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If an MBA Program is not listed, please e-mail and we will be happy to list it. Berkeley / Haas Boston College / Carroll Carnegie Mellon / Tepper Chicago Columbia Concordia Cornell / Johnson Dartmouth / Tuck Duke / Fuqua Emory / Goizueta Harvard HEC Montreal Indiana / Kelley Michigan MIT / Sloan Northwestern / Kellogg New York / Stern North Carolina / Kenan Flagler Notre Dame / Mendoza Pennsylvania / Wharton Queens Stanford Syracuse / Whitman Texas / McCombs Thunderbird Toronto USC / Marshall UCLA / Anderson Vanderbilt / Owen Virginia / Darden Washington University in St. Louis / Olin Western Ontario / Ivey Yale MBA Programs: Rest of the World As there is some variety in the length of international MBA programs, we have denoted the length of the program next to its name (1 = one year; 2 = 2 years). If an MBA Program is not listed, please e-mail and we will be happy to list it. AGSM (Australia) 2 Cambridge / Judge (UK) 1 CIEBS (China) 2 Cheung Kong Graduate School of Business (China) 1 Cranfield School of Mgmt (UK) 1 ESADE (Spain) 1 or 2 HEC (France) 2 Hult (UK) 1 IESE (Spain) 2 IMD (Switzerland) 1 INCAE (Costa Rica) 2 INSEAD (France) 1 IPADE (Mexico) ISB (India) 1 London Business School (UK) 2 Manchester Bus. School (UK) 2 Melbourne (Australia) 2 Oxford / Said (UK) 1 Rotterdam (Netherlands) 1 Tsinghua IMBA (China) 2 University of St. Gallen (Switzerland) 1 Additional Resources Here we link a host of additional resources available across the web. E-mail info@clearadmit.com to have resources added to this list. AACSB International Association of MBAs Beyond Grey Pinstripes EFMD gradschools.com (worldwide) Infozee International Student Loans mba.com (GMAT Scores) MBAInfo mbaleague.blogspot.com MBAzone MBA Jungle TOEFL Top MBA MBA Tipline We encourage admissions officers, students and applicants to alert us of interesting news and developments, please send an email to news@clearadmit.com so we can blog it. Blog Archive
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Get a $10 Amazon.com Gift Card! Contribute your MBA interview reports to the Clear Admit Wiki. Taking the GMAT? Download our free, independent guide to the leading test prep companies - includes coupons for test prep savings! CATEGORY - WORKBOOK WEDNESDAYS Thursday, October 15, 2009 GMAT Practice Question: Difference of Zenzicubes In an effort to meet the needs of our growing audience, we’re pleased to announce that this site will now regularly feature GMAT practice problems, test-taking advice and more. Content will be provided by a range of test preparation companies, many of which were profiled in the Clear Admit Guide to GMAT Test Preparation Companies (a free publication available for immediate download). As such, stay tuned to this site for expanded coverage of the GMAT exam, helpful test-taking strategies and more! Without further ado, today’s sample problem comes from our friends at Manhattan GMAT: Problem: x6 – y6 = (A) (x3 + y3)(x2 + y2)(x – y) (B) (x3 – y3)(x3 – y3) (C) (x2 + y2)(x2 + y2)(x + y)(x – y) (D) (x4 + y4)(x + y)(x – y) (E) (x3 + y3)(x2 + xy + y2)(x – y) Solution: An efficient way to attack this problem is to rephrase the expression in the stem. Since a sixth power is a square, we can look at the difference of sixth powers as the difference of squares, and factor: x6 – y6 = (x3)2 – (y3)2 = (x3 + y3)(x3 – y3) This new expression is not among the answer choices, but as a search strategy, we should focus on these factors. Choice (A) contains (x3 + y3), so we should determine whether the rest of (A) works out to (x3 – y3). The terms (x2 + y2)(x – y) look promising, since they give us +x3 and –y3, but we get cross-terms that don’t cancel: (x2 + y2)(x – y) = x3 – x2y + xy2 – y3. Choice (B) is close but not right. (x3 – y3)(x3 – y3) = (x3 – y3)2, which also gives us cross-terms that don’t cancel: (x3 – y3)2 = x6 – 2×3y3 + y6. (Moreover, the sign is wrong on the y6 term.) Let’s skip to choice (E), since we see (x3 + y3). The remaining terms multiply out as follows: The correct answer is (E). Of course, you can also plug simple numbers. Don’t pick 0 for either x or y, because too many terms will cancel. Also, don’t pick the same number for x and y, because then the result is 0 (and every answer choice gives you 0 as well). But if you pick 2 and 1, you can keep the quantities relatively small and still eliminate wrong answers. 26 – 16 = 64 – 1 = 63. Our target is 63. (A) (8 + 1)(4 + 2)(2 – 1) = 48 Again, the correct answer is (E). Incidentally, a “zenzicube” is a very old word for the sixth power of a number. An even funnier word exists for the eighth power of a number: zenzizenzizenzic. The “zenzi” part means “square,” so a “zenzicube” is the square of a cube, or (x3)2 = x6. Likewise, a “zenzizenzizenzic” is the square of a square of a square, or ((x2)2)2 = x8. Modern exponential notation has its advantages.
Thursday, October 30, 2008 Workbook Wednesdays: Where-4 Art Thou Divisible? Answer Dost thou seek the answer to yesterday’s Challenge Problem? As promised, find it below! Question (1) When p is divided by 8, the remainder is 5. (2) x – y = 3 (A) Statement (1) ALONE is sufficient to answer the question, but statement (2) alone is not. Solution However from all of this, we can infer something else—specifically, that x is NOT a multiple of 4. Here’s why: y2 = (y2 — 1) + 1, which using the quadratic property, yields: y = (y+1)(y—1) +1. Because y is odd, then (y+1)(y—1) is even times even, which is a multiple of 4. However we ALSO know that either y+1 or y—1 is a multiple of 4, because they are consecutive multiples of 2. Therefore, y2 is 1 greater than a multiple of 8. (You can confirm this by thinking about all squared odd numbers: 12 = 1, 32 = 9, 52 = 25, 72 = 49, etc.) Since y2 divided by 8 yields a remainder of 1, from statement (1) we need x2 divided by 8 to yield a remainder of 4. This means that x must be even, but NOT be a multiple of 4, because if it were a multiple of 4, then x2 would be a multiple of 16, and therefore a multiple of 8. Statement (2): INSUFFICIENT. We know that x is even. However, we can come up with two different examples which lead to different answers. For example, if x = 8 and y = 5, then x is a multiple of 4, but if x = 6 and y = 3, then x is not a multiple of 4. The answer is A: Statement (1) is sufficient to answer the question, but statement (2) is insufficient.
Wednesday, October 29, 2008 Workbook Wednesdays: Where-4 Art Thou Divisible? Welcome to this week’s edition of Workbook Wednesdays, featuring the type of problem one would see if scoring above a 700 on the GMAT. As always, thanks to ManhattanGMAT for providing this week’s challenge problem! Be sure to check back tomorrow for the answer! Question (1) When p is divided by 8, the remainder is 5. (2) x – y = 3 (A) Statement (1) ALONE is sufficient to answer the question, but statement (2) alone is not.
Thursday, October 23, 2008 Workbook Wednesdays: X to the Nth Power Answer Check out the answer to yesterday’s Challenge Problem! Question 1) x < 1 2) n is even. (A) Statement (1) ALONE is sufficient to answer the question, but statement (2) alone is not. Solution If n is even, then x^n > 0, no matter what the value of x is (remember that x is nonzero). The reason that we care about the value of x^n is that we can simplify the question by dividing by x^n: After we divide both sides of the inequality by x^n, the question “Is x^n > x^(n+1)?” becomes “Is 1 > x?” ONLY IF x^n > 0, which is true if x > 0 OR if n is even. (Recall that x is nonzero; thus, we are allowed to divide by x^n.) On the other hand, if x^n < 0, then the question rephrases to “Is 1 < x?” Statement 1: INSUFFICIENT. We know that x < 1, but x could be positive or negative. Moreover, we do not know whether n is even or odd. As a result, we do not know the sign of x^n, and thus we do not know the answer to either the rephrased question or to the original question. Alternatively, you can choose positive and negative values of x and an odd n, in order to test the question. If n = 1 and x is positive (but less than 1), then x^n > x^(n+1). But if n = 1 and x is negative, then x^n > x^(n+1). Statement 2: INSUFFICIENT. We know that n is even, so we know that x^n > 0, and therefore we can rephrase the question as “Is 1 > x?” However, we do not know the answer to that question. Statements 1 & 2 TOGETHER: SUFFICIENT. Using Statement (2), we can rephrase the question as “Is 1 > x?”, to which Statement (1) gives us a definitive answer. The answer is C: BOTH statements TOGETHER are sufficient to answer the question, but neither statement alone is sufficient.
Wednesday, October 22, 2008 Workbook Wednesdays: X to the Nth Power Welcome to this week’s edition of Workbook Wednesdays! If you’re ready to sharpen your quantitative skills, check out the Challenge Problem below, courtesy of our friends at ManhattanGMAT. Question 1) x < 1 2) n is even. (A) Statement (1) ALONE is sufficient to answer the question, but statement (2) alone is not.
Thursday, October 16, 2008 Workbook Wednesdays: The Cost of Fuel Answer Check below to see what was paid at the pump in yesterday’s GMAT question! Question 1) Using fuel currently in its tank, the vehicle burned 8 gallons to cover 200 miles. 2) The vehicle can cover 7 and 1/7 miles for every dollar of fuel currently in its tank. (A) Statement (1) ALONE is sufficient to answer the question, but statement (2) alone is not. Solution Statement 1: SUFFICIENT. The statement tells us that the vehicle covered 200 miles on 8 gallons of the fuel mixture; that is, the fuel mixture delivers 25 miles per gallon. As a result, we can find the ratio of Fuel X (20 mpg) to Fuel Y (40 mpg) in the mixture (this ratio turns out to be 3:1, but we do not need to find the exact ratio; we simply need to know that there will be one unique ratio). Finally, we can use this ratio to find the weighted average cost per gallon of the fuel mixture (this weighted average turns out to be $3.50 per gallon, but again, we do not need the exact number). Statement 2: SUFFICIENT. The statement tells us that $1 of fuel “buys” 7 and 1/7 miles. Multiplying through by 7, we can rephrase the ratio as $7 for every 50 miles. Let’s now express the weighted average cost per gallon as $3w + $5(1-w) = 5 – 2w, where w represents the percent of Fuel X in the mixture (and 1-w represents the percent of Fuel Y in the mixture). The number of gallons bought by the $7 is then $7 divided by the average cost per gallon, or 7/(5 – 2w). Likewise, we can express the weighted average fuel efficiency (miles per gallon) as 20w + 40(1-w) = 40 – 20w. The number of gallons burned to cover 50 miles is then 50 divided by the average miles per gallon, or 50/(40 – 20w). We can now set these numbers of gallons equal to each other: 7/(5 – 2w) = 50/(40 – 20w) Knowing that the mixture is 75% Fuel X and 25% Fuel Y, we can now in theory calculate the average cost per gallon of the mixture. (In fact, we should have stopped before calculating w – simply knowing that we could calculate that percentage is sufficient.) The answer is D: EACH statement ALONE is sufficient to answer the question.
Wednesday, October 15, 2008 Workbook Wednesdays: The Cost of Fuel Welcome to this week’s edition of Workbook Wednesdays, featuring the type of problem one would see if scoring above a 700 on the GMAT. As always, thanks to ManhattanGMAT for providing this week’s challenge problem! Be sure to check back tomorrow for the answer! A certain military vehicle can run on pure Fuel X, pure Fuel Y, or any mixture of X and Y. Fuel X costs $3 per gallon; the vehicle can go 20 miles on a gallon of Fuel X. In contrast, Fuel Y costs $5 per gallon, but the vehicle can go 40 miles on a gallon of Fuel Y. What is the cost per gallon of the fuel mixture currently in the vehicle’s tank? 1) Using fuel currently in its tank, the vehicle burned 8 gallons to cover 200 miles. 2) The vehicle can cover 7 and 1/7 miles for every dollar of fuel currently in its tank. (A) Statement (1) ALONE is sufficient to answer the question, but statement (2) alone is not.
Thursday, October 09, 2008 Workbook Wednesdays: Medical Test Answer The results from yesterday’s GMAT question are in! See below for a detailed answer: Question (A) 20% Answer We choose 100 for the total of all tests, since we are only dealing with percents. Filling in the table with the given information and completing the total row and column, we get the following: Since the number of high morning+afternoon exams cannot be less than zero, it must actually be zero. This means that the table fills in this way: Thus, the percentage of evening exams that do NOT result in a high reading is 20/40 x 100%, or 50%. The correct answer is (D).
Wednesday, October 08, 2008 Workbook Wednesdays: Medical Tests Get ready for your quantitative check-up! As always, we’d like to thank ManhattanGMAT for providing this week’s Challenge Question and – more importantly – tomorrow’s answer! Question (A) 20%
Thursday, October 02, 2008 Workbook Wednesdays: A Slippery Slope Answer Check out the answer to yesterday’s Challenge Problem below! Question (1) Line k and the graph of the function f(x) = x^2 – bx, where b is positive, intersect on the x-axis. (2) Line k and the graph of the function g(x) = –x^2 – c, where c is positive, intersect on the y-axis. (A) Statement (1) ALONE is sufficient to answer the question, but statement (2) alone is not. Answer Statement (1) states that the graph of the function f(x) = x^2 – bx, where b is positive, and line k intersect on the x-axis. Since the function is a quadratic, we know that the graph of the function is a parabola. Let’s find the places that this function intersects the x-axis. This is equivalent to finding the values of x for which the function equals zero. Statement (2) states that the graph of the function g(x) and line k intersect on the y-axis. Again, we know that the graph of the function is a parabola, since the function itself is quadratic. To find where the parabola intersects the y-axis, we find the value of the function at x = 0. Statements (1) and (2) together: Line k goes through (0, –c) AND either (0,0) or (b, 0). If line k goes through (0, –c) and (b, 0), where both b and c are positive, then we know that the slope of line k is positive (both the rise and the run are positive). However, line k could go through (0, –c) and (0, 0), which would mean that line k is vertical (it actually would coincide with the y-axis). A vertical line has an undefined slope. Since the slope could either be positive or undefined, the two statements together are INSUFFICIENT. The correct answer is E: Statements (1) and (2) TOGETHER are INSUFFICIENT to answer the question.
Wednesday, October 01, 2008 Workbook Wednesdays: A Slippery Slope Welcome to this week’s edition of Workbook Wednesdays, courtesy of our friends at ManhattanGMAT! Get ready to wrap your brain around an advanced quantitative problem – the kind you would see if you are scoring a 700 or higher on the GMAT – then be sure to check back tomorrow for the answer. Question (1) Line k and the graph of the function f(x) = x^2 – bx, where b is positive, intersect on the x-axis. (2) Line k and the graph of the function g(x) = –x^2 – c, where c is positive, intersect on the y-axis. (A) Statement (1) ALONE is sufficient to answer the question, but statement (2) alone is not.
Thursday, September 25, 2008 Workbook Wednesdays: Solve for N Answer Find the answer to yesterday’s Challenge Problem below! Question (1) 3^n – 1 has three prime factors, not necessarily distinct. (A) Statement (1) ALONE is sufficient to answer the question, but statement (2) alone is not. Answer Statement (1) indicates that 3^n – 1 has three prime factors, not necessarily distinct. There is no way to intuit or derive a general solution for n; we must simply test values of n and see how many prime factors the expression has. If n = 1, then 3^n – 1 = 2, which has just one prime factor. N cannot be 1. Statement (2) indicates that n^2 = 2^n. By testing small numbers, we see that n could only be 2 or 4. For values of n above 4, the right side of the equation grows faster than the left side; for instance, if n = 5, then n^2 = 25, but 2^n = 32. The powers of 2 grow faster than the squares. So n cannot be any larger value. However, we do not know whether n is 2 or 4. INSUFFICIENT. Statements (1) and (2) together: n can only be 2. SUFFICIENT. The correct answer is C: Statements (1) and (2) TOGETHER are sufficient, but neither statement ALONE is sufficient.
Wednesday, September 24, 2008 Workbook Wednesdays: Solving for N Welcome to this week’s edition of Workbook Wednesdays! As always, we’d like to thank our friends at ManhattanGMAT for providing this week’s Challenge Problem. Check back tomorrow for an in-depth look at the answer! Question (1) 3^n – 1 has three prime factors, not necessarily distinct. (A) Statement (1) ALONE is sufficient to answer the question, but statement (2) alone is not.
Thursday, September 18, 2008 Workbook Wednesdays: Learn Your ABC’s Answer Here is the answer to Wednesday’s Challenge Question from Manhattan GMAT. Check back next week for another Workbook Wednesday question! Question If a, b, and c are integers, and the product abc is even, is b even? (A) Statement (1) ALONE is sufficient to answer the question, but statement (2) alone is not. Answer The given information simply guarantees that at least one of the integers a, b, or c is even. Statement (1) indicates that ab = (c)(some even integer). This means that the left side of the equation must be even. However, we could have this result either with an even b or with an odd b (if a is even). INSUFFICIENT. Statement (2) indicates that ac = (b)(some odd integer). If b is even, then we also know that a or c (or both) is even; this scenario fits the constraint that at least one of the variables is even. However, if b is odd, then the right side of the equation is odd. Therefore, BOTH a and c are odd, since ac is odd. This scenario contradicts the constraint that at least one of the variables is even. Thus, we know that b must be even. SUFFICIENT. The correct answer is B: Statement (2) ALONE is sufficient, but statement (1) alone is insufficient.
Wednesday, September 17, 2008 Workbook Wednesdays: Learn Your ABC’s Welcome to this week’s edition of Workbook Wednesdays, courtesy of our friends at ManhattanGMAT! Just like last week, this problem mimics the most advanced quantitative problems on the exam, the type of problem you will see if you are scoring around 700 or higher. Check back tomorrow for the answer! If a, b, and c are integers, and the product abc is even, is b even? (1) (ab)/c is an even integer. (A) Statement (1) ALONE is sufficient to answer the question, but statement (2) alone is not.
Thursday, September 11, 2008 Workbook Wednesdays: Recursive Curse Answer As promised, here is the answer to yesterday’s Challenge Problem! Question What values of A(1) create a sequence such that A(n) = A(n-2) for all n > 2? (I) x < 0 (A) I only Answer Now, you could try various values of A(1) and see whether you get the same value for A(3). A more general approach is to rephrase the question: you are looking for values of x such that f(f(x)) = x. (In other words, when you apply the function to the value twice, you get the same value back.) If f(x) = 1 – 1/(1-x), then f(f(x)) = 1 – 1/(1-[1 – 1/(1-x)], where brackets [] indicate the insertion of f(x) in place of x. This expression looks ugly, but let’s try to simplify it. In general, 1 – [1 – Z] = Z. So 1 – 1/(1-[1 – 1/(1-x)] = 1 – 1/(1/(1-x)). Next, 1/(1/Z) = Z. So 1 – 1/(1/(1-x)) = 1 – (1-x) = x. Amazingly, that ugly expression 1 – 1/(1-[1 – 1/(1-x)] = x, for all legal values of x. Thus, for any value of x besides 1, if you apply the function to it twice, you get the original value back. Try it! The answer is E: I, II, III, and IV.
Wednesday, September 10, 2008 Workbook Wednesdays: Recursive Curse Welcome to this week’s edition of Workbook Wednesdays, brought to you by our friends at ManhattanGMAT. Check back tomorrow for an in-depth look at the answer! Question What values of A(1) create a sequence such that A(n) = A(n-2) for all n > 2? (I) x < 0 (A) I only
Thursday, September 04, 2008 Workbook Wednesdays: Consecutive Divisibility Answer Pencil’s up! Here is the answer to yesterday’s Challenge Question. Question (1) The sum of n consecutive integers, starting at x, is divisible by xn. (A) Statement (1) ALONE is sufficient to answer the question, but statement (2) alone is not. Answer Statement (1) asserts that the sum of n consecutive integers, starting at x, is divisible by xn. First, we can connect this with other facts about consecutive integers. It turns out that the sum of n consecutive integers is divisible by n if and only if n is odd. (The reason is that the average number in a set of consecutive integers is actually the middle integer if you have 3, 5, or some other odd number of integers. But if you have an even number of integers, there is no middle integer, and the average number is not an integer. This matters because the sum of n consecutive integers divided by n IS the average number in a set of consecutive integers.) So we rule out even values of n. However, if we simply let x be 1, then the condition is satisfied by n = 1, 3, 5, or any other positive odd integer. We do not know whether n is equal to 1. INSUFFICIENT. Statement (2) seems even more complicated, but it can be defeated again by a judicious choice of x as 1. If x = 1, then x^n = 1, no matter what n is. Since all positive integers are divisible by 1, then there is no restriction on the value of n, which could be equal to 1 OR to any other positive integer. INSUFFICIENT. Statements (1) and (2) together: Putting what we have learned together, we know that if we let x = 1, then n might equal 1, but it could also equal any other positive odd number. We cannot answer the question definitively. INSUFFICIENT. The correct answer is (E): Statements (1) and (2) TAKEN TOGETHER are NOT sufficient to answer the question.
Wednesday, September 03, 2008 Workbook Wednesdays: Consecutive Divisibility Welcome to the another installment of Workbook Wednesdays. Thanks again to Manhattan GMAT for supplying today’s question and (more importantly) tomorrow’s answer! Just like last week, this problem mimics the most advanced quantitative problems on the exam, the type of problem you will see if you are scoring around 700 or higher. Question (1) The sum of n consecutive integers, starting at x, is divisible by xn. (A) Statement (1) ALONE is sufficient to answer the question, but statement (2) alone is not. Get out your pencils and scratch paper, and be sure to check back tomorrow for an in depth look at the answer!
Thursday, August 28, 2008 Workbook Wednesdays: Boats on the Lake Answer As promised, here is the answer to yesterday’s Manhattan GMAT Challenge Question! The key to this problem is to make the common term in both ratios equal. We should set up a table to display both ratios: Cat Can Kay 4 7 5 9 Now, we can double any of the ratios. In fact, we can multiply any row by any positive-integer factor (x2, x3, x10, etc.). We are constrained to positive-integer factors, though, because the actual number of any boat must be a positive integer. The number of canoes in each row should be the same, so that we can merge the ratios. The least common multiple of 7 and 5 is 35, so we multiply the top row by 5 and the bottom row by 7: Cat Can Kay 20 35 35 63 This means that the “3-way” ratio of catamarans, canoes, and kayaks is 20:35:63. That is, for every 20 catamarans on the lake, there are 35 canoes and 63 kayaks. This ratio is already reduced to lowest integers, because there are no prime factors in common among all 3 integers. The smallest number of boats that can be on the lake is 20 + 35 + 63 = 118. Since the total number of boats must be an integer, any possible number of boats must be a multiple of 118. To check which number is a multiple of 118, we factor 118 into its primes: 2 and 59. This allows us to spot 590, which is 59 x 10 and therefore 118 x 5. The answer is D.
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