Math Overview

liushiyou

美国人对Math的总结，大概看了看，用处不是很大。
不过有时间的战友也可以看看，就算是练阅读了：）

liushiyou

1。Quantitative: intro

This quant "course" serves as supplements to Kaplan Math Workbook, 2/3 edition for GMAT Math prep. I encourage you to purchase this resource. It is a great book designed for self study, on which this course will be based. We are not going to violate the copyright or publish the book material, but rather provide additional comments and alternative solutions not mentioned in the text to get the max output.

The structure of the book is superb for detailed review of the major test points: every chapter has a few pages dealing with the theory, formulae, and rules, and following that, a good number of exercises split into levels: beginner, intermediate, and advanced. It is recommended that you do all the problems in the book, but if you are in a hurry, do at least the intermediate and advanced sections.

Of course, the book is not perfect and is missing some info on statistics, probability, coordinate geometry and other hot subjects (Kaplan may possibly think that those questions are experimental, but it really is not for us to judge). In any case, on this website you will find materials for those sections as well.

Before you move on, we would like to share six basic principles, which will prove crucial in getting maximum results on your test:

1. Do not haste to kill yourself. I am not talking about suicide, but about a general tendency to jump into the math and try to beat it to death. Please, remember that your goal is getting the correct answer, not wearing yourself out with complicated calculations. When you see a river of lava, you do not hurry to jump off a cliff, do you? Even if somebody tells you that you have only 75 minutes to cross it, this 75-minute argument is not really convincing, is it? So why then jump into the GMAT math river?

2. You do not need more than 75 minutes for math. Sometimes people do not wait or even think, but rush to solve, crunch, and calculate. The test rush gets them to be so careless as not to even use a strategy but move ahead as an uncontrollable bunch of barbarians. Perhaps, it is clear that one gunman can take care of enough barbarians, just as long as he has ammo. So, why do so many still use a cudgel instead of M16? 75 is all the time you will need for math, and there is no need to rush or panic on the test. (See our section dedicated to timing). Even if you are a special case and you do need more than 75 mins, you still can't have it.

3. Another point we want to learn today is to take a 5 second pause after reading a hard question. This uncanny technique will allow you to feel more relaxed and in control. When you take this tiny pause, you can look at the problem text and the answers to see possible shortcuts (backsolving, picking numbers, division test, to be discussed later), possible mistakes (units change, tricky questions that do not use all the information, to be discussed later), and just other faster ways to get to the answer. The advantage of spending 5 seconds in the short run is saving minutes in the long run. For all of the 37 questions, you will spend 2-3 minutes for these pauses, but if you are able to spot a faster solution at least twice or simply save yourself from doing faulty calculations, you will gain that time back. Do not hurry to burry yourself in math. GMAT math is easy, and it takes 5 seconds to see that.

4. Another principle you need to remember is that the greatest percentage of errors made are due to inattentiveness. Being attentive should be your main formula, your main rule, and your main principle. Be always on the look out and pay attention: GMAT is not a golf game. Nobody says paying attention is as easy as making pan cakes (the latter is still a challenge for many of us, but still easier). Being careful and paying attention is a hard work and is tiring, but you need to learn to discipline yourself: you will not improve your score unless you are extremely careful. It does not matter how many formulae you know if you mess up in your calculations. If you take this advice seriously, you will score at least 20 more points on your next GMAT.

5. You should know formulae and rules by heart. They should come naturally from extensive practice, not from memory.

6. Do not use your brain for complex calculations. I suggest that you do not trust your brain much during the test . It is primarily because under stress it can make some really silly mistakes and, in addition, it will not be working very fast after you do two essays and a handful of hard math problems. Therefore, use paper, not your brain, for calculations; use real numbers, not variables; use values and do not go into developing formulae; and write down complicated relationships. Try to be as simple as possible – it pays off on the GMAT. We are strongly advocating using paper and not trying to hold things in your head because if you forget something, you will have to go back and recalculate, taking up your time, whereas if you had it on paper, it would be sitting right in front of you. If you catch yourself losing track of the question, you may want to write down the main few keywords, so that you always know what you are after.

Having remembered the main principles of successful GMAT student, you can move on to the first lesson.

liushiyou

Quantitative: Arithmetic: Lesson 1 We are giving suggestions what problems are illustrating some of the common GMAT points, just so that you would not miss them. We are also adding our commentary to the material as to improve the efficiency and provide you with the best techniques. We are thankful to Kaplan for developing such a great self-study book. We are proud that we can recommend it as a great guide that allows getting one’s math level fast and without any external help. Before you move on to arithmetic, please, read the first introductory section pp. 3-12. You should learn about the question format on the GMAT, the way you get your score calculated, etc. you will find more of such questions in our FAQ. Arithmetic is usually one of the most troublesome sections for the majority of students because they often tend to think that they know arithmetic well. In contrast, many of us learnt arithmetic a long time ago and hardly remember anything. Please, read chapter two attentively to brush up those “easy” points: pp. 13-20. You should know the following before moving ahead: integers – can be positive and negative, count numbers (there are a lot of questions on integers) Do not rush to do the math, take a look for an easy way. Know what reciprocal is Let’s see now what you know: Page 23: Do #3, then read on. If you took a look before jumping in, you should have seen that 0.125 can be grouped with 0.375 and 0.25 with 0.75 to make 1.5. Fast and easy. Do # 4, then read on. If you took another look before trying to calculate everything, there would be something you could figure out. There is no clock ticking, so you are probably feeling relaxed and could spotted that some of the answer choices could be grouped. The logic behind is that only ONE of the answer choices will be right, therefore only ONE of them will be less than . Thus, if we compare A, C, and D, D will end up being the smallest (actually = 0.166667 and as soon as we found that D is less than that, we could have stopped, but just to make sure, we will go through all the answers). B is exactly , so it is out and E is a little more than , which is . Therefore, D is the only number less than Know your fractions well. The smaller is the denominator, the greater is the fraction and the smaller is the numerator, the smaller is the fraction. Let’s do #10 now. Read on after you do it. The question is fairly similar to #4 but not the same. Here we can try to represent the values graphically on both sides of 0.40 by locating them on the number line: 0.33 0.4 0.5 At the first glance, we have 2 obvious limits: A, which is 0.33 and E, which is 0.5. You would have noticed that if you took a look and the answer choices before trying to develop a strategy for solving the problem. Now, let’s run through the answer choices and see if any of them approach closer: B is larger than 0.5, so it is out; D is just the same, so it is out too, now we have C, which is larger than 0.33 since it is more than and less than 0.5 since it is less than . We don’t really care how much it is, as long as it is within the interval. As you have noticed, Arithmetic requires that you do calculations fast and that you know some of the common fractions and convert them to decimal values fast. This trick will help you to save time on many questions. Check Arithmetic II section for common fractions Now is #12. Move on after you get tired This is a roman numeral question, as some people call them, and these are of above average difficulty, but they have their own little helper tools too. Before you start, mark 3 numbers on your scratch paper so that you can put a checkmark if the condition is correct or scratch it off if it is wrong, just like below. I II III One of the tricks that we could have used here is reducing the numbers from 160 and 120 to smaller ones since the numerical values don’t really matter. We can take anything like 10 and 1 or 4 and 3, just as long as we follow the rule that the first number is greater than the other. 4 and 3 are convenient since both 160 and 120 can be reduced by 40 to equal 4 and 3. We will have 1,000 – 4 and 1,000 – 3; the latter is greater, YES. and , we will have and ; if you don’t see that the first number is greater than the second, you could have taken 10 and 1 and gotten and , and, clearly, the first value is greater than the second one. Mark it off. If you stayed with the original values, you may have faced the crisis of what is greater or ? In such case you could have, again, taken a set of wild numbers: and or compared and . is closer to 1 than , and therefore larger. NO Ok, you may have attempted to the above calculations in your head, but this one will require you to use paper. Just to mention again, we are strongly advocating using paper and not trying to hold things in your head because if you forget something, you will have to go back and recalculate and use up your time. Anyway, here we will arrive to the following equations: and ; they can be simplified into and ; if you can’t figure out fast which one is larger, try 10 and 2 (1 will not work due to division by 0), and you will have and . YES The last one in this section is # 15. Again, we recommend that you spend 5 seconds looking at the problem and the answer choices. If you do this, you will notice that the answer choices are increasing by 10 times. This will let you know that to miss the right answer, you only need to mess up the ten’s digit, while you can do anything you want with the units, so we can go ahead and round up or down the numbers we will need to calculate. After transformation our task will be the following: = 2 The second sub-section of arithmetic is the most interesting; after studying it, you should know the following: 1. How to manipulate modules | x + 2 | = 4 to solve this equation, we will need to open up the module and solve two equations: x + 2 = 4 and x + 2 = -4; we will have two roots: 2 and -6 If we had an inequality, our solution would slightly change: | x + 2 | > 4 The equations we would get would be: x + 2 > 4 and x + 2 < -4; it this case, we are changing the inequality sign due to the minus that we have added to the 4. Our answer will be: 2>>) 5. Prime factorization The best way to factor out a number is to use a system different from what the book suggests: For example, let’s factor out 264: The advantage of this system is that you can see all the prime factors organized in one line. Also, if you need to find LCM or GCD, you can easily scratch off or circle the factors you need, making sure your numbers are organized and you don’t mess up. 6. Multiple of an integer – an integer that is divisible by another integer; this term is very often used on the GMAT; usually the question says that one number is a multiple of another. E.g. 81 is a multiple of 9, meaning that 81 is divisible by 9. 0 is a multiple of any integer, because 0 divided by any number is 0. See more math terms >>> 7. LCM/GCD. (Least common multiple and greatest common divisor) To find LCM of 2 numbers, you need to find the number that is the product of all the unique factors of 2 numbers. For example, LCM of 10 and 15 is 30. 10 = 2 x 5; 15 = 3 x 5; LCM = 2 x 3 x 5 = 30. For more complex numbers, use the method described above to find the prime factors. GCD, on the other hand, is found by multiplying all the factors that are present in both numbers. For example GCD of 60 and 45 is 15. 60 = 3 x 2 x 2 x 5; 45 = 3 x 3 x 5; therefore, GCD = 3 x 5; these are the only two factors that are common for the two numbers. 8. odd/even. The best strategy is to plug in numbers for odd and even such as 3 and 2. Formula of even numbers is 2n; where n is any integer For odd, 2n+1. Do the following exercise numbers on page 31: 2, 7, 10 and 11 use the system we have suggested, also find the LCM and GCD. 12, 13. 1. Find all the factors of the following numbers: 420 560 286 243 2. Find the GCM (greatest common multiple) and SCM (smallest common multiple) of these integers. On page 34, let’s do #8 This one should be easy for you if you take 5 seconds to look through the answer choices and remember the formula for an even integer: 2n. Do # 11 Find the factors for 130 then we also need to take the unique products of the factors: 2 x 5 =10; 2 x 13 = 26; 5 x 13 = 65; 2 x 5 x 13 x 1 = 130. Therefore, we have 8 integers that satisfy our divisibility condition: 130, 65, 26, 13, 10, 5, 2, 1. Do # 13 This is a pure number picking problem. Pick 2 sets of numbers and check for the greatest integer that will divide the set. You can optionally check another set, just to make sure. Take 0, 2, 4 and 2, 4, 6 and optionally 4, 6, 8. They all divide by 2, 3, and 6. Do # 14 Wherever there are consecutive integers and there is average, sum, or anything like that, you can use one rule: in the consecutive row, the average is the middle number if the row is odd or equidistant from the two middle consecutive numbers if the row is even. Thus, if the sum of 3 consecutive integers is 312, then the middle of them is 312 divided by 3 or 104. Therefore the other two are 103 and 105. The next 3 will be 106, 107, 108. The sum will be 107 x 3 = 321. Consecutive numbers are convenient to operate with. Do # 17 Pick numbers! Do not try to make your brain steam. Again, this is a Roman numerals question, so take 3 sets of numbers that match the conditions specified. I. 3, 3, 3, 3 (12, 81) NO II. 2, 2, 3, 3 (10, 36) YES III. 2, 2, 2, 2 (8, 16) YES We don’t know if the numbers are different or not, so let’s assume they are the same to make it easier to count; it actually does not matter for odd/even.

liushiyou

Quantitative: Arithmetic: Lesson 2

Averages



Moving on with the Arithmetic, we will cover today the chapter to the end, working on the Averages, Ratios, Percents, and Powers.



At the first glance, Averages (also called Arithmetic Mean) seem easy, and they actually are easy and have only one formula:  Average = .

The ways GMAT manages to make things difficult are quite limited. Here are the most common mistakes with averages:

Not understanding what average entails (if you catch yourself stuck on averages problem, write out the average formula, things will get clear)
The sum changes in the course of the problem
The number of values varies and is not clear stated
As always, rusty calculations


We have encountered consecutive numbers and their sum/average in our last session, so you should already know that they are easy to operate and that the average of a set of consecutive numbers is the middle one if the set id odd or the average of the middle two if the set is even.



Once again, we believe that to remember something well, a person needs to make a mistake or at least ponder on something for an extensive period of time, so we give what we believe are good problems first and then provide our explanations and shortcuts for them. This is done to find your weaknesses and try to deal with them. If making too many mistakes is depressing for you (most of the problems we give are above average difficulty), you can read the explanations before doing the problem, but our fear is that you will not remember anything if you don’t commit an error.



To begin, please, do #3 and #9 on page 37, then move on to the explanations.



#3: here we have 5 consecutive numbers, so the average will be the 3rd one, remember?

#9: here you need to learn how to operate with negative number averages. With averages the sign is counted just as well as anywhere, so you will need to subtract if it is a negative number.



pp.40-41



Please, do # 8, then move on to explanations



8: There is a tendency in this question to find the numbers and then add them up, yet by definition of the average, you can find the sum if you have the average and number of values: 12 x 5 = 60. The fact that the numbers are even and consecutive is not of much interest to us.



Please, do #11, then move on to explanations



11: Not a hard problem, but sometimes it confuses. Add n’s and integers separately and the average you’ll get is: n +  = n +1.5.

Please, do # 13, then move on to explanations



13: yet another calculations problem. The trick here is not to get bogged down wit the parameters: 6 x 7.5 = 45 (total sum) 4 x 2 = 8 (sum of 4 numbers) 45 – 8 = 37 (the sum of the other 2 numbers). Average = 18.5



Please do #15 for practice



Please do #17, then move on to explanations



17: the trick here is not to calculate full numbers but take the average, 17.50 as the zero. Thus, we can rewrite the table in the following way:

                        .08

                        .013

-.08

.02

                         x

Total should be 0, since we have assumed that the average is 0.

Therefore: 0.15 + x = 0, x = -0.15

Friday closing price is $17.50 – $0.15 = $17.35









The last problem that we will do on averages will be one we have found on the Princeton Review Forum (the author did not indicate the source):

The average of 3 different positive numbers is 100 and the largest of these 3 integers is 120, what is the least possible value of the smallest of these 3 integers?

a)1

b)10

c)61

d)71

e)80.





Explanations:

You were lucky to solve the problem if you took a moment to read it carefully.

The average is 100, thus the sum is 300. The larges is 120, therefore the sum of the 2 remaining is 180. The trick was that the larges I 120, therefore, the value of the middle number cannot exceed 119. Thus, the smallest possible value for the third number is 61.



RATIOS



Please read what Kaplan has to say about the Ratios first.





Ratios are a form of expressing a numerical relationship between two values. Ratio can be written with a colon 3:5 or as a fraction . Kaplan recommends using the second method because it is better to calculate that way, but there is one very important notice: if you convert from part : part to part : whole ratio, you will need to do a little bit more than just put one on top and the other beneath it and separate them with a line. For example if we know that the ratio of orange juice to apple juice in the refrigerator is 3:5, then we can say that  of the juice is orange and  is apple. The denominator of a fraction ratio needs to be the sum of the ratio quotients. We will meet an example dealing with this. Be alert when converting.



The most common mistakes with ratios:

Ratios need very little information to get an answer. Therefore, very often harder Data Sufficiency questions ask about ratios. We will note to you a few of these.
When a question asks to find a ratio, read carefully what exactly it wants X to Z or Z to X because the test takers will make sure to include both answers.
Also sometimes in the course of the problem, one of the sides of a ratio changes and that needs to be taken into consideration. We will give an example of this further down.
Don’t try to do complicated things; use picking numbers to substitute variables or complex numbers. Sometimes ratios get very, very confusing. We will fully cover picking numbers and other techniques when we get to word problems.
Picking the reverse ratio (reciprocal).






Do #10, then move on to explanations





#10: this is an exercise that will help you evaluate your ability to deal with fractions and compare them





Do #12, then move on to the explanation





#12: The best way to solve this problem is to use a proportion - the rule of the cross that originated in Chemistry See the example below for illustration:

------------------ 30

1 -------------------- X?



Now, using the rule of cross, we get an equation: X =  = 600. Do you see the cross relationship? Now, as you see, there is 600 in the answer choices, but it is a trap for the impatient. 600 is the total number of the contestants, and the number we are looking for is 600-30=570.



Please, do #15, then move on to the explanation









#15: This problem illustrates the rule of the part:part and part:whole ratio. If you had problems getting an answer, try to ask yourself, what is the minimal number of children in the class? 8, so all the possible combinations of children in the class have to be divisible by or multiples of 8.





Please, do #18, then move on to the explanation





#18: This is a rate-related problem. It is harder than most of what we have met as of now, so please, take 5 additional seconds to study it. Immediately from the problem we know that Bob finished the first half of the exam faster than the second, so we can discard D and E. Now, lets try to do some backsolving (we will concentrate on this technique a bit later).  Backsolving is when we try to solve a problem starting from the answer choices, taking it one by one and checking which will work. Usually, one should start with answer choice C because as a rule, the answer choices are lined up in ascending order, so if you hit the wrong one, you will know whether you should move up the line or down. In our case, we eliminated two answer choices, so we can start with the middle of what we have left: 24.

Let’s set up a cross relationship: It takes Bob 2/3 of the second half to do the first, so X is the second half and 2/3x is the first. Now, plug in the answer choice where it says first half.



----------------- first half

1x ------------------- ?



? =  = = 36; Now, we need to make sure that the conditions specified in the question are met: the total time of the test was 1 hour, 60 minutes; 24 + 36 = 60. this is the correct answer choice. We did not have to use X in our equation, we could have just said that it is 2/3 and 1, as percent values, but to keep the math straight, we kept it, otherwise, it is useless.



You can read through Kaplan’s notes to see how they approach this question and it may seem easier to you. In the beginning, it may be a little difficult for you to see the benefit of using backsolving, and it is not surprising, it is a new and unusual for many of us way of getting the answer, but as you practice more, you will see all the uses of backsolving, but we will cover that in full when get to the word problems.





Please, do #22, then move on to the explanation





#22: The reason we have commented this problem is to make sure you know how to equate multiple rations to each other. The possible mistake here was getting a reverse answer: 4:3 instead of 3:4.







Please, do #24, then move on to the explanation





#24: Your success with this problem will depend on your mood and attentiveness. If you took an extra 5 seconds just to stare at the problem, you may have seen the key to the problem. If you did not encounter any difficulties in solving the problem, congratulations, you an exception. The solution is 3 lines, yet there is a difficulty in getting to it, primarily because the test writer chose to compare fractions; this is one way to make a problem more difficult – use cumbersome numbers or fractions.

To solve the problem, you need to switch &frac12; and 1/3 to a whole – 1, and arrive to the following: White mice - of all the mice and grey mice are of the total. Another line and you will get the answer: 3:4.





The last ratio problem was submitted by a club member:

If 5x = 6y and xy ≠ 0, what is the ratio of    to  ?

   (A) 1

   (B)
   (C)
   (D)
   (E)










Explanation:

First, find the ratio of x to y; it will be , we get that from 5x=6y.

The easiest here would be to pick numbers. This will keep the brain in one piece. Let’s say x=12 and y=10; this fits our condition: 5x12=6x10.

Now, let’s combine the ratio and the values we have for x and y:

This is the final answer.



The other way is to solve using the orthodox math way:
First, find the ratio of x to y; it will be , we get that from 5x=6y.

Now, find 1/5x to 1/6y,







The hardest part in the formula above was to realize that we need to divide 6 by 1/6, not multiply it! The reason why we divide is given below, but if we pick numbers, we don’t have to come up with reasons, we just solve, get the answer, and keep moving to harder questions.
Why we need to divide by 1/6 and 1/5:

If we had just  to 1y, we would get 6/25, not 30/5, because now we would need 25 or 5 times as many X’s to fill for each of the Y’s. On the other hand, if X was increased 5 times, we would need only 1X for each of the Y’s (from our original 5x = 6y, we would get x = 6y), so x:y = 6:1. In ratios, when you increase one of the parts, you divide, not multiply because it is a ratio – the smaller, the better.

This is very confusing and the original strategy of picking numbers PAYS OFF! Nothing proves the solution better than real numbers.











Percents:



It is important not waste your time on the Math section on calculations; you need to spend as little time on dividing, multiplying, and adding as possible. Therefore, you need to remember some of the common fractions, powers, square roots, and similar helpful things. They will save you time and prevent from careless errors.

Let’s do a quiz before we move on: you may not come back to a previous answer as you have passed it nor look up a number because some of the values are checked for both fraction and percentage. You have 5 minutes.







=



=



=



=



=



=



=





75% =



20%=



16 % =



83 % =



87&frac12; % =



32 =

33 =

34 =

35 =



12, 24, 36….                                       120



15, 30…                                              120



8, 16, 24, 32, 40, 48, 56, 64, 72…



…………………                                160





25 =



26 =



27 =



43 =



44 =  



54 =



53 =



=



=









In this section try to do #15, 16, 19, 20, 21, 25











#16: this is a good example to use the rule of the cross to balance the equations.





#19: This is an early word problem, an easy one. We will cover solutions and similar things later, but as of now, all you need to do is calculate two proportions: first find out how much pure alcohol is in the original mixture and then calculate it for the new mixture.



#20 very easy





#21: Picking number problem, take 100 as the original: 100 --> 80 + 25% = 100





#25: This is a nice one. Not hard but can cause doubts; $80+25% = $100, now divide 100 by 8 crates = $12.50



















Powers and Roots



There are three major things tested on the GMAT about powers and roots:



For the math and physics majors, = 4 only; the negative root is not used in arithmetic
= ; don’t add/subtract roots with different bases; you can only multiply and divide them: A power of a number is a collection of factors:
214= 21x21x21x21 = 74x34 = 7x7x7x7x3x3x3x3 = 194,481




Consider the following example:



Which of the following is(are) a multiple of 54 x 143 x 65

I. 52 x 142 x 64

II. 35 x 212 x 512

III. 244 x 253

IV. 154 x 28















Your answers should be: I, II, IV



Please, do #5, then move on to the explanation







#5: this tests how easy it is to temp you to subtract powers when you should not. The equation should split into the following:


Please, do #8, then move on to the explanation



#8: This is a attempt to check your ability to read scientific expression of powers



Please, do #12, then move on to the explanation



#12: This is a sneak preview of word problems. The difficulty is to translate the words. If sometimes you can’t understand the problem, try reading it backwards or start solving it backwards; what is a square root of 16?



Please, do #13, then move on to the explanation



This is very similar to our problem, use rule #3 from this page.



Please, do #16, then move on to the explanation



#16: another attempt to check your skills to add/subtract square roots.

liushiyou

DS: enough or not enough



General info:

             Data sufficiency has a unique question format developed especially for the GMAT. Nobody knows why data sufficiency and not data deficiency or something else; perhaps, cause ETS wanted to check decision making skills or ability to act promptly with unfamiliar questions. We don’t know.

             What we know, however, are a few traps that DS poses to a test taker. You will learn most of them when you practice, but we feel we need to warn you about them:



Data Sufficiency Traps



The way answer choices work is the first trap of DS. They seem clear at first, but that’s only the tip of the iceberg. What often confuses people is when enough or not. Thus, an answer choice is enough only and only if using the information provided in it, you can get an exact numerical value such 4, , or similar. However, it is not enough to say that b=a, or to have two values (for example if you have x2 =16, then x is 4 and -4) this is not sufficient, there must be only one value. When one of the statements produces several possible answers (such as 9 and 5 or 3 and -3), the result is undefined.





There are several theories whether the two statements given with the DS question should result in the same numerical answer if either is sufficient. This controversy is illustrated below:



Example 1. Triangle ABC has one angle equal to 90 degrees and AB equal to 5 inches, what is the area of ABC?

(1)  BC =  4

(2)  AC = 12



Apparently both answer choices are sufficient to answer the question, but in two cases, the numeric answer differs. In the first case we get a 3-4-5 triangle and in the second 5-12-13, thus area of ABC1 is 6 and ABC2 is 30. It is still unknown whether this is possible under ETS’s regulations or not. I know for sure that some companies ignore this rule, such as Princeton for example. If we are able to say that ETS does not support the idea of getting two different numeric answers on the same DS question, it will become easier to eliminate some answer choices. Please, respond if you encounter an example of this sort.







To solve or not to solve? Every test taker moves through DS cycles. First, when one encounters GMAT, DS seem to be fairly simple and easy to solve. Then, we realize that we don’t need to solve and DS becomes the easiest thing in the world. The third step happens when Problem Solving score goes up, and DS stays the same or falls down. Then we get back to solving DS, so the pattern goes like this: solve >> not solve >> solve.

There is a general belief that one should not solve DS since all it asks is enough or not enough to answer. In fact, many textbooks tell not to solve.  However, under “do not solve” is implied to do the job of compiling an equation, get it to the final form and stop only when all you have to do is calculations. To get the majority of DS right, you need to solve, but you don’t need to calculate. Don’t try to solve in your head, use paper, don’t stare at the problem trying to hypnotize it. See example below:

Trick 1. What is the volume of a box with dimensions a, b, and c?

                                                      i.      a =
                                                    ii.      b = 2, c = 4



At the first glance both statements are needed to answer the question, but when actually attempted, the problem appears solvable. If you did not compile an equation, you did not see that the volume of a body is ; the first statement is enough to answer. Often, ETS will use fractions and they will cancel out, providing you with an answer; always make sure you write down the formula/equation.



Trick 2. What is the value of x?



1.     x + 2y = 6

2.     4y + 2x = 12



If you solve without a careful look, you will think that since there are two equations and two variables, you will find the solution. Nope. Both statements are masked and appear identical. According to our members, such problems are very common on the real GMAT.

!!! A note to the skeptical test takes: I had two of these tricks appear on my REAL GMAT. It was funny how ridiculously easy they were.





Trick 3.



How many miles is it from George’s house to the groceries store?

                                     1.      If George did not visit a gas station on his way to the groceries store, he would have driven 4 miles less.

                                     2.      The gas station is 8 miles from George’s house





Trick 3.1.



How many children are there in Nancy’s class?

Yesterday there were 14 kids in the class besides Nancy
Usually there are 2 kinds who are sick and not present in the class




Do not assume anything on data sufficiency. Some questions ask you to give them a little of something – Don’t. For question 3, we don’t know how George’s house is situated on the map related to the gas station or the groceries store; it can be a straight line or a triangle, therefore we don’t have enough information to give an answer, E.

In the Example 4, we cannot say how many children are in Nancy’s class since we may not assume that yesterday there was a normal situation and only 2 children were not present. It may have been a big School Play day, so even the sick children came to see it. We don’t know, therefore E.

For more questions like this, see #15, p.327 in Kaplan’s workbook.



Trick 4



Is the sum of six consecutive integers even?

1.     The first integer is odd

2.     The average of six integers is odd

Trick 4.1



Does x equal 3?

x2 = 9
x minus three is negative 6




Watch out for Yes/No data sufficiency questions; they are the hardest and the most misleading. Example 5: The answer to this one is D. (1) Statement says that the sum of the integers is odd, which gives a NO answer to our question, but is SUFFICIENT to give an answer, therefore sufficient. (2) Says that the sum is odd, which is sufficient to give a Yes answer. In both cases it was sufficient to answer the question, except in the first case, the answer was NO and in the other, it was YES. Make sure you don’t confuse No with insufficient because they are not related here.

Example 6: the first statement is not sufficient since a square of x can equal either positive or negative 3, therefore it is not enough. The second statement, however, provides us with a value for x, negative 3. Therefore it is sufficient to answer the question.

The answer is B.







Do not combine answer choices. What ETS often does on harder DS questions, it gives the first piece of info as insufficient and the second being sufficient by itself. Yet, naturally, as one moves on during the test rush, to the second statement that nicely adds to the first and makes it sufficient to answer, he/she misses the possibility that the second choice can be sufficient by itself. Do not eliminate this possibility.



When you solve a medium/hard DS question, it is good to play a game; try to find the trick in the puzzle. If you looked through the both pieces of info and it seems both are needed to be sufficient, try finding a trick why only one could be sufficient or vise versa. This technique pays of with hard DS questions that often have a more complex solution than seems from the first glance.



Make an analysis of your mistakes and see what DS questions cause the most problems.



Make sure you don’t confuse D and C and know the answer choices by heart.



Review trap points on DS:

Always write down the whole formula/equation
Watch out for masked statements; take extra time to check the solution, not just think that you will be able to answer a question since there are two statements; they may cancel out
Do not assume anything on the DS; if you think the author is pushing too much, you are probably right
Yes/No questions – know them
Do not combine answer choices
Play a game with yourself and try to prove yourself wrong; helps with difficult DS
Make sure you know the answer choices and don’t confuse C and D
Make analysis of your errors




Homework: Kaplan Math Workbook 2nd/3rd edition:

             pp. 323, ## 12, 14, 16, 18, 19, 20, 22, 24

pp. 326, ## 3, 7, 9, 10, 15, 18, 21, 24





Review of Tricks:

Make sure you write down the equation for the DS question
Make sure both of the statements are not identical
Do not assume anything: if it is a little fishy, probably it is wrong
Be familiar with Yes/No questions

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SYNOPSIS

The goal of this lesson is to get you accustomed to and comfortable with combinatorics problems on the GMAT. We have incorporated the hardest problems from GMAT that we could find, so this collection should make your experience on the final GMAT less stressful because the real problems are not as difficult. Another goal is to provide you with techniques to solve all combinations problems in time you have – 2-3 minutes.  

Just to start, combinatorics problems are believed to be some of the hardest ones on the GMAT - the elite division of questions that yields only to probability. Yet, even combinatorics questions have several levels of difficulty, making them available through the test:

Easy – one action problems with no complications or limits imposed

Medium – two action problems or one simple limit

Hard – either include extensive calculations or several limits or even both

Still, the hardest thing with problems is to identify the solution method. The test writers, on their part, will make sure to include several almost right, but still wrong answers in order to make sure you pick a wrong answer should you mess up even a bit. Here is the bottom line: the point is to figure out how to solve, the rest is arithmetic. Below you will find several scenarios to approach combinatorics problems that, hopefully, will help you easily identify the right solution for each given problem that you meet on your test day, and at the end we have listed an approximation of a systematic approach to combinatorics problems.

One word on guessing: personally, we don’t like guessing. It is just as hard to guess right as to solve right, so we usually choose solving.

Before we move on, it is recommended that you know by heart two problems from the Kaplan Gmat Guide with CD 2003 or 2002 edition. They are located at the end of the book in the math review section. You have to know them and understand them before continuing.

FORMULAE

There are several rules and formulae to find the number of combinations:

1.     Multiplication rule – when the number of available spaces for combinations matches the number of elements (e.g. we have 5 people for 5 positions – the result will equal to 5x4x3x2x1 or 5!, which is 120.) or when we have several groups within one set; we will need to multiply the results for groups to find the total. (see example 1, 2, and 3).

2.     Addition rule – applies in the more complex situations, primarily when we have a variable number of positions for combinations, and thus have to calculate several different number of combinations for each given number of positions (e.g. we have $1, $5, $10, $20, $50 bills what is the number of sums we can come up with?). (see example 7).

3.     Permutations formula: P(n, k) =  – unique members (example 7, 8).

4.     Combinations formula: C(n, k) =  – non unique members (example 9, 10, 11).



LOGIC

The first kind of the problems that we will consider is not what we usually imply under standard combinatorics problems, but still very interesting. Consider the following example from GMAT Plus:

EXAMPLE 1. Of the science books in a certain supply room, 50 are on botany, 65 are on zoology, 90 are on physics. 50 are on geology, and 110 are on chemistry. If science books are removed randomly from the supply room, how many must be removed to ensure that 80 of the books removed are on the same science?

(A) 81

(B) 159

(C) 166

(D) 285

(E) 324

Such problems invite us to provide a foolproof solution that would work in 100% of the cases. Thus, this means we will need to find a solution for the worst case. In our example, we can say that there is a devil kin sitting in the certain supply room and she hands us the books so that each time we get a new book. So after about 250, we will have removed all of the botany and geology books as well as 50 on zoology, 50 on physics, and 50 on chemistry, but we still don’t have 80 of the same kind. We have 15 zoology, 40 physics, and 60 chemistry books left. So, after another 45 books, we will have removed all zoology books, 15 physics, and 15 chemistry books. Still not enough. We have at most 65 books of one kind. Let’s remove 14 of each kind of the books we have left. So, after removing 14 of physics and chemistry books we will have a total of 323 books removed and we have 5 stacks that are at most 79 books. Now, however, we need to remove only one book because we will know that we have only two kinds of books left (either chemistry or physics) and any of them will give us a set of 50. Of course in reality it would not be that bad, but we have to take the worst situation.

It would be easier, however, if we just took a look at the number of the books that could be left in the room. We know that we need 80 of one kind, so we for sure know that those would not be botany, zoology, or geology books because there too little of them. If we need to guarantee that 80 are removed, we would have 10 physics and 30 chemistry left. However, since we need only 80 of one kind, we can say that either 11 physics and 30 chemistry or 10 physics and 31 chemistry. The trick is to spot see that we need only one stack to be 80, not all. Then, we could subtract 31 from the total number of books (365-41=324).

There are not that many of these remove/remained problems, but they are fun.

Let me know if you come across any, I will include them into the practice section.

LOGIC AND SIMPLE RULES

Usually, on the GMAT you can solve and find the number of combinations or permutations without using any formula or even writing out the combinations, but just by applying your logic. For practice, consider the following example from the Princeton Review:

EXAMPLE 2. Katie must place five stuffed animals--a duck, a goose, a panda, a turtle and a swan in a row in the display window of a toy store. How many different displays can she make if the duck and the goose must be either first or last?

A.      120

B.      60

C.     24

D.     12

E.      6

Here is the explanation by the same company. It works but it is not optimal.

Let’s simplify things. We know either the duck or goose is first or last. Ignore those for the moment. How many ways can we arrange the panda, turtle, and swan in the positions 2, 3 and 4? Be systematic, and list them out: PTS, PST, TPS, TSP, SPT, STP. 6 ways. So if the duck is first and goose is last, there are 6 ways the whole arrangement can work; if the goose is first and the duck is last, that makes 6 more for a total of 12. The answer is (D).

This is an average difficulty problem – we have 5 spots and 5 animals to fill them, so we will just need to run the factorial, but even without knowing that, we can solve the problem. Let’s use the brain for a second. I don’t think you need to write out all the combinations as Princeton Review suggests because it is a pain (yet sometimes it helps when you are not sure about a solution).

Anyway, just think logically: there are 3 spots available for swapping: 2, 3, and 4 (the first and the fifth one are occupied by the duck and goose). So, for the first of the three spots, we have 3 animals; for the second – 2, and for the third only one. This means that for every of the three animals in the spot #1, we have 2 animals in spot #2, and one in spot #3. Therefore, to get the total number of combinations we need to multiply 3x2x1. This falls under the multiplication rule of combinatorics. Since the duck and goose provide us with two options, again, according to the multiplication rule, we need to multiply our final result by 2 to get 12.

Try solving this problem on your own.

EXAMPLE 3. The president of a country and 4 other dignitaries are scheduled to sit in a row on the 5 chairs represented above. If the president must sit in the center chair, how many different seating arrangements are possible for the 5 people?



(A) 4

(B) 5

(C) 20

(D) 24

(E) 120

Let’s consider another example, this time from the official guide:

EXAMPLE 4. In how many arrangements can a teacher seat 3 girls and 3 boys in a row of 6 if the boys are to have the first, third, and fifth seats?

A.      6

B.      9

C.     12

D.     36

E.      720

I don’t know what approach took ETC in this problem, but the most optimal again would be just to straighten things out and then apply logic. This is clearly a unique spots/members situation because otherwise we would have only one combination (girls 2, 4, 6 and boys 1, 3, 5), but it is not in the answers and truly would be stupid to ask.

So let’s devise a plan.

Think for a bit so that you would not waste time doing useless and incorrect calculations.

First of all, we have a limitation on our group that defines the number of combinations for odd and even positions. Usually for problems like this, there are two methods of solution: find the total number of combinations and then subtract the ones that fall under a limitation or count all the possible combinations that respect the limitation and then multiply them (usually the preferable approach). In our case, however, the limitation is fairly large and it will be useless counting the total number of combinations and then subtract a very complicated condition.

So, let’s count the possible number of combinations under our limitation. We know that there are 3 seats for 3 boys, so similarly to the previous example, we get 3! or 3x2=6. The number of girls is the same, which gives us two groups within one set; to find the total number of combinations, we need to multiply the two results for two groups 6x6=36. (because for each combination of girls there are 6 arrangements of boys and vice versa).

If you were entirely at a loss with this question, you could have guessed. First of all, 720 seems just a way too much; actually it would be the correct answer if we did not have limitation, but with the limitation it looks too much, so we are down to 36, 12, and 9. 6 is too little; you could have figured that out too. None of the answers is the product of a factorial. Actually, it is very useful to know the factorial products: 2!=2, 3!=6, 4!=24, 5!=120, 6!=720. Anyway, it would be hard to guess cause we would have 12 (6+6) or 36 (6x6).

Personally I don’t like guessing. I think it is much harder to guess than to solve, so why not do the easiest thing and just solve?

Perhaps a pure multiplication rule will be the following that came from an unknown source:

EXAMPLE 5. If a customer makes exactly 1 selection from each of the 5 categories listed below, what is the greatest number of different ice cream sundaes that a customer can create?

12 ice cream flavors

10 kinds of candy

8 liquid toppings

5 kinds of nuts

With or without whip cream.

(A)9600

(B)4800

(C)2400

(D)800

(E)400

According to the problem, the customer must make 1 selection out of each and it can be only one (if it were different it would much more complex). Basically, we have 5 different ingredients, so after picking one of 12 ice cream flavors, the person has 10 choices of candy, and then for each of 10 choices of candy, he/she has 8 options for toppings, and for each of those 8 toppings – 5 kinds of nuts. Moreover, the person will get either with or without whipped cream. Obviously, this is a multiplication case by all means: the number of positions to fill with combinations equals the number of different ingredients – 5, (in such cases, we can’t use a permutation or combination formulae). Here is what we get after multiplying: 12x10x8x5x2 = 9,600 (don’t forget the whipped cream). This is a one-action problem.

Sometimes, the math may not be very easy. Consider the following example from the Schaum’s Intro to Statistics:

liushiyou

EXAMPLE 6. From a class consisting of 12 computer science majors, 10 mathematics majors, and 9 statistics majors, a committee of 4 computer science majors, 4 mathematics majors, and 3 statistics majors is to be formed. How many distinct committees are there?

To solve the problem, we will need to find the 3 constituting elements – 4 computer majors, 4 math majors, and 3 statistics majors and then, since they are elements of one set – multiply them. Jumping a little ahead, we will use the combinations formula and will get the following results for each group: 495, 210, and 84. Now, since for each computer science major there are a good number of math and stats majors, we multiply. The result we will get is 495x210x84= 8,731,800.
PERMUTATIONS AND COMBINATIONS

Besides the two problems explained in Kaplan, that you should know by heart, about the 3 out of 5 runners and 3 out of 8 committee members, there are few variations.

For example, let’s take a Permutation problem from high school textbook:

EXAMPLE 7. Given a selected committee of 8, in how many ways, can the members of the committee divide the responsibilities of a president, vice president, and secretary?

The solution comes both from the permutations formula and from logic. (Permutations, not combinations formula is used because the order matters since the positions are unique).

Scenario 1 – Formula:

P(8,3)=8!/5!=8x7x6=336

Scenario 2 – Logic:

For the President’s position we have 8 people and for the VP’s – 7, and 6 left for the Secretarial position. Therefore, the total number of permutations equals to 8x7x6=336. Since the position of a person matters (P - Alex, VP- Jen, and S -Sindy is different from Jen, Sindy, Alex), we do not need to divide by anything.

Consider the following more advanced problem from the same textbook (it has a trick to it):

EXAMPLE 8.  How many four-digit numbers can you form using ten numbers

(0, 1, 2, 3, 4, 5, 6, 7, 8, 9) if the numbers can be used only once?

It seems to be easy, we take 10x9x8x7 since we have 4 positions, and get 5040, but there is a trick to this problem because 5040 will include numbers that start with a 0, and in reality we don’t have those. So, we need to subtract the number of the fake 4-digit numbers. There are two ways to arrive to that: either subtract 1/10 out of 5040 since all 10 digits are equally represented (5040-10% = 4536) or use a Permutation formula: P(9,3) = 9x8x7 = 504.

5040-504 = 4536.

However, the problems get more complex by requiring a test taker to make more than one action, so, often, we will need to use addition or multiplication rule along with combination/permutation. Let’s consider an interesting problem:

EXMAPLE 9. A person has the following bills: $1, $5, $10, $20, $50. How many unique sums can one form using any number of these bills only once?

First of all, let’s reason (reasoning is always good!). There are 5 different bills, and we have to make unique sums of money out of them. Basically, as we figure out from the text, we can use either 1 or all 5 bills for our amounts. Good news is that none of the possible combinations seems to overlap, meaning that there is no way to come up with $30, except by taking a $10 and $20 bills. The bad news is that we will need to calculate the possible sums when taking 1, 2, 3, 4, and 5 bills. Again, relying on logic and common sense, taking one bill at a time, we will get 5 unique sums that will equal to the nominal value of each bill. Taking 5 bills altogether, we will get one amount - the max, that equals to $86. Now we need to find the number of sums when taking 2, 3, and 4 bills. Using the combination formula, we will get C(5, 2)=5!/2!3! = 10, for C(5, 3) = 5!/3!2!=10, and for C(5, 4)=5!/4!=5

Total will be: 5+10+10+5+1=31 possible sums.

(You can check by writing all of them out).

Moving towards hard problems, let’s consider a little more complex situation offered again by Princeton:

EXAMPLE 10 A three-person committee must be chosen from a group of 7 professors and 10 graduate students. If at least one of the people on the committee must be a professor, how many different groups of people could be chosen for the committee?

A.      70

B.      560

C.     630

D.     1,260

E.      1,980

This is a hard combinatorics problem that again requires several actions. At first it may be confusing because it has a weird condition that at least one professor needs to be on the committee. One way to solve would be to find a total without a limit and then subtract the number of situations when there are no professors on the team. The other option will be to calculate the number of combinations with 1 professor and 2 students, 2 professors and 1 student and 3 professors.

Scenario 1. We get C(17, 3) =  = 680 total possible committees. Now, the number of teams with students only is, using the same formula C(10,3) =  = 120.

Now, 680-120=560.

Scenario 2. For a committee with 1 professor and 2 students, we will get 7xC(10,2) =  = 45x7=315. (we multiply by 7 because  For 2 professors and 1 student, we will get C(7, 2)x10 = 21x10=210, and for 3 professors, we will get C(7, 3)=35. Adding up the combinations we will get: 315+210+35=560.

Princeton Review suggests using Scenario 2 because it is supposedly simpler to understand, however, taking into consideration that you can make a mistake in the endless calculations and that it requires 3 complex operations in contrast to 2 in the first case, it has a weaker standing. In any case, both ways get the correct answer, but you need to choose the one that appeals more to you—the one easier to use.

Here is another hard problem with some restrictions; again from an unknown source:

EXAMPLE 11. There are 11 top managers that need to form a decision group. How many ways are there to form a group of 5 if the President and Vice President are not to serve on the same team?

Again our options are to solve the problem either to find the total number of committees and then subtracting the number of groups that VP and P would end up together or to find the number of groups with VP and P and None. However, the second method will be lengthy and unnecessary complicated, so the best solution is to find the total and subtract all the cases that fall under the limiting condition.

Here is the best solution:

11!/(6!x5!) = 11x10x9x8x7/5x4x3x2x1 = 11x2x3x7=11x42 = 462 (since the order does not matter, we use the combinations formula).

This means that the total possible number of teams of 5 out of the pool of 11 people is 462, but we have a limiting condition imposed that says that two members of the 11 cannot be on the same team. Therefore, we need to subtract the number of teams where the Vice President and the President serve together. The number of the teams that VP and P would serve together on is perhaps the hardest thing in this problem. Anyway, the trick is to count on how many teams VP and P will be. To do this, we need to imagine the team, and the five chairs: let’s assume that P is chair #1 (since the order does not really matter), VP is #2, and the three other spots are available to the rest (9 total), so for the 3rd chair, we will have 9 managers, for the 4th – 8, and the 5th place will be offered only to the remaining 7. Therefore, the total teams that VP and P would meet is C(9,3)= 9x8x7/3x2=84. (again we divide because the order of the people showing up on the team does not matter).

Final step: 462-84=378

PRACTICE PROBLEMS

1.      There are 9 books on a shelf, 7 hard cover and 2 soft cover. How many different combinations exist in which you choose 4 books from the 9 and have at least one of them be a soft cover book?  (Ans. 91 )

There are 5 married couples and a group of three is to be formed out of them; how many arrangements are there if a husband and wife may not be in the same group? (Ans. 80)
3.      How many different signals can be transmitted by hoisting 3 red, 4 yellow and 2 blue flags on a pole, assuming that in transmitting a signal all nine flags are to be used? (Ans. 1260)

4.      From a group of 8 secretaries, select 3 persons for promotion. How many distinct selections are there? (Ans. 56)

5.      In a certain game, a large container is filled with red, yellow, green, and blue beads worth, respectively, 7, 5, 3, and 2 points each. A number of beads are then removed from the container. If the product of the point values of the removed beads is 147,000, how many red beads were removed? (Ans.…)


(A) 5   (B) 4  (C) 3 (D) 2 (E) 0

6.  There are between 100 and 110 cards in a collection of cards. If they are counted out 3 at a time, there are 2 left over, but if they are counted out 4 at a time, there is 1 left over. How many cards are in the collection? (Ans.…)

(A) 101  (B) 103  (C) 106  (D) 107  (E) 109

7.      The probability that it will rain in NYC on any given day in July is 30%. what is the probability that it will rain on exactly 3 days from July 5 to July 10 ?

There are 4 Fashion magazines and 4 Car magazines. Four magazines are drawn at random, what is the probability that all fashion magazines will be drawn?
a. 1  b. ?   c.   d.   e.  



STRATEGY

Read the problem carefully, trying to see both general and specific details, but do not let the numbers confuse you; try to see the whole picture first.
Choose the best approach for solving the problem: either take a logical approach by drawing the number of members, seats, etc or apply a formula.
If you can’t find a way to solve the problem: it seems to be too complex, try to associate it with one of the problems we did. (It is recommended that you memorize at least two problems that are given in the Kaplan math review section so that you would be able to reproduce their solution on paper in a hard moment of panic). Even the most difficult combinatorics GMAT problems are solved using the simple formulae, so look for a way to apply them. There should be one.
If you are completely at a loss, there is a good way to estimate/guess take the most and then estimate how much less the answer should be. Usually, you will get down to two answer choices and then you can try to “podognat’” your solution to the answers and see which solution makes more sense.
After you have solved the problem, go back to the question and make sure you answered it and make sure you followed all the conditions.
It is also recommended to do a fast check on the questions of such difficulty; try to use the other approach if applicable (formula if you used logic or logic if you used a formula) or just make sure your solution is reasonable. (e.g. you may come up with an answer that there are 150 combinations to sit 5 people into 3 seats, but, in fact, it does not make sense because it is too much since even as much as 5! equals only to 120.)

liushiyou

Note: This essay refers only to the Quant section of the GMAT.

It is not the world’s most organized essay on GMAT timing, but it for sure is the most comprehensive.



Introduction


Almost all the problems that are given on the math section of the GMAT are easy or, actually, very easy. What makes it hard is that there are 2 minutes per problem regardless of the kind and difficulty level. On the other hand, time scarcity adds has numerous negative effects such as stress, rush, fear, analysis paralysis, careless errors, just name it.



The not-so-good news: you will have to solve a 37 various math problems with progressing difficulty level in a short interval.

Still, you can manage these 75 minutes as you wish, deciding how to invest them, seeing for yourself what gives you the highest return on investment. For instance, you can be good in Geometry or Word Problems, and can save some time to deal with hard probability questions, yet if you hit an algebra question that will, most probably, based on your previous experience, take up 4 minutes of your time, you may need to make sure you actually can pay such price.



The good news: The second part of the Math section (the last 20 questions) will be just average difficulty problems with some word problems and DS, but in general nothing special or difficult. Usually, everybody finishes them in 1:35 each, so you will be able to spend a little more time on the first half of quant. section. There is a faulty stereotype that makes everybody think that the difficulty on the GMAT is always increasing if one answers correctly.







It is not so, however, when you make it past the first 15 questions, the difficulty level will settle down somewhere in between and the rest of the test you will be solving average problems that are not hard but time consuming. Here are the reasons why: (1) If the difficulty kept growing, test takers would kill themselves on problems after 20 and nobody would score more than 700; (2) ETS does not want to give away its hard questions that easily. Imagine that a GMAT tutor goes to the testing center and hits 60 on quant. In this case, he would have gotten a good number of hard problems that he/she could tell the students or use for his/her classes. ETS does not want this to happen because good hard questions are not that easy to come up with and as the result are expensive. ETS is a monopoly, so they set the profit maximizing price and then try to cut down on costs. So they save money on us, and instead of giving us more expensive hard questions, stuff in a number of time consuming questions that will require at least 1:40 minutes each to get the right answer. It is the same logic: if you are out of time, you will make a mistake just as well as you would have made it if it were a hard probability question. All they need to make sure is that it takes you at least 1:30 to solve it, and they have all those experimental questions to see if it does*. Here is a testimony of a test taker:


I have spent a lot of time in the first few questions. DON’T DO THAT!! At the end i knew that the problems could be solved but since it involved lots of calculations (and hence a lot of time ) I haad to rush through it. most of the questions after 20 were quite easy.

-- Anonymous, March 02





I still have not made any conclusions as to the last few questions on the GMAT (whether they are weighted the same or if there are worth more points – to penalize those who don’t finish on time). However, I will be glad to hear your suggestions and theories about this on the forum. If you have any theory, just start a new discussion thread and I will join you right away.



Simple question: What does the GMAT test?
It is not high school math or argumentative skills – these are decoys. What’s evaluated is logic, time management, ability to deal with unfamiliar and non-standard questions, perseverance, your ability to focus, and of course concentration and nerve.

In other case, why would an MBA test evaluate your knowledge of high school math?





The Clock


Actually, the clock seems to have a very significant influence on people. One of the usual reactions to that is hurrying as much as possible to finish the test. The stereotype is that there is not a single second to spare. A rush starts: as soon as one is done answering a question, he/she moves to another, and so forth. It is interesting to see how the computer starts devouring the test taker, forcing him/her to give up their freedom and power to choose and control the environment. The GMAT has them. It rushes the test takers to the end of the test, instilling in them a false impression that all they can do is click and be passive. This inhibits efficiency and logic: when such test taker sees an impossible question, he/she does not use logic to figure out that it would better off to leave it alone and save some time for another question or that one of the questions needs to be solved in a slower pace and more time needs to be spent on checking the answers. As the result, more of the precious time is wasted, and even a greater rush starts.

Do NOT let the test intimidate you and force you to give up your usual habits of test taking that you have developed during your practice at home. I realize that you will feel more relaxed and comfortable at home than at the center, but you got to stay calm. Do not concentrate on saving time or making it in time, but on how to solve the problem because if you constantly think about the time, you will not be very attentive to the questions


This idea can very well be illustrated by a person who takes a paper test and times himself/herself with his/her wrist watch. They make it on time and get a nice score, but when they take a very similar test on the computer, the scores is much lower and they don’t make it on time. The difference is that there is a time limit on the computer test and, unlike on the paper test where they could get an extra minute or two if needed, they are more time conscious on the computer tests. Primarily because they no longer control the time, stress starts to build up and, gradually, more and more time is devoted to rushing oneself. The rush, on the other hand, throws in a handful of errors.


Controlling Time


GMAT tests your time management quite closely; of course, the better you are in math, the faster you can do the problems, but being good in math will not deal with all the timing problems. For example, even if you finish a section early, you are not managing your time wisely. Unless you got a perfect score of 60 for the math section, you could probably use some extra time to double check your answers, assuming you checked them once.

It is important to know how you are doing on time in general so that you know what you can afford and what not. Below are some of the ways to learn or improve your timing skills.



Perhaps you know the method of dividing the math part into three sections:

0--12, 13--25, 26--37

75 mins left -- 0 questions done;

50 mins left -- 12 questions done;

25 mins left -- 25 questions done;

0.5 min left -- 37 questions done.







Personally, I have tried to follow it on my test at the test center, but when I had 33 minutes left, I realized that I still had a half of the questions to go; I was like on #15 or 18. I did not seem to have many doubtful questions and had enough time to check my answers in most of the problems, but got a bit nervous when I saw my time (Logically, I was not making it, but practically I did fine). You NEED to be prepared that the timing will not go as you wish; you may get a hard start or a few questions in a row that will get you down. What you need to be is flexible and adjust to the test, just as the test tries to adjust to you. Except you are a human and you should be able to do that much better. I have finished with 30 seconds still left over and 49 on the math section.



One thing that made my timing on the real GMAT different from that of the practice tests was checking my answers. I was too lazy to do that on the practice tests cause I was either tired coming from work or had the whole working day ahead of me. Checking your answers seems to be a nice way to prevent quite a few spontaneous errors - the ones ETS eggs one to make naturally. Therefore, a few weeks before the test, try to start checking your answers even on practice tests or otherwise, finish 2-3 minutes short to compensate for that on the test day.





Pressure
The biggest enemy of time management is pressure. Everybody who takes GMAT always has a thought in the back of their head that tells them that they are not going to make it. Do not rush. This does not mean that you should do everything you can to save time; what we imply here is that you should not let the test rush you. You need to be cold blooded and very attentive, not trying to read the problem as fast as possible, but reading it attentively with the goal of finding the answer and not saving time. Take 5 seconds to give a question another reading: check out the answer choices (how crazy they are, how far apart is the distribution, how precise the calculations need to be, etc.) This short pause will help you concentrate and at the same time relieve pressure of the time race.





Time Savers                                          top
       Try to use the timesaving techniques you’ve learnt in the Math Review -- Backsolving, Picking numbers, Units and Divisibility test. These time savers can also be used to CHECK the results and see whether the solution makes sense. You want to do that on the test day because nobody needs careless mistakes. Another way to save time is to study the common percent values, square roots, powers, and fractions as well as some additional geometry formulae, triangle shortcuts, and rules. This is covered in the Arithmetic 2 section in the Quant course.



What to do when you meet the dead end?              top
   

It is good NOT to panic when meeting the dead end. It is about just as helpful as diving and panicking under water: you immediately run of oxygen and instincts of self-preservation turn your brain off – not very helpful for a meaningful result. I know it is hard to control yourself when time is running out; I just has my Econ final, but, please, don't give up too early on any of the questions (spend at least 1.5 mins, and try different approaches: rewriting the equation, solving for a different variable, simplifying the problem). Some of them sound extremely bizarre and sometimes make NO sense, but often, a second reading reveals an easy answer. Some questions, such as counting, position, last digit, simply require you to do the math. This may not sound too intelligent, but sometimes ETS wants to see how fast you can count, multiply or divide. However, do not waste time on the puzzling questions either. LEARN how to identify ETS tricks fast so that you would know that if you did not figure something out after 2 minutes, perhaps it will not come to you even in 5.

liushiyou

There is one fallacy about the math section many people believe in: if you practice enough, you will meet all the possible math problems and the test won’t be difficult. There are two answers to this belief: one, 90% of the problems will be quite typical, but the remaining 10, usually the hardest, will be unique or unseen by you previously. This supposition gets proven by test takers almost every week. The second response is that it is easier to solve a problem on the test rather than keep solving 20 thousand problems for a year. Be prepared that some of the problems you encounter on your test will not by typical. This is one of the points of GMAT – to test your ability to act in unfamiliar situations and solve untypical problems using unstructured decision making. GMAT rewards people who use multiple approaches trying to solve a problem by applying various techniques and evaluating the result. Finally, if GMAT were a high school math exam, we would not be doing this website and you would not be reading it. It is more than just math.



       Some of the hardest problems you meet on the test day will be experimental questions, so do not worry if you had no idea how to solve a hard one, there is more than 50% chance that it was experimental. About 1/4 or 1/3 of the questions are experimental. These questions are not evaluated and are given just to measure their difficulty level for the future test inclusions. These crazy things are hidden throughout the test and there is no way to identify them. Sometimes they throw off track by being too easy and you start thinking that you have made too many mistakes, and got back to the very basic level of the 50th percentile. Do not worry, sometimes GMAT may ask you to do an addition question such as y=2x+8, y=1+ x. Not that anybody doubts your addition skills after solving that permutation or probability problem, it is just that ETS wants to see how you deal with problems of various difficulty thrown together. GMAT is not only a test of your knowledge of elementary math principles, but also a test of logic, time management, ability to deal with unfamiliar and non-standard questions, and of course a test of concentration and nerve.



             One more point about the dead ends, as we have learnt from a survey, people usually meet only one dead end on Math, and usually it hits around question 12 or so. After that, no one had any doubts or difficulties with questions. Thus, there is a good chance that you will face a very similar situation and only once will you meet an impossible question. (unless you guess it right and the next will be impossible).

Move ON!                                               top
Since you will meet your impossible question only once, it is crucial to advance as far as possible; sometimes it is worth it to spend some extra time, but make sure you don’t take more than 3.5 minutes to solve a problem. You need to analyze your chances for the correct answer: if you spent 3 minutes and the answer will involve complex calculations that you are unsure how to perform, you will not make it in the next 30 seconds, so you will need to move on. Sometimes people, after spending 3 minutes with no result, refuse to move on. They think that since they have invested 3 minutes into the question, they will be better off by investing a few more minutes and getting. They are still doing this after making 2-3 attempts to take the GMAT and getting 550.



You have heard this quote already:

I have spent a lot of time in the first few questions. DON’T DO THAT!! At the end i knew that the problems could be solved but since it involved lots of calculations (and hence a lot of time ) I haad to rush through it. most of the questions after 20 were quite easy.

-- Anonymous, March 02





One thing I can suggest to you guys is try to take an ideal test; when you spend no more time than you have - 2 minutes. I did take one, and what I did was guess when I was overtime on hard questions. My results that day, were the lowest of all and ever, but I felt the timing more deeply. On the other hand, I found that it is also helpful for timing to try to solve all the problems without letting them go and see where you get with that and compare your normal and “ideal” results. Good luck on the test.

Bryant, May 02



Using the GMAT official guide 10th edition I found I was getting all the questions correct but was taking far too long. So I used a stopwatch and recorded my time for each question individually. Although I made a point not to rush through the process of working through the question for the sake of getting a low time, I found that the simple fact of recording and being aware of the time it took for every question helped me drastically improve my speed. For verbal I got to the point where I was able to read a RC passage in about 3 minutes and answer each RC question in 30 seconds. For CR I took between :40 and 1:20 per question. SC was highly variable taking anywhere from :25 to 2:00. DS and PS usually were between :20 and 1:30. Of course, all these times were WAY better than they were when I started timing. You may be wondering to yourself "okay, I still don't get how just tracking the time it takes to do questions will help me do them faster", well, I'm sure it wasn't the only reason and I can't explain the process behind it, but shortly after I started timing the individual questions my times improved drastically. There is definately something to it. I think it's well worth the additional effort to time how long it takes you to do each question.

Joe800, April 02



Morale and Mood
       It is recommended that during your GMAT test you not think about school rankings and the schedules of classes that you will have in your MBA program. Try not to worry about your overall standing or your chances for a scholarship, but concentrate on questions. You should totally forget about the last question you worked on and not think about what will come up, rather do your work on the problem that you see on the screen. Another helpful thing is to turn your fear and emotions off for the test; just for 4 hours. Try to be a logical machine that has no personality because if emotions mattered for the GMAT, my former girlfriend would have easily scored 810.

A big, big point: you won't be able to tell the score until you see it. I have known people who made approximations and were wrong more than 50 points. Some guessed their score correctly, but that was because they took a powerprep test just before the real GMAT, and therefore knew their potential score. The nature of the GMAT is that it keeps you on the edge of you abilities, giving you questions that you can hardly answer; therefore, it will always feel as though you are falling (this is more true on verbal than on Math. Read what one of the old dogs has to say:



Remember the test really made me feel like a fool. i thought that i would get somewhere around 650 but got 710 . so keep ur cool and dont worry too much if u cant solve a couple of questions. just dont spend too much time on any one question. Also my test was at 12:30 so i warmed up in the morning by doing 10DS questions( beacuse i found that generally in the tests i consistently made a mistake in the first 5 q).

Anonymous, March 02



If you think that since you made too many mistakes, and now you need to concentrate more on the problems, then, sorry, your strategy is wrong! You need to concentrate all the time -- squeeze all the juices out of yourself. You should always work at the top of you abilities, using all your power, strength, and logic. If you want to be in the top school, you gotta be the top man. Stick the earplugs as far as you can, and let the rest of the world stop existing for a while. Everybody knows that GMAT is an important part of the application process and that it is the only thing left, but you should forget that for the testing day because thoughts like that only build up pressure and worries.

tongxun

感谢liushiyou！