The post Quarter Wit, Quarter Wisdom: Solving the Pouring Water Puzzle appeared first on Veritas Prep Blog.

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

Here is the puzzle:

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

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

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

Now think about it:

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

*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!*

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]]>The post 4 Predictions for Test Prep and Admissions in 2017 appeared first on Veritas Prep Blog.

]]>**One-year MBA programs will reach a tipping point in the United States.**

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

**Video prompts will become much more common in business school applications.**

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

**An Asia-scale cheating scandal will hit the SAT or ACT in the United States.**

News articles about standardized test cheating scandals like this one and this one seem to come out nearly every month. Much of the blame lies with the pressure that students — and especially their families — put on themselves to do well on these exams.

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

**Community colleges will gain a lot more recognition.**

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

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

*By Scott Shrum*

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]]>The post How to Answer GMAT Questions That are About an Unfamiliar Topic appeared first on Veritas Prep Blog.

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

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

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

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

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

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

*(B) Patrons who are under financial pressure from their credit-card obligations tend to tip less when presented with a restaurant bill on a tray with credit-card logo than when the tray has no logo.*

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

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

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

Let’s break down the argument:

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

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

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

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

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

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

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

A: After 12 hours of night time sleep, I can’t study.

B: Yeah, because your sleep pattern is linked to your level of concentration. After a long sleep, your mind is still muddled and lazy so you can’t study.

A: After only 4 hrs of night time sleep, I can’t study either.

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

Let’s now review answer choice E since it tends to confuse people:

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

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

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

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

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

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

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

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

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

Therefore, our answer is B.

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

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

*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!*

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]]>The post GMAT Tip of the Week: Taking the Least Amount of Time to Solve “At Least” Probability Problems appeared first on Veritas Prep Blog.

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

To illustrate, let’s consider an example:

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

*(A) 5/17*

* (B) 12/17*

* (C) 25/81*

* (D) 56/81*

* (E) 4/9*

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

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

First marble: Yellow; Second: Not Yellow

First: Not Yellow; Second: Yellow

First: Yellow; Second: Yellow

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

PROBABILITY OF NO YELLOW, FIRST DRAW: 10 non-yellow / 18 total

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

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

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

Yellow Yellow Yellow

Yellow Not-Yellow Not-Yellow

Yellow Not-Yellow Yellow

Yellow Yellow Not-Yellow

Not-Yellow Yellow Yellow

…

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

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

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

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

*By Brian Galvin.*

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]]>The post Solving GMAT Geometry Problems That Involve Infinite Figures appeared first on Veritas Prep Blog.

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

*(A) 18*

*(B) 32*

*(C) 36*

*(D) 64*

*(E) None*

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

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

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

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

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

Hypotenuse = *s*/√(2)

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

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

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

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

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

Total Sum = *a*/(1-*r*)

*a:* First Term

*r:* Common Ratio

Sum of areas of all squares = *s*^2 + *s*^2/2 + *s*^2/4 + *s*^2/8 + …

Sum of areas of all squares = *s*^2/(1 – 1/2)

Sum of areas of all squares = 2*s*^2

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

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

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

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

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]]>The post How to Find the Maximum Distance Between Points on a 3D Object appeared first on Veritas Prep Blog.

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

*(A) 15*

*(B) 20*

*(C) 25*

*(D) 10 * √(2)*

*(E) 10 * √(3)*

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

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

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

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

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

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

In our question we know that:

l = 10 inches

w = 10 inches

h = 5 inches

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

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

DB^2 = DC^2 + BC^2

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

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

BH^2 = HD^2 + DB^2

BH^2 = 10^2 + 125

BH = √(225) = 15

Thus, our answer to this question is A.

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

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

*(A) 5 * √2*

*(B) 5 * √3*

*(C) 5 * √5*

*(D) 10*

*(E) 15*

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

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

Diameter^2 + Height^2 = Distance^2

10^2 + 5^2 = Distance^2

Distance = 5 * √5

Thus, our answer is C.

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

**free online GMAT seminars** running all the time. And, be sure to follow us on **Facebook**, **YouTube**, **Google+**, and **Twitter**!

*GMAT** for Veritas Prep and regularly participates in content development projects such as this blog!*

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]]>The post GMAT Sentence Correction: How to Tackle Inverted Sentence Structures appeared first on Veritas Prep Blog.

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

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

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

*(A) the Aramaic script with it, from which was derived both northern and*

*(B) the Aramaic script with it, and from which deriving both the northern and the*

*(C) with it the Aramaic script, from which derive both the northern and the*

*(D) with it the Aramaic script, from which derives both northern and*

*(E) with it the Aramaic script, and deriving from it both the northern and *

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

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

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

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

or

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

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

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

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

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

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]]>The post Advanced Number Properties on the GMAT – Part VII appeared first on Veritas Prep Blog.

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

*1) When p is divided by 8, the remainder is 5.*

*2) x – y = 3*

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

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

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

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

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

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

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

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

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

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

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

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

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

So x = 2 * √(an odd number)

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

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

Now let’s look at the next statement:

*Statement 2: x – y = 3*

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

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

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

See you next week!

(For more advanced number properties on the GMAT, check out Parts **I**, **II**, **III**, **IV,** **V **and** VI** of this series.)

**free online GMAT seminars** running all the time. And, be sure to follow us on **Facebook**, **YouTube**, **Google+**, and **Twitter**!

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The post Advanced Number Properties on the GMAT – Part VII appeared first on Veritas Prep Blog.

]]>The post GMAT Tip of the Week: 3 Guiding Principles for Exponent Problems appeared first on Veritas Prep Blog.

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

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

**1) Find Common Bases**

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

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

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

*(A) 3x = y*

* (B) x = 3y*

* (C) y = (3/5)x*

* (D) x = (3/5)y*

* (E) 2x^2 = y*

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

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

Which simplifies to:

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

Now you can add together the exponents on the left:

2^5x = 2^3y

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

5x = 3y

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

**2) Factor to Create Multiplication**

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

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

*(A) 18*

* (B) 20*

* (C) 21*

* (D) 22*

* (E) 24*

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

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

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

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

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

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

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

**3) Test Small Numbers and Look For Patterns**

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

*What is the tens digit of 11^13?*

*(A) 1*

* (B) 2*

* (C) 3*

* (D) 4*

* (E) 5*

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

11^1 = 11 –> tens digit of 1

11^2 = 121 –> tens digit of 2

11^3 = 1331 –> tens digit of 3

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

11^4 = 14641 –> tens digit of 4

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

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

1) Find Common Bases

2) Factor to Create Multiplication

3) Test Small Numbers to Find a Pattern

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

*By Brian Galvin.*

The post GMAT Tip of the Week: 3 Guiding Principles for Exponent Problems appeared first on Veritas Prep Blog.

]]>The post How NOT to Write the Equation of a Line on the GMAT appeared first on Veritas Prep Blog.

]]>*y* = *mx* + *c* (where *m* is the slope and *c* is the *y*-intercept)

and

*y* – *y*_{1} = *m* * (*x* – *x*_{1}) [where *m* is the slope and (*x*_{1},*y*_{1}) is a point on the line]

We also know that *m* = (*y*_{2} – *y*_{1})/(*x*_{2} – *x*_{1}) – this is how we find the slope given two points that lie on a line. The variables are *x*_{1}, *y*_{1} and *x*_{2}, *y*_{2,} and they represent specific values.

But think about it, is *m* = (*y*_{2} – *y*)/(*x* – *x*_{1}) really the equation of a line? Let’s further clarify this idea using a GMAT practice question:

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

*(A) 3.5*

*(B) 7 *

*(C) 8*

*(D) 10*

*(E) 14*

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

*y* = *mx* + *c*

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

*y* = 2*x*

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

(4 – *y*)/(*x* – 3) = 2

(4 – *y*) = 2*(*x* – 3)

Thus, 2*x *+ *y* = 10

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

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

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

To get the second equation, we have used *m* = (*y*_{2} – *y*)/(*x* – *x*_{1}). But is this really the equation of a line? No. This formula doesn’t have *y* and *x*, the generic variables for the *x*– and *y*-coordinates in the equation of a line.

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

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

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

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

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

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

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

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

**free online GMAT seminars** running all the time. And, be sure to follow us on **Facebook**, **YouTube**, **Google+**, and **Twitter**!

*GMAT** for Veritas Prep and regularly participates in content development projects such as this blog!*

The post How NOT to Write the Equation of a Line on the GMAT appeared first on Veritas Prep Blog.

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