The post GMAT Math: The Uses and Abuses of Formulas appeared first on Magoosh GMAT Blog.
]]>1) In the figure above, AC = BC = 8, angle C = 90°, and the circular arc has its center at point C. Find the area of the shaded region.
2) Employees at a company will vote for an executive team of five people from eight qualified candidates. The executive team consists of a president, a treasurer, and three warrant officers. If an executive team is considered different if any of the same people hold different offices, then how many possible executive teams could be selected from the eight candidates?
3) In the above diagram, the 16 dots are in rows and columns, and are equally spaced in both the horizontal & vertical direction. How many triangles, of absolutely any shape, can be created from three dots in this diagram? Different orientations (reflections, rotations, etc.) and/or positions count as different triangles. (Notice that three points all on the same line cannot form a triangle; in other words, a triangle must have some area.)
What all three of these problems have in common is: relying on formula alone will not help you. Complete explanations will come at the end of this article.
Clearly, it’s a good idea to know some GMAT math formulas —– many of the best formulas to know are listed on that blog. Formulas can be wonderful shortcuts, very efficient time-savers. Clearly it would be a mistake to walk into the GMAT Math section without knowing a single formula.
Formulas are useful to a point, but one of the worst mistakes folks make when they study for GMAT Math is to think that all they have to do to master the GMAT Math section is to memorize all the formulas. Similarly, the worst way to approach a GMAT math section is to approach each question asking only, “What’s the best formula to use for this question?” That’s roughly the same as thinking that all I need to be the world’s best lion tamer is a fancy hat that says “lion tamer”! In fact, the GMAT regularly designs math questions that predictably punish folks who blindly memorize formulas. Most of the harder math problems on the GMAT have this property: if you approach the problem with the perspective of “what’s the best formula to use to solve this question?”, then you very likely will miss what the question is really about and get it wrong. Here are some tips you can use to escape these traps.
What do I mean by this? Yes, you have to remember some formulas for the GMAT. One option would be to memorize each one separately as an isolate mathematical factoid. Some people are exceptionally good memorizers, but for most people, this is not too helpful: in particular, if, in the stress of the real GMAT, you forget what you memorized, then you are out of luck. Memorizing allows you no recourse if you forget, and it gives you no insight into mathematical logic.
Instead, I would recommend: understand the derivation of every formula: that is, understand the logical argument that underlies the formula and from which the formula arises. Think in terms of “why is it true?”: don’t stop at simply “what is true?” For example, both the permutation and combination formulas can be derived directly from the Fundamental Counting Principle. In another blog, I talked about the derivation of the formula for the area of an equilateral triangle. For most of the formulas you need to know for the GMAT, there is a context of logical argument that is also helpful to know. One exception is Archimedes‘ formula for the area of a circle:
Yes, the great master Archimedes had a brilliant argument for why that formula is true, but that’s well beyond the kind of logic you need to employ on the GMAT. For this one, you can take a pass and simply memorize the formula, if you haven’t already. For most formulas, though, you should explore the logical of the context in depth.
Problem-solving in mathematics is a subtle and sophisticated subject, and even in the limited context of the GMAT math section, the harder questions sometimes demand insightful approaches. Formulas are very cookie-cutter: if you have a formula to do something, it’s very good at doing that one thing, but that one thing is all it can do. That’s quite different from the flexible analyses characteristic of mathematical thinking.
Think about it this way. When you approach the GMAT math section, you should have a well-equipped tool box. The individual formulas you know are tools, but each one is a highly specific tool, useful for only one thing. Other more widely applicable tools are skills such as estimation or backsolving or picking numbers. It’s important to have in your “tool box” a mix of tools, some more general and some more specific; what’s more important, though, it knowing when and how to use them.
In the post How to do GMAT Math Faster, I talk in depth about left-brain vs. right-brain thinkers. Left-brain dominant thinkers want to know “what to do,” and they love formulas, because it’s always very clear what to do with a formula. A more right-brain perspective focuses on “how to look at the problem“, on the question of the best way to frame a problem, the best problem-solving perspective to adopt. Often, on a challenging GMAT problem, when one adopts the best perspective, what to do becomes obvious. Given the correct perspective, many test-takers could solve the problem, but most folks get that problem wrong because it’s hard to come up with that right perspective on one’s own. Coming up with the best perspective for a problem, the best way to frame a problem, takes time. It’s a right-brain pattern-matching skill: these always take time & experience to master. It’s very important to read solutions carefully: left-brain thinkers will want to jump ahead to the formulas, to “what to do“, and may be frustrated because they knew all those parts already. To get the most from solutions, it’s very important to study the very beginning, any reasons given for making one perceptual choice rather than another. Sometimes, it may be helpful to solve the problem in more than one way, to experience the difference in different solution routes. Sometimes, the perceptual choices are hard to get from the solution, and the only way to find an answer is to post a question to an expert in the forums, asking them: “why do we get the solution this way but not in that way?” Through asking questions and investigating, over time one develops more of a sense for problem-solving perspective.
Yes, know the GMAT formulas, but don’t think they are a magic bullet. You will get considerably more from understanding the logic behind each formula, than you will from blind memorization of only the formula. Finally, it’s most important to develop the problem-solving perspective, which will help you to see which formulas to use when. Here’s another practice question from inside the Magoosh product:
4) http://gmat.magoosh.com/questions/124
If you would like to share your own experience with formulas, or would like to clarify something I have said, let us know in the comments section.
1) There’s not a single formula we can use to get the answer, but by combining a few formulas, we can calculate this.
First, think about the circle. The circle has radius r = 8, so its total area would be
This entire figure is a quarter of the circle, so that area would be
Now, the shaded area (technically known as a circular segment), would have an area of
(circular segment) = (quarter circle) – (triangle ABC)
Well, we already have the area of the quarter circle. The triangle would have an area of (1/2)bh = 32. Therefore, the area of the segment is
Answer = (B)
2) We can use a quick formula for this. We have to go back to the Fundamental Counting Principle and think this through.
Choice #1: for the president, we have eight choices
Choice #2: for the treasurer, we have seven remaining choices
Choice #3: we have to pick 3 warrant officers from the remaining 6. This would be
The FCP tells us to multiply the number of choices in each selection:
8*7*20 = 56*20 = 1120 choices — Answer = (D)
3) This is a very subtle counting problem. This post on difficult counting problems may provide some insights.
The basic strategy will be to calculate the total number of sets of three points we can select in this diagram, and then subtract the small number of sets of three collinear points.
The first step: in how many ways can we select three points from a set of 16 points?
There are 560 sets of three points we can select. Right away, notice, we can eliminate answer choices (C), (D), and (E) as too big.
How many sets of three points are collinear? Well, first look at the column. In any column, we have four collinear points, so we could eliminate any one point, and we would be left with a unique set of three collinear points. That means we have four collinear sets in each column, or 16 vertical sets of collinear points. By symmetry, we must have 16 horizontal sets of collinear point. That’s 32 together.
But that’s not all. Notice there are two long diagonals, of four points each: much in the same way as in the argument above, there must be four unique sets of three collinear points along each diagonal. For the two diagonals, that’s 8 more. We up to 40 now.
Now, notice the short diagonals of only three points:
That’s the last four. There are a total of 44 unique sets of three collinear points. Since any of these would not constitute valid triangles, we have to subtract this from the total number of sets of three points.
560 – 44 = 516 —- answer = (A)
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]]>The post Compound Interest on the GMAT appeared first on Magoosh GMAT Blog.
]]>1) If $5,000,000 is the initial amount placed in an account that collects 7% annual interest, which of the following compounding rates would produce the largest total amount after two years?
(A) compounding annually
(B) compounding quarterly
(C) compounding monthly
(D) compounding daily
(E) All four of these would produce the same total
2) If A is the initial amount put into an account, R is the annual percentage of interest written as a decimal, and the interest compounds annually, then which of the following would be an expression, in terms of A and R, for the interest accrued in three years?
3) At the beginning of January 2003, Elizabeth invested money in an account that collected interest, compounding more frequently than a year. Assume the annual percentage rate of interest remained constant. What is the total amount she has invested after seven years?
Statement #1: her initial investment was $20,000
Statement #2: the account accrued 7% annual interest
4) Sarah invested $38,700 in an account that paid 6.2% annual interest, compounding monthly. She left the money in this account, collecting interest for a full three-year period. Approximately how much interest did she earn in the last month of this period?
(A) $239.47
(B) $714.73
(C) $2793.80
(D) $7,888.83
(E) $15,529.61
Solutions to these will be given at the end of the article.
In grade school, you learn about simple interest, largely because we want to teach little kids something about the idea of interest, and that’s the only kind of interest that children can understand. No one anywhere in the real world actually uses simple interest: it’s a pure mathematical fiction.
Here’s how it works. There’s an initial amount A, and an annual percentage P. At the end of each year, you get interest in the amount of P percent of A — the same amount every year. Suppose the initial amount is $1000, and the annual percentage is 5%. Well, 5% of $1000 is $50, so each year, you would get the fixed value of $50 in interest. If you plotted the value of the account (principle + interest) vs. time, you would get a straight line.
Again, this is a fiction we teach children, the mathematical equivalent of Santa Claus. This never takes place in the real world.
With compound interest, in each successive year or period, you collect more interest not merely on the principle but on all the interest you have accrued up to that point in time. Interest on interest: that’s the big idea of compound interest.
Here’s how it plays out. Again, there’s an initial amount A, and an annual percentage P, and we also have to know how frequently we are compounding. For starters, let’s just say that we are compounding annually, once a year at the end of the year. In fact, let’s say we have $1000, and the annual interest rate is 5%. Well, in the first year, we would earn five percent on $1000, and gain $50 in interest. The first year is exactly the same as the simple interest scenario. After that first year, we now have $1050 in the account, so at the end of the second year, we gain 5% of $1050, or $52.50, for a new total of $1102.50. Now, that’s our new total, so at the end of the third year, we gain 5% of $1102.50, or $55.12 (rounded down to the nearest penny), for a new total of $1157.62. At the end of three years, the simple interest scenario would give us $1150, so the compound interest gains us an extra $7.62 —- not much, but then again, $1000 is not a lot to have invested. You can see that, with millions or billions of dollars, this would be a significant difference.
Here, I was demonstrating everything step-by-step for clarity, but if we wanted to calculate the total amount after a large number of years, we would just use a formula. We know that each year, the amount increased by 5%, and we know that 1.05 is the multiplier for a 5% increase. After 20 years, the amount in the account would have experienced twenty 5% increases, so the total amount would be
We don’t have to do it step-by-step: we can just jump to the answer we need, using multipliers. Of course, for this exact value, we would need a calculator, and you don’t get a calculator on the GMAT Quantitative section. Sometimes, though, the GMAT lists some answers in “formula form”, and you would just have to recognize this particular expression, , as the right formula for this amount.
If we graph compound interest against time, we get an upward curving graph (purple), which curves away from the simple interest straight line (green):
The curve of the graph, that is to say, the multiplying effect of the interest, gets more pronounces as time goes on.
BIG IDEA #1: as long as there is more than one compounding period, then compound interest always earns more than simple interest.
A year is a long time to wait to get any interest. Historically, some banks have compounded over shortened compounding period. Here is a table of common compounding periods:
Technically, the fraction for “compounding daily” would be 1/365 in a non-leap year and 1/366 in a leap year; alternatively, one could use 1/365.25 for every year.
Now, how does this work? Let’s say the bank gives 5% annual, compounding quarterly. It would be splendid if the bank wanted to give you another 5% each and every quarter, but that’s not how it works. The bank takes the percentage rate of interest and multiplies it by the corresponding fraction. For 5% annual, compounding quarterly, we would multiply (5%)*(1/4) = 1.25%. That’s the percentage increase we get each quarter. The multiplier for a 1.25% increase is 1.0125. Suppose we invest $1000 initially and keep the money in this account for seven years: that would be 7*4 = 28 compounding periods, so there are twenty-eight times in that period in which the account experiences a 1.25% increase. Thus, the formula would be
For compounding quarterly, we divide the annual rate by four and compound four times each year. For compounding monthly, we divide the annual rate by twelve and compound twelve times a year. Similarly, for daily or any other conceivable compounding period.
How do the amounts of interest accrued compare for different compounding periods? To compare this, let’s pick a larger initial value, $1,000,000, and collect over a longer period, 20 years. Below are the total amounts, after twenty years, on an initial deposit of one million dollars compound at 5% annual:
As we go down that list, notice the values keep increasing as we decrease the size of the compounding period (and, hence, increase the total number of compound periods). This leads to:
BIG IDEA #2: We always get more interest, and larger account value overall, when the compounding period decreases; the more compounding periods we have, the more interest we earn.
Admittedly, the difference between “compounding daily” and “compounding hourly” only turn out to be a measly $178 on a million dollar investment over a twenty year period. A infinitesimally small difference, but technically, it is still an increase to move from “compounding daily” to “compounding hourly.”
Notice that so far, we have been talking about investments and interest that you earn. All of this, everything in this article, works equally well for debt and interest that you have to pay. Just as the compounding effect, over time, magically multiplies an investment, so the same compounding effect will sink you deeper and deeper into debt. This, in a nutshell, is the lose-lose proposition of credit card debt.
You may have noticed that, as the compounding periods get smaller and smaller, the amount increase in every diminishing steps. In fact, as you may suspect, the total amount you possibly could earn from decreasing the compound period reaches a ceiling, a limit. This limit is called “compounding continuously.” The mathematics of this involves the special irrational number e, named for Leonard Euler (1707 – 1783), who is often considered the single greatest mathematician of all times.
e = 2.71828182845904523536028747135266249775724709369995 …
The formula for calculating continuously compound interest involves e, and is more complicated than anything you need to understand for the GMAT. I will simply point out, in the example above, with one million dollars invested at 5% annual for 20 years, the limit of continuously compounding would be 1 million times e, which is $2,718,281.83. You do not need to understand why that is or how this was calculated.
Most banks use monthly compounding interest, for accounts and for mortgages: this make sense for accounts with monthly statement or payments. Credit cards tend to use continually compounding interest, because charges or payments could occur at any point, at any time of any day of the month. In addition to understanding the mathematics of compound interest, it’s good to have a general idea of how it works in the real world: after all, the history or logic of compound interest would be a very apt topic for a Reading Comprehension passage or a Critical Reasoning prompt on the GMAT!
If you had any “aha’s” while reading this article, you may want to go back a take another look at the four practice problems above. If you would like to express anything on these themes, or if you have a question about anything I said in this article, please let us know in the comments section.
1) The smaller the compounding period is, the greater the number of times the interest will be compounded. Of course, if we compound monthly instead of quarterly, then we are compounding by 1/12 of the annual rate each time, instead of 1/4. The number of times we compound goes up, but the percentage by which we compound each time goes down. Naively, you may think that those two would cancel out, but they don’t. As discussed above, as the compounding period gets smaller, the total amount of interest earned goes up. Therefore, we will get the most with the smallest compound period, daily. Answer = (D)
2) Notice that, since R is the annual percent as a decimal, we can form a multiplier simply by adding one: (1 + R). That’s very handy! We will explore two different methods to get the answer.
Method One: Step-by-step
Starting amount = A
After one year, we multiply by the multiplier once
That’s the total amount at the end of the first year. The amount A is the original principle, and AR is the interest earned.
At the end of the second year, that entire amount is multiplied by the multiplier. We need to FOIL.
That’s the total amount at the end of the second year. The amount A is the original principle, and the rest is the interest earned.
At the end of the third year, this entire amount is again multiplied by the multiplier.
That’s the total amount at the end of the third year. The amount A is the original principle, and the rest is the interest earned.
Answer = (C)
Method Two: some fancy algebra
Over the course of three years, the initial amount A is multiplied by the multiplier (1 + R) three times. Thus, after three years,
Now, if you happen to know it offhand, we can use the cube of a sum formula:
Thus,
and
Answer = (C)
3) In order to determine the total amount at the end of an investment, we would need to know three things: (a) the initial deposit; (b) the annual percentage rate; and (c) the compounding period.
Statement #1 tells us the initial deposit but not the annual percentage rate. Insufficient.
Statement #2 tells us the annual percentage rate but not the initial deposit. Insufficient.
Together, we know both the initial deposit and the annual percentage rate, but we still don’t know the compounding period. All we know is that it’s less than a year, but quarterly compounding vs. monthly compounding vs. daily compounding would produce different total amounts at the end. Without knowing the exact compounding period, we cannot calculate a precise answer. Even together, the statements are insufficient.
Answer = (E)
4) Without a calculator available, this is a problem screaming for estimation. The problem even uses the magic word “approximately” to indicate that estimation is a good idea, and the answer choices are spread far apart, making it easier to estimate an individual answer.
Let’s round the deposit up to $40,000, and the percentage down to 6% annual. Compounding monthly means each month, Sarah will accrue 6/12 = 0.5% in interest. Well, 1% of $40,000 is $400. Divide by 2: then 0.5% of $40,000 would be $200. That would be the simple interest amount, as well as the interest in the first month. We expect the amount in the last month to be a little more than this, but certain not even as large as double this amount. The only possible answer is (A).
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]]>The post GMAT Quant: Difficult Units Digits Questions appeared first on Magoosh GMAT Blog.
]]>1) The units digit of is:
(A) 1
(B) 3
(C) 5
(D) 7
(E) 9
2) The units digit of is:
(A) 2
(B) 4
(C) 6
(D) 8
(E) 0
3) The units digit of is:
(A) 2
(B) 4
(C) 6
(D) 8
(E) 0
Admittedly, these problems are probably a notch harder than anything you are likely to see on the GMAT. If you understand these, you definitely will understand anything of this variety that the GMAT throws at you!
All of those problems above involve numbers with hundreds of decimal places. No one can calculate those answers without a calculator: in fact, no calculator would be sufficient to do the calculation, because no calculator can accommodate that many digits. If one needed the exact answer, one could always use that most extraordinary web computing tool, Wolfram Alpha. Of course, one will not have access to the web or a calculator or anything other than one’s owns wits when confronting a question such as this on the GMAT. How do we proceed?
It turns out, what appears as a ridiculously hard calculation is actually quite easier. No part of the calculation we are going to do will involve anything beyond single-digit arithmetic!
The units digits of large numbers are special: they form a kind of elite and exclusive club. The big idea: only units digits affect units digits. What do I mean by that? Well, first of all, suppose you add or subtract two large numbers —- the units digit of the sum or the difference will depend only on the units digits of the two input numbers. For example, 3 + 5 = 8 —- this means that any number ending in 3 plus any number ending in 5 will be a number ending in 8. If you remember your “column addition” processes from grade school, this one might make intuitive sense.
The one that can be a little harder for folks to swallow is multiplication. If you multiply two large numbers, the unit digit of the product will have the same units digit as the product of the units digit of the two factors. That’s a mouthful! In other words, let’s take 3*7 = 21, so a units digit of a 3, times a unit digit of a 7, equals a units digit of a 1. That means, we could take any large number ending in 3, times any large number ending in 7, and the product absolutely will have to have a units digit of 1. If this is new idea to you, I strongly recommend: sit down with a calculator and multiply ridiculously large numbers together, with all combinations of units digits, until you are 100% satisfied that this pattern works.
This part will be a recap of an earlier post on powers of units digits. When we raise to a power, of course, that’s iterated multiplication, so we just follow the multiplication rule above. As it turns out, a simple pattern will always emerge.
Suppose we were considering powers of 253 — first of all, only the units digit, 3, matters, for the units digit of the powers. Any number ending in three will have the same sequence of units digits for the powers.
= 3
= 9
9*3 = 27, so has a units digit of 7
7*3 = 21, so has a units digit of 1
1*3 = 3, so has a units digit of 3
3*3 = 9, so has a units digit of 9
9*3 = 27, so has a units digit of 7
7*3 = 21, so has a units digit of 1
Notice a pattern has emerged — 3, 9, 7, 1, 3, 9, 7, 1, 3, 9, 7, 1, … It repeats like mathematical wallpaper. The pattern has a period of 4 — that is to say, it take four steps to repeat. This means, 3 to the power of any multiple of 4 has a units digit of 1: , , and all have a units digit of one. If I want to find power that’s not a multiple of 4, that’ easy: I just go to the nearest multiple of 4 and follow the wallpaper from there. For example, if I wanted —–
has a units digit of 1
has a units digit of 3
has a units digit of 9
As it happens, is a number that has 213 digits, but the units digit must be a 9.
The really expansive idea: everything I have just said about powers of 3 also applies to any larger number that happens to have a units digit of 3. Thus,
has a units digit of 9
That number has over a thousand digits (you don’t need to know how to figure that out!) but we know for sure that the units digit of this gargantuan number is 9.
If reading this article gave you any insights, you may want to give the questions above a second try. Here’s another problem, slightly easier and more GMAT-like, of this genre:
4) http://gmat.magoosh.com/questions/648
If there is anything you would like to say on this topic, or if you have any questions, please let me know in the comments section.
Finally, on a totally gratuitous note, here, in it’s the full thousand-plus-digit glory, is , courtesy of Wolfram Alpha:
= 6190880832531899190821690500833264378796558621138440416684569816896956548548008694
5501637120599244459832600741641042834529231770919235178215551804306285786976623543
72380800482154741117883167160446214356675850891464689434900627878445424968534812213
04515041521698298553935861735825956938171070649017829532068323699186643091574519875
641384511684642401730622089635116285314952964987090639595147866795944410916642166093
585848058327971863158257619930226042698661632905146850162960633520155118628867911625
239725418415604877699453370534194837316774432534349898272185517986005836675979188507
704257742239368667474408667895362250511057160490511029003928521905584001998500412272
300652930331121107733643816582958394189572596322595033481338694429893546070448926193
272806103607662243076062238492673013489615386273692928543218155895489937520257687664
947027647847750945509362588170852889875925160078794611182855714905968753089053225774
863189760920183769031795458368827168630624310066175033265292467587132663811805301906
7641362643313498166787213628751583911774745199740840719668395714479929
Notice, of course, that the units digit is 9.
1) First of all, all we need is the last digit of the base, not 137, but just 7. Here’s the power sequence of the units of 7
has a units digit of 7
has a units digit of 9 (e.g. 7*7 = 49)
has a units digit of 3 (e.g. 7*9 = 63)
has a units digit of 1 (e.g. 7*3 = 21)
has a units digit of 7
has a units digit of 9
has a units digit of 3
has a units digit of 1
etc.
The period is 4, so 7 to the power of any multiple of 4 has a units digit of 1
has a units digit of 1
has a units digit of 7
So the inner parenthesis is a number with a units digit of 7.
Now, for the outer exponent, we are following the same pattern — starting with a units digit of 7. The period is still 4.
has a units digit of 1
has a units digit of 7
has a units digit of 9
has a units digit of 3
So the unit digit of the final output is 3. Answer = B
BTW, this number is the great-granddaddy, the biggest number of all the big numbers mentioned in this post. The number clocks in with over 1300 digits!
2) We have to figure out each piece separately, and then add them. The first piece is remarkably easy — any power of anything ending in 5 always has a units digit of 5. So the first term has a units digit of 5. Done.
The second term takes a little more work. We can ignore the tens digit, and just treat this base as 3. Here is the units digit patter for the powers of 3.
has a units digit of 3
has a units digit of 9
has a units digit of 7 (e.g. 3*9 = 27)
has a units digit of 1 (e.g. 3*7 = 21)
has a units digit of 3
has a units digit of 9
has a units digit of 7
has a units digit of 1
The period is 4. This means, 3 to the power of any multiple of 4 will have a units digit of 1.
has a units digit of 1
has a units digit of 3
has a units digit of 9
Therefore, the second term has a units digit of 9.
Of course 5 + 9 = 14, so something with a units digit of 5 plus something with a units digit of 9 will have a units digit of 4. Answer = B
3) We have to figure out each piece separately, and then multiply them. The powers of 4 are particularly easy.
has a units digit of 4
has a units digit of 6 (e.g. 4*4 = 16)
has a units digit of 4 (e.g. 4*6 = 24)
has a units digit of 6
has a units digit of 4
has a units digit of 6
Four to any odd power will have a units digit of 4. Thus, any number with a units digit of four, raised to an odd power, will also have a units digit of 4. The first factor, , has a units digit of 4.
Now, the base in the second factor ends in a 3 (we can ignore the tens digit). Here is the pattern for powers of three.
has a units digit of 3
has a units digit of 9
has a units digit of 7 (e.g. 3*9 = 27)
has a units digit of 1 (e.g. 3*7 = 21)
has a units digit of 3
has a units digit of 9
has a units digit of 7
has a units digit of 1
The period is 4. This means, 3 to the power of any multiple of 4 will have a units digit of 1.
has a units digit of 1
has a units digit of 3
Thus, any number with a units digit of 7, when raised to the power of 37, will have a units digit of 3. The second factor, , has a units digit of 3.
Of course, 4*3 = 12, so any number with a units digit of 4 times any number with a units digit of 3 will yield a product with a units digit of 2.
Answer = A
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]]>The post Three Algebra Formulas Essential for the GMAT appeared first on Magoosh GMAT Blog.
]]>1) The numbers a, b, and c are all positive. If , then what is the value of ?
Statement #1: a – b = 3
Statement #2: = 7
2) Given that (P + 2Q) is a positive number, what is the value of (P + 2Q)?
Statement #1: Q = 2
Statement #2:
3) In the diagram above, O is the center of the circle, DC = a and DO = b. What is the area of the circle?
Statement #1:
Statement #2: a + b = 22
4) ABCD is a square with a side y, and JKLM is a side x. If Rectangle S (not shown) with length (x + y) has the same area as the shaded region above, what is the width of Rectangle S?
(A) x
(B) y
(C) y + x
(D) y – x
(E)
Doing math involve both following procedures and recognizing patterns. Three important patterns for algebra on the GMAT are as follows:
Pattern #1: The Difference of Two Squares
Pattern #2: The Squares of a Sum
Pattern #3: The Squares of a Difference
For GMAT Quant success, you need to know these patterns cold. You need to know them as well as you know your own phone number or address. The GMAT will throw question after question at you in which you simply will be expected to recognize these patterns. In such a question, if you recognize the relevant formula, it will enormously simplify the problem. If you don’t recognize the relevant formula, you are likely to be stymied by such a question.
You might think I would say: memorize them. Instead, I will ask you to remember them. What’s the difference? Memorization implies a rote process, simply trying to stuff an isolated and disconnected factoid into your head. By contrast, you strengthen you capacity to remember a math formula when you understand all the logic that underlies it.
Here, the logic behind these formulas is the logic of FOILing and factoring. You should review those patterns until you can follow each both ways — until you can FOIL the product out, or factor it back into components. If you can do that, you really understand these, and are much more likely to remember them in an integrated way.
If these patterns are relatively new to you, you may want to revisit the problems at the top with the list handy: see if you can reason your way through them, before reading the explanations below. Here’s another practice problem from inside Magoosh:
5) http://gmat.magoosh.com/questions/129
Do you have questions? Is there anything you would like to say? Let us know in the comment section at the bottom!
1) Let X = , the value we are seeking. Notice if we subtract the first equation in the prompt from this equation, we get = X – 117. In other words, if we could find the value of , then we could find the value of X.
Statement #1: a – b = 3
From this statement alone, we cannot calculate , so we can’t find the value of X. Statement #1, alone and by itself, is insufficient.
Statement #2: = 7
From this statement alone, we cannot calculate , so we can’t find the value of X. Statement #2, alone and by itself, is insufficient.
Statements #1 & #2 combined: Now, if we know both statements are true, then we could multiple these two equations, which cancel the denominator, and result in the simple equation a + b = 21. Now, we have the numerical value of both (a – b) and (a + b), so from the difference of two squares formula, we can figure out , and if we know the numerical value of that, we can calculate X and answer the prompt. Combined, the statements are sufficient.
Answer = C
2) The prompt tells us that (P + 2Q) is a positive number, and we want to know the value of P. Remember number properties! We don’t know that (P + 2Q) is a positive integer, just a positive number of some kind.
Statement #1: Q = 2
Obvious, by itself, this tells us zilch about P. Alone and by itself, this statement is completely insufficient.
Statement #2:
Now, this may be a pattern-recognition stretch for some folks, but this is simply the “Square of a Sum” pattern. It may be clearer if we re-write it like this:
This is now the “Square of a Sum” pattern, with P in the role of A and 2Q in the role of B. Of course, this should equal the square of the sum:
All we have to do is take a square root. Normally, we would have to consider both the positive and the negative square root, but since the prompt guarantees that (P + 2Q) is a positive number, we need only consider the positive root:
This statement allows us to determine the unique value of (P + 2Q), so this statement, alone and by itself, is sufficient.
Answer = B
3) To find the area of the circle, we need to use Archimedes’ formula, . For that we need the radius, OC. We are not given this directly, but notice that r = OC = DC – OD = a – b. If we knew that, we could find the area of the circle.
Statement #1:
A major pattern-matching hit! This, as written, is the “Square of a Difference” pattern.
In fact, this statement already gives us , so we just have to multiply by pi and we have the area. This statement, alone and by itself, is sufficient.
Statement #2: a + b = 22
We need a – b, and this statement gives us a value of a + b. If we had more information, perhaps we could use this in combination with other information to find what we want, but since this is all we have, it’s simply not enough to find a – b. This statement, alone and by itself, is insufficient.
Answer = A
4) A tricky one. First of all, notice that the shaded area, quite literally and visually, is the difference of two square — Area = . We know from the Difference of Two Squares pattern, this factors into:
Area = = (y + x)(y – x)
Well, if a rectangle had this same area, and it had a length of (y + x), it would have to have to have a width of (y – x) — that would make the area the same. The width has to be (y – x). Answer = D
To find out where algebra sits in the “big picture” of GMAT Quant, and what other Quant concepts you should study, check out our post entitled:
What Kind of Math is on the GMAT? Breakdown of Quant Concepts by Frequency
The post Three Algebra Formulas Essential for the GMAT appeared first on Magoosh GMAT Blog.
]]>The post GMAT Math Equations appeared first on Magoosh GMAT Blog.
]]>
It’s not really an equation, but you should know how to add & subtract & multiply & divide fractions. Percents are all over the GMAT, and you should understand converting among decimals & fractions & percents, as well as the crucial topic of percent changes. Among the big arithmetic equations are:
= the “remainder” equation (a.k.a. “rebuilding the dividend”)
= the doubling and halving method
= the average formula (esp. for finding the sum)
= the probability “AND” and “OR” rules
= the Fundamental Counting Principle
= the formulas for permutations & combinations
Notice, I do not list standard deviation as an equation you need to know — you need to understand the properties of standard deviation, but not the exact formula.
It’s not really an equation, but you should understand the procedures for solving for x, or solving for two variables with two equations. You should understand the rules for inequalities. Ironically, I will not recommend perhaps the most famous high school algebra equation, the Quadratic Formula — it’s much more important on the GMAT to know how to solve quadratics by factoring. Algebra equations important on the GMAT include:
= absolute values and negatives produce befuddling equations
The beautiful realm of geometry entails several important formulas, starting with the great granddaddy of them all:
others include:
= area of a triangle (A = 0.5bh)
= area of a rectangle (A = bh)
=
= Archimedes’ formula:
= Volume of a box = (height)*(width)*(depth)
= slope in the x-y plane
= slopes for parallel & perpendicular lines
= the special properties of the line y = x
Oddly, I am going to strongly recommend NOT to memorize a formula for finding distance in the x-y plane — I explain what to do instead at that link. 🙂
= rate, distance, and time
= (profit) = (revenue) – (cost)
Problems of interest and discount just involve the same percent equations discussed above.
Mathematical thinking on the GMAT
If you have questions or feedback about the GMAT math equations you’ll need to know, please let us know in the comments section below.
The post GMAT Math Equations appeared first on Magoosh GMAT Blog.
]]>The post Pythagorean Triplets to Memorize for the GMAT appeared first on Magoosh GMAT Blog.
]]>There aren’t many numbers you need to memorize for success on the GMAT Quantitative section, but knowing a few key Pythagorean triplets will save you a ton of time. First, try these GMAT practice question: remember: no calculator!
1) In right triangle ABC, BC = 48 and AB = 60. Find AC
2) In the x-y plane, what is the distance between (-4, -2) and (11, 6)?
3) In the diagram above, ∠L = ∠M = 90°, KL = 4, LM = 8, MN = 10, and JN = JK = 13. What is the area of JKLMN?
In almost all cases, I will recommend that GMAT student learn to remember mathematical facts without memorizing them. This is one of the few cases in which I will unapologetically recommend memorizing. There are certain sets of positive integers that satisfy the Pythagorean Theorem: these sets of three integers are called Pythagorean triplets. Some of them are very obscure, but some of them are extremely common. The most common by far is the triplet (3, 4, 5). In all of these, I am listing a set (a, b, c) that satisfies , so the largest number would be the hypotenuse of the triangle. Two other common sets of Pythagorean triplets are (5, 12, 13) and (8, 15, 17). Right there, BAM! Memorize those three sets, and you will spare yourself many stressful moments of lengthy calculations on the GMAT Quantitative section, when you have no calculator.
First of all, if you encounter a right triangle with legs 8 & 15, you won’t have to square and add things up: rather, you will just know that the hypotenuse is 17. The benefits, though, of that wee bit of memorizing expend wildly when you realize: you can multiply any of those three sets by any integer and get a new set of Pythagorean Triplets. The first set, (3, 4, 5), is the most common to see in multiplies —- (6, 8, 10), (9, 12, 15), (12, 16, 20), etc. —- but once in a while you may see one of the other two multiplied by something small, like 2 or 3.
Imagine on the GMAT, you see this:
That’s the diagram, and you are asked to find the length of YZ — you are given five answer choices, any of which looks like it could be plausible at first blush. Well, the unskilled GMAT taker will consume a great deal of time squaring 24, then squaring 45, then adding those together and — Gadzooks! — trying to figure out the square root of the four-digit number that results. 🙁 It’s MUCH easier simply to recognize that 24 = 3*8 and 45 = 3*15, so we are clearly dealing with 3 times the (8, 15, 17) triplet, which we conveniently have memorized. That means the answer must be YZ = 3*17 = 51. 🙂
Armed with these tricks, take another look at the practice problems again, before reading the explanations below. Also, here’s a Magoosh practice GMAT question that uses one of these Pythagorean triplets:
4) http://gmat.magoosh.com/questions/100
1) Clearly, the wrong approach would be to square 48 and 60, subtract, and without calculator try to take the square root of the resultant four-digit number. Not fun! A little GCF exploration reveals: 48 and 60 have a GCF = 12. More to the point, 48 = 4*12, and 60 = 5*12, so clearly we have the (3, 4, 5) triplet multiplied by 12. That means the missing side must be AC = 3*12 = 36. Answer = B
2) For those who would like the visual, here’s a diagram:
The purple line is the actual distance we want to find. One very important trick to know: when you need to find a diagonal distance in the x-y plane, always use the Pythagorean Theorem. We draw the connecting horizontal and vertical lines, shown in green here, to create a right triangle, the hypotenuse of which is the distance we want. Horizontal & vertical distances are very easy in the x-y plane: we simply count, or subtract the corresponding coordinates. These x- and y-distances are:
These are the lengths of the two green lines. Lo and behold: our old friend, the (8, 15, 17) triplet! Without any further calculations, we see immediately that the distance between these two points must be 17 units. Answer =
3) We will subdivide the figure as shown.
Notice, we constructed KR, such that KR is perpendicular to MN. This makes KLMR a rectangle and KRN a right triangle. S is the midpoint of KN, so that JS is the median & altitude of isosceles triangle JKN. (Since it’s easy to find the area of rectangles and right triangles, those make particularly good choices for subdivision units when you have to find area.)
Now, we will figure things out piece-by-piece. First of all, a particularly easy piece is rectangle KLMR: Area = bh = 4*8 = 32.
The problem gives that MN = 10, and we know MR = 4, so NR must equal 6, and KR must equal 8. These are two legs of the Pythagorean triplet (6, 8, 10), which is the (3, 4, 5) triplet times two. This means we know KN = 10. We also know the area of triangle KRN is: Area = 0.5*bh = 0.5*6*8 = 24.
Since we know KN = 10, we know the midpoint divides that in half, so KS = SN = 5. Notice, we now have two right triangles, KJS and NJS, each with a leg of 5 and a hypotenuse of 13. This is another one of the triplets, (5, 12, 13)! Immediately, without further calculation, we know JS = 12. Now we can find the area of the big isosceles triangle JKN: Area = 0.5*bh = 0.5*12*10 = 60.
Total area = (area of rect. KLMR) + (area of triangle KRN) + (area of triangle JKN)
= 32 + 24 + 60 = 116
Answer = E
The post Pythagorean Triplets to Memorize for the GMAT appeared first on Magoosh GMAT Blog.
]]>The post GMAT Coordinate Geometry: Distance Between Two Points appeared first on Magoosh GMAT Blog.
]]>First, try these practice questions.
1) What is the distance from (–7, 2) to (5, –3)?
2) Consider the three points in the x-y plane: P = (8, 4), Q = (6, 7), and R = (9, 0). Rank these three points from closest to the origin, (0, 0), to furthest from the origin.
3) Trapezoid JKLM in the x-y plane has coordinates J = (–2, –4), K = (–2, 1), L = (6, 7), and M = (6, –4). What is its perimeter?
The x-y plane is sometimes called the Cartesian plane, in honor of its creator, the multitalented philosopher and mathematician René Descartes (1596 – 1650). In the x-y plane, the x-coordinate of any point gives its horizontal position, how far right or left of the y-axis it is; the y-coordinate of any points gives its vertical position, how far above or below the x-axis it is. Notice that, on any horizontal line, all the y-coordinates would have to be the same, and on any vertical line, all the x-coordinates would have to be the same. Also notice: if any two points have the same x-coordinate, the segment that connects them must be vertical; and if any two points have the same y-coordinate, the segment that connects them must be horizontal. Those are very handy facts to know for the GMAT.
Big Idea #1: Horizontal & vertical distances are very easy to find!!
If two points have the same y-coordinate, then the segment connected them is horizontal. Simply subtract their x-coordinates to find the length of the segment between them.
If two points have the same x-coordinate, then the segment connected them is vertical. Simply subtract their y-coordinates to find the length of the segment between them.
BTW, whenever the GMAT gives you a geometric figure in the x-y plane and one or more of the sides are horizontal and/or vertical, the test is handing you a gift on a silver platter. Never fail to appreciate that.
Big Idea #2: For any oblique distance, use the Pythagorean Theorem
Perhaps in school you encountered something called the Distance Formula. If you already have that memorized and use it flawlessly, then good for you. If you don’t have it memorized, then I want you to blot out of your mind any trace of its existence. I don’t want you to memorize a needlessly complicated formula: instead, I want you to remember a simple procedure that accomplishes exactly the same thing.
The big idea is: use that most miraculously amazing theorem of mathematics, the Pythagorean Theorem. Combine this insight with Big Idea #1: we know horizontal & vertical distances are very easy to find, so construct a right triangle so that the distance you want to find is the hypotenuse, one leg is horizontal, and the other leg is vertical. Thus, you automatically get a triangle in which two of the sides are very simple to find, and the triangle is also automatically a right triangle, which guarantees that we can use the Pythagorean Theorem. For example, if we need to find the distance between A = (–1, 1) and B = (3, 3) —-
— then we simply draw a vertical segment going down from B, and a horizontal segment going to the right from A, and choose the third vertex, C, at the intersection:
Well, clearly AC = 4 and BC = 2, so then we are set up to find AB with the Pythagorean Theorem.
If you understand these two “big ideas”, then that’s all you need to know about distance in the x-y plane for the GMAT. If you had trouble with the practice questions, you may want to give them another look before reading the solutions below. Here’s another question from inside the Magoosh product:
4) http://gmat.magoosh.com/questions/820
1) Suppose we made a right triangle, as in the example above. The difference in the x-coordinates is (–7) – 5 = –12, so the horizontal leg would have a length of 12 (length has to be positive). The difference in the y-coordinates is 2 – (–3) = 5, so this is the length of the vertical leg. We recognize that we have a common Pythagorean triplet, so we don’t even need to perform a calculation: we know the length of the hypotenuse, and hence, the distance between the two points, is 13. Answer = D
2) Of course, the easiest distance to find is that of point R: point R is a distance of 9 from the origin. To find the distance to P, we construct a triangle with a horizontal leg of 8 and a vertical leg of 4.
Similarly,
Therefore P is the closest to the origin, and Q is the furthest. The order is P, R, Q. Answer = B
3) This one may look challenging initially, but all kinds of simplifying tricks. First of all, J & K have the same x-coordinate, so that’s a vertical segment, and the length is JK = 1 – (–4) = 5. L & M also have the same x-coordinate, so that’s also a vertical segment, and the length is LM = 7 – (–4) = 11. J & M have the same y-coordinate, so that’s also a horizontal segment, and the length is JM = 6 – (–2) = 8. Right there, we have the lengths of three of the four sides.
The last side to find is KL. From K = (–2, 1), we have to go over 8 and up 6 to get to L = (6, 7), so 6 and 8 the legs of the right triangle, and KL is the hypotenuse. We recognize another common Pythagorean triplet — (6, 8, 10), which is a multiple of the (3, 4, 5) triplet. Thus, without further calculations, we immediately know KL = 10.
Perimeter = JK + KL + LM + JM = 5 + 10 + 11 + 8 = 34
Answer = A
For those who would like a visual, here’s the graph of this trapezoid.
The post GMAT Coordinate Geometry: Distance Between Two Points appeared first on Magoosh GMAT Blog.
]]>The post GMAT Quant: Difference of Two Squares appeared first on Magoosh GMAT Blog.
]]>This formula, called “the difference of two squares” formula, is a favorite of standardized test writers. A simple enough pattern: see if you can detect where it shows up in the following challenging problems.
1) =
(A)
(B)
(C)
(D)
(E)
2) What is the sum of a and b?
(1)
(2)
3) In the diagram above, ∠A = ∠ABC, ∠CBD = ∠BDC, and ∠CBE =90°. If AE = 16 and DE = 4, what is the length of BE?
(A) 7
(B) 8
(C) 9
(D) 10
(E) 11
1) This involves a relatively sophisticated trick known as “multiplying by the conjugate.” When we have an expression of the form a+sqrt(b), the “conjugate” of this is a-sqrt(b). When we multiply a radical expression by its conjugate, we employ the difference of two squares. For example:
This is answer B. BTW, the trick of multiplying by the conjugate is at the very outside edge of what you might be expected to do the hardest GMAT math problems.
2) The prompt of this DS problem is straightforward.
Statement #1 tells us a = 4, but we have no idea of b, so, by itself, this is not sufficient for finding the sum.
Statement #2 gives us a value for an algebraic expression that lends itself nicely to simplification.
Thus, the sum is 7. Statement #2, by itself is sufficient. Answer = B.
3) This is a tricky one. Remember, it’s a diagram drawn to scale, so if all else fails, you can estimate (see this post). But, let’s solve this with math. The fact that ∠A = ∠ABC tells us triangle ABC is isosceles, with AC = BC. The fact that ∠CBD = ∠BDC tells us triangle BCD is isosceles, with BC = CD. The fact that ∠CBE = 90° means that (BC)^{2} + (BE)^{2} = (CE)^{2}. This means
(substitutions from the two isosceles triangles)
Answer = B
Here’s a further practice question.
http://gmat.magoosh.com/questions/118
The post GMAT Quant: Difference of Two Squares appeared first on Magoosh GMAT Blog.
]]>The post Breakdown of GMAT Quant Concepts by Frequency appeared first on Magoosh GMAT Blog.
]]>Here are the samples of Official Material he used to figure out the GMAT Quant breakdown:
1.GMAT Official Guide (12th Edition) Problem Solving Practice Questions (Pg. 152-265)*
2. GMAT Official Guide (12th Edition) Data Sufficiency Practice Questions (Pg. 272-351)*
3. GMATPrep Test #1
4. GMATPrep Test #2
5. Released past exam from GMAC, test code 14, from the 90s/early 2000’s
Total | Percentage | ||
Arithmetic | Percents | 21 | 10.71% |
Arithmetic | Properties of Integers | 18 | 9.18% |
Arithmetic | Descriptive Statistics | 12 | 6.12% |
Arithmetic | Fractions | 11 | 5.61% |
Algebra | Linear Equations, Two Unknowns | 11 | 5.61% |
Algebra | Simplifying Algebraic Expressions | 9 | 4.59% |
Arithmetic | Powers & Roots of Numbers | 8 | 4.08% |
Arithmetic | Counting Methods | 7 | 3.57% |
Algebra | Linear Equations, One Unknown | 7 | 3.57% |
Algebra | Functions/Series | 7 | 3.57% |
Geometry | Coordinate Geometry | 7 | 3.57% |
Word Problems | Rate Problems | 7 | 3.57% |
Arithmetic | Ratio & Proportions | 6 | 3.06% |
Arithmetic | Decimals | 5 | 2.55% |
Arithmetic | Discrete Probability | 5 | 2.55% |
Algebra | Exponents | 5 | 2.55% |
Algebra | Inequalities | 5 | 2.55% |
Geometry | Quadrilaterals | 5 | 2.55% |
Geometry | Circles | 5 | 2.55% |
Word Problems | Interest Problems | 4 | 2.04% |
Algebra | Solving Quadratic Equations | 3 | 1.53% |
Geometry | Triangles | 3 | 1.53% |
Geometry | Rectangular Solids & Cylinders | 3 | 1.53% |
Word Problems | Work Problems | 3 | 1.53% |
Word Problems | Mixture Problems | 3 | 1.53% |
Geometry | Intersecting Angles and Lines | 2 | 1.02% |
Word Problems | Profit | 2 | 1.02% |
Word Problems | Sets | 2 | 1.02% |
Word Problems | Measurement Problems | 2 | 1.02% |
Word Problems | Data Interpretation | 2 | 1.02% |
Arithmetic | Real Numbers | 1 | 0.51% |
Arithmetic | Sets | 1 | 0.51% |
Algebra | Solving by Factoring | 1 | 0.51% |
Geometry | Polygons | 1 | 0.51% |
Word Problems | Discount | 1 | 0.51% |
Word Problems | Geometry Problems | 1 | 0.51% |
Algebra | Equations | 0 | 0.00% |
Algebra | Absolute Value | 0 | 0.00% |
Geometry | Lines | 0 | 0.00% |
Geometry | Perpendicular Lines | 0 | 0.00% |
Geometry | Parallel Lines | 0 | 0.00% |
Total | Percentage | ||
Arithmetic | Properties of Integers | 30 | 19.23% |
Arithmetic | Descriptive Statistics | 13 | 8.33% |
Arithmetic | Percents | 11 | 7.05% |
Algebra | Linear Equations, Two Unknowns | 11 | 7.05% |
Word Problems | Rate Problems | 9 | 5.77% |
Algebra | Inequalities | 8 | 5.13% |
Arithmetic | Sets | 7 | 4.49% |
Arithmetic | Ratio & Proportions | 6 | 3.85% |
Geometry | Triangles | 6 | 3.85% |
Geometry | Circles | 6 | 3.85% |
Arithmetic | Decimals | 5 | 3.21% |
Geometry | Coordinate Geometry | 5 | 3.21% |
Arithmetic | Fractions | 4 | 2.56% |
Algebra | Linear Equations, One Unknown | 4 | 2.56% |
Algebra | Exponents | 4 | 2.56% |
Geometry | Rectangular Solids & Cylinders | 4 | 2.56% |
Arithmetic | Discrete Probability | 3 | 1.92% |
Algebra | Functions/Series | 3 | 1.92% |
Word Problems | Interest Problems | 3 | 1.92% |
Geometry | Lines | 2 | 1.28% |
Word Problems | Work Problems | 2 | 1.28% |
Word Problems | Discount | 2 | 1.28% |
Word Problems | Profit | 2 | 1.28% |
Arithmetic | Real Numbers | 1 | 0.64% |
Arithmetic | Counting Methods | 1 | 0.64% |
Algebra | Simplifying Algebraic Expressions | 1 | 0.64% |
Algebra | Absolute Value | 1 | 0.64% |
Geometry | Quadrilaterals | 1 | 0.64% |
Word Problems | Measurement Problems | 1 | 0.64% |
Arithmetic | Powers & Roots of Numbers | 0 | 0.00% |
Algebra | Equations | 0 | 0.00% |
Algebra | Solving by Factoring | 0 | 0.00% |
Algebra | Solving Quadratic Equations | 0 | 0.00% |
Geometry | Intersecting Angles and Lines | 0 | 0.00% |
Geometry | Perpendicular Lines | 0 | 0.00% |
Geometry | Parallel Lines | 0 | 0.00% |
Geometry | Polygons | 0 | 0.00% |
Word Problems | Mixture Problems | 0 | 0.00% |
Word Problems | Sets | 0 | 0.00% |
Word Problems | Geometry Problems | 0 | 0.00% |
Word Problems | Data Interpretation | 0 | 0.00% |
Total | Percentage | ||
Arithmetic | Properties of Integers | 6 | 17.14% |
Arithmetic | Ratio & Proportions | 4 | 11.43% |
Arithmetic | Descriptive Statistics | 4 | 11.43% |
Algebra | Exponents | 3 | 8.57% |
Algebra | Functions/Series | 3 | 8.57% |
Arithmetic | Percents | 2 | 5.71% |
Arithmetic | Powers & Roots of Numbers | 2 | 5.71% |
Word Problems | Interest Problems | 2 | 5.71% |
Arithmetic | Fractions | 1 | 2.86% |
Arithmetic | Counting Methods | 1 | 2.86% |
Algebra | Equations | 1 | 2.86% |
Geometry | Intersecting Angles and Lines | 1 | 2.86% |
Geometry | Triangles | 1 | 2.86% |
Geometry | Circles | 1 | 2.86% |
Word Problems | Work Problems | 1 | 2.86% |
Word Problems | Sets | 1 | 2.86% |
Word Problems | Measurement Problems | 1 | 2.86% |
Arithmetic | Decimals | 0 | 0.00% |
Arithmetic | Real Numbers | 0 | 0.00% |
Arithmetic | Sets | 0 | 0.00% |
Arithmetic | Discrete Probability | 0 | 0.00% |
Algebra | Simplifying Algebraic Expressions | 0 | 0.00% |
Algebra | Linear Equations, One Unknown | 0 | 0.00% |
Algebra | Linear Equations, Two Unknowns | 0 | 0.00% |
Algebra | Solving by Factoring | 0 | 0.00% |
Algebra | Solving Quadratic Equations | 0 | 0.00% |
Algebra | Inequalities | 0 | 0.00% |
Algebra | Absolute Value | 0 | 0.00% |
Geometry | Lines | 0 | 0.00% |
Geometry | Perpendicular Lines | 0 | 0.00% |
Geometry | Parallel Lines | 0 | 0.00% |
Geometry | Polygons | 0 | 0.00% |
Geometry | Quadrilaterals | 0 | 0.00% |
Geometry | Rectangular Solids & Cylinders | 0 | 0.00% |
Geometry | Coordinate Geometry | 0 | 0.00% |
Word Problems | Rate Problems | 0 | 0.00% |
Word Problems | Mixture Problems | 0 | 0.00% |
Word Problems | Discount | 0 | 0.00% |
Word Problems | Profit | 0 | 0.00% |
Word Problems | Geometry Problems | 0 | 0.00% |
Word Problems | Data Interpretation | 0 | 0.00% |
Total | Percentage | ||
Arithmetic | Descriptive Statistics | 6 | 18.75% |
Arithmetic | Properties of Integers | 4 | 12.50% |
Arithmetic | Ratio & Proportions | 2 | 6.25% |
Arithmetic | Percents | 2 | 6.25% |
Algebra | Simplifying Algebraic Expressions | 2 | 6.25% |
Algebra | Exponents | 2 | 6.25% |
Geometry | Coordinate Geometry | 2 | 6.25% |
Arithmetic | Fractions | 1 | 3.13% |
Arithmetic | Powers & Roots of Numbers | 1 | 3.13% |
Arithmetic | Counting Methods | 1 | 3.13% |
Algebra | Equations | 1 | 3.13% |
Algebra | Linear Equations, Two Unknowns | 1 | 3.13% |
Algebra | Inequalities | 1 | 3.13% |
Algebra | Functions/Series | 1 | 3.13% |
Geometry | Triangles | 1 | 3.13% |
Geometry | Circles | 1 | 3.13% |
Word Problems | Rate Problems | 1 | 3.13% |
Word Problems | Discount | 1 | 3.13% |
Word Problems | Measurement Problems | 1 | 3.13% |
Arithmetic | Decimals | 0 | 0.00% |
Arithmetic | Real Numbers | 0 | 0.00% |
Arithmetic | Sets | 0 | 0.00% |
Arithmetic | Discrete Probability | 0 | 0.00% |
Algebra | Linear Equations, One Unknown | 0 | 0.00% |
Algebra | Solving by Factoring | 0 | 0.00% |
Algebra | Solving Quadratic Equations | 0 | 0.00% |
Algebra | Absolute Value | 0 | 0.00% |
Geometry | Lines | 0 | 0.00% |
Geometry | Intersecting Angles and Lines | 0 | 0.00% |
Geometry | Perpendicular Lines | 0 | 0.00% |
Geometry | Parallel Lines | 0 | 0.00% |
Geometry | Polygons | 0 | 0.00% |
Geometry | Quadrilaterals | 0 | 0.00% |
Geometry | Rectangular Solids & Cylinders | 0 | 0.00% |
Word Problems | Work Problems | 0 | 0.00% |
Word Problems | Mixture Problems | 0 | 0.00% |
Word Problems | Interest Problems | 0 | 0.00% |
Word Problems | Profit | 0 | 0.00% |
Word Problems | Sets | 0 | 0.00% |
Word Problems | Geometry Problems | 0 | 0.00% |
Word Problems | Data Interpretation | 0 | 0.00% |
Total | Percentage | ||
Arithmetic | Properties of Integers | 7 | 13.73% |
Arithmetic | Ratio & Proportions | 5 | 9.80% |
Arithmetic | Percents | 5 | 9.80% |
Arithmetic | Descriptive Statistics | 4 | 7.84% |
Word Problems | Sets | 4 | 7.84% |
Arithmetic | Powers & Roots of Numbers | 3 | 5.88% |
Algebra | Linear Equations, Two Unknowns | 3 | 5.88% |
Word Problems | Rate Problems | 3 | 5.88% |
Arithmetic | Fractions | 2 | 3.92% |
Arithmetic | Counting Methods | 2 | 3.92% |
Algebra | Exponents | 2 | 3.92% |
Algebra | Inequalities | 2 | 3.92% |
Geometry | Triangles | 2 | 3.92% |
Algebra | Simplifying Algebraic Expressions | 1 | 1.96% |
Algebra | Equations | 1 | 1.96% |
Algebra | Functions/Series | 1 | 1.96% |
Geometry | Quadrilaterals | 1 | 1.96% |
Geometry | Circles | 1 | 1.96% |
Geometry | Rectangular Solids & Cylinders | 1 | 1.96% |
Geometry | Coordinate Geometry | 1 | 1.96% |
Arithmetic | Decimals | 0 | 0.00% |
Arithmetic | Real Numbers | 0 | 0.00% |
Arithmetic | Sets | 0 | 0.00% |
Arithmetic | Discrete Probability | 0 | 0.00% |
Algebra | Linear Equations, One Unknown | 0 | 0.00% |
Algebra | Solving by Factoring | 0 | 0.00% |
Algebra | Solving Quadratic Equations | 0 | 0.00% |
Algebra | Absolute Value | 0 | 0.00% |
Geometry | Lines | 0 | 0.00% |
Geometry | Intersecting Angles and Lines | 0 | 0.00% |
Geometry | Perpendicular Lines | 0 | 0.00% |
Geometry | Parallel Lines | 0 | 0.00% |
Geometry | Polygons | 0 | 0.00% |
Word Problems | Work Problems | 0 | 0.00% |
Word Problems | Mixture Problems | 0 | 0.00% |
Word Problems | Interest Problems | 0 | 0.00% |
Word Problems | Discount | 0 | 0.00% |
Word Problems | Profit | 0 | 0.00% |
Word Problems | Geometry Problems | 0 | 0.00% |
Word Problems | Measurement Problems | 0 | 0.00% |
Word Problems | Data Interpretation | 0 | 0.00% |
Total | Percentage | ||
Arithmetic | Properties of Integers | 65 | 13.83% |
Fractions | 19 | 4.04% | |
Decimals | 10 | 2.13% | |
Real Numbers | 2 | 0.43% | |
Ratio & Proportions | 23 | 4.89% | |
Percents | 41 | 8.72% | |
Powers & Roots of Numbers | 14 | 2.98% | |
Descriptive Statistics | 39 | 8.30% | |
Sets | 8 | 1.70% | |
Counting Methods | 12 | 2.55% | |
Discrete Probability | 8 | 1.70% | |
Algebra | Simplifying Algebraic Expressions | 13 | 2.77% |
Equations | 3 | 0.64% | |
Linear Equations, One Unknown | 11 | 2.34% | |
Linear Equations, Two Unknowns | 26 | 5.53% | |
Solving by Factoring | 1 | 0.21% | |
Solving Quadratic Equations | 3 | 0.64% | |
Exponents | 16 | 3.40% | |
Inequalities | 16 | 3.40% | |
Absolute Value | 1 | 0.21% | |
Functions/Series | 15 | 3.19% | |
Geometry | Lines | 2 | 0.43% |
Intersecting Angles and Lines | 3 | 0.64% | |
Perpendicular Lines | 0 | 0.00% | |
Parallel Lines | 0 | 0.00% | |
Polygons | 1 | 0.21% | |
Triangles | 13 | 2.77% | |
Quadrilaterals | 7 | 1.49% | |
Circles | 14 | 2.98% | |
Rectangular Solids & Cylinders | 8 | 1.70% | |
Coordinate Geometry | 15 | 3.19% | |
Word Problems | Rate Problems | 20 | 4.26% |
Work Problems | 6 | 1.28% | |
Mixture Problems | 3 | 0.64% | |
Interest Problems | 9 | 1.91% | |
Discount | 4 | 0.85% | |
Profit | 4 | 0.85% | |
Sets | 7 | 1.49% | |
Geometry Problems | 1 | 0.21% | |
Measurement Problems | 5 | 1.06% | |
Data Interpretation | 2 | 0.43% | |
Total | 470 | 100.00% |
Total | Percentage | ||
Arithmetic | Properties of Integers | 65 | 13.83% |
Arithmetic | Percents | 41 | 8.72% |
Arithmetic | Descriptive Statistics | 39 | 8.30% |
Algebra | Linear Equations, Two Unknowns | 26 | 5.53% |
Arithmetic | Ratio & Proportions | 23 | 4.89% |
Word Problems | Rate Problems | 20 | 4.26% |
Arithmetic | Fractions | 19 | 4.04% |
Algebra | Exponents | 16 | 3.40% |
Algebra | Inequalities | 16 | 3.40% |
Algebra | Functions/Series | 15 | 3.19% |
Geometry | Coordinate Geometry | 15 | 3.19% |
Arithmetic | Powers & Roots of Numbers | 14 | 2.98% |
Geometry | Circles | 14 | 2.98% |
Algebra | Simplifying Algebraic Expressions | 13 | 2.77% |
Geometry | Triangles | 13 | 2.77% |
Arithmetic | Counting Methods | 12 | 2.55% |
Algebra | Linear Equations, One Unknown | 11 | 2.34% |
Arithmetic | Decimals | 10 | 2.13% |
Word Problems | Interest Problems | 9 | 1.91% |
Arithmetic | Sets | 8 | 1.70% |
Arithmetic | Discrete Probability | 8 | 1.70% |
Geometry | Rectangular Solids & Cylinders | 8 | 1.70% |
Geometry | Quadrilaterals | 7 | 1.49% |
Word Problems | Sets | 7 | 1.49% |
Word Problems | Work Problems | 6 | 1.28% |
Word Problems | Measurement Problems | 5 | 1.06% |
Word Problems | Discount | 4 | 0.85% |
Word Problems | Profit | 4 | 0.85% |
Algebra | Equations | 3 | 0.64% |
Algebra | Solving Quadratic Equations | 3 | 0.64% |
Geometry | Intersecting Angles and Lines | 3 | 0.64% |
Word Problems | Mixture Problems | 3 | 0.64% |
Arithmetic | Real Numbers | 2 | 0.43% |
Geometry | Lines | 2 | 0.43% |
Word Problems | Data Interpretation | 2 | 0.43% |
Algebra | Solving by Factoring | 1 | 0.21% |
Algebra | Absolute Value | 1 | 0.21% |
Geometry | Polygons | 1 | 0.21% |
Word Problems | Geometry Problems | 1 | 0.21% |
Geometry | Perpendicular Lines | 0 | 0.00% |
Geometry | Parallel Lines | 0 | 0.00% |
The list of concepts tested on the quantitative section is from GMAC, which you can see on page 107 of the Official Guide (either 12th or 13th edition). It isn’t perfect– “Integer Properties” is a wide area of knowledge, whereas something like “Circles” is very specific.
NO. For the sake of simplicity and accuracy in reporting absolute frequency, we’ve only assigned each question to one concept. This means that even though GMAC lists “Perpendicular lines” as a topic tested on the GMAT, and we have 0 questions marked as pertaining to that topic, that certainly doesn’t mean the idea of perpendicular lines did not come up at all on all of the exams. It certainly appeared, but often in questions that were better categorized, overall, as “Coordinate Geometry”, or “Intersecting Angles and Lines”.
We hope this serves as a guideline for the relative frequency of math topics tested on the GMAT to help you decide how to focus your time! In Magoosh practice, you can set up customized practice sessions to focus on specific concepts, as well as review your performance on individual concepts to identify your weak spots using our Review tool.
Let us know whether you find this type of breakdown helpful, and whether you have any questions about any of the information above! 🙂
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Euclid first proved this theorem over 2200 years ago. The theorem says:
If the two sides are equal, then the opposite angles are equal
and
if the two angles are equal, then the opposite sides are equal.
It can be used for deductions in both directions (logically, this is called a “biconditional” statement”). Another way to say this: a statement about equal angles is sufficient to conclude equal sides; conversely, a statement about equal sides is sufficient to conclude equal angles. Remember that on GMAT Data Sufficiency!
Another favorite GMAT geometry fact is that the sum of all three angles in a triangle − any triangle − is 180º. This is particularly fruitful if combined with the Isosceles Triangle Theorem.
Suppose you are told that Triangle ABC is isosceles, and one of the bottom equal angles (called a “base angle”) is 50º. Then immediately you know that the measure of the other base angle is also 50º, and that means the top angle (the “vertex angle”) must be 80º. Knowing the measure of one base angle is sufficient to find the measures of all three angles of an isosceles triangle.
Suppose you are told that Triangle ABC is isosceles, and the vertex angle is 50º. Well, you don’t know the measures of the base angle, but you know they’re equal. Let x be the degrees of the base angle; then . So, each base angle is 65º. Knowing the measure of the vertex angle is sufficient to find the measures of all three angles of an isosceles triangle.
BUT, if you are told that Triangle ABC is isosceles, and one of angles is 50º, but you don’t know whether that 50º is a base angle or a vertex angle, then you cannot conclude anything about the other angles in the isosceles triangle without more information. That’s a subtle but important distinction to remember on GMAT Data Sufficiency.
http://gmat.magoosh.com/questions/1019
http://gmat.magoosh.com/questions/1024
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