Hello!

Guess what? It’s Tuesday! And on this Tuesday, we’ll discuss how to build a GMAT syllabus that will successfully prepare you for the test!

Here are a series of links to the resources that I mention in the video:

- Magoosh Study Schedules
- GMATPrep
- Manhattan GMAT Practice Tests
- Magoosh GMAT eBook
- Magoosh IR book
- Idiom Flashcards
- Math Flashcards

With that combination of resources, you should be off to a good start! Let me know if you have any questions in the comment box below.

]]>Tune in to Accepted’s webinar, *Round 3 vs. Next Year: The MBA Admissions Debate*, for professional advice that will clarify the two options and help you come to a conclusion that will increase your chances of getting accepted.

During the webinar, Linda Abraham, founder & president of Accepted, will discuss:

• The **differences** between R3 and earlier rounds.

• The **pros and cons** of applying R3.

• 6 reasons why **some people should wait** until next year.

…and more!

The webinar will take place on Wednesday, January 21, 2015 at 10:00 AM PST/1:00 PM EST. Reserve your spot by registering for free now!

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OK, from this solved Venn diagram, there’s a ton we know:

total in AP Chemistry = 50 + 30 = 80

total in AP Literature = 30 + 80 = 110

taking both AP Chemistry and AP Literature = 30

taking AP Chemistry, and not AP Literature = 50

taking AP Literature, and not AP Chemistry = 80

taking neither = 40

All that is unambiguous, and most folks looking at this Venn Diagram would not have much trouble with all that. The problem comes with the question: “How many students are in AP Literature **OR** AP Chemistry?” Do we include the folks in both or not? There seem to be two readings of this question, and for some students, both reading might make sense. What’s going on?

It turns out, in mathematical logic, there are two completely different kinds of OR: the **inclusive OR** and the **exclusive OR**. The inclusive OR always includes the “both” case. For the inclusive OR, the phrase “A or B” includes the possibilities:

(a) A by itself

(b) B by itself

(c) A and B together

By contrast, the exclusive OR always excludes the “both” case. For the exclusive OR, the phrase “A or B” includes (a) & (b) above, but definitely NOT (c).

In colloquial language, often context makes it clear which OR is intended. For example, if you imagine when you were, say, eight years old, and it was around dessert time, and your mother said emphatically, “*You can have cake or ice cream*!!!” At that moment, your mother most certainly was not expatiating on the merits of the inclusive OR. In fact, in many scenarios of ordinary life, the word OR implicitly means the exclusive OR—especially when any emotional emphasis lands on the word itself!

Things change when we look at the use within mathematics. First of all, the word OR is very important: it is part of the logical apparatus in a problem. For example, the solution of quadratic usually yields two solutions, and these solutions are related by the word OR.

In mathematics, wherever the word OR appears, it is always the inclusive OR, 100% of the time. If a mathematician says “A or B,” she always implicitly includes the case of A and B together. If the mathematician wants to specific an exclusive OR, she would have to go out of her way to say something wordier: “*A alone or B alone but not both*.” (In Symbolic Logic and in the Computer Science world, folks sometimes use the abbreviation XOR for the exclusive OR.)

This means, on the GMAT Quant section, wherever the word OR appears, 100% of the time it will be the inclusive OR. Thus, in the Venn Diagram above, it would be absolutely clear and unambiguous: if the test asked, “How many students are in AP Literature OR AP Chemistry?”, the answer would have to be 160.

The GMAT Quant section presents exceptionally high quality questions, and these are always devoid of ambiguity. Even when the word OR appears, it will have a precisely defined mathematical meaning, leaving absolutely no room for ambiguity. The word OR always, 100% of the time, means the inclusive OR. If you can remember this piece of advice, it will save you trouble in counting and probability problems. In fact, you can apply these ideas in these Practice Probability Questions.

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If medical researchers are correct, then the human microbiome, made up of the microorganisms in our body, may hold the cure to diseases that have long plagued __humanity, amounting to a major oversight in Western medicine that has, until recently, all but__ ignored any such role of the microbiome.

(A) humanity, amounting to a major oversight in Western medicine that has, until recently, all but

(B) humanity, a discovery that amounts to a major oversight in Western medicine, which, until recently, had all but

(C) humanity, a discovery amounting to a major oversight by Western medicine: until recently, Western medicine all but

(D) humanity, amounting to a major oversight made by Western medicine, one that, until recently, all but

(E) humanity, which amounts to a major oversight in Western medicine and until recently it all but

A very subtle point that will come up on the harder GMAT SC questions is the idea of a summative modifier: a word that “encapsulates” the action of the preceding clause. To illustrate: the original sentence states that “microbiome…diseases…humanity, amounting…”. In this case, what does “amounting” refer to? Whichever of the three you argue for—assuming you argue for any—none amount to a major oversight in Western medicine. “The microbiome amounts to a major oversight” is odd, though the least odd of the three.

In order to clearly state what is doing the “amounting”, we use a word that “encapsulates” or captures the preceding phrase. This word is known as a summative modifier. In other words, it sums up what is being said. The good news here is you do not need to come up with the summative modifier itself, but you will have to find an answer choice that uses one.

Side note: the GMAT will never have multiple possible summative modifiers, asking you to pick the word that best encapsulates the preceding clause.

In this question, “a discovery” is a perfect summative modifier since it captures the idea of the preceding clause: the micribiome may hold cures to disease, a discovery…”

Just like that we can eliminate (A), (D) and (E).

The difference between (B) and (C) is somewhat similar. But instead of using another summative modifier, which is stylistically off, the correct answer, (C), uses a colon and cleans up the error by clearly stating who has been doing the ignoring: Western medicine.

As for (B), the “which” refers to the phrase “a major oversight”. It is not the “major oversight” that has until recently been ignored.

Finally, the “until recently + verb tense” is one of those faux errors that test takers can get hung up. There is no hard and fast rule here. Either works.

What the GMAT *does* want you to know is how to correctly modify clauses. And knowing how a summative modifier works will go a long way.

Oh, if you missed it, the answer to the question above is **(C)**.

Hello!

It’s the perfect day: you wake up and realize you have 3 free hours to use for GMAT prep! In addition to that, you also have a Magoosh premium account–lucky you! Now, how do you make the most of this study session? Where should you begin to maximize the time you have?

That’s what I’m addressing in this week’s GMAT Tuesday! Watch the video to get tips on how to mix things up during your study sessions.

And after you’ve watched the video, you can check out this sample of a three-hour routine below that I’d recommend to my students. Feel free to start here, but adapt it to fit your needs:

1. 20 minutes: study weekly verbal concept

2. 20 minutes: study weekly math concept

3. 20 minutes: do timed verbal practice

4. 10 minutes: break time

5. 20 minutes: review verbal practice, taking notes

6. 20 minutes: review math concepts from prior study session

7. 20 minutes: do timed math practice

8. 10 minutes: break time

9. 20 minutes: review math practice, taking notes

10. 20 minutes: review verbal concepts from prior study session

Good luck studying! Leave me any comments or questions you have below in the comment box.

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(A) 720

(B) 1,512

(C) 2,520

(D) 6,400

(E) 12,600

2) A librarian has a set of ten books, including four different books about Abraham Lincoln. The librarian wants to put the ten books on a shelf with the four Lincoln books next to each other, somewhere on the shelf among the other six books. How many different arrangements of the ten books are possible?

3) Over the course of a full-day seminar, 66 students must give short presentations. Most of the presentations are independent, but Sam’s presentation refers to something in Ruth’s, so Ruth must go before Sam; similarly, Matt’s presentation refers to something in Ruth’s and something in Sam’s, so Matt’s must come after both of those. These three presentations need not be consecutive with each other: whatever the order is, Ruth’s must come before the other two but not necessarily first of the 66, and sometime later, Matt must come after the other two but not necessarily last of all. The other 63 participants can present in any order. How many orders obey these constraints?

(A) 22!

(B) 63!

(C) 63!/3!

(D) 66!/3

(E) 66!/3!

4) A bag contains ten marbles of the same size: 3 are identical green marbles, 2 are identical red marbles, and the other 5 are five distinct colors. If 5 marbles are selected at random, how many distinct combinations of five marbles could be drawn?

(A) 41

(B) 51

(C) 62

(D) 72

(E) 82

5) In a bag, there are five 6-sided dice (numbered 1 to 6), three 12-sided dice (numbered 1 to 12), and two 20-sided dice (numbered 1 to 20). If four of these dice are selected at random from the bag, and then the four are rolled and we find the sum of numbers showing on the four dice, how many different possible totals are there for this sum?

(A) 61

(B) 106

(C) 424

(D) 840

(E) 960

6) A chessboard is an 8×8 array of identically sized squares. Each square has a particular designation, depending on its row and column. An L-shaped card, exactly the size of four squares on the chessboard, is laid on the chessboard as shown, covering exactly four squares. This L-shaped card can be moved around, rotated, and even picked up and turned over to give the mirror-image of an L. In how many different ways can this L-shaped card cover exactly four squares on the chessboard?

(A) 256

(B) 336

(C) 424

(D) 512

(E) 672

7) A shipping company has four empty trucks that will head out in the morning, all four to the same destination. The clerk has four different boxes to ship to that same destination. All four boxes could go on any one of the trucks, or the boxes could be split up into any groupings and given to the trucks in any combinations (ie. two to one truck, one to another, and one to another). In how many different ways could the boxes be put on the four trucks?

(A) 16

(B) 64

(C) 256

(D) 576

(E) 4096

8) Suppose we have six marbles: 3 blue marbles, 2 red marbles, and one green marble. Suppose we are going to put them into three cups: a black cup, a white cup, and a purple cup. We could put all six in any cup and leave two cups empty; or we could put marbles in two cups and leave one cup empty; or we could put some marbles in each of the three cups. How many combinations are possible?

(A) 90

(B) 180

(C) 360

(D) 540

(E) 720

9) Suppose we have six marbles: 3 blue marbles, 2 red marbles, and one green marble. Suppose we are going to put them into three cups: a black cup, a white cup, and a purple cup. The only restriction is that the two red marbles can’t be in the same cup. We could put as many as five (all except one of the reds) in any cup. We could leave one cup empty, or put some in each of the three cups. All combinations are allowed that don’t involve the two red marbles in the same cup. How many combinations are possible?

(A) 90

(B) 180

(C) 360

(D) 540

(E) 720

10) In a certain mathematical activity, we start with seven cards, each with a different prime number written on it. These seven cards are randomly put into three boxes in the following way: one box must get four cards, one must get two, and one gets only one. Then, for each box, we find the product of all the cards in the box, and that’s the “number” of the box. Then, we put those three numbers in order, from lowest to highest, and that is our set. How many different sets can be created from this process?

(A) 35

(B) 105

(C) 210

(D) 420

(E) 630

11) In a certain mathematical activity, we have five cards with five different prime numbers on them. We will distribute these five cards among three envelope: all could go in any envelope, or they could be broken up in any way among the envelopes. Then in each envelop, we find the product of all the cards in that envelope: that is the “number” of the envelope. An envelope containing no cards has the number 1. We then put the three envelope numbers in order, from lowest to highest, and that is our set. How many different sets can be produced by this process?

(A) 51

(B) 89

(C) 125

(D) 243

(E) 512

12) In the diagram below, points A and B are on opposite corners of a lattice consisting of 12 segments. A “true path” from A to B is a path on which no segment is traversed more than once. How many “true paths” from A to B are there?

(A) 6

(B) 10

(C) 12

(D) 14

(E) 18

Solutions will appear at the end of this article.

The following blogs cover some of the basic ideas of counting problems:

1) The Fundamental Counting Principle

2) Permutations and Combinations

4) Difficult Counting Problems

The first three just cover the basic rules. The fourth, in addition to four hard practice problems, has a discussion of some of the reasons why counting is such a challenging topic. I would refer you to that discussion, and would simply add: if you found the problems above challenging, then read the solutions very careful. The hardest part of a counting problem is often how to frame the problem, how to parse the situation, how to imagine it in do-able stages: once these choices are made, the application of the rules is often quite straightforward. How do you learn how to frame a counting problem? Well, first of all, pay attention to how solutions do this, starting with the solutions below. For more on right brain skills, see this post.

If the related blogs and the solutions here gave you some new insights, please let us know about this in the comments section!

1) If the ten books were all different, the arrangements would be (10!), a very big number. Because we have repeats of identical copies, not all 10! arrangements will be unique. For a subset of k identical members, those k members could be interchanged in k! orders, and the resulting arrangements would be the same, so we have to divide by k! to remove repetitions. In this example, we need to divide 10! by 4! and 3! and 2!:

Answer = **(E)**

2) Start out by thinking of the four books about Lincoln as one big book. This is one book, and we have six others, so we start by figuring out the arrangements of these seven books. That’s 7!

Now, for each one of those 7! arrangements, we can put the four Lincoln books in four different orders. Thus, for each 7! arrangements of the Lincoln slot with the six other books, we have 4! variants because of the order of the Lincoln books. That’s a total of

N = (7!)(4!)

Answer = **(C)**

3) If the constraints with these three individuals didn’t matter, there would be 66! arrangements. Any one arrangement will have those three in some particular order. If we kept the other 63 participants in the same places, and just re-arranged these three, there would be 3! = 6 possibilities:

We could group all (66!) arrangements into groups of 6 like this, and in each case, only 1 of the 6 would have the correct order for these three individuals. Thus, we need to divide (66!) by 3! = 6.

N = (66!)/(3!)

Answer = **(E)**

4) Here, we have to do a combination of listing and calculating. We have to think about several individual scenarios. First, the “three green” scenarios. By “other color”, I mean the colors other than green and red; there are five of these.

Scenario #1: 3 green, 2 red = 1 possibility

Scenario #2: 3 green, 1 red, 1 other color = 5 possibilities

Scenario #3: 3 green, 2 other colors = 5C2 = 10 possibilities

Now, the “two green” scenarios:

Scenario #4: 2 green, 2 red, 1 other color = 5 possibilities

Scenario #5: 2 green, 1 red, 2 other colors = 5C2 = 10 possibilities

Scenario #6: 2 green, 3 other colors = 5C3 = 10 possibilities

Now, the “one green” scenarios:

Scenario #7: 1 green, 2 red, 2 other colors = 5C2 = 10 possibilities

Scenario #8: 1 green, 1 red, 3 other colors = 5C3 = 10 possibilities

Scenario #9: 1 green, 4 other colors = 5C4 = 5 possibilities

Now, the “no green” scenarios:

Scenario #10: 2 red, 3 other colors = 5C3 = 10 possibilities

Scenario #11: 1 red, 4 other colors = 5C4 = 5 possibilities

Scenario #12: 5 other colors = 1 possibility

Now add these.

(10 + 10 + 10 + 10 + 10 + 10) + (5 + 5 + 5 + 5) + 1 + 1 = 60 + 20 + 2 = 82

Answer = **(E)**

5) This is not really a counting question, in that it doesn’t involves any of the standard counting techniques. We just have think about this logically. No matter what four dice we pick, the lowest roll we could get is a “1” on each of the four dice, for a total of 4. We could get any integer value from 4 up to the highest value. The highest value would occur if we picked the two 20-sided dice and two of the 12-sided dice, and got the highest value on each die: 20 + 20 + 12 + 12 = 64. We could get any integer from 4 to 64, inclusive. For this, we simply need inclusive counting. 64 – 4 + 1 = 61.

Answer = **(A)**

6) Consider the L in its current orientation, and start with it in the lower left corner.

We can move this up so that the top square is in any of the five empty squares above it: those, plus this original, is 6 positions. Now, we can move any of these six to the right one space, then again, then again, until we have moved it to the sixth new space, when the right side of the L will come up flush against the rightmost boundary. That’s seven total columns, each with 6 positions, for a total of 42 while it is in this orientation.

Clearly, we can rotate by 90° clockwise, and we would have 42 new positions for that orientation. Then we can rotate by 90° clockwise again, and again. Four orientations, each with 42 positions, for 42*4 = 168 positions.

All of this is for the “forward L.” Now, if we pick up the card, and put it down flipped over, to get a “mirror image L,” this again will have 42 positions in each of 4 rotated orientations, for another 168 position.

The total is 2*168 = 336

Answer = **(B)**

7) Where we put one box has absolutely no bearing on where we put any of the other boxes. The placement of the four boxes is completely independent of one another. For each box, we have four choices.

N = 4*4*4*4 = 16*16 = 256

Answer = **(C)**

8) First, consider possibilities the 3 blue marbles.

Blue Case I: all three in one cup = 3 possibilities

Blue Case II: two in one cup, one in another è three choices for the cup with two, then two choices for the cup with one = 6 possibilities

Blue Case III: one in each cup = 1 possibility

total = 3 + 6 + 1 = 10 possibilities for the blue marbles

Now, the 2 red marbles.

Red Case I: two in one cup = 3 possibilities

Red Case II: one in cup, one in another = 3 possibilities for which cup is empty

total = 6 possibilities for the red marbles

Now, the green marble can simply go in one of the three cups = 3 possibilities.

By the Fundamental Counting Principle, multiple the blue & red & green possibilities:

N = 10*6*3 = 180

Answer = **(B)**

9) First, consider possibilities the 3 blue marbles.

Blue Case I: all three in one cup = 3 possibilities

Blue Case II: two in one cup, one in another è three choices for the cup with two, then two choices for the cup with one = 6 possibilities

Blue Case III: one in each cup = 1 possibility

total = 3 + 6 + 1 = 10 possibilities for the blue marbles

Now, the 2 red marbles.

Red Case I: two in one cup = FORBIDDEN!

Red Case II: one in cup, one in another = 3 possibilities for which cup is empty

total = 3 legal possibilities for the red marbles

Now, the green marble can simply go in one of the three cups = 3 possibilities.

By the Fundamental Counting Principle, multiple the blue & red & green possibilities:

N = 10*3*3 = 90

Answer = **(A)**

10) Part of the logic of the problem involves recognizing that every unique combination of prime numbers produces a unique product. There is no way that two different groupings cards will produce the same set of numbers.

First of all, the cards to go into the cup holding 4 cards.

;

Once we have placed 4 card in the first cup, we have three cards left, which means we have three choices of a single card to put into the third cup. Three choices. Once we place that, the remaining two cards must go into the second cup: no choice there.

N = 35*3 = 105

Answer = **(B)**

11) We have to consider different groupings. First, all the cards together.

Case I: all five cards in one envelope (2 envelopes empty) = one possibility

Now, two envelopes used, one empty.

Case II: four in one, one in another, one empty = five choices for the one by itself = 5 possibilities

Case III: three in one, two in another, one empty = 5C3 = 10 possibilities

Now, no empty envelopes, all three used:

Case IV: 3-1-1 split = 5C3 = 10 possibilities

Case V: 2-2-1 split = 5C2 = 10 for the first pair, then 3 possibilities for the single one, leaving the other two for the pair. This would be 10*3 = 30, but that double-counts the two pairs, so we need to divide by two. 15 possibilities.

N = 1 + 5 + 10 + 10 + 10 + 15 = 51

Answer = **(A)**

12) First of all, consider the minimum paths, the paths of just four segments from A to B. Each one of them will consist of, in some order, two horizontal segments and two vertical segments. How many of these are there? Well, how many ways can we distribute two horizontal segments among four slots? 4C2 = 6. We put the horizontal segments in two slots, and then the two vertical segments must go in the remaining slots. There are six minimum slots.

Now, we have to consider paths that take more than four segments to get from A to B. Consider the possibilities that begin: (down) then (right) then (up). The two possibilities are

- a) (down)(right)(up)(right)(down)(down)
- b) (down)(right)(up)(right)(down)(left)(down)(right)

Now, consider the possibilities that begin (down)(down) then (right) then (up). The two possibilities are:

- c) (down)(down)(right)(up)(right)(down)
- d) (down)(down)(right)(up)(up)(right)(down)(down)

Those are all the possibilities for path that start (down). We also could start with a first step of (right), but those possibilities are just the reflections of these over an imaginary mirror line from A to B. There are four path that start (down), and four mirror images that start (right). That’s 8 longer-than-necessary paths, to add to the 6 minimum paths, for a total of 14 paths.

Answer = **(A)**

- Using Diagrams to Solve Rate Problems: Part 1
- Using Diagrams to Solve Rate Problems: Part 2
- A Different Use of the RTD Table: Part 1
- A Different Use of the RTD Table: Part 2
- Using the RTD Table for a Complicated Problem
- One More RTD Table Problem: Average Rates Part 1

Yesterday we solved this problem:

*Div’s bicycle tour consists of three legs of equal length. For the first leg Div averaged 16 kilometers per hour. For the second leg he averaged 24 kilometers per hour. What speed must Div average for the final leg in order to average 24 kilometers per hour for the entire tour? *

*A) 20 kilometers per hour*

*B) 28 kilometers per hour*

*C) 32 kilometers per hour*

*D) 40 kilometers per hour*

*E) 48 kilometers per hour*

We made a pretty good use of the RTD table, but we still took quite a while to arrive at our answer, and we did a lot of computation and algebra, every step inviting some sort of error.

There are two shortcuts that could save us some time and energy while reducing opportunities for computational error.

Notice that this problem doesn’t specify the lengths of the legs, but that every answer is a constant. That implies that *so long as the legs are all of the same length* as required by the problem, *any leg length will yield the same answer.*

Let’s take advantage of that fact to avoid some unnecessary algebra. Let’s stipulate an easy leg length. What leg length would be easiest to work with? 48 kilometers, since 48 is the least common multiple of 16 and 24.

Now we can put constants in the cells for leg lengths as well as in some of the cells for rates.

We can also determine a time for the bottom row—a combined time for all three legs—by dividing the combined distance by the average rate.

That allows us to determine the time for the third leg, since the sum of the times for the three legs is the combined time for the tour.

Finally, we can determine the rate for the third leg by dividing the distance by the time.

The little shortcut isn’t as powerful as the big one, and it’s peculiar to average problems, so you won’t get much use out of it, but here it is.

Since the rate for the second leg is equal to the average rate, we can ignore that leg. Because Div averages 24 kph for the entire tour and 24 kph for the second leg, he must also average 24 kph for the balance of the tour. So we could leave the second leg out altogether.

What would the RTD table for that amended problem look like? Well, if we use the big shortcut, we’ll still probably make each leg 48 kilometers, since 48 is the least common denominator of the rate in kph of the first leg and the rate in kph of the entire tour.

We can complete the top and bottom rows by dividing distance by the time.

Next, we can complete the “time” column.

Finally, we can divide the length of the third leg, 48 kilometers, by the time for the third leg, 1 hour, to determine the rate for the third leg.

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- Using Diagrams to Solve Rate Problems: Part 1
- Using Diagrams to Solve Rate Problems: Part 2
- A Different Use of the RTD Table: Part 1
- A Different Use of the RTD Table: Part 2
- Using the RTD Table for a Complicated Problem

Here’s today’s problem:

*Div’s bicycle tour consists of three legs of equal length. For the first leg Div averaged 16 kilometers per hour. For the second leg he averaged 24 kilometers per hour. What speed must Div average for the final leg in order to average 24 kilometers per hour for the entire tour? *

*A) 20 kilometers per hour
*

I usually start with the bare outline of an RTD table, and then fill it in bit-by-bit. Today, though, I’m going to start with a warning. Every problem on average rates has at least one attractive wrong answer that is intuitively appealing and seems promise a quick solution. Can you spot it here?

You might have said 20 kilometers per hour, because that is the average of the given rates, 16 and 24 kilometers per hour. Fair enough. That is wrong and will attract some people who read too quickly. I’m more concerned about a different trap, though.

Average-rate problems encourage you to assume that spending the same *distance* at two (or more) different rates allows you simply average those rates. For instance, in this problem, we might suppose that traveling one leg at 16 kph, one at 24 kph, and one at 32 kph would yield an average speed for the whole tour of 24 kph, because 24 is the simple mean of 16, 24, and 32 . And there’s 32 kilometers per hour waiting for you!

In fact, though, speeds are weighted according to how much *time* you spend at each, not how much *distance* you spend at each. This means that average speeds are generally less than you’d guess based on their component speeds, since it takes you *more time* to travel a given distance when you travel it slowly. (This neat fact deserves a post of its own, and I’ll write one soon.)

As a practical matter, what this means for average speed problems is that you will usually need to approach them *not* as simple averages or even as weighted averages. Instead, you need to view average speed problems through this formula:

Let’s devote one row of the table to each leg, and a fourth row to the total.

For an average-rate problem, you’ll need a wall between the rates of the various legs and the rate for the whole trip. I’ve represented that wall by drawing a heavy black line above the bottom-left cell. The only way to fill in that cell is from the other information in the bottom row, not directly from the cells above it in the same column.

The red lines below show the direction in you will usually *add each of the columns* and *divide the bottom row*.

Today’s problem is a bit strange though, so we’ll do things in a slightly different way.

Okay, let’s get rid of those red lines and add the rate information we’ve been given: the rates for the first two legs and the average rate.

Those are the only constant values we’ve been given, but we can add a little more information even so. We don’t know the distance, but we know that it’s the same for every leg, so let’s call that distance *d*, and then sum the legs to get a combined distance.

Now that we’ve represented both rate and distance for three of our four rows, we can determine times for those rows by division;

Notice that the column for “time” is almost complete. Since the total time, d/8, must be the sum of the other times, we can determine the missing time for the third leg.

Multiply through by 48, the least common denominator.

Transpose.

Divide.

Let’s add that to the table.

Now that we have the distance and time for the third leg, we can divide to determine the rate;

Okay, even using the RTD table, we still had to do an awful lot of work. It turns out, though, that a couple of shortcuts could have saved us most of that work. Come back tomorrow to see how.

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Hello!

Ever wonder who the man behind Magoosh is? On this edition of GMAT Tuesday’s with Kevin, I’m treating you to an interview with Magoosh CEO Bhavin!

Watch this week’s video to learn about Bhavin’s life leading up to business school, his advice for how to thrive in b-school, and of course, his spirit animal.

Be sure to let us know what you think in the comment box below! We’ll be back to our regular GMAT Tuesday format next week, so see you then.

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If you’d like to review before moving on, check out my previous posts on the topic:

- Using Diagrams to Solve Rate Problems: Part 1
- Using Diagrams to Solve Rate Problems: Part 2
- A Different Use of the RTD Table: Part 1
- A Different Use of the RTD Table: Part 2

We used the RTD (Rate, Time, and Distance) table to manage these problems, but some test-takers probably found that method to be overkill. Those readers judged that they could translate the problems directly from English to algebra without benefit of any table.

Today I’m going to give you a more complicated problem, one in which the travelers don’t move simultaneously:

*Jan’s house and Cindy’s house are joined by a straight road 24 miles long. Jan and Cindy agree to meet at a restaurant along that road, twice as far from Jan’s house as from Cindy’s. Beginning at noon, Cindy walks to the restaurant at a constant speed of 3 miles per hour. Later, Jan drives to the restaurant at a constant speed of 30 miles per hour. If they arrive at the restaurant simultaneously, at what time did Jan begin her drive?*

*A) 12:32
*

This is likely to be a frustrating problem. The algebra—if we can get to the algebra—probably won’t be too complicated, but the translation *into* algebra looks daunting, if only because we haven’t likely seen a very close model of this problem before. Let’s try our good friend the RTD table, and see if it eases the translation.

We’ll set the table up as we have before, with columns for rate, time, and distance, and with rows for Jan, Cindy, and their combined distance:

We’re given Jan’s and Cindy’s rates directly, so let’s put those in.

We are given the combined distances for Jan and Cindy, 24 miles, so we can include that. We’re told that the restaurant is “twice as far from Jan’s house as from Cindy’s.” If we call the distance from Cindy’s house *d*, then the distance from Jan’s house must be 2*d*.

*d*+2*d*=24

3*d*=24

*d*=8.

So the restaurant is 8 miles from Cindy’s house and 16 miles from Jan’s.

Let’s complete the rows for Cindy and Jan. Since we know that Cindy’s rate is 3 mph and that her distance is 8 miles, we can conclude that her time was 8/3 hours. Since we know that Jan’s rate is 30 mph and that her distance is 16 miles, we can conclude that her time was 16/30 hours.

But now what? Well, if Cindy took 8/3 of an hour, then she took hours, or 2 hours and 40 minutes. Since she started at noon, she arrived at the restaurant at 2:40.

Since Cindy and Jan arrived simultaneously, Jan too arrived at 2:40. At what time did she leave? Well, she took 16/30 of an hour. That’s 32/60, or 32 minutes. Since she arrived at 2:40 after 32 minutes of travel, she must have started her drive at 2:08.

If you’ve read my other blogs on the use of this table, then you know that the answer is “yes,” at least for some people. The table is an aid to translation, and a few people are able to translate even very complicated rate problems directly into algebra without benefit of the table.

Still, I think that the RTD table is especially useful for this problem, even though as we used it here it turned out to just be two equations stacked one on the other, and even though it left us with a fair bit of work still to do. Disciplined well-practiced use of the RTD table allows us to chip away at the familiar parts of the problem and to reserve our cognitive resources for the less familiar parts.

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