The post GMAT Tip of the Week: Cam Newton’s GMAT Success Strategy appeared first on Veritas Prep Blog.

]]>**Why is Cam dancing and smiling so much?**

The answer? Because smiling may very well be the secret to success, both in the Super Bowl and on the GMAT.

Note: this won’t be the most mathematically tactical GMAT tip post you read, and it’s not something you’ll really be able to practice on Sunday afternoon while you hit the *Official Guide for GMAT Review* before your Super Bowl party starts. But it may very well be the tip that most impacts your score on test day, because managing stress and optimizing performance are major keys for GMAT examinees. And smiling is a great way to do that.

First, there’s science: the act of smiling itself is *known to release endorphins*, relaxing your mind and giving you a more positive outlook. And this happens regardless of whether you’re actually happy or optimistic – you can literally “fake it till you make it” by smiling through a stressful or unpleasant experience.

(Plus there’s the fact that smiling puts OTHER people in a better mood, too, which won’t really help you on the GMAT since it’s you against a computer, but for your b-school and job interviews, a smile can go a long way toward an upbeat experience for both you and the interviewer.)

There are plenty of ways to force yourself to smile. One is the obvious: just do it. Write it down on the top of your noteboard in all caps: **SMILE!** And force yourself to do it, even when it doesn’t feel natural.

But you can also laugh/smile at yourself more naturally: when Question 1 is a permutations problem and you were dreading the idea of a permutations problem, you can laugh at your bad luck but also at the fact that at least you’re getting it over with while you still have plenty of time to recover. When you blank on a rule and have to test small numbers to prove it, you can laugh at the fact that had you not been so fascinated with the video games on your calculator in middle school you’d know that cold. You can smile when you see a friend’s name in a word problem or a Sentence Correction reference to a place you want to visit someday.

And the tactical rationale there: when you can smile in relation to the subject matter on the test, you can remind yourself that, at least on some level, you enjoy learning and problem-solving and striving for achievement. The biggest difference between “good test takers” and “good students, but bad test takers” is in the way that each approaches problems: the latter group says, “I don’t know,” and feels doubt, while the former says, “I don’t know…yet,” and starts from a position of confidence and strength. Then when you apply that confidence and figure out a problem that for a second had you totally stumped, you’ve earned that next smile and the positive energy snowballs.

As you watch Cam Newton on Sunday (For you brand management hopefuls, he’ll be playing football between those commercials you’re so excited to see!), pay attention to that megawatt smile that’s been the topic of so much talk radio controversy the last few weeks. Cam smiles because he’s having fun out there, and then that smile leads to big plays, which is even more fun, and then he’s smiling again. Apply that Cam Newton “smile your way to success” philosophy on test day and maybe you’ll be the next one getting paid hundreds of thousands of dollars to go to school for two years… (We kid, Cam – we kid!)

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

*By Brian Galvin.*

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]]>The post Why Logic is More Important Than Algebra on the GMAT appeared first on Veritas Prep Blog.

]]>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?”

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

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

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

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]]>The post How to Choose the Right Number for a GMAT Variable Problem appeared first on Veritas Prep Blog.

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

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

*(A) 2 *

*(B) 3 *

*(C) 4 *

*(D) 6 *

*(E) 8 *

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

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

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

*(A) **x + 10*

*(B) **10x + 1*

*(C) **100(x + 10)*

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

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

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

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

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

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

Start: 100 liters

After adding water: 120 liters

After removing water: 90 liters

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

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

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

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

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

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

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

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

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

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

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]]>The post Quarter Wit, Quarter Wisdom: Cyclicity in GMAT Remainder Questions appeared first on Veritas Prep Blog.

]]>First let’s look at a pattern:

**20/10** gives us a remainder of** 0** (as 20 is exactly divisible by 10)

**21/10** gives a remainder of **1**

**22/10** gives a remainder of **2**

**23/10** gives a remainder of **3**

**24/10** gives a remainder of **4**

**25/10** gives a remainder of **5**

and so on…

In the case of this pattern, 20 is the closest multiple of 10 that goes completely into all these numbers and you are left with the units digit as the remainder. Whenever you divide a number by 10, the units digit will be the remainder. Of course, if the units digit of a number is 0, the remainder will be 0 and that number will be divisible by 10 — but we already know that. So remainder when 467,639 is divided by 10 is 9. The remainder when 100,238 is divided by 10 is 8 and so on…

Along the same lines, we also know that every number that ends in 0 or 5 is a multiple of 5 and every multiple of 5 must end in either 0 or 5. So if the units digit of a number is 1, it gives a remainder of 1 when divided by 5. If the units digit of a number is 2, it gives a remainder of 2 when divided by 5. If the units digit of a number is 6, it gives a remainder of 1 when divided by 5 (as it is 1 more than the previous multiple of 5).

With this in mind:

**20/5** gives a remainder of** 0** (as 20 is exactly divisible by 5)

**21/5** gives a remainder of **1**

**22/5** gives a remainder of **2**

**23/5** gives a remainder of **3**

**24/5** gives a remainder of **4**

**25/5** gives a remainder of **0** (as 25 is exactly divisible by 5)

**26/5** gives a remainder of **1**

**27/5** gives a remainder of **2**

**28/5** gives a remainder of **3**

**29/5** gives a remainder of **4**

**30/5** gives a remainder of** 0** (as 30 is exactly divisible by 5)

and so on…

So the units digit is all that matters when trying to get the remainder of a division by 5 or by 10.

Let’s take a few questions now:

*What is the remainder when 86^(183) is divided by 10?*

Here, we need to find the last digit of 86^(183) to get the remainder. Whenever the units digit is 6, it remains 6 no matter what the positive integer exponent is (**previously discussed in this post**).

So the units digit of 86^(183) will be 6. So when we divide this by 10, the remainder will also be 6.

Next question:

*What is the remainder when 487^(191) is divided by 5?*

Again, when considering division by 5, the units digit can help us.

The units digit of 487 is 7.

7 has a cyclicity of 7, 9, 3, 1.

Divide 191 by 4 to get a quotient of 47 and a remainder of 3. This means that we will have 47 full cycles of “7, 9, 3, 1” and then a new cycle will start and continue until the third term.

7, 9, 3, 1

7, 9, 3, 1

7, 9, 3, 1

7, 9, 3, 1

…

7, 9, 3

So the units digit of 487^(191) is 3, and the number would look something like ……………..3

As discussed, the number ……………..0 would be divisible by 5 and ……………..3 would be 3 more, so it will also give a remainder of 3 when divided by 5.

Therefore, the remainder of 487^(191) divided by 5 is 3.

Last question:

*If x is a positive integer, what is the remainder when 488^(6x) is divided by 2?*

Take a minute to review the question first. If you start by analyzing the expression 488^(6x), you will waste a lot of time. This is a trick question! The divisor is 2, and we know that every even number is divisible by 2, and every odd number gives a remainder 1 when divided by 2. Therefore, we just need to determine whether 488^(6x) is odd or even.

488^(6x) will be even no matter what x is (as long as it is a positive integer), because 488 is even and we know even*even*even……(any number of terms) = even.

So 488^(6x) is even and will give remainder 0 when it is divided by 2.

That is all for today. We will look at some GMAT remainders-cyclicity questions next week!

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 Quarter Wit, Quarter Wisdom: Cyclicity of Units Digits on the GMAT (Part 2) appeared first on Veritas Prep Blog.

]]>**Cyclicity of 2:** 2, 4, 8, 6

**Cyclicity of 3:** 3, 9, 7, 1

**Cyclicity of 7:** 7, 9, 3, 1

**Cyclicity of 8:** 8, 4, 2, 6

Digits 4 and 9 have a cyclicity of 2, i.e. the units digit repeats itself every 2 digits:

**Cyclicity of 4:** 4, 6

**Cyclicity of 9:** 9, 1

Digits 0, 1, 5 and 6 have a cyclicity of 1, i.e. the units digit is 0, 1, 5, or 6 respectively.

Now let’s take a look at how to apply these fundamentals:

*What is the units digit of 813^(27)?*

To get the desired units digit here, all we need to worry about is the units digit of the base, which is 3.

Remember, our cyclicity of 3 is 3, 9, 7, 1 (four numbers total).

We need the units digit of 3^(27). How many full cycles of 4 will be there in 27? There will be 6 full cycles because 27 divided by 4 gives 6 as quotient and 3 will be the remainder. So after 6 full cycles of 4 are complete, a new cycle will start:

3, 9, 7, 1

3, 9, 7, 1

… (6 full cycles)

3, 9, 7 (new cycle for remainder of 3)

7 will be the units digit of 3^(27), so 7 will be the units digit of 813^(27).

Let’s try another question:

*What is the units digit of 24^(1098)?*

To get the desired units digit here, all we need to worry about is the units digit of the base, which is 4.

Remember, our cyclicity of 4 is 4 and 6 (this time, only 2 numbers).

We need the units digit of 24^(1098) – every odd power of 24 will end in 4 and every even power of 24 will end in 6.

Since 1098 is even, the units digit of 24^(1098) is 6.

Not too bad; let’s try something a little harder:

*What is the units digit of 75^(25)^5*

Note here that you have 75 raised to power 25 which is further raised to the power of 5.

25^5 is not the same as 25*5 – it is 25*25*25*25*25 which is far more complicated. However, the simplifying element of this question is that the last digit of the base 75 is 5, so it doesn’t matter what the positive integer exponent is, the last digit of the expression will always be 5.

Now let’s take a look at a Data Sufficiency question:

*Given that x and y are positive integers, what is the units digit of (5*x*y)^(289)?*

*Statement 1: x is odd.*

*Statement 2: y is even.*

Here there is a new complication – we don’t know what the base is exactly because the base depends on the value of x and y. As such, the real question should be can we figure out the units digit of the base? That is all we need to find the units digit of this expression.

When 5 is multiplied by an even integer, the product ends in 0.

When 5 is multiplied by an odd integer, the product ends in 5.

These are the only two possible cases: The units digit must be either 0 or 5.

With Statement 1, we do not know whether y is odd or even, we only know that x is odd. If y is odd, x*y will be odd. If y is even, x*y will be even. Since we don’t know whether x*y is odd or even, we don’t know whether 5*x*y will end in 5 or 0, so this statement alone is not sufficient.

With Statement 2, if y is even, x*y will certainly be even because an even * any integer will equal an even integer. Therefore, it doesn’t matter whether x is odd or even – regardless, 5*x*y will be even, hence, it will certainly end in 0.

As we know from our patterns of cyclicity, 0 has a cyclicity of 1, i.e. no matter what the positive integer exponent, the units digit will be 0. Therefore, this statement alone is sufficient and the answer is B (Statement 2 alone is sufficient but Statement 1 alone is not sufficient).

Finally, let’s take a question from our own book:

**If n and a are positive integers, what is the units digit of n^(4a+2) – n^(8a)?**

**Statement 1: n = 3**

**Statement 2: a is odd.**

We know that the cyclicity of every digit is either 1, 2 or 4. So to know the units digit of n^{4a+2} – n^{8a}, we need to know the units digit of n. This will tell us what the cyclicity of n is and what the units digit of each expression will be individually.

Statement 1: n = 3

As we know from our patterns of cyclicity, the cyclicity of 3 is 3, 9, 7, 1

Plugging 3 into “n”, n^{4a+2} = 3^{4a+2}

In the exponent, 4a accounts for “a” full cycles of 4, and then a new cycle begins to account for 2.

3, 9, 7, 1

3, 9, 7, 1

…

3, 9

The units digit here will be 9.

Again, plugging 3 into “n”, n^{8a} = 3^{8a}

8a is a multiple of 4, so there will be full cycles of 4 only. This means the units digit of 3^{8a} will be 1.

3, 9, 7, 1

3, 9, 7, 1

3, 9, 7, 1

…

3, 9, 7, 1

Plugging these answers back into our equation: n^{4a+2} – n^{8a} = 9 – 1

The units digit of the combined expression will be 9 – 1 = 8.

Therefore, this statement alone is sufficient.

In Statement 2, we are given what the exponents are but not what the value of n, the base, is. Therefore, this statement alone is not sufficient, and our answer is A (Statement 1 alone is sufficient but Statement 2 alone is not sufficient).

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!*

The post Quarter Wit, Quarter Wisdom: Cyclicity of Units Digits on the GMAT (Part 2) appeared first on Veritas Prep Blog.

]]>The post Quarter Wit Quarter Wisdom: Cyclicity of Units Digits on the GMAT appeared first on Veritas Prep Blog.

]]>The first thing you need to understand is that when we multiply two integers together, the last digit of the result depends only on the last digits of the two integers.

For example:

24 * 12 = 288

Note here: …4 * …2 = …8

So when we are looking at the units digit of the result of an integer raised to a certain exponent, all we need to worry about is the units digit of the integer.

Let’s look at the pattern when the units digit of a number is 2.

**Units digit 2:**

2^1 = 2

2^2 = 4

2^3 = 8

2^4 = 1__6__

2^5 = 3__2__

2^6 = 6__4__

2^7 = 12__8__

2^8 = 25__6__

2^9 = 51__2__

2^10 = 102__4__

…

Note the units digits. Do you see a pattern? 2, 4, 8, 6, 2, 4, 8, 6, 2, 4 … and so on

So what will 2^11 end with? The pattern tells us that two full cycles of 2-4-8-6 will take us to 2^8, and then a new cycle starts at 2^9.

2-4-8-6

2-4-8-6

2-4

The next digit in the pattern will be 8, which will belong to 2^11.

In fact, any integer that ends with 2 and is raised to the power 11 will end in 8 because the last digit will depend only on the last digit of the base.

So 652^(11) will end in 8,1896782^(11) will end in 8, and so on…

A similar pattern exists for all units digits. Let’s find out what the pattern is for the rest of the 9 digits.

**Units digit 3:**

3^1 = 3

3^2 = 9

3^3 = 2__7__

3^4 = 8__1__

3^5 = 24__3__

3^6 = 72__9__

The pattern here is 3, 9, 7, 1, 3, 9, 7, 1, and so on…

**Units digit 4:**

4^1 = 4

4^2 = 16

4^3 = 64

4^4 = 256

The pattern here is 4, 6, 4, 6, 4, 6, and so on…

Integers ending in digits 0, 1, 5 or 6 have the same units digit (0, 1, 5 or 6 respectively), whatever the positive integer exponent. That is:

1545^23 = ……..5

1650^19 = ……..0

161^28 = ………1

Hope you get the point.

**Units digit 7:**

7^1 = 7

7^2 = 4__9__

7^3 = 34__3__

7^4 = ….__1 __(Just multiply the last digit of 343 i.e. 3 by another 7 and you get 21 and hence 1 as the units digit)

7^5 = ….__7 __(Now multiply 1 from above by 7 to get 7 as the units digit)

7^6 = ….__9__

The pattern here is 7, 9, 3, 1, 7, 9, 3, 1, and so on…

**Units digit 8:**

8^1 = 8

8^2 = 6__4__

8^3 = …__2__

8^4 = …__6__

8^5 = …__8__

8^6 = …__4__

The pattern here is 8, 4, 2, 6, 8, 4, 2, 6, and so on…

**Units digit 9: **

9^1 = 9

9^2 = 81

9^3 = 729

9^4 = …1

The pattern here is 9, 1, 9, 1, 9, 1, and so on…

Summing it all up:

1) Digits 2, 3, 7 and 8 have a cyclicity of 4; i.e. the units digit repeats itself every 4 digits.

**Cyclicity of 2:** 2, 4, 8, 6

**Cyclicity of 3:** 3, 9, 7, 1

**Cyclicity of 7:** 7, 9, 3, 1

**Cyclicity of 8: **8, 4, 2, 6

2) Digits 4 and 9 have a cyclicity of 2; i.e. the units digit repeats itself every 2 digits.

**Cyclicity of 4:** 4, 6

**Cyclicity of 9:** 9, 1

3) Digits 0, 1, 5 and 6 have a cyclicity of 1.

**Cyclicity of 0:** 0

**Cyclicity of 1:** 1

**Cyclicity of 5:** 5

**Cyclicity of 6:** 6

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 Is Technology Costing You Your GMAT Score? appeared first on Veritas Prep Blog.

]]>The book’s core thesis – that our smartphones and tablets are fragmenting our concentration and robbing us of a fundamental part of what it means to be human – isn’t a terribly original one. The difference between Turkle’s work and less effective screeds about the evils of technology is the scope of the research she provides in demonstrating how the overuse of our devices is eroding the quality of our education, our personal relationships, and our mental health.

What’s amazing is that these costs are, to some extent, quantifiable. Ever wonder what the impact is of having most of our conversations mediated through screens rather than through hoary old things like facial expressions? College students in the age of smartphones score 40% lower on tests measuring indicators of empathy than college students from a generation ago. In polls, respondents who had access to smartphones by the time they were adolescents reported heightened anxiety about the prospect of face-to-face conversations in general.

Okay, you say. Disturbing as that is, those findings have to do with interpersonal relationships, not education. Can’t technology be used to enhance the learning environment as well? Though it would be silly to condemn any technology as wholly corrosive, particularly in light of the fact that most schools are making a concerted effort to incorporate laptops and tablets in the classroom, Turkle makes a persuasive case that the overall costs outweigh the benefits.

In one study conducted by Pam Mueller and Daniel Oppenheimer, the researchers compared the retention rates of students who took notes on their laptops versus those who took notes by hand. The researchers’ assumption had always been that taking notes on a laptop would be more beneficial, as most of us can type faster than we can write longhand. Much to their surprise, the students who took notes by hand did significantly better than those who took notes on their laptops when tested on the contents of a lecture a week later.

The reason, Mueller and Oppenheimer speculate, is that because the students writing longhand couldn’t transcribe fast enough to record everything, they had to work harder to filter the information they were provided, and this additional cognitive effort allowed them to retain more. The ease of transcription – what we perceive as a benefit of technology – actually proved to be a cost. Even more disturbing, another study indicated that the mere presence of a smartphone – even if the phone is off – will cause everyone in its presence to retain less of a lecture, not just the phone’s owner.

I’ve been teaching long enough that when I first started, it was basically unheard of for a student’s attention to wander because he’d been distracted by a device. Smartphones didn’t exist yet. No one brought laptops to class. Now, if I were to take a poll, I’d be surprised if there were a single student in class who didn’t at least glance at a smartphone during the course of a lesson. One imagines that the same is true when students are studying on their own – a phone is nearby, just in case something important comes up. I’d always assumed the presence of these devices was relatively harmless, but if a phone that’s off can degrade the quality of our study sessions, just imagine the impact of a phone that continually pings and buzzes as fresh texts, emails and notifications come in.

The GMAT is a four-hour test that requires intense focus and concentration, so anything that hampers our ability to focus is a potential drag on our scores. There’s no easy solution here. I’m certainly not advocating that anyone throw away their smartphone – the fact that certain technology has costs associated with it is hardly a reason to discard that technology altogether. There are plenty of well-documented educational benefits: one can use a long train ride as an opportunity to do practice problems or watch a lecture. We can easily store data that can shed light on where we need to focus our attention in future study sessions. So the answer isn’t a draconian one in which we have to dramatically alter our lifestyles. Technology isn’t going anywhere – it’s a question of moderation.

Takeaways: No rant about the costs of technology is going to be terribly helpful without an action plan, so here’s what I suggest:

**Put the devices away in class and take notes longhand.**Whether you’re in a GMAT prep class, or an accounting class in your MBA program, this will benefit both you and your classmates.

- If you aren’t using your device to study,
**turn it off, and make sure it’s out of sight when you work**. The mere visual presence of a smartphone will cause you to retain less.

**Give yourself at least 2 hours of device-free time each day.**This need not be when you’re studying. It can also be when you’re out to dinner with friends or spending time with family. In addition to improving your interpersonal relationships, conversation actually makes you smarter.

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 by him here.*

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]]>The post What to Do if You’re Struggling with GMAT Solutions appeared first on Veritas Prep Blog.

]]>The short answer is that they struggled just like you did, but like anybody else, they wanted to make it look easy. (Think of all the time some people spend preening their LinkedIn or their Instagram: you only ever see the flashy corporate name and the glamour shot, never the 5 AM wake up call or the 6 AM look in the mirror.) Solution writers, particularly those who work for the GMAC, never seem to tell you that problem solving is mostly about blundering through a lot of guesswork before hitting upon a pattern, but that’s really what it is. Your willingness to blunder around until you hit something promising is a huge part of what’s being tested on the GMAT; after all, as depressing as it sounds, that’s basically how life works.

Here’s a great example:

*I haven’t laid eyes on it in thirty years, but I still remember that the rope ladder to my childhood treehouse had exactly ten rungs. I was a lot shorter then, and a born lummox, so I could only climb the ladder one or two rungs at a time. I also had more than a touch of childhood OCD, so I had to climb the ladder a different way every time. After how many trips up did my OCD prevent me from ever climbing it again? (In other words, how many different ways was I able to climb the ladder?) *

*A) 55 *

*B) 63 *

* C) 72 *

*D) 81 *

*E) 89*

Just the thought of trying 55 to 89 different permutations of climbing the ladder has my OCD going off like a car alarm, so I’m going to look for an easier way of doing this. It’s a GMAT problem, albeit one on the level of a Google interview question, so it must have a simple solution. There has to be a pattern here, or the problem wouldn’t be tested. Maybe I could find that pattern, or at least get an idea of how the process works, if I tried some shorter ladders.

Suppose the ladder had one rung. That’d be easy: there’s only one way to climb it.

Now suppose the ladder had two rungs. OK, two ways: I could go 0-1 then 1-2, or straight from 0-2 in a single two step, so there are two ways to climb the ladder.

Now suppose that ladder had three rungs. 0-1, 1-2, 2-3 is one way; 0-2, 2-3 is another; 0-1, 1-3 is the third. So the pattern is looking like 1, 2, 3 … ? That can’t be right! Doubt is gnawing at me, but I’m going to give it one last shot.

Suppose that the ladder had four rungs. I could do [0-1-2-3-4] or [0-1-3-4] or [0-1-2-4] or [0-2-4] or [0-2-3-4]. So there are five ways to climb it … wait, that’s it!

While I was mucking through the ways to climb my four-rung ladder, I hit upon something. When I take my first step onto the ladder, I either climb one rung or two. If I climb one rung, then there are 3 rungs left: in other words, I have a 3-rung ladder, which I can climb in 3 ways, as I saw earlier. If my step is a two-rung step instead, then there are 2 rungs left: in other words, a 2-rung ladder, which I can climb in 2 ways. Making sense?

By the same logic, if I want to climb a 5-rung ladder, I can start with one rung, then have a 4-rung ladder to go, or start with two rungs, then have a 3-rung ladder to go. So the number of ways to climb a 5-rung ladder = (the number of ways to climb a 3-rung ladder) + (the number of ways to climb a 4-rung ladder). Aha!

My pattern starts 1, 2, 3, so from there I can find the number of ways to climb each ladder by summing the previous two. This gives me a 1-, 2-, 3-, … rung ladder list of 1, 2, 3, 5, 8, 13, 21, 34, 55, and 89, so a 10-rung ladder would have 89 possible climbing permutations, and we’re done.

And the lesson? Much like a kid on a rope ladder, for a GMAT examinee on an abstract problem there’s often no “one way” to do the problem, at least not one that you can readily identify from the first instant you start. Very often you have to take a few small steps so that in doing so, you learn what the problem is all about. When all else fails in a “big-number” problem, try testing the relationship with small numbers so that you can either find a pattern or learn more about how you can better attack the bigger numbers. Sometimes your biggest test-day blunder is not allowing yourself to blunder around enough to figure the problem out.

Congratulations: that’s the hardest GMAT problem you’ve solved yet! (And bonus points if you noticed that the answer choices differed by 8, 9, 9, and 8. I still have OCD, and a terrible sense of humor.)

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]]>The post You Can Do It! How to Work on GMAT Work Problems appeared first on Veritas Prep Blog.

]]>When dealing with a complex work question there are typically only two things we need to keep in mind, aside from our standard “rate * time = work” equation. First, we know that rates are additive. If I can do 1 job in 4 hours, my rate is 1/4. If you can do 1 job in 3 hours, your rate is 1/3. Therefore, our combined rate is 1/4 + 1/3, or 7/12. So we can do 7 jobs in 12 hours.

The second thing we need to bear in mind is that rate and time have a reciprocal relationship. If our rate is 7/12, then the time it would take us to complete a job is 12/7 hours. Not so complex. What’s interesting is that these simple ideas can unlock seemingly complex questions. Take this official question, for example:

*Pumps A, B, and C operate at their respective constant rates. Pumps A and B, operating simultaneously, can fill a certain tank in 6/5 hours; pumps A and C, operating simultaneously, can fill the tank in 3/2 hours; and pumps B and C, operating simultaneously, can fill the tank in 2 hours. How many hours does it take pumps A, B, and C, operating simultaneously, to fill the tank. *

*A) **1/3*

*B) **1/2*

*C) **2/3*

*D) **5/6*

*E) **1*

So let’s start by assigning some variables. We’ll call the rate for p ump A, R_{a. }Similarly, we’ll designate the rate for pump B as R_{b,}and the rate for pump C as R_{c.}

If the time for A and B together to fill the tank is 6/5 hours, then we know that their combined rate is 5/6, because again, time and rate have a reciprocal relationship. So this first piece of information yields the following equation:

R_{a }+ R_{b} = 5/6.

If A and C can fill the tank in 3/2 hours, then, employing identical logic, their combined rate will be 2/3, and we’ll get:

R_{a }+ R_{c} = 2/3.

Last, if B and C can fill tank in 2 hours, then their combined rate will be ½, and we’ll have:

R_{b}+ R_{c} = 1/2.

Ultimately, what we want here is the time it would take all three pumps working together to fill the tank. If we can find the combined rate, or R_{a }+ R_{b }+ R_{c}, then all we need to do is take the reciprocal of that number, and we’ll have our time to full the pump. So now, looking at the above equations, how can we get R_{a }+ R_{b }+ R_{c} on one side of an equation? First, let’s line our equations up vertically:

_{ }R_{a }+ R_{b} = 5/6.

R_{a }+ R_{c} = 2/3.

R_{b}+ R_{c} = 1/2.

_{ }Now, if we sum those equations, we’ll get the following:

2R_{a }+ 2R_{b }+ 2R_{c} = 5/6 + 2/3 + 1/2. This simplifies to:

2R_{a }+ 2R_{b }+ 2R_{c} = 5/6 + 4/6 + 3/6 = 12/6 or 2R_{a }+ 2R_{b }+ 2R_{c } = 2.

Dividing both sides by 2, we’ll get: R_{a }+ R_{b }+ R_{c } = 1.

This tells us that the pumps, all working together can do one tank in one hour. Well, if the rate is 1, and the time is the reciprocal of the rate, it’s pretty obvious that the time to complete the task is also 1. The answer, therefore, is E.

Takeaway: the most persistent myth we have about our academic limitations is that we’re simply not good at a certain subset of problems when, in truth, we just never properly learned how to do this type of question. Like every other topic on the GMAT, rate/work questions can be mastered rapidly with a sound framework and a little practice. So file away the notion that rates can be added in work questions and that time and rate have a reciprocal relationship. Then do a few practice questions, move on to the next topic, and know that you’re one step closer to mastering the skills that will lead you to your desired GMAT score.

**GMATPrep question courtesy of the Graduate Management Admissions Council.*

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]]>The post Don’t Panic on the GMAT! appeared first on Veritas Prep Blog.

]]>*A palindrome is a number that reads the same front-to-back as it does back-to-front (e.g. 202, 575, 1991, etc.) p is the smallest integer greater than 200 that is both a prime and a palindrome. What is the sum of the digits of p? *

*A) 3*

*B) 4*

*C) 5*

*D) 6 *

*E) 7*

Thud.

I don’t know about you, but I’m petrified. I mean, yeah, I know what you’re saying — I’m the bozo who just dreamed up that question — but I don’t know where it came from, and I’m sort of thinking I might need to summon an exorcist, because I must be possessed by a math demon. What does that question even say? How the heck are we going to solve it?

This is such a common GMAT predicament to be in that I’m willing to bet that 99% of test takers experience it: the feeling that you don’t even know what the question is saying, and the sense of creeping terror that maybe you don’t know what any of these questions are saying. This is by design, of course. The test writers love these sort of “gut check” questions that test your ability to calmly unpack and reason out a cruel and unusual prompt. So many students take themselves out of the game by panicking, but like any GMAT question, once we get past the intimidation factor, the problem is simple at heart. Let’s try to model the process.

We’ll start by clarifying our terms. Palindrome, palindrome … what on earth is a palindrome!? Is that some sort of hovercraft where Sarah Palin lives? Where are our flash cards? Maybe we should just go to law school or open a food truck or something, this test is absurd.

Wait, the answer is right in front of us, in the very first line! “A palindrome is a number that reads the same back-to-front as it does front-to-back.” Phew, OK, and there are even some examples. So a palindrome is a number like 101, 111, 121, etc. Alright, got that. And it’s prime … prime, prime … OK, right, that WAS on a flashcard: a prime number is a number with exactly two factors, such as 2, or 3, or 5, or 7. So if we were to make lists of each of these numbers, primes and palindromes, we’d have

Primes: 2, 3, 5, 7, 11, 13, 17, 19, …

Palindromes: 101, 111, 121, 131, …

and we want the first number that’s greater than 200 that appears on both lists. OK!

Now let’s think of where to start. We know our number is greater than 200, so 202 seems promising. But that can’t be prime: it’s even, so it has at least three factors (1, itself, and 2). Great! We can skip everything that begins/ends with 2, and fast forward to 303. That looks prime, but what was it that Brian kept telling us about divisibility by 3 … ah, yes, test the sum of the digits! 3 + 0 + 3 = 6, and 6 divides by 3, so 303 also divides by 3.

Our next candidate is 313. This seems to be our final hurdle: a lot of quick arithmetic. That’s what the question is testing, after all, right? How quickly can you factor 313?

It sure seems that way, but take one last look at the answers. The GMAT tests efficiency as much as anything else, and it has a way of hiding easter eggs for the observant. Our largest answer is 7, and what’s 3+1+3? 7! So this MUST be the answer, and any time spent factoring 313 is wasted time.

We made it! In hindsight, that didn’t really feel like a math problem, did it? It was testing our ability to:

1) Remember a definition (“prime”)

2) Actually read the question stem (“a palindrome is…”)

3) Not panic, and try a few numbers (“202”? “303”?)

4) Realize that heavy calculation is for suckers, and that the answer might be right in front of us (“check the answers”)

So we just had to remember, actually read the directions, have the courage to try something to see where it leads, and look for clues directly around us. I don’t know about you, but if I were running a business, those are exactly the sort of skills I’d want my employees to have; maybe these test writers are on to something after all!

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