \$360,000 = 0.8(Total), then the total will be \$450,000. That means that he’ll spend \$90,000 in New York City. Which is answer choice D…but that’s not the question!

The question asked for how much of his campaign money would be left over, so the calculation you need to focus on is the \$520,000 he started with minus the \$450,000 he spent for a total of \$70,000, answer choice C. And in a larger context, you can learn a major lesson from Wharton’s most famous alumnus: it’s not enough for your answer to be related to the question. On the GMAT, you must answer the question directly! So make sure that you:

1. Double check which portion of a word problem the question asked for. Don’t be relieved when your algebra spits out “a” number. Make sure it’s “the” number.
2. Be careful with Strengthen/Weaken Critical Reasoning problems. A well-written Strengthen problem will likely have a good Weaken answer choice, and vice-versa.
3. In algebra problems, make sure to identify the proper variable (or combination of variables if they ask, for example “What is 6x – y?”).
4. With Data Sufficiency problems, pay attention to the exact values being asked for. One of the most common mistakes that people make is saying that a statement is insufficient because they’re looking to fill in all variables, when actually it is sufficient to answer the exact combination that the test asked for.

As you watch the debate this weekend, notice (How could you not?) how absurd it is that the candidates just about never directly answer the question…and then vow to not make the same mistake on your GMAT exam.