- Black coffee
- Kimchi
- Black metal
- Grammar
- Liver
- Absinthe
- Gertrude Stein
- Very dark chocolate
- Kombucha
- Distance running
- Free jazz
- Math problems

Alright, so that’s a pretty weird list. Some of them are common acquired tastes (almost nobody likes coffee the first time they drink it, yet it’s one of the most traded commodities in the world). Meanwhile, some are a bit more idiosyncratic, and two of those, you probably noticed, are actually things that are on the GMAT (if only more of them were…).

But they’re put into one list for a reason: there was a time in my life, for each of them, that I actively disliked them.

So let’s take a step back and think about that process in general. Why is it, and *how* is it, that something we dislike can become more palatable? What changes, and what drives that change? The easiest answer is exposure—no acquired taste can be learned until you’ve spent enough time with it. But there’s more to it than that. If I forced you to listen to Ornette Coleman every day for a month, you might still think it sounds like a somebody put Miles Davis’s music into a food processor.

The second part of the recipe is just as important. You have to *want* to like it. Not just that, but you need to **want to be the type of person **who likes it. If you start identifying as that type of person, start *acting* that way, then you’re primed to approach whatever bitter thing you’re learning to like with a bit more positivity. On the other hand, if you tell yourself (and other people) that you’re just “not a math person,” or anything like that, then you’re doing yourself a *huge* disservice, because that cuts of the opportunity to act as if you do like it and start forming positive associations.

Alright, so let’s bring this back to the realm of GMAT prep. Maybe you think grammar is kind of dull or math is frustrating. If *any* part of the test is a chore to study, then you’re setting yourself up to learn less—and therefore improve less—than you could with a different attitude. You learn the most when you’re seriously *engaged* with the material. And that’s all about mentality. Every practice question is a puzzle, and you’re a puzzle enthusiast. It doesn’t matter if you weren’t before; you are **now.**

When you talk about your prep, talk about parts you like. When you hit a particularly challenging question, share it with a friend in the same way you might share a puzzle or riddle. Do your prep in a comfortable place, at your own speed, and enjoy yourself. At the very least, *act* like you enjoy it. In time, that act becomes reality.

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Hello!

Today, we’re diving into another installment of “Real World Matters”. It’s my favorite subject–politics–and it might appear on your GMAT test, so watch the video to get yourself ready!

If you have any questions about this, be sure to leave them in the comments below.

]]>1) A car, moving at a constant speed, covers 100 m in 5 seconds. What is the car’s speed in km/hr? (1 km = 1000 m)

- 20 km/hr
- 54 km/hr
- 60 km/hr
- 72 km/hr
- 90 km/hr

2) Frank and Georgia started traveling from A to B at the same time. Georgia’s constant speed was 1.5 times Frank’s constant speed. When Georgia arrived at B, she turned right around and returned by the same route. She cross paths with Frank, who was coming toward B, when they were 60 miles away from B. How far away are A and B?

- 72 mi
- 120 mi
- 144 mi
- 240 mi
- 300 mi

3) Kevin drove from A to B at a constant speed of 60 mph, turned right around, and returned at a constant speed of 80 mph. Exactly 4 hours before the end of his trip, he was still approaching B, only 15 miles away from it. What is the distance between A and B?

- 275 mi
- 300 mi
- 320 mi
- 350 mi
- 390 mi

4) Car A and B are traveling from Town X to Town Y on the same route at constant speeds. Car A is initially behind Car B, and Car A’s speed is 1.25 times Car B’s speed. Car A passes Car B at 1:30 pm. At 3:15 pm, Car A reaches Town Y, and at that moment, Car B is still 35 miles away from Town Y. What is the speed of Car A?

- 60 mph
- 75 mph
- 80 mph
- 96 mph
- 100 mph

5) Cars P & Q are approaching each other on the same highway. Car P is moving at 49 mph northbound and Car P is moving at 61 mph southbound. At 2:00 pm, they are approaching each other and 120 mi apart. Eventually they pass each other. At what clock time are they moving away from each other and 45 miles apart?

- 3:06 pm
- 3:30 pm
- 3:54 pm
- 5:21 pm
- 6:15 pm

Solutions will be given at the end of this article.

The test loves motion problems. See this introductory article about rate questions for the basics. In particular, that article talks about the tricky issue of “average speed” or “average velocity,” and has some practice problems involve those topics.

Some problems, such as the ones above, specify distances between two cars — how far apart they are. Just as D = RT is true for each individual moving item, it is also true for a gap, and often, thinking about the D = RT of the gap is an enormous shortcut in a problem.

Fact: when two cars are moving in **opposite** directions, either approaching or receding, we **add** the individual speeds to get the rate for the gap. When two cars are moving the **same** direction, we subtract the individual speeds to get the rate for the gap.

Keep in mind, at any time, the gap may be expanding or shrinking, and so the rate may be how fast the gap is getting bigger or how fast the gap is getting smaller.

Here are four cases to keep in mind.

**Case I: cars in opposite directions, approach**

Here, the gap is getting smaller, and the sum of the two individual speeds of the cars is the rate at which the gap is shrinking.

**Case II: cars in opposite directions, receding**

Here, the gap is getting larger, and the sum of the two individual speeds of the cars is the rate at which the gap is expanding.

**Case III: cars in same directions, faster car behind slower car**

Here, the gap is getting smaller, and the difference of the two individual speeds of the cars is the rate at which the gap is shrinking.

**Case IV: cars in same directions, faster car ahead of slower car**

Here, the gap is getting larger, and the difference of the two individual speeds of the cars is the rate at which the gap is expanding.

Using these four cases, if we figure out the rate at which the gap is getting larger or smaller, then it would take a simple sum or difference to related one speed to the other. Even if both individual speeds are unknown, and we need to set up simultaneous equations, and the “gap” equation can be one of those two equations.

If you had any insights while reading this, you may want to revisit the practice problem above. If you have any insights about these problems you want to share, please let us know in the comments section below.

1) This question is more about pure unit conversion than about motion *per se*. In an hour, there are 60 minute, each with 60 seconds, so 1 hr = 60*60 = 3600 seconds.

The car’s speed is (100 m)/(5 s) = 20 m/s, so the question is: what is 20 m/s in km per hour.

Well, in one hour (i.e. 3600 second), a car moving 20 m/s would cover

D = RT = (20 m/s)(3600 s) = 72,000 m

That’s 72 km, or 72 km/hr. Answer = **(D)**.

2) Let F = Frank’s speed, and G = Georgia’s speed. We know G = 1.5*F. Let T be the time from start until they pass one another, and let D be the distance between the two locations. At time T, Frank has not gotten to destination B yet, and so has traveled a distance of (D – 60); meanwhile, by time T, Georgia has gone the entire distance of D and 60 miles beyond that, for a total distance of (D + 60).

Thus, our D = RT equations for the two cars are:

Frank: D – 60 = F*T

Georgia: D + 60 = G*T = 1.5F*T

Since the first equation, Frank’s equation, gives us an expression for F*T, and we don’t need the values for either of those variables, substitute that in to the second equation:

D + 60 = 1.5(D – 60)

D + 60 = 1.5D – 90

60 = 0.5D – 90

150 = 0.5D

300 = D

Answer = **(E)**

3) In the last 15 miles of his approach to B, Kevin was traveling at 60 mph, so he traveled that distance in ¼ hr, or 15 minutes. That means, when he arrived at B, 15 minutes had elapsed, and he took (4 hr) – (15 min) = 3.75 hr to drive the distance D at 80 mph. It will be easier to leave that time in the form (4 hr) – (15 min).

D = RT = (80 mph)[ (4 hr) – (15 min)] = 320 mi – 20 mi = 300 mi

Answer = **(B)**

4) Let A = speed of Car A, and B = speed of Car B. We know A = 1.25*B. Let D be the distance from where the cars pass each other at 1:30 pm to Town Y. Car A covers the distance D in 1 hr 45 min, or 1.75 hr; in that same time period, B covers (D – 35).

Formula for A: D = A*1.75

Formula for B: D – 35 = B*175

Substitute the former into the latter, to eliminate D.

A*1.75 – 35 = B*1.75

7A – 140 = 7B

A – 20 = B

Now, substitute A = 1.25*B (it’s easier to solve for B, then find A).

1.25*B – 20 = B

0.25*B – 20 = 0

B = 80 mph

A = 1.25*80 = 100 mph

Answer =** (E) **

5) Rather than focus on the motion of the individual cars, it will be much easier in this problem to think about the gap between the cards, as discussed above. When the cars are approaching, the rate at which the gap is shrinking is the sum of the two velocities, 110 mph, and that time, the time until meeting, would be the T = D/R = 120 mi/110 mph. Don’t calculate yet: just hold that thought. Once the cars pass, the gap is expanding, and the rate at which it expands is again the sum of the velocities, 110 mph, so we would again divide that distance by 110 mph.

Once we have those two fractions for the time, we will add them, and of course, they have a common denominator, so we can simply combine them by adding the numerators. In essence, if we think about the gap, there is a total distance of (120 mi + 45 mi) = 165 mi that is traversed at 110 mph. That would take a time of:

T = (165 mi)/(110 mph) = 15/10 = 1.5 hours

Therefore, the cars would be 45 mile apart and receding 1.5 hours later, at 3:30 pm

Answer = **(B)**

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14, 23, 32, 41, 50, 59, …

1) In the sequence above, each term is 9 more than the previous term. What is the 41^{st} term of the sequence?

- 360
- 365
- 369
- 374
- 383

2) What is the difference between the fourth and third terms of the sequence defined by

- 18
- 23
- 47
- 65
- 83

3) Which of the following could be true of at least some of the terms of the sequence defined by

I. divisible by 15

II. divisible by 18

III. divisible by 27

- I only
- II only
- I and II only
- I and III only
- I, II, III

4) Let S be the set of all positive integers that, when divided by 8, have a remainder of 5. What is the 76^{th} number in this set?

- 605
- 608
- 613
- 616
- 621

5) Let T be a sequence of the form . If and , find .

- 37
- 38
- 39
- 40
- 41

6) What is the sum of all the multiples of 20 from 160 to 840?

- 14,000
- 17,500
- 18,060
- 28,000
- 35,000

7) A sequence is defined by for n > 2, and it has the starting values of and . Find the value of .

- 25
- 32
- 36
- 93
- 279

9) In the set of positive integers from 1 to 500, what is the sum of all the odd multiples of 5?

- 10,000
- 12,500
- 17,500
- 22,500
- 25,000

10) If =, , , , and , what is the value of ?

- 7
- 11
- 19
- 42
- 130

Solutions will follow this article.

Sequences are a tricky topic on the GMAT Quant section. In that linked article, I discuss sequence notation, which is a variant of function notation, and I discuss recursive sequences, that is, sequences in which each term is determined by the previous term or terms.

An **arithmetic sequence** is a sequence in which we add some fixed amount to each term to get the next term. Another way to say that is that, if we subtract any term from the following term, the difference will always be the same: this difference is called the **common difference** of the arithmetic sequence. Let d be the common difference. Then, in algebraic form, the terms of the arithmetic sequence would be

Recall that, say, for the 3^{rd} term, the little subscript 3 is the **index**, that is, the position on the list. For this particular sequence, every term equals the first term plus a factor times **d**, and that factor is always one less than the index. This means we can write the general term as:

That is a very important formula, although, as always, don’t just memorize it; instead, remember the logic of the argument that leads up to it.

This is an important formula because any evenly spaced list is an arithmetic sequence. The consecutive multiples of any factor form an arithmetic sequence.

The sum of the first n terms in an arithmetic sequence is given by the formula

That’s the sum of the first term and the last term, times half the number of items on the list. You can also thing of that as the average of the first & last terms times the number of items on the list. One special case is the sum of the first n integer, given by

If you had some insights while reading that first article on sequences or the section on arithmetic sequences, then take another look at the problems above before looking at the solutions below. If you found this article helpful, or if you have an alternative solution for solving any of these problems, please let us know in the comments section below!

1) This is an arithmetic sequence, with

= 14 and d = 9

Using the formula for the nth term, we find that:

= 14 + 9*40 = 14 + 360 = 374.

Answer = **(D)**.

2) This is general sequence, with an explicitly defined nth term.

Answer = **(C)**.

3) First of all, if n = 8, then

= (16 – 1)(16 + 3) = 15*19

We don’t have to calculate that: clearly, whatever it is, it is divisible by 15. Similarly, if n = 12, then

= (24 – 1)(24 + 3) = 23*27

Whatever that equals, it must be divisible by 27. Thus, I & III are true. Notice that, for any integer, 2n must be even, so both (2n – 1) and (2n + 3) are odd numbers, and their product must be odd. Every term in this sequence is an odd number. Now, no odd number can be divisible by an even number, because there is no factor of 2 in the odd number. Therefore, no terms could possibly be divisible by 18. Statement II is absolutely not true.

Answer = **(D)**.

4) Think about the first few numbers on this list:

5, 13, 21, 29, 37, 45, …

Notice that 5 is the first number in S, because when 5 is divided by 8, the quotient is zero and the remainder is 5.

This is an arithmetic sequence with

= 5 and d = 8

Using the general formula for the nth term of an arithmetic sequence, we have

= 5 + 8*(75) = 5 + 4*(150) = 5 + 600 = 605

Answer = **(A)**.

5) The formula in the first sentence tells us that this is an arithmetic sequence. The first term and the common difference are unknown, but we can generate two equations from the values of the two terms given.

Subtract the first equation from the second, and we get 16d = 48, which means d = 3. From the value of third term, we can see that first term must equal 11. Therefore,

Answer = **(B)**.

6) This is arithmetic sequence, and we have the first and last terms already. How many terms are there? Well, 160 = 20*(8) and 840 = 20*(42); we have to use inclusive counting to see that there are 42 – 8 + 1 = 35 terms.

Answer = **(B)**.

7) This is a recursive sequence, and we have to find it term by term.

Answer = **(E)**.

8) This is a recursive sequence, so we have to work backwards term by term.

Answer = **(A)**.

9) Let’s think about the terms in this sequence:

5, 15, 25, 35, …., 485, 495

The first term is 5 and the last is 495. There are 100 multiples of 5 from 1 to 500, so there are 50 odd multiples and 50 even multiple. The sum is:

Answer = **(B)**.

10) This recursive sequence is probably more difficult than anything that the GMAT is going to throw you, but solving this problem is not too bad. As with any recursive sequence, we have to go term by term.

Answer = **(C)**.

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Hello!

Today, it’s fraction time! I am answering a question that one of your fellow GMAT-studiers asked about comparing fractions. Watch the video to learn more on this topic, and check out Mike McGarry’s blog post for even more tips!

If you have any questions about this, be sure to leave them in the comments below.

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*Analytical Writing Assignment:*One essay about the weaknesses in an argument that you read*Integrated Reasoning:*Math-based questions on reading tables, charts, and diagrams*Quantitative:*Math*Verbal:*Reading comprehension multiple choice, grammar, critical reasoning

*Reading:*Reading comprehension multiple choice*Listening:*Listening comprehension multiple choice*Speaking:*Open responses (speaking into a microphone), including summarizing recordings and text*Writing:*One essay summarizing a lecture, another essay explaining your opinion

So although two of the sections on each test are measure the same basic skills (verbal and writing), there are very few questions that are similar. In fact, the only part of either test that’s actually comparable is the reading comprehension.

Both the GMAT and the TOEFL focus on advanced, academic English. Neither test includes fiction, for example, and both are likely to include some text about science or history. However, that is the only real similarity between the two. In fact, the passages are written in a very different style between the two tests, and the questions are not very similar at all.

That’s because they have different goals. The GMAT wants to test your logical abilities and how well you can make concrete inferences from a text without assuming too much. In other words, your goal during GMAT reading comprehension is to understand the author’s purpose and show that you can use information from that text correctly to create your own conclusions.

The TOEFL, on the other hand, only wants you to understand the language. Very much of TOEFL reading comprehension relies on vocabulary. Even questions that are not directly about vocabulary require you to understand difficult sentence structures and idioms, but do not require you to analyze the information you read. You only need to understand it and recognize other sentences that give the same information.

Besides that, TOEFL reading passages, at around 700 words, are twice as long as “long” GMAT passages, which rarely pass 350 words, and there are many more questions per passage (12-14 on the TOEFL, 3-4 on the GMAT) so the strategies for answering questions are different.

These two parts of the tests have some things in common, but only very little.

Basically, the GMAT tests grammar and written style extensively. The TOEFL doesn’t test grammar directly, but if you write and speak with correct grammar, then you’ll score better on those sections. However, the GMAT’s grammar is mostly **very** advanced and often extremely subtle. If you understand English grammar well enough to score well on GMAT sentence correction questions, you’ll probably be able to write well.

And finally, the GMAT and the TOEFL both require you to structure an essay well. Using transition phrases (such as “on the other hand,”) and advanced vocabulary helps on both tests. The GMAT essay is about much, much more than that—your logical analysis of an argument, specifically—but the fundamentals of good English writing apply to both tests.

So even if the question types are different between the two tests, improving your written grammar can help for both the GMAT and the TOEFL.

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*Hear from a Yale grad about his experience with the GMAT. Thanks for sharing, Brendan!*

* About Me:* My name is Brendan and I’m an Electrical Engineering Major from Yale University (2010). I have been a submarine officer in the US Navy for the past four years. My job is primarily concerned with operating the reactor on board, so I am looking to pursue energy marketing and infrastructure at business school.

* Biggest Challenge:* I won’t lie, despite my engineering background, I was still weak on many quantitative skills required by the GMAT, and was also very weak on sentence correction (not surprising considering years of short hand engineering prose). I attacked this initially by skimming the lessons and really attacking practice problems with volume. In the end I realized two things:

1. Many of the lessons that seem simple have very valuable parts crucial to getting top GMAT scores. It wasn’t enough to skip the areas I thought I already understood. **Take the time to expose yourself to all the information in the lessons.**

2. Practice problems are only effective with a **thorough review**. Taking the time to understand why I got something incorrect (or often why I got something correct) was far more valuable than continuing to be incorrect on similar questions. Magoosh’s explanation videos are terrific and proved to be a great asset once I decided to go through them more thoroughly.

As my quantitative improved, I shifted my focus to my weaker verbal section. I poured in time with lessons and practice questions. In the end, I did very well in verbal (99 percentile), but actually a little worse than I expected in math (74 percentile). My first practice test with Magoosh was a 630, so I was happy with my overall score of 760 – but it was clear that I neglected my math studies during my last big push in verbal.

* Tips for Others:* Remember: if an engineer like me can beat the verbal section, so can you, but

Hello!

Welcome to another GMAT Tuesday installment, complete with a 2-second guitar solo! This week, learn about two tricky exceptions to the “empty it” rule that might trip you up on test day.

If you have any questions about this, please leave them in the comments below.

]]>1) At a certain school of 200 students, the students can study French, Spanish, both or neither. Just as many study both as study neither. One quarter of those who study Spanish also study French. The total number who study French is 10 fewer than those who study Spanish only. How many students study French only?

- 30
- 50
- 70
- 90
- 120

2) In a company of 300 employees, 120 are females. A total of 200 employees have advanced degrees, and the rest have a college degree only. If 80 employees are males with college degree only, how many are females with advanced degrees?

- 60
- 80
- 100
- 120
- 160

3) In a certain school, there are 80 Freshmen, 100 Sophomores, and 220 Upperclassmen, drawn from three cities: A, B, and C. Sixty percent of students are from A, 30% from B, the rest from C, and all these students from C are freshmen. Half the student from B are upperclassman, and the rest are split evenly between the other two grades. How many sophomores are from A?

- 60
- 70
- 80
- 90
- 100

4) There are a total of 400 students at a school, which offers a chorus, baseball, and Italian. This year, 120 students are in the chorus, 40 students in both chorus & Italian, 45 students in both chorus & baseball, and 15 students do all three activities. If 220 students are in either Italian or baseball, then how many student are in none of the three activities?

- 40
- 60
- 70
- 100
- 130

Solutions will be given at the end of the article.

For two or more overlapping categories, in scenarios in which any given member can belong to all, some, or none of the categories, Venn Diagrams can be helpful. If there are two separate variables, and each member is classified according to each variable, then the Double Matrix Method can be helpful. If reading the articles at either of those links gives you insights on any of these problems, you may want to give the problems a second look before reading the solutions below.

1) Let x be the number of folks studying both, which means it is also the number of folks studying neither.

“One quarter of those who study Spanish also study French.” If the Spanish students studying French are x, then all Spanish students are 4x, and those who do not study French are 3x. Also, let y be the number of students who study French but not Spanish.

“The total number who study French is 10 fewer than those who study Spanish only.” In other words,

x + y = 3x – 10

10 = 2x – y

Also, notice that the total number of students is 200:

3x + x + y + x = 200

5x + y = 200

We have two equations with two unknowns. Add the equations (2x – y = 10) and (5x + y = 200), and we get

7x = 210

x = 30

y = 50

And the number who study French is x + y = 80. Answer = **(B)**

2) This problem is handled best with a double-matrix solution.

Because there are 120 females, we know the rest are males. Because there are 200 with advanced degrees, we know there are 100 with college degrees only.

At this point, we could go either way: we could figure out all the numbers in the “college” row or all the numbers in the “male” column.

Either way, it is easy to fill in the last box:

Thus, there are 100 employees who are females with advanced degrees. Answer = **(C)**.

3) This problem calls for a 3×3 double matrix. Here are the numbers for the grade levels, with the column totaled:

Obviously, 10% of 400 is 40, so 30% is 3*40 = 120, and 60% = 6*40 = 240. This allows us to complete the bottom row.

All 40 from C are freshmen, so we can put zeros in the other two grade level slots:

Half of the 120 from B, 60, are upperclassmen, and the other 60 are evenly split, 30 in each of the other two slots.

Now, we can complete the sophomore row:

There are 70 sophomores from A. Answer = **(B)**.

4) This problem calls for a 3-way Venn Diagram. Here’s the diagram with no numbers filled in.

We know that C = 15. If 40 students are in both chorus & Italian, B + C = 40, and because C = 15, B = 25. If 45 students in both chorus & baseball, C + F = 45, and F = 30. We know that there are 120 in chorus, and B + C + F = 70, so E = 50.

Now, we are told that 220 student are in either Italian or baseball. Think about that region, Italian or baseball:

That entire purple region, A + B + C + D + F + G, is 220. If we add E = 50, that’s a total of 270 inside all three circles, which means that the outside of the circle, H, must equal 400 – 270 = 130. Answer = **(E)**.

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