May 23, 2012
Sometimes, you’ll come across a GMAT problem that gives you a lot of information. The relationship between the pieces of information may not even be clear at first. In my experience as a GMAT teacher, I’ve found that these types of questions tend to confuse students, especially when students try to immediately write one big equation to solve. Instead of attempting to reach a solution all at once, take a step back and follow a few standard steps.
First, do not panic. The question may look like it has too much data to consider in the two minute time frame you are allotted per question. You need to remind yourself that GMAT questions are written to be solved in about two minutes, so a strategy must exist to complete the problem in a timely manner.
Second, in order to find that strategy, start translating the question stem piece by piece. Do not expect to find one big equation; many complex word problems require a system of equations to solve. Even if you end up with four or five equations to start, you will be ok as long as those equations are relatively straightforward.
Third, solve any one variable equations. Use those results to solve two variable equations. Keep the variable you ultimately want to solve for in mind. Doing so may allow you to bypass some of the equations you initially wrote down.
By following the steps outlined here, you will avoid getting the various pieces of information confused, and you will be less likely to put variables in the wrong place. Additionally, you will not rush through the question, which leads to mistakes and ends up taking longer than the method described above.
Try the problem below. As you do, be sure to follow the steps we discussed.
A truck driver drove for 2 days. On the second day, he drove 3 hours longer and at an average speed of 15 miles per hour faster than he drove on the first day. If he drove a total of 1,020 miles and spent 21 hours driving during the 2 days, what was his average speed on the first day, in miles per hour?
Step 1: Analyze the Question
Another two-stage journey. So despite the intimidating presentation, we know that we will transfer the data from the question stem into this chart:
We’re solving for speed on the first day, which is the top left box of the chart.
Step 3: Approach Strategically
Before you get too worried about what your solution will be, plug the data into the chart to help organize your thinking. The first thing we read is “On the second day, he drove 3 hours longer . . . than he did on the first day.” We know the total time will be 21 hours, so we can’t just pick a number. Let’s use t for time on the first day. That makes time on the second day t + 3.
Similarly, “On the second day, he drove . . . at an average speed of 15 miles per hour faster than he drove on the first day,” allows us to say that if r is speed on the first day, then r +15 is speed on the second day. The rest of the data is simply numerical:
Since the total of the two days’ times will be the time for the entire trip, we can say:
t + (t + 3) = 21
2t + 3 = 21
2t = 18
t = 9
We can put that into the chart:
That allows us to find distance for each day by multiplying (Rate 3 Time = Distance).
Now we have an equation that will allow us to solve for r, which is what we’re looking for—speed on Day 1. Since the total of the two days’ distances will be the distance for the entire trip, we can say:
9r + 12(r + 15) = 1,020
9r + 12r + 180 = 1,020
21r = 840
r = 40
Step 4: Confirm Your Answer
Reread the question stem, making sure that you didn’t miss anything about the problem.
May 21, 2012
The trickiest question type in the quantitative section of the GMAT for most students is yes/no data sufficiency questions. When approaching these problems, it is imperative that you keep in mind the purpose of data sufficiency.
Let’s start with a review of data sufficiency. On these questions, your goal is not to find the answer. Rather, it is to determine if you have enough information to find the answer, regardless of what the answer is. On value questions, this is fairly straightforward. If you are asked for the value of x and you know it is 4, that’s sufficient, but if it could be 4 or 6, that’s not sufficient. In the former case we could narrow down the possibilities to one answer. In the latter, we could not.
Now let’s see how this same concept applies to yes/no questions. If a question asks us if x is positive and we know the answer is “yes,” that’s sufficient, because we only have one possible answer. Likewise, if we know that the answer is “no”, that’s also sufficient, because we still have only one possible answer. Since this is data sufficiency, it does not matter if the answer is “yes” or “no”, it only matters that we have one possible answer, just as was the case with a value question. If you must answer the question “sometimes yes, sometimes no”, that will be insufficient, because you have two possible solutions.
Thus, when you encounter a yes/no data sufficiency question on test day, see if you can answer it with an “always yes” or “always no.” If so, remind yourself that, since you are able to give a definitive answer to the question, the statement is sufficient.
Is the sum of x , y , and z equal to 3?
(1) xyz = 1
(2) x , y , and z are each greater than zero.
The stem does not tell you anything, so move on to the statements.
Statement 1: Don’t assume that x, y, and z are positive integers! Pick numbers:
If x = y = z = 1, then x × y × z = 1 and x + y + z = 3.
If x = 1 and y = z = – 1, then x × y × z = 1 but x + y + z = – 1.
Statement 1 gives a “sometimes yes, sometimes no” answer. Statement 1 is insufficient.
Statement 2: x, y, and z can be any positive numbers.
This yields a “sometimes yes, sometimes no” answer. Statement 2 is insufficient.
Statements 1 & 2: Pick numbers.
If x = y = z = 1, then x × y × z = 1 and x + y + z = 3.
If x = 1, , and z = 2, then x × y × z = 1, but .
Combined, we still get an insufficient answer, “sometimes yes, sometimes no.”
The answer is E.
Did you have trouble with the above question? Post your questions in the comments area so we can help you work through it.
May 17, 2012
Ever since I started teaching GMAT classes, I have taken note of any references to standardized tests I come across in television shows and movies. In the six years of doing so, I have found that these references almost always follow the same pattern. One of the characters needs to take a standardized test that they find difficult or boring. In order to illustrate this to the other characters, they will read an example of one of the questions on the exam. Invariably, the question they read involves two trains leaving two different stations at two different times and traveling towards each other.
Because of this, rate problems that feature two trains (or cars or people or anything else) have a bit of a bum rap. These questions are seen, unjustly, as difficult, time consuming and complicated. However, by learning only a few basic rules, you can handle these questions quickly and correctly.
The first step is to make sure the trains leave at the same time. If one train leaves earlier than the other one, calculate the distance the earlier train will have travelled by the time the later train leaves. Subtract that distance from the distance originally separating the trains, and use the that new distance as the total distance.
The second step will depend on the exact type of problem. If the trains are coming towards each other or going away from each other, add their speeds. If one train is catching up to the other, subtract their speeds. Use this result as the total speed.
Once you have calculated the total distance and total speed, you can solve for the time as you would on any other rate question. You just plug these numbers into the same formula you used back in step one to find the earlier trains distance, which is distance = rate x time.
The problem below is a perfect example of this type of question. While it looks complicated at first, draw a diagram and then follow the steps outlined above to reach the correct answer.
Train A left Centerville Station, heading towards Dale City Station, at 3:00 p.m. Train B left Dale City Station, heading toward Centerville Station, at 3:20 p.m. on the same day. The trains rode on straight tracks that were parallel to each other. If train A traveled at a constant speed of 30 miles per hour and Train B traveled at a constant speed of 10 miles per hour, and the distance between the Centerville Station and Dale City Station is 90 miles, when did the trains pass each other?
(A) 4:45 p.m.
(B) 5:00 p.m.
(C) 5:20 p.m.
(D) 5:35 p.m.
(E) 6:00 p.m.
Begin as you do for any word problem, by understanding the basic situation. Two trains, 90 miles apart, start moving toward each other at different times. One train moves at 30 mph, the other at 10 mph. Our task is to determine the times at which the trains pass each other, which is to say when they will be at the same point on these 90-mile tracks.
Our story begins at 3:00 pm when train A leaves. It is going 30 mph. The next event happens at 3:20 pm when Train B leaves its station going 10 mph. In the 20 minutes before train B leaves, train A has travelled 10 miles. This leaves 80 miles of track between them when train B starts at 3:20. This is now our total distance.
The question is then, “How fast will the two trains close that distance?” Here we add the speeds to get total speed: 30 + 10 = 40.
So, at a combined rate of 40 mph, how long will it take them to close an 80-mile gap? Time = distance/speed. Thus, 80/40 = 2 hours to close the gap. It will be 5:20 at that point. Answer (C) is correct.
For about a year, I always used the same method to solve the following GMAT problem:
How many liters of water must be evaporated from 50 liters of a 3 percent sugar solution to get a 5 percent sugar solution?
“This is simple percentages,” I would say. “Just start by taking 3% of 50 liters, which is 3 over 100 times 50, which comes out to 1.5 liters sugar…”
But one day, teaching this same quantitative problem, a student’s hand shot straight up. “Yes, James?” I said. (That wasn’t his real name, by the way, but it will do.)
“Eli, who cares about the sugar?”
I paused. “Well, the sugar will help us figure out the solution.”
“But you don’t need it!” James explained. “I’ve been a chemical engineer for years, so I do this problem all the time. The sugar is a constant. The amount of sugar doesn’t change, and that amount is always equal to the concentration times the volume. So just do CV = CV; 50 times 3 is equal to the final volume times 5!”
I paused, impressed, and amazed—and have taught his timesaving shortcut ever since.
However, there is a bigger lesson here than simple mixture problems. I had approached that problem uncritically. I “knew” how to find the right answer, so I never gave it a second thought. I spent far more time prepping the combination and probability problems given their complexities and hidden challenges. As a result of my complacency, I made extra work for myself.
In other words, the takeaway is this: don’t focus exclusively on your mistakes. Review your correct answers as carefully as your errors. There might be a time-saving shortcut that you missed the first time through. Or better yet, you might have used such a technique that you can notice, reconstruct, and repeat! It’s very tempting to spend all of your time studying those red X’s on your quizzes and CATs. But avoid that trap; looking at your right answers as well as your errors will help you stay positive, and more importantly, will give you many opportunities to raise your score.
May 6, 2012
I want you to think back to when you were in grammar school (strange how often the test for business school has us thinking about grammar school). Specifically, I want you to think about gym class and every time you needed to pick teams. For purposes of today’s discussion, let’s say we are picking a baseball team from the students in class. Now I realize that for some of you this was a traumatic experience, and GMAT teachers are no exception to that – we are not particularly notable for our athletic prowess. However, this scenario can help you understand a difficult question type on the GMAT – permutation problems.
A permutation problem will ask you to determine the number of ordered subsets of a certain size that exist in a group. If we move away from GMAT speak, this means permutations will occur when you must perform two actions. First, you will be creating groups. In our example of a baseball team in a gym class, the class may have 30 students, but you need to select nine. Second, you will put the selected entities in an order. On our baseball team, you can think of this as the batting order. If A is first, B is second, and C is third you have a different outcome than if C is first, B is second, and A is third. However, A, B, and C may not even be on the team – you will have lots of other outcomes in which they are left out.
Thus, we need to both select nine players to be on the team AND put those players in an order. On GMAT test day, when we must perform both of these actions, we will have a permutations problem. The formula to determine the number of permutations is n!/(n – k)!, where n is the number of total entities, in our example 30, and k is the number of entities we want to select, in our example nine.
See how you do on this next problem, and ask questions if you have them…
Kim has four trophies which she wishes to display in a cabinet with five shelves, with only one trophy to a shelf. How many different ways are there to arrange the trophies?
This is an arrangement question. Often you can simply count the different possible arrangements, making sure to do so systematically, but it is usually easier to remember that the number of ways of arranging x objects is x !
There are too many arrangements here to count by simply writing them out, so you will have to use factorials to count the arrangements.
In this case you have 4 objects, but 5 spaces. So you have five possibilities for each shelf-it could contain one of the four trophies, or it could be empty. So the number of possible arrangements is 5! = 120. The answer is (E)
May 2, 2012
Not surprisingly, most GMAT test takers have a background in the business world. As such, many readers have worked on a committee formed from a larger group of employees. Every time a committee is formed in this fashion, you are, in fact, doing a GMAT problem. More specifically, you are attempting one of the most dreaded question types on the GMAT quantitative section – combinations.
While these questions can be tough, by thinking about the real life experience of forming a committee, you can more easily understand exactly what a combinations question is asking you to do. Let’s say that your business has ten employees and needs to create a committee of four people. If you want to determine how many different possible committees you could create, you would use the combinations formula, n!/[k!(n - k)!], where n is the number of people with which you start (in this case 10) and k is the number of people you want in your group (in this case 4).
Now, let’s take a step back and see how we know that this is the appropriate formula to use. A combinations question will ask you how many different groups you can create. For example, if our people are A, B, C, D, E, F, G, H, I, and J, one group we could make is A, B, C, and D and another group we could create is B, C, H, and I. However, in combinations problems, order will not matter; that is, changing the order of the entities in the group will not make it a different outcome. In our example, A, B, C, and D is the same group as D, C, B, and A. Thus, we are grouping, but not arranging.
On GMAT test day, when this occurs, you know that you are dealing with a combinations question and should apply the formula outlined above.
Corporation Z has 2 locations, one in New York and one in Los Angeles. The New York location has 6 executives and the Los Angeles location has 4 executives. If a meeting of 4 executives is to take place, and exactly 2 executives from New York must attend, how many different groups of 4 executives are possible?
The question asks for the number of combinations of 4 executives, 2 from New York and 2 from Los Angeles, which can be formed starting with a larger group in each city.
Use the combination formula to find the number of pairs of executives from each city.
New York has 6 executives.
Los Angeles has 4 executives.
Then multiply the two results.
15 × 6 = 90 – answer choice (A)
April 28, 2012
Remember, the quantitative section of the GMAT asks you to complete 37 questions in 75 minutes. This means that you have only two minutes per question. That’s not much time as I am sure many of you already know. Thus, it is important that you use your time in the most efficient manner possible.
While many test-takers feel they will not be able to complete the test in time, you should also keep in mind that the questions are designed so that it is possible to complete each one in an average of two minutes. While some questions will take a little more or a little less time than this, you should never be spending over five minutes on an individual problem. If you do find you are reaching the five minute mark on a question, even if you think you are approaching it correctly, you need to move to the next problem. For someone who knows the correct approach, no GMAT question will ever take five minutes. The time you are taking is itself the clue that you do not know how to reach the answer correctly.
No matter how high of a score you are trying to achieve, you will miss some questions on the GMAT. You are better off identifying those questions quickly and providing yourself with more time to handle the questions you will be able to get right. By following this strategy, you will maximize your GMAT score, which is always the end goal.
April 21, 2012
One of the big GMAT skills that is often overlooked by students is translation. Any time you decide approach a word problem using algebra, you will need to translate the English in the question stem into an algebraic equation. While this seems as if it would usually be fairly straightforward, the GMAT will often find ways to make it more difficult. A translation error will often lead to a trap answer, so it is essential that you learn how to translate difficult statements before test day.
To understand why translation can be more difficult than it seems, think about translating a foreign language. If you only need to translate one word, you can usually just find the equivalent word in English. Similarly, if a GMAT problem uses the phrase “more than” you know that it must translate to addition.
However, when you try to translate an entire sentence from a foreign language to English, the process becomes much more complicated. You do not need to only determine the matching English words, but also need to consider the structure and meaning of the sentence as whole, since the foreign language might construct sentences differently than English. Along the same lines, when you translate a word problem to math, you must consider the relationships between the different parts of the sentence and check to see if any words are used in unusual ways. You might get lucky and have a problem that uses the word “equals”, but you are just as likely to see the words “costs”, “weighs”, and “sells” used. Identifying when these other words translate to “equals” will be key to writing a correct equation.c
Below, you will find a word problem that can be solved algebraically. Be careful on your translation and then check out the solution to see how you did.
Jacob is now 12 years younger than Michael. If 9 years from now Michael will be twice as old as Jacob, how old will Jacob be in 4 years?
Step 1: Analyze the Question
We must first translate the question into algebraic equations, then apply the techniques of combination or substitution to solve for Jacob’s age.
Step 2: State the Task
Translate the question stem, then solve for Jacob’s age using substitution.
Step 3: Approach Strategically
Equation for Michael’s and Jacob’s current ages: M = J + 12.
Equation for Michael’s and Jacob’s ages 9 years from now: M + 9 = 2(J + 9)
We now have two distinct equations with two unknowns that we can solve. We can substitute the first equation into the second for M, yielding the equation:
( J + 12) + 9 = 2 (J + 9)
J + 12 + 9 = 2J + 18
J + 21 = 2J + 18
Thus, J = 3. So if Jacob is 3 years old now, then in 4 years, he will be 7 years old. Don’t forget to add these 4 years at the end.
Step 4: Confirm Your Answer
We can plug in J = 3 into the original question stem to confirm our answer. Be careful with your calculations, since (A) is a trap for those who don’t add 4 years to his current age.
April 19, 2012
Most students learn that absolute value is the positive version of a number. Thus, the absolute value of 7 is 7 and the absolute value of -7 is also 7. While these absolute values are correct, many GMAT problems will be more straightforward if you learn the true definition of absolute value, which is the distance a number is from zero on a number line. Thus, the absolute values of 7 and -7 are 7 because both numbers are 7 away from zero on a number line.
To understand how absolute value works, imagine you live in a house right in the middle of a block. The street has 5 houses to the left of your house and 5 houses to the right of your house. Whether you walk two houses to the left or two houses to the right you will be 2 houses away from your home. Now, instead of picturing a block, consider the same concept as a number line in which your house represents zero. If you walk 2 houses to the left, you are at -2 and if you walk 2 houses to the right, you are at 2. Normally, we use positive and negative numbers to distinguish these as different spots. However, if we are concerned with the distance from us rather than the location relative to us, we no longer need to distinguish between positive and negative numbers. Since that distance cannot be negative – it is either 2 to the right or 2 to the left – we must define the distance in positive terms. Thus, absolute value will always be positive.
Keeping the concept of absolute value in mind, try the problem below and then check out the solution to see how you did.
Which of the following could be the value of x, if |4x – 2| = 10?
Analyze: We see an equation with an absolute value sign in the Q-stem. Also, what important GMAT wording do we see? “could be the value”—that means that there are at least two possible solutions for x in the equation and the absolute value sign means the expression could equal 10 or –10.
Task: Pick the answer that makes the equation true. There may be another solution, too, but only one of the answer choices will work.
Approach Strategically: To solve a normal algebraic equation, we have to do the same thing to both sides in order to isolate the variable.
We’ll do the same here, but we’ll set up two equations because of the 2 possible answers that the expression can equal, like so: 4x – 2 =10 and 4x – 2 = –10.
Solving for 4x – 2 = 10 gives us x = 3.
That’s not one of our answer choices.
Solving for 4x – 2 = – 10 gives us x = -2 which is answer choice B.
You could also backsolve here but it’s important to know the rules of absolute value on Test Day.
April 9, 2012
Recently, I took a trip to the Metropolitan Museum of Art in New York. As I perused the galleries, I noticed that not all of the painting were equally easy to understand. In the works by Da Vinci, that were made to look realistic, I could clearly tell what was depicted. The Monet’s were a bit tougher, but with the explanation provided by the museum, I could clearly see the subject. At the Picasso’s the subject was a bit harder to find, even with the explanations. Finally, I came across the Pollock’s, which required me to depend entirely on the explanation to understand what was happening on the canvas.
In art, I realized, just as is the case on the GMAT, the more abstract the presentation of a concept is, the harder it is to understand. Luckily, the curator at the museum had written blurbs for each painting – blurbs I relied upon more and more as the works became more abstract. Unfortunately, on the GMAT no such blurbs are provided, so we need to find other ways to make abstract problems more concrete.
One major way to accomplish this is to pick numbers. When you encounter a problem that features unknowns or variables, but does not ask you to solve for these variables directly, it can be helpful to make up numbers for each unknown and solve using those numbers. Then you can plug the numbers you selected into each answer choice and see which choice produces the same result. The option that does will be the correct answer. Keep in mind that you should almost always check every answer when you use this strategy, in case more than one choice produces your result with the particular numbers you picked. When this happens, you will need to pick new numbers and test those choices again.
The problem below can be pretty tough to solve algebraically, so see if you can solve it by picking numbers.
Each writer for the local newspaper is paid as follows: a dollars for each of the first n stories each month, and a + b dollars for each story thereafter, where a > b. How many more dollars will a writer who submits n + a stories in a month earn than a writer who submits n + b stories?
(A) (a – b)(a + b + n)
(B) a – b
(C) a2 – b2
(D) n(a – b)
(E) an + bn – an
A couple ways to think about this problem.
Author 1’s earnings – Author 2’s earnings
an + a(a+b) - [an + b(a+b)]
an + a^2 + ab – [an + ab + b^2]
an + a^2 + ab – an – ab – b2
a^2 – b^2 (Answer Choice C)
Picking Numbers: When you have variables in the answer choices, this usually signals an opportunity to pick numbers if setting up the equation proves difficult. Pick some permissible, manageable numbers such as n =10, a=3 and b= 2. Remember that the only rule was a must be greater than b. Read the original question stem with the numbers in it and realize that they’re asking how much more does a writer who writes 13 stores make than one who writes 12. Since he’s only getting paid for one more story, that’s a+b more dollars which in our case is 5 more dollars.
Plus the values back into the answer choices and find which equals 5. It might take some extra time but we’ve reduced some abstract algebra into simple arithmetic which can be done quickly.