How to Simplify Percent Questions on the GMAT

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

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

Take the following question, for example:

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

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

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

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

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

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

24 min. 18 sec. = 0.90t

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

22A + 22B + 22C = 1100

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

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

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

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

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

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

By Brian Galvin.

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

By Brian Galvin.

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

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

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

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

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

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

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

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

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

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

Here’s a quick recap with some examples:

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

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

If x is not 0, then √(x^2)/x =

(A) -1
(B) 0
(C) 1
(D) x
(E) |x|/x

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

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

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

Is √[(x – 3)^2] = (3 – x)?

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

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

This means that our question is basically:

Is |x – 3| = 3 – x?

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

Is |x – 3| = -(x – 3)?

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

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

So our question now boils down to:

Is (x – 3) negative (or 0)?

Statement 1: x is not 3

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

Statement 2: -x * |x| > 0

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

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

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

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

GMAT Rate Questions: Tackling Problems with Multiple Components

stopwatch-620A few posts ago, I tackled rate/work questions, which are invariably a source of consternation for GMAT test-takers. On the latest official practice tests that GMAC has released, these questions showed up with surprising frequency, so I thought it might be worthwhile to tackle a challenging incarnation of this question type: one in which a single machine begins a project and then multiple machines complete the partially-finished work.

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

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

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

And that’s it!

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

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

A) 6
B) 12
C) 15
D) 18
E) 24

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

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

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

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

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

1/8 + 1/24 + z = ¼

Multiply everything by 24, and we get:

3 + 1 + 24z = 6

24z = 2

z = 1/12.

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

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

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

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

Avoid Obtaining the Wrong Values in Percent Increase Questions

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

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

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

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

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

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

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

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

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

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

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

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

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

How to Simplify Sequences on the GMAT

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

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

Take this question for example:

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

I. 4
II. -1
III. -5

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

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

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

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

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

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

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

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

1) n is odd

2) an is positive

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

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

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

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

Let’s keep going.

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

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

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

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

We’re asked about the sum:

After one term, we have 1.

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

And the answer is A.

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

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

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

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

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

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

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

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

*GMATPrep questions courtesy of the Graduate Management Admissions Council.

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

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

How to Avoid Trap Answers On GMAT Data Sufficiency Questions

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

*Official Guide question courtesy of the Graduate Management Admissions Council.

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

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

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

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

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

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

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

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

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

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

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

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

Let’s move on to the other three options:

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

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

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

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

Let’s try another one:

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

 

 

 

 

 

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

DG blog pic 2

 

 

 

 

 

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

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

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

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

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

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

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

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

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

*GMATPrep question courtesy of the Graduate Management Admissions Council.

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

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

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

Quarter Wit, Quarter WisdomLet’s continue our discussion on coordinate geometry today.

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

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

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

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

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

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

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

Method 1: The Traditional Approach
Traditionally, one would solve this question like this:

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

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

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

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

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

Thus, the equations of the 2 lines become:

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

QWQW pic 1

 

 

 

 

 

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

QWQW pic 2

 

 

 

 

 

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

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

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

Now, we can easily solve this official question:

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

Statement 1: b not equal to 0

Statement 2: ab > 0

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

Statement 1: b not equal to 0

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

Statement 2: ab>0

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

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

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

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

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

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

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

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

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

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

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

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

Reading Comprehension

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

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

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

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

By Brian Galvin.

What Makes GMAT Quant Questions So Hard?

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

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

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

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

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

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

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

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

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

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

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

450 = 2 * 3^2 * 5^2

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

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

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

Bringing the given equation back, we get:

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

Statement 1: a is 1.

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

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

1 + c = 2

c = 1

Statement 2:  b is 2.

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

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

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

1 + c = 2

c = 1

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

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

3^b * 5^c = 45

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

If b  = 3, then 3^3 * 5^c = 45.

5^c = 45/27

C will take a non-integer value here.

c = .3174

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

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

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

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

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

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

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

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

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

Quarter Wit, Quarter Wisdom: Dealing with Tangents on the GMAT

Quarter Wit, Quarter WisdomConsidering a two dimensional figure, a tangent is a line that touches a curve at a single point.  Here are some examples of tangents:
QWQW 1

 


 

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

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

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

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

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

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

QWQW 2

 

 

 

 

 

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

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

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

tx = 2ab = (2√3)x * √5

t = 2√15

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

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

f(x) = (√3)x – (√5)^2

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

(√3)x – (√5)^2  = 0

x = (√5)/(√3)

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

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

Therefore, our answer is A.

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

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

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

2 Tips to Make GMAT Remainder Questions Easy

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

*GMATPrep questions courtesy of the Graduate Management Admissions Council.

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

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

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

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

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

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

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

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

(A) 16
(B) 18
(C) 20
(D) 24
(E) 30

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

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

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

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

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

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

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

(x^6)^3 = 2^3

x^18 = 8

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

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

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

*GMATPrep question courtesy of the Graduate Management Admissions Council.

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

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

Quarter Wit, Quarter Wisdom: Ratios in GMAT Data Sufficiency

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

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

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

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

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

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

From the Question stem:

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

Question: What is n?

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

2, 3, and 4,

or 4, 6, and 8,

or 6, 9, and 12,

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

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

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

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

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

They could be

2, 3, 4 and 9

or 6, 9, 12 and 3

Therefore, Statement 2 alone is also not sufficient.

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

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

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

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

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

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

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

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

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

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

Aero_img084The ancient Greeks were, to put it mildly, really neat. They created or helped to create the foundations of philosophy, theater, science, democracy, and mathematics – no small accomplishment for a small war-torn civilization from over two millennia ago. Many of our contemporary ideas, beliefs, and traditions are rooted in contributions made by Greek thinkers, and the GMAT is no exception.

A few months ago, I wrote about this difficult Data Sufficiency question.

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

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

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

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

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

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

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

(A) Between 1 and 10

(B) Between 11 and 15

(C) Between 15 and 20

(D) Between 20 and 25

(E) Greater than 25

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

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

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

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

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

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

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

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

Hit this kk and become yourself.

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

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

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

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

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

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

(1) a = 1

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

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

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

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

(1) -2 < a < 2

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

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

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

By Brian Galvin.

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

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

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

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

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

With 2016, for example:

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

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

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

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

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

(A) 18

(B) 36

(C) 42

(D) 54

(E) 56

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

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

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

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

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

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

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

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

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

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

By Brian Galvin.

How to Choose the Right Number for a GMAT Variable Problem

Pi to the 36th digitWhen you begin studying for the GMAT, you will quickly discover that most of the strategies are, on the surface, fairly simple. It will not come as a terribly big surprise that selecting numbers and doing arithmetic is often an easier way of attacking a problem than attempting to perform complex algebra. There is, however, a big difference between understanding a strategy in the abstract and having honed that strategy to the point that it can be implemented effectively under pressure.

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

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

(A) 2

(B) 3

(C) 4

(D) 6 

(E) 8 

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

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

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

(A) x + 10

(B) 10x + 1

(C) 100(x + 10)

(D) 100 * (x+10)/(x+100)

(E) 100 * (10x + 1)/(10x+10)

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

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

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

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

Start: 100 liters

After adding water: 120 liters

After removing water: 90 liters

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

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

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

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

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

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

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

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

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

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

How to Use Difference of Squares to Beat the GMAT

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

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

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

3600 – 1 = 602 – 12

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

60– 1= (60 + 1)(60 – 1) = 61 * 59

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

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

First:

999,9992 – 1 = 

A) 1010 – 2

B) (106 – 2) 2   

C) 105 (106 -2)

D) 106 (105 -2)

E) 106 (106 -2)

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

Using our equation for difference of squares, we get:

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

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

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

Let’s try one more:

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

A) 97

B) 65

C) 35

D) 13

E) 5

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

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

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

Takeaways:

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

*Official Guide question courtesy of the Graduate Management Admissions Council.

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

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

Quarter Wit, Quarter Wisdom: Cyclicity in GMAT Remainder Questions (Part 2)

Quarter Wit, Quarter WisdomLast week, we reviewed the concepts of cyclicity and remainders and looked at some basic questions. Today, let’s jump right into some GMAT-relevant questions on these topics:

 

 

If n is a positive integer, what is the remainder when 3^(8n+3) + 2 is divided by 5?

(A) 0

(B) 1

(C) 2

(D) 3

(E) 4

In this problem, we are looking for the remainder when the divisor is 5. We know from last week that if we get the last digit of the dividend, we will be able to find the remainder, so let’s focus on finding the units digit of 3^(8n + 3) + 2.

The units digit of 3 in a positive integer power has a cyclicity of: 3, 9, 7, 1

So the units digit of 3^(8n + 3) = 3^(4*2n + 3) will have 2n full cycles of 3, 9, 7, 1 and then a new cycle will start:

3, 9, 7, 1

3, 9, 7, 1

3, 9, 7, 1

3, 9, 7, 1

3, 9, 7

Since the exponent a remainder of 3, the new cycle ends at 3, 9, 7. Therefore, the units digit of 3^(8n + 3) is 7. When you add another 2 to this expression, the units digit becomes 7+2 = 9.

This means the units digit of 3^(8n+3) + 2 is 9. When we divide this by 5, the remainder will be 4, therefore, our answer is E.

Not so bad; let’s try a data sufficiency problem:

If k is a positive integer, what is the remainder when 2^k is divided by 10?

Statement 1: k is divisible by 10

Statement 2: k is divisible by 4

With this problem, we know that the remainder of a division by 10 can be easily obtained by getting the units digit of the number. Let’s try to find the units digit of 2^k.

The cyclicity of 2 is: 2, 4, 8, 6. Depending on the value of k is, the units digit of 2^k will change:

If k is a multiple of 4, it will end after one cycle and hence the units digit will be 6.

If k is 1 more than a multiple of 4, it will start a new cycle and the units digit of 2^k will be 2.

If k is 2 more than a multiple of 4, it will be second digit of a new cycle, and the units digit of 2^k will be 4.

If k is 3 more than a multiple of 4, it will be the third digit of a new cycle and the units digit of 2^k will be 8.

If k is 4 more than a multiple of 4, it will again be a multiple of 4 and will end a cycle. The units digit of 2^k will be 6 in this case.

and so on…

So what we really need to find out is whether k is a multiple of 4, one more than a multiple of 4, two more than a multiple of 4, or three more than a multiple of 4.

Statement 1: k is divisible by 10

With this statement, k could be 10 or 20 or 30 etc. In some cases, such as when k is 10 or 30, k will be two more than a multiple of 4. In other cases, such as when k is 20 or 40, k will be a multiple of 4. So for different values of k, the units digit will be different and hence the remainder on division by 10 will take multiple values. This statement alone is not sufficient.

Statement 2: k is divisible by 4

This statement tells you directly that k is divisible by 4. This means that the last digit of 2^k is 6, so when divided by 10, it will give a remainder of 6. This statement alone is sufficient. therefore our answer is B.

Now, to cap it all off, we will look at one final question. It is debatable whether it is within the scope of the GMAT but it is based on the same concepts and is a great exercise for intellectual purposes. You are free to ignore it if you are short on time or would not like to go an iota beyond the scope of the GMAT:

What is the remainder of (3^7^11) divided by 5?

(A) 0

(B) 1

(C) 2

(D) 3

(E) 4

For this problem, we need the remainder of a division by 5, so our first step is to get the units digit of 3^7^{11}. Now this is the tricky part – it is 3 to the power of 7, which itself is to the power of 11. Let’s simplify this a bit; we need to find the units digit of 3^a such that a = 7^{11}.

We know that 3 has a cyclicity of 3, 9, 7, 1. Therefore (similar to our last problem) to get the units digit of 3^a, we need to find whether a is a multiple of 4, one more than a multiple of 4, two more than a multiple of 4 or three more than a multiple of 4.

We need a to equal 7^{11}, so first we need to find the remainder when a is divided by 4; i.e. when 7^{11} is divided by 4.

For this, we need to use the binomial theorem we learned earlier in this post (or we can use the method of “pattern recognition”):

The remainder of 7^{11} divided by 4

= The remainder of (4 + 3)^{11} divided by 4

= The remainder of 3^{11} divided by 4

= The remainder of 3*3^{10} divided by 4

= The remainder of 3*9^5 divided by 4

= The remainder of 3*(8+1)^5 divided by 4

= The remainder of 3*1^5 divided by 4

= The remainder of 3 divided by 4, which itself = 3

So when 7^{11} is divided by 4, the remainder is 3. This means 7^{11} is 3 more than a multiple of 4; i.e. a is 3 more than a multiple of 4.

Now we go back to 3^a. We found that a is 3 more than a multiple of 4. So there will be full cycles (we don’t need to know the exact number of cycles) and then a new cycle with start with three digits remaining:

3, 9, 7, 1

3, 9, 7, 1

3, 9, 7, 1

3, 9, 7, 1

3, 9, 7

With this pattern, we see the last digit of 3^7^11 is 7. When this 7 is divided by 5, remainder will be 2 – therefore, our answer is C.

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

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

Rounding Up Some Official GMAT Questions!

Quarter Wit, Quarter WisdomLast week we looked at some rounding rules. Today, let’s go over some official questions on rounding. They are quite simple and if we just keep the “Slip to the side and look for a 5” rule in mind, they can be easily solved.

Question 1: If n = 2.0453 and n* is the decimal obtained by rounding n to the nearest hundredth, what is the value of n* – n?

(A) -0.0053
(B) -0.0003
(C) 0.0007
(D) 0.0047
(E) 0.0153

Solution: A quick note on place value nomenclature:

Given a decimal 345.789, we know that 5 represents the units digit, 4 the tens digit and 3 the hundreds digit. Also, 7 represents the tenths digit, 8 the hundredths digit and 9 the thousandths digit and so on…

Now let’s go back to this question:

n = 2.0453

We need to round n to the nearest hundredth which means we will retain 2 digits after the decimal. The third digit after the decimal is 5 so 2.0453 rounded to the nearest hundredth is 2.05.

Thus n* – n = 2.05 – 2.0453 = 0.0047

Answer (D)

Question 2: If digit h is the hundredths digit in the decimal n = 0.2h6, what is the value of n, rounded to the nearest tenth?

Statement 1: n < 1/4

Statement 2: h < 5

Solution: Given that n = 0.2h6

We need to find the value of n rounded to the nearest tenth i.e. we need to keep only one digit after the decimal.

Statement 1: n < 1/4

In decimal form, it means n < 0.25

If h were 5 or greater, n would become 0.256 or 0.266 or higher. All these values would be more than 0.25 so h must be less than 5 such as 0.246 or 0.236 etc. In all such cases, n would be rounded to 0.2

This statement alone is sufficient.

Statement 2: h < 5

This is even simpler. Since we have been given that h is less than 5, when we round n to the tenths digit, we will get 0.2

This statement alone is also sufficient.

Answer (D)

Question 3: If d denotes a decimal number, is d >= 0.5?

Statement 1: When d is rounded to the nearest tenth, the result is 0.5.

Statement 2: When d is rounded to the nearest integer, the result is 1.

Solution: Again, a simple question!

We need to find whether d is greater than or equal to 0.5 or not.

Statement 1: When d is rounded to the nearest tenth, the result is 0.5.

This means that whatever d is, when we round it to the nearest tenth, we get 0.5. What are the possible values of d? If d is anywhere from 0.450 to 0.5499999…, it will be rounded to 0.5

Some of these numbers are less than 0.5 and others are greater than 0.5 so this statement alone is not sufficient.

Statement 2: When d is rounded to the nearest integer, the result is 1.

In this case d must be at least 0.5; only then can it be rounded to 1.

d can be anything from 0.50 to 1.499999… In any case, d will be greater than or equal to 0.5.

This statement alone is sufficient to answer the question.

Answer (B)

We hope you see that if we just remember the rules, we can solve most rounding questions very quickly and efficiently.

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!

Rounding Rules on the GMAT: Slip to the Side and Look for a Five!

Quarter Wit, Quarter WisdomThe famous rounding song by Joe Crone is pretty much all you need to solve the trickiest of rounding questions on GMAT:

You just slip to the side, and you look for a five.

Well if the number that you see is a five or more, 

You gotta round up now, that’s for sure.

If the number that you see is a four or less, 

   You gotta round down to avoid a mess.

To put it in our own words, when we round a decimal, we drop the extra decimal places and apply certain rules:

–          If the first dropped digit is 5 or greater, we round up the last digit that we keep.

–          If the first dropped digit is 4 or smaller, we keep the last digit that we keep, the same.

For Example, we need to round the following decimals to two digits after decimal:

(a) 3.857

We drop 7. Since 7 is ‘5 or greater’, we are left with 3.86

(b) 12.983

We drop 3. Since 3 is ‘4 or smaller’, we are left with 12.98

(c) 26.75463

We drop 463. Since 4 is ‘4 or smaller’, we are left with 26.75

(d) 8.9675

We drop 75. Since 7 is ‘5 or greater’, we are left with 8.97

Note example (c) carefully:

When we round 26.75463 to two decimal places, we do not start rounding from the rightmost digit i.e. this is incorrect: 26.75463 becomes 26.7546 which becomes 26.755 which further becomes 26.76 – this is not correct. .00463 is less than .005 and hence should be ignored. You only need to worry about the digit right next to the digit you are keeping. Just slip to the side, and look for a five!

A logical question arises: what happens when we have, say, 2.5 and we need to round it to the nearest integer? 2.5 is midway between 2 and 3. In that case, why do we round the number up, as the rule suggests? Note that a 2.5 is a tie and we have many tie breaking rules that can be used. They are ‘Round half to odd’, ‘Round half to even’, ‘Round up’, ‘Round down’, ‘Round towards 0’, ‘Round away from 0’ etc. We don’t need to worry about all these since GMAT uses only Round up i.e. 2.5 will be rounded up to 3.

Let’s take a look at a question now which uses these fundamentals.

Question: The exact cost price to make each unit of a widget is $7.6xy7, where x and y represent single digits. What is the value of y?

Statement 1: When the cost is rounded to the nearest cent, it becomes $7.65.

Statement 2: When the cost is rounded to the nearest tenth of a cent, it becomes $7.65.

Solution: The question is based on rounding. We need to figure out the value of y given some rounding scenarios. Let’s look at them one by one.

Statement 1: When the cost is rounded to the nearest cent, it becomes $7.65.

When rounded to the nearest cent, the cost becomes 7 dollars and 65 cents. 6xy7 cents got rounded to 65 cents. When will .6xy7 get rounded to .65? When .6xy7 lies anywhere in the range .6457 to .6547. Note that in all these cases, when you round the number to 2 digits, it will become .65.

Say price is 7.6468. We need to drop 68 but since 6 is ‘5 or greater’, 4 gets rounded up to 5.

Similarly, say the price is 7.6543. We need to drop 43. Since 4 is ‘4 or smaller’, 5 stays as it is.

So x and y can take various different values. This statement alone is not sufficient.

Statement 2: When the cost is rounded to the nearest tenth of a cent, it becomes $7.65

Now the cost is rounded to the tenth of a cent which means 3 places after the decimal. But the cost is given to us as $7.65. Since we need 3 places, the cost must be $7.650 (which will be written as $7.65)

When will 7.6xy7 get rounded to 7.650? Now this is the tricky part of the question – from 7.6xy7, you need to drop the 7 and round up y. When you do that, you get 7.650. This means 7.6xy7 must have been 7.6497. Only in this case, when we drop the 7, we round up the 9 to make 10, carry the 1 over to 4 and make it 5. This is the only way to get 7.650 on rounding 7.6xy7 to the tenth of a cent. Hence x must be 4 and y must be 9. This statement alone is sufficient to answer the question.

Answer (B)

Hope you see that a few simple rules can make rounding questions quite easy.

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!