# What to Do When You Find a Weighted Average Question In the Verbal Section of the GMAT

Weighted averages show up everywhere on the GMAT. Most test-takers are prepared to see them on the Quantitative Section, but they’ll show up on the Integrated Reasoning and Verbal Sections, as well.  Because it is such an exam staple, we want to make sure that we have a thorough, intuitive understanding of the concept.

In class, I’ll typically start with a simple example. Say you have two solutions, A and B. A is 10% salt and B is 20% salt. If we combine these two solutions to get a composite solution that is 14% salt, do we have more A or B in this composite solution? Most students eventually see that we’ll have more of solution A, but it doesn’t always feel instinctive. If we had had equal quantities of both solutions, the combined solution would have been 15% salt – equidistant from 10% and 20%. So, if there is 14% salt, the average skews closer to A than B, and thus, there must be more of solution A.

I’ll then give another example. Say that there is an intergalactic party in which both humans and aliens are present. The humans, on average, are 6 feet tall. The aliens, on average, are 100 feet tall. If the average height at the party is 99 feet, who is dominating the party? It isn’t so hard to see that this party is packed with aliens and that the few humans present would likely spend the evening cowering in some distant corner of the room. The upshot is that it’s easier to feel the intuition behind a weighted average question when the numbers are extreme.

Take this tough Critical Reasoning argument for example:

To be considered for inclusion in the Barbizon Film Festival, a film must belong either to the category of drama or of comedy. Dramas always receive more submissions but have a lower acceptance rate than comedy. All of the films are either foreign or domestic. This year, the overall acceptance rate for domestic films was significantly higher than that for foreign films. Within each category, drama and comedy, however, the acceptance rate for domestic films was the same as that for foreign films.

From the cited facts it can be properly concluded that

(A) significantly fewer foreign films than domestic films were accepted.
(B) a higher proportion of the foreign than of the domestic films submitted were submitted as dramas.
(C) the rate of acceptance of foreign films submitted was the same for drama as it was for comedies.
(D) the majority of the domestic films submitted were submitted as comedies.
(E) the majority of the foreign films submitted were submitted as dramas.

Okay. We know that dramas had a lower acceptance rate than comedies, and we know that the overall acceptance rate for domestic films was significantly higher than the acceptance rate for foreign films. So, let’s assign some easy numbers to try and get a handle on this information:

Say that the acceptance rate for dramas was 1% and the acceptance rate for comedies was 99%.

We’ll also say that the acceptance rate for domestic films was 98% and the acceptance rate for foreign films was 2%.

The acceptance rate within both domestic and foreign films is a weighted average of comedies and dramas. If only dramas were submitted, clearly the acceptance rate would be 1%. If only comedies were submitted, the acceptance rate would be 99%. If equal amounts of both were submitted, the acceptance rate would be 50%.

What do our numbers tell us? Well, if the acceptance rate for domestic films was 98%, then almost all of these films must have been comedies, and if the acceptance rate for foreign films was 2%, then nearly all of these films must have been dramas. So, domestic films were weighted towards comedies and foreign films were weighted towards drama. (An unfair stereotype, perhaps, but this is GMAC’s question, not mine.)

We can see that answer choice A is out, as we only have information regarding rates of acceptance, not absolute numbers. C is also out, as it violates a crucial premise of the question stem – we know that the acceptance rate for dramas is lower than for comedies, irrespective of whether we’re talking about foreign or domestic films.

That leaves us with answer choices B, D and E. So now what?

Let’s pick another round of values, but see if we can invalidate two of the three remaining options.

What if the acceptance rate for domestic films was 3% and the acceptance rate for foreign films was still 2%? (We’ll keep the acceptance rate for dramas at 1% and the acceptance rate for comedies at 99%.) Now domestic films would be mostly dramas, so option D is out – the majority of domestic films would not be comedies, as this  answer choice states.

Similarly, what if the acceptance rate for domestic films was 98% and the rate for foreign films was 97%? Now the foreign films would be mostly comedies, so option E is also out – the majority of foreign films would not be dramas, as this answer choice states.

Because the acceptance rate is lower for dramas than it is for comedies, and foreign films have a lower acceptance rate than do domestic films, the foreign films must be weighted more heavily towards dramas than domestic films are. This analysis is perfectly captured in option B, which is, in fact, the correct answer.

Takeaway: certain concepts, such as weighted averages, are such exam staples that will appear in both Quant and Verbal questions. If you see one of these examples in the Verbal Section, assigning extreme values to the information you are given can help you get a handle on the underlying logic being tested.

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

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

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

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

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

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

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

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

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

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

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

Here’s a quick recap with some examples:

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

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

If x is not 0, then √(x^2)/x =

(A) -1
(B) 0
(C) 1
(D) x
(E) |x|/x

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

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

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

Is √[(x – 3)^2] = (3 – x)?

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

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

This means that our question is basically:

Is |x – 3| = 3 – x?

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

Is |x – 3| = -(x – 3)?

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

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

So our question now boils down to:

Is (x – 3) negative (or 0)?

Statement 1: x is not 3

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

Statement 2: -x * |x| > 0

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

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

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

# GMAT Rate Questions: Tackling Problems with Multiple Components

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

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

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

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

And that’s it!

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

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

A) 6
B) 12
C) 15
D) 18
E) 24

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

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

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

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

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

1/8 + 1/24 + z = ¼

Multiply everything by 24, and we get:

3 + 1 + 24z = 6

24z = 2

z = 1/12.

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

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

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

# Avoid Obtaining the Wrong Values in Percent Increase Questions

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

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

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

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

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

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

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

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

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

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

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

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

# How to Simplify Sequences on the GMAT

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

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

Take this question for example:

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

I. 4
II. -1
III. -5

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

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

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

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

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

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

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

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

1) n is odd

2) an is positive

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

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

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

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

Let’s keep going.

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

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

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

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

After one term, we have 1.

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

# Quarter Wit, Quarter Wisdom: Using Visual Symmetry to Solve GMAT Probability Problems

Today, let’s take a look at an official GMAT question involving visual skills. It takes a moment to understand the given diagram, but at close inspection, we’ll find that this question is just a simple probability question – the trick is in understanding the symmetry of the figure:

The figure shown represents a board with 4 rows of pegs, and at the bottom of the board are 4 cells numbered 1 to 4. Whenever the ball shown passes through the opening between two adjacent pegs in the same row, it will hit the peg directly beneath the opening. The ball then has the probability 1/2 of passing through the opening immediately to the left of that peg and probability 1/2 of passing through the opening immediately to the right. What is the probability that when the ball passes through the first two pegs at the top it will end in cell 2?

(A) 1/16
(B) 1/8
(C) 1/4
(D) 3/8
(E) 1/2

First, understand the diagram. There are small pegs arranged in rows and columns. The ball falls between two adjacent pegs and hits the peg directly below. When it does, there are two ways it can go – either to the opening on the left or to the opening on the right. The probability of each move is equal, i.e. 1/2.

The arrow show the first path the ball takes. It is dropped between the top two pegs, hits the peg directly below it, and then either drops to the left side or to the right. The same process will be repeated until the ball falls into one of the four cells – 1, 2, 3 or 4.

Method 1: Using Symmetry
Now that we understand this process, let’s examine the symmetry in this diagram.

Say we flip the image along the vertical axis – what do we get? The figure is still exactly the same, but now the order of cells is reversed to be 4, 3, 2, 1. The pathways in which you could reach Cell 1 are now the pathways in which you can use to reach Cell 4.

OR think about it like this:

To reach Cell 1, the ball needs to turn left-left-left.

To reach Cell 4, the ball needs to turn right-right-right.

Since the probability of turning left or right is the same, the situations are symmetrical. This will be the same case for Cells 2 and 3. Therefore, by symmetry, we see that:

The probability of reaching Cell 1 = the probability of reaching Cell 4.

Similarly:

The probability of reaching Cell 2 = the probability of reaching Cell 3. (There will be multiple ways to reach Cell 2, but the ways of reaching Cell 3 will be similar, too.)

The total probability = the probability of reaching Cell 1 + the probability of reaching Cell 2 + the probability of reaching Cell 3 + the probability of reaching Cell 4 = 1

Because we know the probability of reaching Cells 1 and 4 are the same, and the probabilities of reaching Cells 2 and 3 are the same, this equation can be written as:

2*(the probability of reaching Cell 1) + 2*(the probability of reaching Cell 2) = 1

Let’s find the probability of reaching Cell 1:

After the first opening (not the peg, but the opening between pegs 1 and 2 in the first row), the ball moves left (between pegs 1 and 2 in second row) or right (between pegs 2 and 3 in second row). It must move left to reach Cell 1, and the probability of this = 1/2.

After that, the ball must move left again – the probability of this occurring is also 1/2, since probability of moving left or right is equal. Finally, the ball must turn left again to reach Cell 1 – the probability of this occurring is, again, 1/2. This means that the total probability of the ball reaching Cell 1 = (1/2)*(1/2)*(1/2) = 1/8

Plugging this value into the equation above:

2*(1/8) + 2 * probability of reaching Cell 2 = 1

Therefore, the probability of reaching Cell 2 = 3/8

Method 2: Enumerating the Cases
You can also answer this question by simply enumerating the cases.

At every step after the first drop between pegs 1 and 2 in the first row, there are two different paths available to the ball – either it can go left or it can go right. This happens three times and, hence, the total number of ways in which the ball can travel is 2*2*2 = 8

The ways in which the ball can reach Cell 2 are:

Left-Left-Right

Left-Right-Left

Right-Left-Left

So, the probability of the ball reaching Cell 2 is 3/8.

Note that here there is a chance that we might miss some case(s), especially in problems that involve many different probability options. Hence, enumerating should be the last option you use when tackling these types of questions on the GMAT.

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: Mother Knows Best

This weekend is Mother’s Day here in the United States, and also, as the first full weekend in May, a weekend that will kick off a sense of study urgency for those intent on the September Round 1 MBA admissions deadlines. (If your mother were here she’d tell you why: if you want two full months to study for the GMAT and two full months to work on your applications, you have to start studying now!)

In honor of mothers everywhere and in preparation for your GMAT, let’s consider one of the things that makes mothers so great. Even today as an adult, you’ll likely find that if you live a flight or lengthy drive/train from home, when you leave your hometown, your mother loads you up with snacks for the plane, bottled water for the drive, hand sanitizer for the airport, etc. Why is that? When it comes to their children – no matter how old or independent – mothers are prepared for every possible situation.

What if you get hungry on the plane, or you’re delayed at your connecting airport and your credit card registers fraud because of the strange location and you’re unable to purchase a meal?! She doesn’t want you getting sick after touching the railing on an escalator, so she found a Purell bottle that’s well less than the liquid limit at security (and also packed a clear plastic bag for you and your toiletries). Moms do not want their children caught in a unique and harmful (or inconvenient) situation, so they plan for all possible occurrences.

And that’s how you should approach Data Sufficiency questions on the GMAT.

When a novice test-taker sees the problem:

What is the value of x?

(1) x^2 = 25

(2) 8 < 2x < 12

He may quickly say “oh it’s 5” to both of them. 5 is the square root of 25, and the second equation simplifies to 4 < x < 6, and what number is between 4 and 6? It’s 5.

But your mother would give you caution, particularly because her mission is to avoid *negative* outcomes for you. She’d be prepared for a negative value of x (-5 satisfies Statement 1) and for nonintegers (x could be 4.00001 or 5.9999 given Statement 2). Knowing those contingencies, she’d wisely recognize that you need both statements to guarantee one exact answer (5) for x.

Just like she’d tie notes to your mittens or pin them on your shirt when you were a kid so that you wouldn’t forget (and like now she’ll text you reminders for your grandmother’s birthday or to RSVP to your cousin’s wedding), your mom would suggest that you keep these unique occurrences written down at the top of your noteboard on test day: Negative, Zero, Noninteger, Infinity, Biggest/Smallest Value. That way, you’ll always check for those unique situations before you submit your answer, and you’ll have a much better shot at a challenge-level problem like this:

The product of consecutive integers a, b, c, and d is 5,040. What is the value of d?

(1) d is prime

(2) d < c < b < a

So where does mom come in?

Searching for consecutive integers, you’ll likely factor 5,040 to 7, 8, 9, and 10 (the 10 is obvious because 5,040 ends in a 0, and then when you see that the rest is 504 and know that’s divisible by 9, and you’re just about done). And so with Statement 1, you’ll see that the only prime number in the bunch is 7, meaning that d = 7 and Statement 1 is sufficient. And Statement 2 seems to support that exact same conclusion – as the smallest of the 4 integers, d is, again, 7.

Right?

Enter mom’s notes: did you consider zero? (irrelevant) Did you consider nonintegers? (they specified integers, so irrelevant) Did you consider negative numbers?

That’s the key. The four consecutive integers could be -10, -9, -8, and -7 meaning that d could also be -10. That wasn’t an option for Statement 1 (only positives are prime) and so since you did the “hard work” of factoring 5,040 and then finally got to where Statement 2 was helpful, there’s a high likelihood that you were ready to be finished and saw 7 as the only option for Statement 2.

This is why mom’s reminders are so helpful: on harder problems, the “special circumstances” numbers that mom wants to make sure you’re always prepared for tend to be afterthoughts, having taken a backseat to the larger challenges of math. But mother knows best – you may not be stranded in a foreign airport without a snack and your car might not stall in the desert when you don’t have water, but in the rare event that such a situation occurs she wants you to be prepared. Keep mom’s list handy at the top of your noteboard (alas, the Pearson/Vue center won’t allow you to pin it to your shirt) and you, like mom, will be prepared for all situations.

By Brian Galvin.

# How to Avoid Trap Answers On GMAT Data Sufficiency Questions

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

Let’s move on to the other three options:

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

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

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

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

Let’s try another one:

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

Let’s continue our discussion on coordinate geometry today.

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

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

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

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

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

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

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

Traditionally, one would solve this question like this:

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

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

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

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

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

Thus, the equations of the 2 lines become:

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

Now, we can easily solve this official question:

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

Statement 1: b not equal to 0

Statement 2: ab > 0

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

Statement 1: b not equal to 0

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

Statement 2: ab>0

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

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

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

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

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

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

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

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

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

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

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

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

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

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

By Brian Galvin.

# Use This Tip to Avoid Critical Reasoning Traps on the GMAT

When you’ve been teaching test prep for a while you begin to be able to anticipate the types of questions that will give your students fits. The reason isn’t necessarily because these questions are unusually hard in a conventional sense, but because embedded within these problems is a form of misdirection that is nearly impossible to resist. It’s often worthwhile to dissect these problems in greater detail to reveal some deeper truths about how the test works.

Here is a problem I knew I’d be asked about often the moment I saw it:

W, X, Y, and Z represent distinct digits such that WX * YZ = 1995. What is the value of W?

1. X is a prime number
2. Z is not a prime number

The first instinct for most students I work with is, “I’m told nothing about W in either statement. There have to be many possibilities, so each statement alone is not sufficient.” When this thought occurs to you during the test, it’s important to resist it. By this, I don’t mean that you should simply assume that you’re wrong – there likely will be times when your first instincts are correct. Instead, what I mean is that you should take a bit more time to prove your assumptions to yourself. If there really are many workable scenarios, it won’t take much time to find them.

First, whenever there is an unusually large number and we’re dealing with multiplication, we want to take the prime factorization of that large number so that we can work with that figure’s basic building blocks and make it more manageable. In this case, the prime factorization of 1995 is 3 * 5 * 7 * 19. (First we see that five is a factor of 1995 because 1995 = 5*399. Next, we see that 3 is a factor of 399, because the digits of 399 sum to a multiple of 3. Now we have 5 * 3 * 133. Last, we know that 133 = 7 * 19, because if there are twenty 7’s in 140, there must be nineteen 7’s in 133.)

Now we can use these building blocks to form two-digit numbers that multiply to 1995. Here is a list of two-digit numbers we can assemble from those prime factors:

3 * 5 = 15

3 * 7 = 21

3 * 19 = 57

5 * 7 = 35

5 * 19 = 95

These are our candidates for WX and YZ. There aren’t many possibilities for multiplying two of these two-digit numbers and still getting a product of 1995. In fact, there are only two: 95*21 = 1995 and 35*57 = 1995. But we’re told that each digit must be unique, so 35*57 can’t work, as two of our variables would equal 5. This means that we know, before we even look at the statements, that our two two-digit numbers are 95 and 21 – we just need to know which is which.

It’s possible that WX = 95 and YZ = 21, or WX = 21 and YZ = 95. That’s it. What at first appeared to be a very open-ended question actually has very few workable solutions. Now that we’ve established our sample space of possibilities, let’s examine the statements:

Statement 1: If we know X is prime, we know that WX cannot be 21, as X would be 1 in this scenario and 1 is not a prime number. This means that WX has to be 95, and thus we know for a fact that W = 9. This statement alone is sufficient to answer the question.

Statement 2: If we know that Z is not prime, we know that YZ cannot be 95, as Z would be 5 in this scenario and 5 is, of course, prime. Thus, YZ is 21 and WX is 95, and again, we know for a fact that W is 9, so this statement alone is also sufficient.

The answer is D, either statement alone is sufficient to answer the question, a result very much at odds with most test-taker’s initial instincts.

Takeaway: the GMAT is engineered to wrong-foot test-takers, using our instincts against us.  Rather than simply assuming our instincts are wrong – they won’t always be – we want to be methodical about proving our intuitions one way or another by confirming them in some instances, refuting them in others. By being thorough and methodical, we reduce the odds that we’ll step into one of the traps the question-writer has set for us and increase the odds that we’ll answer the question correctly.

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

# Quarter Wit, Quarter Wisdom: Using a Venn Diagram vs. a Double Set Matrix on the GMAT

Critics may have given a rotten rating to the recently released “Batman v. Superman” movie, but we sure can use it to learn a valuable GMAT lesson. A difficult decision point for GMAT test takers is picking the probable winner between Venn diagrams and Double Set Matrices for complicated sets questions. If that is true for you too, then the onscreen rivalry between Batman and Superman will help you remember this trick:

Venn diagrams are like Superman – all powerful. They can help you solve almost all questions involving either 2 or 3 overlapping sets. But then, there are some situations in which double set matrix method (aka Batman with his amazing weaponry) might be easier to use. It is possible to solve these questions using Venn diagrams, too, but it is more convenient to solve them using a Double Set Matrix.

We have discussed solving three overlapping sets using Venn diagrams.

Today, we will look at the case in which using a Double Set Matrix is easier than using a Venn diagram – in instances where we have two sets of variables, such as English/Math and Middle School/High School, or Cake/Ice cream and Boys/Girls, etc.

Eventually, we will solve our question again using a Venn diagram, for those who like to use a single method for all similar questions. First, take a look at our question:

A business school event invites all of its graduate and undergraduate students to attend. Of the students who attend, male graduate students outnumber male undergraduates by a ratio of 7 to 2, and females constitute 70% of the group. If undergraduate students make up 1/6 of the group, which of the following CANNOT represent the number of female graduate students at the event?

(A) 18
(B) 27
(C) 36
(D) 72
(E) 180

To solve this problem using a Double Set Matrix, first jot down one set of variables as the row headings and the other as the column headings, as well as a row and column for “totals.” Now all you need to do is add in the information line by line as you read through the question.

“…male graduate students outnumber male undergraduates by a ratio of 7 to 2…

“…females constitute 70% of the group.

Female students make up 70% of the group, which implies that male students (total of 9x) make up 30% of the group.

9x = (30/100)*Total Students

Total Students = 30x

Since 9x is the total number of male students while 30x is the total number of all students, the total number of female students must be 30x – 9x = 21x.

If undergraduate students make up 1/6 of the group…

Undergrad students make 1/6 of the group, i.e. (1/6)*30x = 5x

If the total number of undergrad students is 5x and the number of male undergrad students is 2x, the number of female undergrad students must be 5x – 2x = 3x.

This implies that the number of graduate females must be 18x, since the total number of females is 21x.

Therefore, the number of graduate females must be a multiple of 18. 27 is the only answer choice that is not a multiple of 18, so it cannot be the number of graduate females – therefore, our answer must be B.

Now, here is how Superman can rescue us in this question. An analysis similar to the one above will give us a Venn diagram which looks like this:

Of course, we will get the same answer: the number of graduate females must be a multiple of 18. We know 27 is not a multiple of 18, so it cannot be the number of graduate females and therefore, our answer is still B.

Hopefully, next time you come across an overlapping sets question, you will know exactly who your superhero is!

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: Don’t Be the April Fool with Trap Answers!

Today, people across the world are viewing news stories and emails with a skeptical eye, on guard to ensure that they don’t get April fooled. Your company just released a press release about a new initiative that would dramatically change your workload? Don’t react just yet…it could be an April Fool’s joke.

But in case your goal is to leave that job for the greener pastures of business school, anyway, keep that April Fool’s Day spirit with you throughout your GMAT preparation. Read skeptically and beware of the way too tempting, way too easy answer.

First let’s talk about how the GMAT “fools” you. At Veritas Prep we’ve spent years teaching people to “Think Like the Testmaker,” and the only pushback we’ve ever gotten whilethemselves has been, “Hey! We’re not deliberately trying to fool people.”

So what are they trying to do? They’re trying to reward critical thinkers, and by doing so, there need to be traps there for those not thinking as critically. And that’s an important way to look at trap answers – the trap isn’t set in a “gotcha” fashion to be cruel, but rather to reward the test-taker who sees the too-good-to-be-true answer as an invitation to dig a little deeper and think a little more critically. One man’s trash is another man’s treasure, and one examinee’s trap answer is another examinee’s opportunity to showcase the reasoning skills that business schools crave.

With that in mind, consider an example, and try not to get April fooled:

What is the greatest prime factor of 12!11! + 11!10! ?

(A) 2
(B) 7
(C) 11
(D) 19
(E) 23

If you’re like many – more than half of respondents in the – you went straight for the April Fool’s answer. And what’s even more worrisome is that most of those test-takers who choose trap answer C don’t spend very long on this problem. They see that 11 appears in both additive terms, see it in the answer choice, and pick it quickly. But that’s exactly how the GMAT fools you – the trap answers are there for those who don’t dig deeper and think critically. If 11 were such an obvious answer, why are 19 and 23 (numbers greater than any value listed in the expanded versions of those factorials 12*11*10*9…) even choices? Who are they fooling with those?

If you get an answer quickly it doesn’t necessarily mean that you’re wrong, but it should at least raise the question, “Am I going for the fool’s answer here?”. And that should encourage you to put some work in. Here, the operative verb even appears in the question stem – you have to factor the addition into multiplication, since factors are all about multiplication/division and not addition/subtraction. When you factor out the common 11!:

11!(12! + 10!)

Then factor out the common 10! (12! is 12*11*10*9*8… so it can be expressed as 12*11*10!):

11!10!(12*11 + 1)

You end up with 11!*10!(133). And that’s where you can check 19 and 23 and see if they’re factors of that giant multiplication problem. And since 133 = 19*7, 19 is the largest prime factor and D is, in fact, the correct answer.

So what’s the lesson? When an answer comes a little too quickly to you or seems a little too obvious, take some time to make sure you’re not going for the trap answer.

Consider this – there are only four real reasons that you’ll see an easy problem in the middle of the GMAT:

1) It’s easy. The test is adaptive and you’re not doing very well so they’re lobbing you softballs. But don’t fear! This is only one of four reasons so it’s probably not this!

2) Statistically it’s fairly difficult, but it’s just easy to you because it’s something you studied well for, or for which you had a great junior high teacher. You’re just that good.

3) It’s not easy – you’re just falling for the trap answer.

4) It’s easy but it’s experimental. The GMAT has several problems in each section called “pretest items” that do not count towards your final score. These appear for research purposes (they’re checking to ensure that it’s a valid, bias-free problem and to gauge its difficulty), and they appear at random, so even a 780 scorer will likely see a handful of below-average difficulty problems.

Look back at that list and consider which are the most important. If it’s #1, you’re in trouble and probably cancelling your score or retaking the test anyway. And for #4 it doesn’t matter – that item doesn’t count. So really, the distinction that ultimately matters for your business school future is whether a problem like the example above fits #2 or #3.

If you find an answer a lot more quickly than you think you should, use some of that extra time to make sure you haven’t fallen for the trap. Engage those critical thinking skills that the GMAT is, after all, testing, and make sure that you’re not being duped while your competition is being rewarded. Avoid being the April Fool, and in a not-too-distant September you’ll be starting classes at a great school.

By.

# What Makes GMAT Quant Questions So Hard?

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

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

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

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

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

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

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

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

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

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

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

450 = 2 * 3^2 * 5^2

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

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

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

Bringing the given equation back, we get:

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

Statement 1: a is 1.

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

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

1 + c = 2

c = 1

Statement 2:  b is 2.

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

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

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

1 + c = 2

c = 1

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

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

3^b * 5^c = 45

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

If b  = 3, then 3^3 * 5^c = 45.

5^c = 45/27

C will take a non-integer value here.

c = .3174

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

tx = 2ab = (2√3)x * √5

t = 2√15

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

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

f(x) = (√3)x – (√5)^2

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

(√3)x – (√5)^2  = 0

x = (√5)/(√3)

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

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

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

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

# 2 Tips to Make GMAT Remainder Questions Easy

Several months ago, I wrote an article aboutBecause the concepts of division, quotients, and remainders show up so often on the GMAT, I thought it would be useful to revisit this frequently-tested topic. (After all, there always remains at least something unsaid about remainders, right?) When you encounter quotient/remainder problems on the GMAT, at times, it will be helpful to know the kind of division terminology we’re taught in grade school – in particular the quotient + remainder formula you’ll see detailed below – while at other times, you will simply want to select simple numbers that satisfy the parameters of a Data Sufficiency statement.

To ensure that you’re prepared for all types of quotient/remainder problems on the GMAT, let’s explore each of these division-related scenarios in a little more detail. A simple example can illustrate the important division terminology: if we divide 7 by 4, we’ll have 7/4 = 1 + 3/4.

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 even if the terminology takes a little reminding to come back to you.

In the abstract, the classic remainder formula is: Dividend/Divisor = Quotient + Remainder/Divisor. If we multiply through by the Divisor, we get another helpful variant of the remainder formula: Dividend = Quotient*Divisor + Remainder.

Simply knowing this terminology and the remainder equation will be sufficient to answer the following official GMAT 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 division 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 remainder 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 remainder formula, 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.

Consider, for example, the following question: 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 that emerges with our x values: x = 4 or 9 or 14, or 19… In essence, the first allowable value of x is the remainder. Afterwards, we’re simply adding the divisor, 5, over and over. Without much math at all, you could continue this cycle indefinitely: 4, 9, 14, 19, 24, 29, etc. 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

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

In this problem, Statement 1 gives us potential values for x – y. Remember the pattern discussed above: x – y must be 1 greater than a multiple of 5. 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 by itself.

Statement 2 gives us potential values for x + y. Once again, let’s use that pattern: x + y must be 2 greater than a multiple of 5. 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 on its own.

Now that we find ourselves in the classic C or E scenario, let’s 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.

Now remember this: for us to pick a non-E answer on Data Sufficiency, we must know that the value will stay the same in any scenario allowed by the question and statements. To be safe, let’s 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 which values we select, this will be the case – we can prove definitively that the remainder is 0. Together, the statements are sufficient, so the correct answer is C.

Now let’s summarize some important takeaways regarding GMAT quotient/remainder problems and the ever-important remainder formula. 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 remainder formula: 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.

Remember, also, that division is something you were once quite good at as a primary school student, so do not let the terminology intimidate you as an adult! The GMAT thrives on abstraction in these problems, so if you find yourself distracted by terminology or abstraction, simply try using small numbers to remind yourself how the operation works. The remainder equation Dividend = Divisor*Quotient + Remainder is an important one, but if you blank on it you can reconstruct it. Try, as we did at the beginning, 7 divided by 4. The result of that is 1, remainder 3. And the quotient is 1 because 4 goes into 7 one time, leaving 3 left over. So you can reconstruct the equation: to get back to 7, multiply the divisor (4) by the 1 time it went in to 7, and then add back the remaining 3: 7 = 4(1) + 3. Once you’ve stripped away the abstraction in quotient/remainder problems, the remainder is a concept you’ve been quite adept with your whole life!

Interested in more practice with GMAT division problems and the remainder equation? Check out some of our other articles on this frequently-tested GMAT topic, or try your hand at some practice questions via the Veritas Prep GMAT Question Bank or practice tests.

# Understanding Absolute Values with Two Variables

We have looked at quite a few absolute value and inequality concepts. (Check out our discussion on the basics of absolute values and inequalities, here, and our discussion on how to handle inequalities with multiple absolute value terms in a single variable, here.) Today let’s look at an absolute value concept involving two variables. It is unlikely that you will see such a question on the actual GMAT, since it involves multiple steps, but it will help you understand absolute values better.

Recall the definition of absolute value:

|x| = x if x ≥ 0

|x| = -x if x < 0

So, to remove the absolute value sign, you will need to consider two cases – one when x is positive or 0, and another when it is negative.

Say, you are given an inequality, such as |x – y| < |x|. Here, you have two absolute value expressions: |x – y| and |x|. You need to get rid of the absolute value signs, but how will you do that?

You know that to remove the absolute value sign, you need to consider the two cases. Therefore:

|x – y| = (x – y) if (x – y) ≥ 0

|x – y| = – (x – y) if (x – y) < 0

But don’t forget, we also need to remove the absolute value sign that |x| has. Therefore:

|x| = x if x ≥ 0

|x| = -x if x < 0

In all we will get four cases to consider:

Case 1: (x – y) ≥ 0 and x ≥ 0

Case 2: (x – y) < 0 and x ≥ 0

Case 3: (x – y) ≥ 0 and x < 0

Case 4: (x – y) < 0 and x < 0

Let’s look at each case separately:

Case 1: (x – y) ≥ 0 (which implies x ≥ y) and x ≥ 0

|x – y| < |x|

(x – y) < x

-y < 0

Multiply by -1 to get:

y > 0

In this case, we will get 0 < y ≤ x.

Case 2: (x – y) < 0 (which implies x < y) and x ≥ 0

|x – y| < |x|

-(x – y) < x

2x > y

x > y/2

In this case, we will get 0 < y/2 < x < y.

Case 3: (x – y) ≥ 0 (which implies x ≥ y) and x < 0

|x – y| < |x|

(x – y) < -x

2x < y

x < y/2

In this case, we will get y ≤ x < y/2 < 0.

Case 4: (x – y) < 0 (which implies x < y) and x < 0

|x – y| < |x|

-(x – y) < -x

-x + y < -x

y < 0

In this case, we will get x < y < 0.

Considering all four cases, we get that both x and y are either positive or both are negative. Case 1 and Case 2 imply that if both x and y are positive, then x > y/2, and Case 3 and Case 4 imply that if both x and y are negative, then x < y/2. With these in mind, there is a range of values in which the inequality will hold. Both x and y should have the same sign – if they are both positive, x > y/2, and if they are both negative, x < y/2.

Here are some examples of values for which the inequality will hold:

x = 4, y = 5

x = 8, y = 2

x = -2, y = -1

x = -5, y = -6

etc.

Here are some examples of values for which the inequality will not hold:

x = 4, y = -5 (x and y have opposite signs)

x = 5, y = 15 (x is not greater than y/2)

x = -5, y = 9 (x and y have opposite signs)

x = -6, y = -14 (x is not less than y/2)

etc.

As said before, don’t worry about going through this method during the actual GMAT exam – if you do get a similar question, some strategies such as plugging in values and/or using answer choices to your advantage will work. Overall, this example hopefully helped you understand absolute values a little better.

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: The Case of a Correct Answer Despite Incorrect Logic!

It is common for GMAT test-takers to think in the right direction, understand what a question gives and what it is asking to be found out, but still get the wrong answer. Mistakes made during the execution of a problem are common on the GMAT, but what is rather rare is going with incorrect logic and still getting the correct answer! If only life was this rosy so often!

Today, we will look at a question in which exactly this phenomenon occurs – we will find the flaw in the logic that test-takers often come up with and then learn how to correct that flaw:

If a motorist had driven 1 hour longer on a certain day and at an average rate of 5 miles per hour faster, he would have covered 70 more miles than he actually did. How many more miles would he have covered than he actually did had he driven 2 hours longer and at an average rate of 10 miles per hour faster on that day?

(A) 100

(B) 120

(C) 140

(D) 150

(E) 160

This little gem (and it’s detailed algebra solution) is from our Advanced Word Problems book. We will post its solution here, too, for the sake of a comprehensive discussion:

Method 1: Algebra

D = R*T ……….(I)

From here, the first theoretical trip can be represented as D + 70 = (R + 5)(T + 1), (the motorist travels for 1 extra hour at a rate of 5 mph faster), which can be expanded to D + 70 = RT + R + 5T +5.

We can then eliminate “D” by plugging in the value of “D” from our equation (I):

RT + 70 = RT + R + 5T + 5, which simplifies to 70 = R + 5T + 5 and then to 65 = R + 5T ……….. (II)

The second theoretical trip can be represented as (R+10)(T+2), which expands to RT + 2R + 10T + 20 (not that we only have an expression since we don’t know what the distance is).

The two middle terms (2R + 10T) can be factored to 2(R+5T), which allows us to use equation (II) here:

RT + 2(R+5T) + 20 = RT + 2(65) + 20 = RT + 150.

Since the original distance was RT, the additional distance is 150 more miles, or answer choice D.

We totally understand that this solution is a bit convoluted – algebra often is. So, understandably, students often look for a more direct logical solution.

Here is one they sometimes employ:

Method 2: Logic (Incorrect)
If the motorist had driven 1 hour longer at a rate 5 mph faster, then his original speed would be 70 miles subtracted by the extra 5 miles he drove in that hour to get 70 – 5 = 65 mph. If he drives at a rate 10 mph faster (i.e. at 65 + 10 = 75) * 2 for the extra hours, he/she would have driven 150 miles extra.

But here is the catch in this logic:

The motorist drove for an average rate of 5 mph extra. So the 70 includes not only the extra distance covered in the last hour, but also the extra 5 miles covered every hour for which he drove. Hence, his original speed is not 65. Now, let’s see the correct logical method of solving this:

Method 3: Logic (Correct)
Let’s review the original problem first. Say, speed is “S” mph – we don’t know the number of hours for which this speed was maintained.

STEP 1:

S + S + S + … + S + S = TOTAL DISTANCE COVERED

In the first hypothetical case, the motorist drove for an extra hour at a speed of 5 mph faster. This means he covered 5 extra miles every hour and then covered another S + 5 miles in the last hour. The underlined distances are the extra ones which all add up to 70.

STEP 2:

S + S + S + … + S + S = TOTAL DISTANCE COVERED

+5 +5 +5 + … + 5 + 5 = +70

In the second hypothetical case, in which the motorist drove for two hours longer at a speed of 10 mph faster,  he adds another 5 mph to his hourly speed and covers yet another distance of “S” in the second extra hour. In addition to S, he also covers another 10 miles in the second extra hour. The additional distances are shown in red  in the third case – every hour, the speed is 10 mph faster and he drove for two extra hours in this case (compared with Step 1).

STEP 3:

S + S + S + … + S + S + S + S = TOTAL DISTANCE COVERED

+5  +5  +5 + …  +5  +5  +5 = +70

+5  +5  +5 + …  +5  +5  +5 + 10 = +70 + 10

Note that the +5s and the S all add up to 70 (as seen in Step 2). We also separately add the extra 10 from the last hour. This is the logic of getting the additional distance of 70 + 70 + 10 = 150. It involves no calculations, but does require you to understand the logic. Therefore, our answer is still D.

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

# GMAT Tip of the Week: The Biggie Smalls Sufficiency Strategy

If it’s March, it must be Hip Hop Month in the Veritas Prep GMAT Tip of the Week space, where this week we’ll tackle the most notorious GMAT question type – Data Sufficiency – with some help from hip hop’s most notorious rapper – Biggie Smalls.

Biggie’s lyrics – and his name itself – provide a terrific template for you to use when picking numbers to test whether a statement is sufficient or not. So let’s begin with a classic lyric from “Big Poppa” – you may think Big is describing how he’s approach a young lady in a nightclub, but if you listen closely he’s actually talking directly to you as you attack Data Sufficiency:

“Ask you what your interests are, who you be with. Things to make you smile; what numbers to dial.”

“What numbers to dial” tends to be one of the biggest challenges that face GMAT examinees, so let’s examine the strategies that can take your score from “it was all a dream” to sipping champagne when you’re thirsty.

Biggie Smalls Strategy #1: Biggie Smalls
Consider this Data Sufficiency problem:

What is the value of integer z?

1) z is the remainder when positive integer y is divided by positive integer (y – 1)

2) y is not a prime number

Statistically, more than 50% of respondents in the Veritas Prep practice tests incorrectly choose answer choice A, that Statement 1 alone is sufficient but Statement 2 alone is not sufficient. Why? Because they’re not quite sure “what numbers to dial.” People know that they need to test numbers – Statement 1 is very abstract and difficult to visualize with variables – so they test a few numbers that come to mind:

If y = 5, y – 1 = 4, and the problem is then 5/4 which leads to 1, remainder 1.

If y = 10, y – 1 = 9, so the problem is then 10/9 which also leads to 1, remainder 1.

If they keep choosing random integers that happen to come to mind, they’ll see that pattern hold – the answer is ALMOST always 1 remainder 1, with exactly one exception. If y = 2, then y – 1 = 1, and 2 divided by 1 is 2 with no remainder. This is the only case where z does not equal 1, but that one exception shows that Statement 1 is not sufficient.

The question then becomes, “If there’s only one exception, how the heck does the GMAT expect me to stumble on that needle in a haystack?” And the answer comes directly from the Notorious BIG himself:

You need to test “Biggie Smalls,” meaning that you need to test the biggest number they’ll let you use (here it can be infinite, so just test a couple of really big numbers like 1,000 and 1,000,000) and you need to test the smallest number they’ll let you use. Here, that’s y = 2 and y – 1 = 1, since y – 1 must be a positive integer, and the smallest of those is 1.

The problem is that people tend to simply test numbers that come to mind (again, over half of all respondents think that Statement 1 is sufficient, which means that they very likely never considered the pairing of 2 and 1) and don’t push the limits. Data Sufficiency tends to play to the edge cases – if you get a statement like 5 < x < 12, you can’t just test 8, 9, and 10 – you’ll want to consider 5.00001 and 11.9999. When the GMAT gives you a range, use the entire range – and a good way to remind yourself of that is to just remember “Biggie Smalls.”

Biggie Smalls Strategy #2:  Juicy
In arguably his most famous song, “Juicy”, Biggie spits the line, “Damn right I like the life I live, because I went from negative to positive and it’s all…it’s all good (and if you don’t know, now you know).”

There, of course, Biggie is reminding you that you have to consider both negative and positive numbers in Data Sufficiency problems. Consider this example:

a, b, c, and d are consecutive integers such that the product abcd = 5,040. What is the value of d?

1) d is prime

2) a>b>c>d

This problem exemplifies why keeping Big’s words top of mind is so crucial – difficult problems will often “satisfy your intellect” with interesting math…and then beat you with negative/positive ideology. Here it takes some time to factor 5040 into the consecutive integers 7 x 8 x 9 x 10, but once you do, you can see that Statement 1 is sufficient: 7 is the only prime number.

But then when you carry that over to Statement 2, it’s very, very easy to see 7, 8, 9, and 10 as the only choices and again see that d = 7. But wait! If d doesn’t have to be prime – primes can only be positive – that allows for a possibility of negative numbers: -10, -9, -8, and -7. In that case, d could be either 7 or -10, so Statement 2 is actually not sufficient.

So heed Biggie’s logic: you’ll like the life you live much better if you go from negative to positive (or in most cases, vice versa since your mind usually thinks positive first), and if you don’t know (is that sufficient?) now, after checking for both positive and negative and for the biggest and smallest numbers they’ll let you pick, now you know.

By Brian Galvin.

# Quarter Wit, Quarter Wisdom: How to Find Composite Numbers on the GMAT

We love to talk about prime numbers and their various properties for GMAT preparation, but composite numbers usually aren’t mentioned. Composite numbers are often viewed as whatever is leftover after prime numbers are removed from a set of positive integers (except 1 because 1 is neither prime, nor composite), but it is important to understand how these numbers are made, what makes them special and what should come to mind when we read “composite numbers.”

Principle: Every composite number is made up of 2 or more prime numbers. The prime numbers could be the same or they could be distinct.

For example:

2*2 = 4 (Composite number)

2*3*11 = 66 (Composite number)

5*23 = 115 (Composite number)

and so on…

Look at any composite number. You will always be able to split it into 2 or more prime numbers (not necessarily distinct). For example:

72 = 2*2*2*3*3

140 = 2*2*5*7

166 = 2*83

and so on…

This principle does look quite simple and intuitive at first, but when tested, we could face problems because we don’t think much about it. Let’s look at it with the help of one of our 700+ level GMAT questions:

x is the smallest integer greater than 1000 that is not prime and that has only one factor in common with 30!. What is x?

(A) 1009

(B) 1021

(C) 1147

(D) 1273

(E) 50! + 1

If we start with the answer choices, the way we often do when dealing with prime/composite numbers, we will get stuck. If we were looking for a prime number, we would use the method of elimination – we would find factors of all other numbers and the number that was left over would be the prime number.

But in this question, we are instead looking for a composite number – a specific composite number – and some of the answer choices are probably prime. Try as we might, we will not find a factor for them, and by the time we realize that it is prime, we will have wasted a lot of precious time. Let’s start from the question stem, instead.

We need a composite number that has only one factor in common with 30!. Every positive integer will have 1 as a factor, as will 30!, hence the only factor our answer and 30! will have in common is 1.

30! = 1*2*3*…*28*29*30

30! is the product of all integers from 1 to 30, so all prime numbers less than 30 are factors of 30!.

To make a composite number which has no prime factor in common with 30!, we must use prime numbers greater than 30. The first prime number greater than 30 is 31.

(As an aside, note that if we were looking for the smallest number with no factor other than 1 in common with 31!, we would skip to 37. All integers between 31 and 37 are composite and hence, would have factors lying between 1 and 31. Similarly, if we were looking for the smallest number with no factor other than 1 in common with 50!, 53 would be the answer.)

Let’s get back to our question. If we want to make a composite number without using any primes until 30, we must use two or more prime numbers greater than 30, and the smallest prime greater than 30 is 31. If we use two 31’s to get the smallest composite number, we get 31*31 = 961 But 961 is not greater than 1000, so it cannot be our answer.

So, let’s find the next prime number after 31 – it is 37. Multiplying 31 and 37, we get 31*37 = 1147. This is the smallest composite number greater than 1000 with no prime factors in common with 30! – the only factor it has in common with 30! is 1. Therefore, our answer is (C).

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!

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

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

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

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

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

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

(A) 16
(B) 18
(C) 20
(D) 24
(E) 30

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

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

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

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

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

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

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

(x^6)^3 = 2^3

x^18 = 8

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

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

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

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

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

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

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

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

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

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

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

From the Question stem:

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

Question: What is n?

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

2, 3, and 4,

or 4, 6, and 8,

or 6, 9, and 12,

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

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

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

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

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

They could be

2, 3, 4 and 9

or 6, 9, 12 and 3

Therefore, Statement 2 alone is also not sufficient.

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

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

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

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

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

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

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

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

# A 750+ Level GMAT Geometry Question

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

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

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

(A) Sqrt(3) – 1

(B) Sqrt(3) + 2

(C) (Sqrt(3) – 1)/2

(D) (Sqrt(3) + 1)/2

(E) 2*(Sqrt(3) + 1)

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

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

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

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

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

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

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

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

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

Let’s use the pythagorean theorem on triangle BDC:

BD^2 + DC^2 = BC^2

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

X^4/32 + 2 = X^2

X^4 – 32*X^2 + 64 = 0

X^4 – 16X^2 + 8^2 – 16X^2 = 0

(X^2 – 8)^2 – (4X)^2 = 0

(X^2 -8 + 4X) * (X^2 – 8 – 4X) = 0

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

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

x = [-B +- Sqrt(B^2 – 4AC)]/2A

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

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

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

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

To recap:

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

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

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

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

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

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

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

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

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

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

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

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

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

(A) Between 1 and 10

(B) Between 11 and 15

(C) Between 15 and 20

(D) Between 20 and 25

(E) Greater than 25

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

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

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

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

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

# Quarter Wit, Quarter Wisdom: Solving GMAT Critical Reasoning Questions Involving Rates

In our “Quarter Wit, Quarter Wisdom” series, we have seen how to solve various rates questions – the basic ones as well as the complicated ones. But we haven’t considered critical reasoning questions involving rates, yet. In fact, the concept of rates makes these problems very difficult to both understand and explain. First, let’s look at what “rate” is.

Say my average driving speed is 60 miles/hr. Does it matter whether I drive for 2 hours or 4 hours? Will my average speed change if I drive more (theoretically speaking)? No, right? When I drive for more hours, the distance I cover is more. When I drive for fewer hours, the distance I cover is less. If I travel for a longer time, does it mean my average speed has decreased? No. For that, I need to know  what happened to the distance covered. If the distance covered is the same while time taken has increased, only then can I say that my speed was reduced.

Now we will look at an official question and hopefully convince you of the right answer:

The faster a car is traveling, the less time the driver has to avoid a potential accident, and if a car does crash, higher speeds increase the risk of a fatality. Between 1995 and 2000, average highway speeds increased significantly in the United States, yet, over that time, there was a drop in the number of car-crash fatalities per highway mile driven by cars.

Which of the following, if true about the United States between 1995 and 2000, most helps to explain why the fatality rate decreased in spite of the increase in average highway speeds?

(A) The average number of passengers per car on highways increased.

(B) There were increases in both the proportion of people who wore seat belts and the proportion of cars that were equipped with airbags as safety devices.

(C) The increase in average highway speeds occurred as legal speed limits were raised on one highway after another.

(D) The average mileage driven on highways per car increased.

(E) In most locations on the highways, the density of vehicles on the highway did not decrease, although individual vehicles, on average, made their trips more quickly.

Let’s break down the given argument:

• The faster a car, the higher the risk of fatality.
• In a span of 5 years, the average highway speed has increased.
• In the same time, the number of car crash fatalities per highway mile driven by cars has reduced.

This is a paradox question. In last 5 years, the average highway speed has increased. This would have increased the risk of fatality, so we would expect the number of car crash fatalities per highway mile to go up. Instead, it actually goes down. We need to find an answer choice that explains why this happened.

(A) The average number of passengers per car on highways increased.

If there are more people in each car, the risk of fatality increases, if anything. More people are exposed to the possibility of a crash, and if a vehicle is in fact involved in an accident, more people are at risk. It certainly doesn’t explain why the rate of fatality actually decreases.

(B) There were increases in both the proportion of people who wore seat belts and the proportion of cars that were equipped with airbags as safety devices.

This option tells us that the safety features in the cars have been enhanced. That certainly explains why the fatality rate has gone down. If the cars are safer now, the risk of fatality would have reduced, hence this option does help us in explaining the paradox. This is the answer, but let’s double-check by looking at the other options too.

(C) The increase in average highway speeds occurred as legal speed limits were raised on one highway after another.

This option is irrelevant – why the average speed increased is not our concern at all. Our only concern is that average speed has, in fact, increased. This should logically increase the risk of fatality, and hence, our paradox still stands.

(D) The average mileage driven on highways per car increased.

This is the answer choice that troubles us the most. The rate we are concerned about is number of fatalities/highway mile driven, and this option tells us that mileage driven by cars has increased.

Now, let’s consider the parallel with our previous distance-rate-time example:

Rate = Distance/Time

We know that if I drive for more time, it doesn’t mean that my rate changes. Here, however:

Rate = Number of fatalities/highway mile driven

In this case, if more highway miles are driven, it doesn’t mean that the rate will change. It actually has no impact on the rate; we would need to know what happened to the number of fatalities to find out what happened to the rate. Hence this option does not explain the paradox.

(E) In most locations on the highways, the density of vehicles on the highway did not decrease, although individual vehicles, on average, made their trips more quickly.

This answer choice tells us that on average, the trips were made more quickly, i.e. the speed increased. The given argument already tells us that, so this option does not help resolve the paradox.

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: Kanye, Wiz Khalifa, Twitter Beef…and GMAT Variables

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

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

Hit this kk and become yourself.

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

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

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

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

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

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

(1) a = 1

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

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

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

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

(1) -2 < a < 2

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

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

By Brian Galvin.

# Why Logic is More Important Than Algebra on the GMAT

One common complaint I get from students is that their algebra skills aren’t where they need to be to excel on the GMAT. This complaint, invariably, is followed by a request for additional algebra drills.

If you’ve followed this blog for any length of time, you know that one of the themes we stress is that Quantitative Reasoning is not, primarily, a math test. Though math is certainly involved – How could it not be? – logic and reasoning are far more important factors than conventional mathematical facility. I stress this in every class I teach. So why the misconception that we need to hone our algebra chops?

I suspect that the culprit here is the explanations that often accompany official GMAC questions. On the whole, they tend to be biased in favor of purely algebraic solutions.  They’re always technically correct, but often suboptimal for the test-taker who needs to arrive at a solution within two minutes. Consequently, many students, after reviewing these solutions and arriving at the conclusion that they would not have been capable of the hairy algebra proffered in the official solution, think they need to work on this aspect of their prep. And for the most part it isn’t true.

Here’s a good example:

If x, y, and k are positive numbers such that [x/(x+y)]*10 + [y/(x+y)]*20 = k and if x < y, which of the following could be the value of k?

A) 10
B) 12
C) 15
D) 18
E) 30

A large percentage of test-takers see this question, rub their hands together, and dive into the algebra. The solution offered in the Official Guide does the same – it is about fifteen steps, few of them intuitive. If you were fortunate enough to possess the algebraic virtuosity to solve the question in this manner, you’d likely chew up 5 or 6 minutes, a disastrous scenario on a test that requires you to average 2 minutes per problem.

The upshot is that it’s important for test-takers, when they peruse the official solution, not to arrive at the conclusion that they need to solve this question the same way the solution-writer did. Instead, we can use the same simple strategies we’re always preaching on this blog: pick some simple numbers.

We’re told that x<y, but for my first set of numbers, I like to make x and y the same value – this way, I can see what effect the restriction has on the problem. So let’s say x = 1 and y = 1. Plugging those values into the equation, we get:

(1/2) * 10 + (1/2) * 20  = k

5 + 10 = k

15 = k

Well, we know this isn’t the answer, because x should be less than y. So scratch off C. And now let’s see what the effect is when x is, in fact, less than y. Say x = 1 and y = 2. Now we get:

(1/3) * 10 + (2/3) * 20  = k

10/3 + 40/3 = k

50/3 = k

50/3 is about 17. So when we honor the restriction, k becomes larger than 15. The answer therefore must be D or E. Now we could pick another set of numbers and pay attention to the trend, or we can employ a bit of logic and common sense. The first term in the equation x/(x+y)*10 is some fraction multiplied by 10. So this term, logically, is some value that’s less than 10.

The second term in the equation is y/(x+y)*20, is some fraction multiplied by 20, this term must be less than 20. If we add a number that’s less than 10 to a number that’s less than 20, we’re pretty clearly not going to get a sum of 30. That leaves us with an answer of 18, or D.

(Note that if you’re really savvy, you’ll recognize that the equation is a weighted average. The coefficients in the weighted average are 10 and 20. If x and y were equal, we’d end up at the midway point, 15. Because 20 is multiplied by y, and y is greater than x, we’ll be pulled towards the high end of the range, leading to a k that must fall between 15 and 20 – only 18 is in that range.)

Takeaway: Never take a formal solution to a problem at face value. All you’re seeing is one way to solve a given question. If that approach doesn’t resonate for you, or seems so challenging that your conclusion is that you must purchase a host of textbooks in order to improve your formal math skills, then you haven’t absorbed what the GMAT is really about. Often, the relevant question isn’t, “Can you do the math?” It’s, “Can you reason your way to the answer without actually doing the math?”

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

# Quarter Wit, Quarter Wisdom: Should You Use the Permutation or Combination Formula?

A recurring question from many students who are preparing for GMAT is this: When should one use the permutation formula and when should one use the combination formula?

People have tried to answer this question in various ways, but some students still remain unsure. So we will give you a rule of thumb to follow in all permutation/combination questions:

You never NEED to use the permutation formula! You can always use the combination formula quite conveniently. First let’s look at what these formulas do:

Permutation: nPr = n!/(n-r)!
Out of n items, select r and arrange them in r! ways.

Combination: nCr = n!/[(n-r)!*r!]
Out of n items, select r.

So the only difference between the two formulas is that nCr has an additional r! in the denominator (that is the number of ways in which you can arrange r elements in a row). So you can very well use the combinations formula in place of the permutation formula like this:

nPr = nCr * r!

The nCr formula is far more versatile than nPr, so if the two formulas confuse you, just forget about nPr.

Whenever you need to “select,” “pick,” or “choose” r things/people/letters… out of n, it’s straightaway nCr. What you do next depends on what the question asks of you. Do you need to arrange the r people in a row? Multiply by r!. Do you need to arrange them in a circle? Multiply by (r-1)!. Do you need to distribute them among m groups? Do that! You don’t need to think about whether it is a permutation problem or a combination problem at all. Let’s look at this concept more in depth with the use of a few examples.

There are 8 teachers in a school of which 3 need to give a presentation each. In how many ways can the presenters be chosen?

In this question, you simply have to choose 3 of the 8 teachers, and you know that you can do that in 8C3 ways. That is all that is required.

8C3 = 8*7*6/3*2*1 = 56 ways

Not too bad, right? Let’s look at another question:

There are 8 teachers in a school of which 3 need to give a presentation each. In how many ways can all three presentations be done?

This question is a little different. You need to find the ways in which the presentations can be done. Here the presentations will be different if the same three teachers give presentations in different order. Say Teacher 1 presents, then Teacher 2 and finally Teacher 3 — this will be different from Teacher 2 presenting first, then Teacher 3 and finally Teacher 1. So, not only do we need to select the three teachers, but we also need to arrange them in an order. Select 3 teachers out of 8 in 8C3 ways and then arrange them in 3! ways:

We get 8C3 * 3! = 56 * 6 = 336 ways

Let’s try another one:

Alex took a trip with his three best friends and there he clicked 7 photographs. He wants to put 3 of the 7 photographs on Facebook. How many groups of photographs are possible?

For this problem, out of 7 photographs, we just have to select 3 to make a group. This can be done in 7C3 ways:

7C3 = 7*6*5/3*2*1 = 35 ways

Here’s another variation:

Alex took a trip with his three best friends and there he clicked 7 photographs. He wants to put 3 of the 7 photographs on Facebook, 1 each on the walls of his three best friends. In how many ways can he do that?

Here, out of 7 photographs, we have to first select 3 photographs. This can be done in 7C3 ways. Thereafter, we need to put the photographs on the walls of his three chosen friends. In how many ways can he do that? Now there are three distinct spots in which he will put up the photographs, so basically, he needs to arrange the 3 photographs in 3 distinct spots, which that can be done in 3! ways:

Total number of ways = 7C3 * 3! = (7*6*5/3*2*1) * 6= 35 * 6 = 210 ways

Finally, our last problem:

12 athletes will run in a race. In how many ways can the gold, silver and bronze medals be awarded at the end of the race?

We will start with selecting 3 of the 12 athletes who will win some position in the race. This can be done in 12C3 ways. But just selecting 3 athletes is not enough — they will be awarded 3 distinct medals of gold, silver, and bronze. Athlete 1 getting gold, Athlete 2 getting silver, and Athlete 3 getting bronze is not the same as Athlete 1 getting silver, Athlete 2 getting gold and Athlete 3 getting bronze. So, the three athletes need to be arranged in 3 distinct spots (first, second and third) in 3! ways:

Total number of ways = 12C3 * 3! ways

Note that some of the questions above were permutation questions and some were combination questions, but remember, we don’t need to worry about which is which. All we need to think about is how to solve the question, which is usually by starting with nCr and then doing any other required steps. Break the question down — select people and then arrange if required. This will help you get rid of the “permutation or combination” puzzle once and for all.

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

# GMAT Tip of the Week: Your MLK Study Challenge (Remove Your Biases)

As we celebrate Martin Luther King, Jr. this weekend, you may take some of your free time to study for the GMAT. And if you do, make sure to heed the lessons of Dr. King, particularly as you study Data Sufficiency.

If Dr. King were alive today, he would certainly be proud of the legislation he inspired to end much of the explicit bias – you can’t eat here, vote there, etc. – that was part of the American legal code until the 1960s. But he would undoubtedly be dismayed by the implicit bias that still runs rampant across society.

This implicit bias is harder to detect and even harder to “fix.” It’s the kind of bias that, for example, the movie Freaknomics shows; often when the name at the top of a resume connotes some sort of stereotype, it subconsciously colors the way that the reader of that resume processes the rest of the information on it.

While that kind of subconscious bias is a topic for a different blog to cover, it has an incredible degree of relevance to the way that you attack GMAT Data Sufficiency problems. If you’re serious about studying for the GMAT, you’ll probably have long enacted your own versions of the Voting Rights Act and Civil Rights Act well before you get to test day – that is to say, you’ll have figured out how to eliminate the kind of explicit bias that comes from reading a question like:

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

1) x > 0

2) y is a 3 digit number

Here, you’ll likely see very quickly that Statement 1 is not sufficient, and come back to Statement 2 with fresh eyes. You don’t know that x is positive, so you’ll quickly see that y could be 111 and x could be 2, or that y could be -111 and x could be -2, so Statement 2 is clearly also not sufficient. The explicit bias that came from seeing “x is positive” is relatively easy to avoid – you know not to carry over that explicit information from Statement 1 to Statement 2.

But you also need to be just as aware of implicit bias. Try this question, as it is more likely to appear on the actual GMAT:

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

1) x is a prime number

2) y is a 3 digit number

On this version of the problem, people become extremely susceptible to implicit bias. You no longer get to quickly rule out the obvious “x is positive.” Here, the first statement serves to pollute your mind – it is, on its own merit, sufficient (if y is odd and the product of x and y is even, the only prime number x could be is 2, the only even prime), but it also serves to get you thinking about positive numbers (only positive numbers can be prime) and integers (only integers are prime). But those aren’t explicitly stated; they’re just inferences that your mind quickly makes, and then has trouble getting rid of. So as you assess Statement 2, it’s harder for you to even think of the possibilities that:

x could be -2 and y could be -111: You’re not thinking about negatives!

x could be 2/3 and y could be 333: You’re not thinking about non-integers!

On this problem, over 50% of users say that Statement 2 is sufficient (and less than 25% correctly answer A, that Statement 1 alone is sufficient), because they fall victim to that implicit bias that comes from Statement 1 whispering – not shouting – “positive integers.”

Harder problems will generally prey on your more subtle bias, so you need to make sure you’re giving each statement a fresh set of available options. So this Martin Luther King, Jr. weekend, applaud the progress that you have made in removing explicit bias from your Data Sufficiency regimen – you now know not to include Statement 1 directly in your assessment of Statement 2 ALONE – but remember that implicit bias is just as dangerous to your score. Pay attention to the times that implicit bias draws you to a poor decision, and be steadfast in your mission to give each statement its deserved, unbiased attention.

By Brian Galvin.

# Quarter Wit, Quarter Wisdom: An Interesting Property of Exponents

Today, let’s take a look at an interesting number property. Once we discuss it, you might think, “I always knew that!” and “Really, what’s new here?” So let me give you a question beforehand:

For integers x and y, 2^x + 2^y = 2^(36). What is the value of x + y?

Think about it for a few seconds – could you come up with the answer in the blink of an eye? If yes, great! Close this window and wait for the next week’s post. If no, then read on. There is much to learn today and it is an eye-opener!

Let’s start by jotting down some powers of numbers:

Power of 2: 1, 2, 4, 8, 16, 32 …

Power of 3: 1, 3, 9, 27, 81, 243 …

Power of 4: 1, 4, 16, 64, 256, 1024 …

Power of 5: 1, 5, 25, 125, 625, 3125 …

and so on.

Obviously, for every power of 2, when you multiply the previous power by 2, you get the next power (4*2 = 8).

For every power of 3, when you multiply the previous power by 3, you get the next power (27*3 = 81), and so on.

Also, let’s recall that multiplication is basically repeated addition, so 4*2 is basically 4 + 4.

This leads us to the following conclusion using the power of 2:

4 * 2 = 8

4 + 4 = 8

2^2 + 2^2 = 2^3

(2 times 2^2 gives 2^3)

Similarly, for the power of 3:

27 * 3 = 81

27 + 27 + 27 = 81

3^3 + 3^3 + 3^3 = 3^4

(3 times 3^3 gives 3^4)

And for the power of 4:

4 * 4 = 16

4 + 4 + 4 + 4 = 16

4^1 + 4^1 + 4^1 + 4^1 = 4^2

(4 times 4^1 gives 4^2)

Finally, for the power of 5:

125 * 5 = 625

125 + 125 + 125 + 125 + 125 = 625

5^3 + 5^3 + 5^3 + 5^3 + 5^3 = 5^4

(5 times 5^3 gives 5^4)

Quite natural and intuitive, isn’t it? Take a look at the previous question again now.

For integers x and y, 2^x + 2^y = 2^(36). What is the value of x + y?

A) 18

(B) 32

(C) 35

(D) 64

(E) 70

Which two powers when added will give 2^(36)?

From our discussion above, we know they are 2^(35) and 2^(35).

2^(35) + 2^(35) = 2^(36)

So x = 35 and y = 35 will satisfy this equation.

x + y = 35 + 35 = 70

One question arises here: Is this the only possible sum of x and y? Can x and y take some other integer values such that the sum of 2^x and 2^y will be 2^(36)?

Well, we know that no matter which integer values x and y take, 2^x and 2^y will always be positive, which means both x and y must be less than 36. Now note that no matter which two powers of 2 you add, their sum will always be less than 2^(36). For example:

2^(35) + 2^(34) < 2^(35) + 2^(35)

2^(2) + 2^(35) < 2^(35) + 2^(35)

etc.

So if x and y are both integers, the only possible values that they can take are 35 and 35.

How about something like this: 2^x + 2^y + 2^z = 2^36? What integer values can x, y and z take here?

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: Make 2016 The Year Of Number Fluency

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

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

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

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

With 2016, for example:

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

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

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

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

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

(A) 18

(B) 36

(C) 42

(D) 54

(E) 56

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

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

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

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

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

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

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

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

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

By Brian Galvin.

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

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

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

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

(A) 2

(B) 3

(C) 4

(D) 6

(E) 8

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

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

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

(A) x + 10

(B) 10x + 1

(C) 100(x + 10)

(D) 100 * (x+10)/(x+100)

(E) 100 * (10x + 1)/(10x+10)

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

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

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

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

Start: 100 liters

After removing water: 90 liters

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

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

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

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

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

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

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

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

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

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

# Quarter Wit, Quarter Wisdom: Calculating the Probability of Intersecting Events

We know our basic probability formulas (for two events), which are very similar to the formulas for sets:

P(A or B) = P(A) + P(B) – P(A and B)

P(A) is the probability that event A will occur.

P(B) is the probability that event B will occur.

P(A or B) gives us the union; i.e. the probability that at least one of the two events will occur.

P(A and B) gives us the intersection; i.e. the probability that both events will occur.

Now, how do you find the value of P(A and B)? The value of P(A and B) depends on the relation between event A and event B. Let’s discuss three cases:

1) A and B are independent events

If A and B are independent events such as “the teacher will give math homework,” and “the temperature will exceed 30 degrees celsius,” the probability that both will occur is the product of their individual probabilities.

Say, P(A) = P(the teacher will give math homework) = 0.4

P(B) = P(the temperature will exceed 30 degrees celsius) = 0.3

P(A and B will occur) = 0.4 * 0.3 = 0.12

2) A and B are mutually exclusive events

If A and B are mutually exclusive events, this means they are events that cannot take place at the same time, such as “flipping a coin and getting heads” and “flipping a coin and getting tails.” You cannot get both heads and tails at the same time when you flip a coin. Similarly, “It will rain today” and “It will not rain today” are mutually exclusive events – only one of the two will happen.

In these cases, P(A and B will occur) = 0

3) A and B are related in some other way

Events A and B could be related but not in either of the two ways discussed above – “The stock market will rise by 100 points” and “Stock S will rise by 10 points” could be two related events, but are not independent or mutually exclusive. Here, the probability that both occur would need to be given to you. What we can find here is the range in which this probability must lie.

Maximum value of P(A and B):

The maximum value of P(A and B) is the lower of the two probabilities, P(A) and P(B).

Say P(A) = 0.4 and P(B) = 0.7

The maximum probability of intersection can be 0.4 because P(A) = 0.4. If probability of one event is 0.4, probability of both occurring can certainly not be more than 0.4.

Minimum value of P(A and B):

To find the minimum value of P(A and B), consider that any probability cannot exceed 1, so the maximum P(A or B) is 1.

Remember, P(A or B) = P(A) + P(B) – P(A and B)

1 = 0.4 + 0.7 – P(A and B)

P(A and B) = 0.1 (at least)

Therefore, the actual value of P(A and B) will lie somewhere between 0.1 and 0.4 (both inclusive).

Now let’s take a look at a GMAT question using these fundamentals:

There is a 10% chance that Tigers will not win at all during the whole season. There is a 20% chance that Federer will not play at all in the whole season. What is the greatest possible probability that the Tigers will win and Federer will play during the season?

(A) 55%

(B) 60%

(C) 70%

(D) 72%

(E) 80%

Let’s review what we are given.

P(Tigers will not win at all) = 0.1

P(Tigers will win) = 1 – 0.1 = 0.9

P(Federer will not play at all) = 0.2

P(Federer will play) = 1 – 0.2 = 0.8

Do we know the relation between the two events “Tigers will win” (A) and “Federer will play” (B)? No. They are not mutually exclusive and we do not know whether they are independent.

If they are independent, then the P(A and B) = 0.9 * 0.8 = 0.72

If the relation between the two events is unknown, then the maximum value of P(A and B) will be 0.8 because P(B), the lesser of the two given probabilities, is 0.8.

Since 0.8, or 80%, is the greater value, the greatest possibility that the Tigers will win and Federer will play during the season is 80%. Therefore, our answer is E.

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

# Quarter Wit, Quarter Wisdom: Basic Operations for GMAT Inequalities

We know that we can perform all basic operations of addition, subtraction, multiplication and division on two equations:

a = b

c = d

When these numbers are equal, we know that:

a + c = b + d (Valid)

a – c = b – d (Valid)

a * c = b * d (Valid)

a / c = b / d (Valid assuming c and d are not 0)

When can we add, subtract, multiply or divide two inequalities? There are rules that we need to follow for those. Today let’s discuss those rules and the concepts behind them.

We can add two inequalities when they have the same inequality sign.

a < b

c < d

a + c < b + d (Valid)

Conceptually, it makes sense, right? If a is less than b and c is less than d, then the sum of a and c will be less than the sum of b and d.

On the same lines:

a > b

c > d

a + c > b + d (Valid)

Case 2: What happens when the inequalities have opposite signs?

a > b

c < d

We need to multiply one inequality by -1 to get the two to have the same inequality sign.

-c > -d

a – c > b – d

Subtraction:

We can subtract two inequalities when they have opposite signs:

a > b

c < d

a – c > b – d (The result will take the sign of the first inequality)

Conceptually, think about it like this: from a greater number (a is greater than b), if we subtract a smaller number (c is smaller than d), the result (a – c) will be greater than the result obtained when we subtract the greater number from the smaller number (b – d).

Note that this result is the same as that obtained when we added the two inequalities after changing the sign (see Case 2 above). We cannot subtract inequalities if they have the same sign, so it is better to always stick to addition. If the inequalities have the same sign, we simply add them. If the inequalities have opposite signs, we multiply one of them by -1 (to get the same sign) and then add them (in effect, we subtract them).

Why can we not subtract two inequalities when they have the same inequality sign, such as when a > b and c > d?

Say, we have 3 > 1 and 5 > 1.

If we subtract these two, we get 3 – 5 > 1 – 1, or -2 > 0 which is not valid.

If instead it were 3 > 1 and 2 > 1, we would get 1 > 0 which is valid.

We don’t know how much greater one term is from the other and hence we cannot subtract inequalities when their inequality signs are the same.

Multiplication:

Here, the constraint is the same as that in addition (the inequality signs should be the same) with an extra constraint: both sides of both inequalities should be non-negative. If we do not know whether both sides are non-negative or not, we cannot multiply the inequalities.

If a, b, c and d are all non negative,

a < b

c < d

a*c < b*d (Valid)

When two greater numbers are multiplied together, the result will be greater.

Take some examples to see what happens in Case 1, or more numbers are negative:

-2 < -1

10 < 30

Multiply to get: -20 < -30 (Not valid)

-2 < 7

-8 < 1

Multiply to get: 16 < 7 (Not valid)

Division:

Here, the constraint is the same as that in subtraction (the inequality signs should be opposite) with an extra constraint: both sides of both inequalities should be non-negative (obviously, 0 should not be in the denominator). If we do not know whether both sides are positive or not, we cannot divide the inequalities.

a < b

c > d

a/c < b/d (given all a, b, c and d are positive)

The final inequality takes the sign of the numerator.

Think of it conceptually: a smaller number is divided by a greater number, so the result will be a smaller number.

Take some examples to see what happens in Case 1, or more numbers are negative.

1 < 2

10 > -30

Divide to get 1/10 < -2/30 (Not valid)

Takeaways:

Addition: We can add two inequalities when they have the same inequality signs.

Subtraction: We can subtract two inequalities when they have opposite inequality signs.

Multiplication: We can multiply two inequalities when they have the same inequality signs and both sides of both inequalities are non-negative.

Division: We can divide two inequalities when they have opposite inequality signs and both sides of both inequalities are non-negative (0 should not be in the denominator).

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

# How to Make Rate Questions Easy on the GMAT

I recently wrote about the reciprocal relationship between rate and time in “rate” questions. Occasionally, students will ask why it’s important to understand this particular rule, given that it’s possible to solve most questions without employing it.

There are two reasons: the first is that knowledge of this relationship can convert incredibly laborious arithmetic into a very straightforward calculation. And the second is that this same logic can be applied to other types of questions. The goal, when preparing for the GMAT, isn’t to internalize hundreds of strategies; it’s to absorb a handful that will prove helpful on a variety of questions.

The other night, I had a tutoring student present me with the following question:

It takes Carlos 9 minutes to drive from home to work at an average rate of 22 miles per hour.  How many minutes will it take Carlos to cycle from home to work along the same route at an average rate of 6 miles per hour?

(A) 26

(B) 33

(C) 36

(D) 44

(E) 48

This question doesn’t seem that hard, conceptually speaking, but here is how my student attempted to do it: first, he saw that the time to complete the trip was given in minutes and the rate of the trip was given in hours so he did a simple unit conversion, and determined that it took Carlos (9/60) hours to complete his trip.

He then computed the distance of the trip using the following equation: (9/60) hours * 22 miles/hour = (198/60) miles. He then set up a second equation: 6miles/hour * T = (198/60) miles. At this point, he gave up, not wanting to wrestle with the hairy arithmetic. I don’t blame him.

Watch how much easier it is if we remember our reciprocal relationship between rate and time. We have two scenarios here. In Scenario 1, the time is 9 minutes and the rate is 22 mph. In Scenario 2, the rate is 6 mph, and we want the time, which we’ll call ‘T.” The ratio of the rates of the two scenarios is 22/6. Well, if the times have a reciprocal relationship, we know the ratio of the times must be 6/22. So we know that 9/T = 6/22.

Cross-multiply to get 6T = 9*22.

Divide both sides by 6 to get T = 9*22/6.

We can rewrite this as T = (9*22)/(3*2) = 3*11 = 33, so the answer is B.

The other point I want to stress here is that there isn’t anything magical about rate questions. In any equation that takes the form a*b = c, a and b will have a reciprocal relationship, provided that we hold c constant. Take “quantity * unit price = total cost”, for example. We can see intuitively that if we double the price, we’ll cut the quantity of items we can afford in half. Again, this relationship can be exploited to save time.

Take the following data sufficiency question:

Pat bought 5 lbs. of apples. How many pounds of pears could Pat have bought for the same amount of money?

(1) One pound of pears costs \$0.50 more than one pound of apples.

(2) One pound of pears costs 1 1/2 times as much as one pound of apples.

Statement 1 can be tested by picking numbers. Say apples cost \$1/pound. The total cost of 5 pounds of apples would be \$5.  If one pound of pears cost \$.50 more than one pound of apples, then one pound of pears would cost \$1.50. The number of pounds of pears that could be purchased for \$5 would be 5/1.5 = 10/3. So that’s one possibility.

Now say apples cost \$2/pound. The total cost of 5 pounds of apples would be \$10. If one pound of pears cost \$.50 more than one pound of apples, then one pound of pears would cost \$2.50. The number of pounds of pears that could be purchased for \$10 would be 10/2.5 = 4. Because we get different results, this Statement alone is not sufficient to answer the question.

Statement 2 tells us that one pound of pears costs 1 ½ times (or 3/2 times) as much as one pound of apples. Remember that reciprocal relationship! If the ratio of the price per pound for pears and the price per pound for apples is 3/2, then the ratio of their respective quantities must be 2/3. If we could buy five pounds of apples for a given cost, then we must be able to buy (2/3) * 5 = (10/3) pounds of pears for that same cost. Because we can find a single unique value, Statement 2 alone is sufficient to answer the question, and we know our answer must be B.

Takeaway: Remember that in “rate” questions, time and rate will have a reciprocal relationship, and that in “total cost” questions, quantity and unit price will have a reciprocal relationship. Now the time you save on these problem-types can be allocated to other questions, creating a virtuous cycle in which your time management, your accuracy, and your confidence all improve in turn.

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

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

The GMAT asks a fair number of questions about the properties of two-digit numbers whose tens and units digits have been reversed. Because these questions pop up so frequently, it’s worth spending a little time to gain a deeper understanding of the properties of such pairs of numbers. Like much of the content on the GMAT, we can gain understanding of these problems by simply selecting random examples of such numbers and analyzing and dissecting them algebraically.

Let’s do both.

First, we’ll list out some random pairs of two-digit numbers whose tens and units digits have been reversed: {34, 43}; {17, 71}; {18, 81.} Now we’ll see if we can recognize a pattern when we add or subtract these figures. First, let’s try addition: 34 + 43 = 77; 17 + 71 = 88; 18 + 81 = 99. Interesting. Each of these sums turns out to be a multiple of 11. This will be true for the sum of any two two-digit numbers whose tens and units digits are reversed. Next, we’ll try subtraction: 43 – 34 = 9; 71 – 17 = 54; 81 – 18 = 63; Again, there’s a pattern. The difference of each pair turns out to be a multiple of 9.

Algebraically, this is easy enough to demonstrate. Say we have a two-digit number with a tens digit of “a” and a units digit of “b”. The number can be depicted as 10a + b. (If that isn’t clear, use a concrete number to illustrate it to yourself. Let’s reuse “34”. In this case a = 3 and b = 4. 10a + b = 10*3 + 4 = 34. This makes sense. The number in the “tens” place should be multiplied by 10.) If the original number is 10a + b, then swapping the tens and units digits would give us 10b + a. The sum of the two terms would be (10a + b) + (10b + a) = 11a + 11b = 11(a + b.) Because “a” and “b” are integers, this sum must be a multiple of 11. The difference of the two terms would be  (10a + b) – (10b + a) = 9a – 9b = 9(a – b) and this number will be a multiple of 9.

Now watch how easy certain official GMAT questions become once we’ve internalized these properties:

The positive two-digit integers x and y have the same digits, but in reverse order. Which of the following must be a factor of x + y?

A) 6

B) 9

C) 10

D) 11

E) 14

If you followed the above discussion, you barely need to be conscious to answer this question correctly. We just proved that the sum of two-digit numbers whose units and tens digits have been reversed is 11! No need to do anything here. The answer is D. Pretty nice.

Let’s try another, slightly tougher one:

If a two-digit positive integer has its digits reversed, the resulting integer differs from the original by 27. By how much do the two digits differ?

A) 3

B) 4

C) 5

D) 6

E) 7

This one is a little more indicative of what we’re likely to encounter on the actual GMAT. It’s testing us on a concept we’re expected to know, but doing so in a way that precludes us from simply relying on rote memorization. So let’s try a couple of approaches.

First, we’ll try picking some numbers. Let’s use the answer choices to steer us. Say we try B – we’ll want two digits that differ by 4. So let’s use the numbers 84 and 48. Okay, we can see that the difference is 84 – 48 = 36. That difference is too big, it should be 27. So we know that the digits are closer together. This means that the answer must be less than 4. We’re done. The answer is A. (And if you were feeling paranoid that it couldn’t possibly be that simple, you could test two numbers whose digits were 3 apart, say, 14 and 41. 41-14 = 27. Proof!)

Alternatively, we can do this one algebraically. We know that if the original two-digit numbers were 10a +b, that the new number, whose digits are reversed, would be 10b + a. If the difference between the two numbers were 27, we’d derive the following equation: (10a + b) – (10b + a) = 27. Simplifying, we get 9a – 9b = 27. Thus, 9(a – b) = 27, and a – b = 3. Also not so bad.

Takeaway: Once you’ve completed a few hundred practice questions, you’ll begin to notice that a few GMAT strategies are applicable to a huge swath of different question types. You’re constantly picking numbers, testing answer choices, doing simple algebra, or applying a basic number property that you’ve internalized. In this case, the relevant number property to remember is that the sum of two two-digit numbers whose units and tens digits have been reversed is always a multiple of 11, and the difference of such numbers is always a multiple of 9. Generally speaking, if you encounter a particular question type more than once in the Official Guide, it’s always worth spending a little more time familiarizing yourself with it.