Data Sufficiency Flashcards

Question

If n is not equal to 0, is |n| < 4 ? (1) n2 > 16 (2) 1/|n| > n

Answer 1

CAT 1, Q 31. Answer A. Inequalities. Lines The statement can be rephrased to: (i) if n > 0, is n < 4? (ii) if n < 0, is n > -4? Combing the two questions yields: is -4 < n < 4? (1) SUFFICIENT. We are able to conclude that n is definitely not between -4 and 4. Note that an absolute no is sufficient! n^2 > 16 n > 4 OR n < -4 (2) INSUFFICIENT. Test 1 > |-4| * -4. 1 > -16. Thus, n can be any negative value since |n|*n will always be negative and therefore less than 1. This statement is insufficient because n might not be between -4 and 4. 1/|n| > n 1 > |n| * n

Answer 2

CAT 1, Q 33. Answer A. Lines & Angles. Note that a = e because they are vertical angles, and e + f + g = 180. We can rewrite this equation to be e = 180 – f – g. Thus, we are looking for f + g = e = a ``` (1) SUFFICIENT. This statement allows us to calculate f + g. More specifically: b + c = 287 (b + f) + (c + g) = 180 + 180 287 + (f + g) = 360 (f + g) = 73 = e = a ``` (2) INSUFFICIENT. d + e = 269 OR d + a = 269. There are many combinations that satisfy this constrain. We are unable to determine a unique value for a

Answer 3

CAT 1, Q 34. Answer C. Combinatorics. 3 women + 2 men = panel. Chosen from x women and y men. In order to determine how many different panels that can be formed with these constraints, we need to know the values of x and y. The number of different panels that can be formed = xC3 * yC2 (1) INSUFFICIENT. Choosing 3 from x + 2 yields 56 groups. Thus, (x+2)C3 = 56. One concept that you need to know for the exam is that when dealing with combinations and permutations, each result corresponds to a unique set of circumstances. Because we know the number of groups yielded (in this case 56), then we know that there is only one possible of x (which is 6). However, we still do not know the value of y (2) INSUFFICIENT. knowing that x = y + 1 is only helpful if we know one of the values. With only this information, there are infinite possible values for x and y. (1) and (2). SUFFICIENT Together. We are able to determine x from statement (1), which allows us to calculate the value of y with statement (2) NOTE: with data sufficiency problems, we do not need to actually be able to calculate exact values; it is enough to know that we CAN calculate them if we wanted to do so

Answer 4

CAT 1, Q 37. Answer B. Algebraic Translations. Number of antelope in the herd today = 500 x = annual growth factor 1,000 2 We need to determine minimum n to yield a population of 1,000 or more. In order to solve this problem, we need to know the value of x. (1) INSUFFICIENT. There are infinite values of x that make this inequality true 500x^10 > 5,000 x^10 > 10 (2) SUFFICIENT. Let y = the growth factor from the statement (Note that y does not necessarily equal 2x because x is a growth factor. 500y^2 = 980 y = Sqrt(980/500) = Sqrt(49/25) = 7/5 = 1.4 x = (1.4 – 1)/2 + 1 = 1.2

Answer 5

Lecture # 1. Answer D. (1) SUFFICIENT. x can be 9, 18, 27, 36, etc. With this information, we can conclude that x does NOT fall between 20 and 26, inclusive, because multiples of 9 do not fall in this range (2) SUFFICIENT. We are told that x is unequivocally equal to 18, which does not fall in the given range. Thus, we can conclude that x does NOT fall between 20 and 26, inclusive.

Answer 6

OG13, Lecture # 1, DS #172. Answer B. (1) INSUFFICIENT. We can rephrase this question to say, “is the square root of an integer also an integer?” The answer is no, because if n is equal to the square of some prime number, then n^2 will be an integer, but n itself is not actually an integer. For example, if n = Sqrt(11), then n^2 = 11, which is an integer. However, n itself is not an integer. Thus, we do not have enough information to conclude that n is an integer (2) SUFFICIENT. This can be rephrased to state that n = (integer)^2. For all values of n, Sqrt(n) is an integer AND n is an integer.

Answer 7

``` Lecture # 1. Answer D. First we must detangle the complicated inequality in the question. a – b + c > a + b – c - b + c > b – c -2b > -2c b < c ``` (1) INSUFFICIENT. With this information, we know that b is negative, but the value of c could feasibly be even more negative than the value of b (2) INSUFFICIENT. With this information, we know that c is positive, but the value of b could feasibly be even more positive than the value of c (1) and (2) SUFFICIENT. If b is negative and c is positive, we can conclude that the original inequality is true because a negative will always be less than a positive

Answer 8

OG11, Lecture # 1, DS # 153. Answer B. The question can be rephrased to state, “is k prime?” because only prime numbers only have factors of 1 and themselves. (1) INSUFFICIENT. k > 4*3*2*1 or k > 24. k could be prime (e.g. 31), in which case the inequality in the question does not hold true. k could also not be prime (e.g. 27), in which case the inequality does hold true. (2) SUFFICIENT. None of the numbers that fall between 13! + 2 and 13! + 13 are prime (to test this, draw a number line). As a result, we know that the inequality in the question holds true.

Answer 9

FDP Strategy, Ch 6, Q 7. Answer A. Total Revenue in 2009 = C + M. Total Revenue in 2010 = (1+k)C + (1-k)M Since you already know that k is a positive integer, you know that C increases by the same percent by which M decreases. So, you do not actually need to know k. You already know that k percent of whichever number is bigger (C or M) will constitute a bigger change to the overall revenue. If C starts off bigger, then a k% increase in C means more new dollars coming in than you would lose due to k% decrease in the smaller number, M. The question can thus be rephrased, “Which is bigger, C or M?” (1) SUFFICIENT. this statement tells you that C is bigger than M. Thus, k% increase in c is larger than k% decrease in M, and the overall revenue went up (2) INSUFFICIENT. Knwoing the percent change doesn’t help, since you do not know whether C or M is bigger

Answer 10

FDP Strategy, Ch 6, Q 9. Answer C. For this problem, you do not necessarily need to know the value of x, y or z. You simply need to know the ratio of x:y:z (in other words, the value of x/y AND y/z). You must manipulate the information given to see whether you can determine this ratio ``` (1) INSUFFICIENT. There is no way to manipulate this equation to solve for a ratio. x + y = 2z x = 2z – y x/y = (2z – y)/y = 2z/y – 1 (2) INSUFFICIENT. Same as above 2x + 3y = z 2x = z – 3y x/y = (z – 3y)/2y = z/2y – 3/2 (1) and (2) SUFFICIENT. Use substitution to combine the equations x + y = 2(2x + 3y) x/y = -5/3 ``` 2(2z – y) + 3y = z y/z = -3 Therefore, the ratio x:y:z = 5:-3:1

Answer 11

OG13, HW1, Q6. Word Problem. Answer E. In this Number Line question, we have to figure out something about the relationship between K and L (1) INSUFFICIENT. Does not provide enough information to calculate the distance between K and L (2) INSUFFICIENT. Does not provide enough information to calculate the distance between K and L (1) and (2) INSUFFICIENT. Even with both statements, there are multiple possible values for the distance from K to L. In essence, the fixed distances (21km and 26km) can slide past each other, changing the K-to-L distance. Below are two examples JK = 11, KL = 10, LM =16 JK = 16, KL = 5, LM =21

Answer 12

OG13, HW1, Q17. Linear Equations. Answer B. We are trying to find the value of, k = m – n. It would be sufficient to know either k or m – n (1) INSUFFICIENT. k = m – 10. There are an infinite number of values for m, thus changing the value of k. (2) SUFFICIENT. m + 10 = n can be rewritten to say m – n = -10, which is the value of k

Answer 13

OG13, HW1, Q40. FDP. Answer A. Let S = the selling price of the TV, C = the cost of the TV, and M = markup of the TV. Thus, we can write the following equation: S = C + M. We are trying to determine M/S. (1) SUFFICIENT. Using substitution, we can determine M/S. We are told that M = 0.25C, which can be substituted into the original equation: S = C + 0.25C S = 1.25C M/S = 0.25C/1.25C = 1/5 (2) INSUFFICIENT. This statement tells us that the sum of the cost and markup is $250, but the cost and markup could be any two numbers summing to that amount. Thus, the markup could be anywhere from 0% to 100% of the cost. The markup-price ratio is not fixed.

Answer 14

OG13, HW1, Q72. Number Properties. Answer B. If m is an integer, is m odd? No rephrasing is necessary (1) INSUFFICIENT. m/2 is NOT an even integer. This does NOT mean that m/2 is an odd integer! Rather, it may be that m/2 is not an integer at all. Set up a table to test scenarios. Since both examples conform to the condition given in statement 1 yet give different answers to the question, the statement is not sufficient. m m/2 = not even m = odd int? 1 1/2 Yes 2 1 No (2) SUFFICIENT. Apply the properties of odds and evens. If m – 3 = even integer, then m = even integer + 3. m = even + odd is always odd!

Answer 15

OG13, HW1, Q103. Word Problems. Answer D. This is Sudoku! This means each digit must occur exactly once in each column and each row. This also means that there will be only three 1’s, three 2’s and three 3’s. We are asked for the value of r. (1) SUFFICIENT. If v + z = 6, then both v and z must equal 3, since 3 + 3 is the only way to achieve 6 if only 1’s, 2’s and 3’s can be used. If v and z are 3, then we know the other numbers sharing either a row or a column with v or z cannot be 3. The only remaining possibility is r = 3. (2) SUFFICIENT. We know that r = 6 – s – t and that r = 6 – u – x. Using substitution: 6 – s – t = 6 – u – x s + t = u + x Using the information given in statement 2 and substituting: s + t = 6 – u – x 6 – u – x = u + x 6 = 2u + 2x The only possibility is: u = 2 or 1, x = 2 or 1. Thus, r must be equal to 3

Answer 16

OG13, HW1, Q111. Number Properties. Answer D. In this odds and evens problem, we are asked whether the product of xy is even. In order for this to be true, just one of the two integers needs to be even. By the properties of odds and evens, an even integer multiplied by any other integer produces an even product. Therefore, we can rephrase the question as “Is x OR y even?” (1) SUFFICIENT. If x = y + 1, then we are dealing with two consecutive integers. In any pair of consecutive integers, one of the integers must be even and one must be odd. The answer to the rephrased question is yes. (2) SUFFICIENT. If x/y is an even integer, then x must be even because an odd number contains only odd prime factors. Thus, if x were odd and divided by y, the quotient would either be an odd integer or a fraction. Since x is definitely even, the answer to the rephrased question is yes.

Answer 17

OG13, HW1, Q131. FDP. Answer B. Let S = total value of all six shipments, Truck1 = value of shipments on truck 1 and Truck2 = value of shipments on truck 2. Given that Truck1 > 1/2S, we know that Truck2

Answer 18

OG13, HW1, Q141. Word Problem (Algebraic Translation). Answer B. Translate the words into equations Profit = P*Q – 100,000 – 0.05(P*Q) = 0.95(P*Q) – 100,000 = (19/20)(P*Q) – 100,000 Profit > 0 (19/20)(P*Q) – 100,000 > 0 R = P*Q Is Q > 21,000? We need to know price in order to determine Q. (1) INSUFFICIENT. R > 110,000. Without knowing the price per unit, it will be impossible to tell how many units were sold. (2) SUFFICIENT. Incorporating P = 5 into the equality allows us to solve for Q. After doing the first few steps of long division, we are able to determine that Q > 21,000 (19/20)(5Q) > 100,000 Q > 100,000/(19/4) Q > 400,000/19

Answer 19

OG13, HW1, Q144. Word Problem. Answer B. The question cannot be usefully rephrased. (1) INSUFFICIENT. The sum of 15 numbers is 60. Each of the numbers could be 4, but there are many other possibilities. (2) SUFFICIENT. The sum of ANY 3 numbers is 12. At first glance, it appears that each number must equal 4. In order to prove this fact, test scenarios. Make a list of numbers of which any 3 numbers sum to 12. It will quickly become apparent that this is impossible unless each number is equal to 4. 4,4,4,4,4,4,4,4,4,4,4,4,4,5,3 = 60 Example 1: 4 + 5 + 3 = 12 Example 2: 4 + 4 + 5 = 13

Answer 20

FDP Strategy Guide, Ch 7, Q6. FDP. Answer B. We can write two equations with the given information where x is the tens digit and y is the units digit for A. We can use these equations to rewrite the given equation for Q and determine that Q is only dependent on y. Thus, we can rephrase the question as: “what is the value of the units digit of A or the tens unit of B?” A = 10x + y B = 10y + x Q = 10(10y + x) – (10x + y) = 100y + 10x – 10x – y = 99y (1) INSUFFICIENT. Tens digit of A is 7 means that x = 7. This is insufficient to provide the value of Q. (2) SUFFICIENT. Tens digit of B is 6 means that y = 6. Thus Q = 99(6)

Answer 21

``` OG13, Lecture # 2, Q169. Exponents & Roots. Answer D. We can rephrase the question by manipulating the inequality (1/10)^n < (1/100) (1/10)^n < 1/(10^2) 1/(10^n) < 1/(10^2) n > 2? ``` (1) SUFFICIENT. The statement tells us that n is greater than 2, which answers our rephrased question. (2) SUFFICIENT. We need to rephrase this as well. (1/10)^(n-1) < 1/10 1/(10^(n-1)) < 1/10 n – 1 > 1 n > 2

Answer 22

Algebra Strategy Guide, Ch 9. (1) INSUFFICIENT. Does not tell you enough to solve for x (2) INSUFFICIENT. Does not tell you enough to solve for x (1) and (2) SUFFICIENT. Substitute y = 25 – x. Since x must be an integer, x must equal 4 280 < 20x + 10y < 300 280 < 20x + 10(25 – x) < 300 280 < 20x + 250 – 10x < 300 30 < 10x < 50 3 < x < 5

Answer 23

OG13, HW2, Q55. Percent Change = $ Change in Net Income / $ Original Net Income Net Income = Gross Income – Deductions (1) INSUFFICIENT. Guy’s gross income increased by 4% on the specific date in question. This does not allow us to calculated either the change in net income or the starting net income. (2) INSUFFICIENT. Guy’s deductions increased by 15% on the specific date in question. This does not allow us to calculated either the change in net income or the starting net income. (C) INSUFFICIENT. We know that gross income increases by 4% and that deductions increased by 15%, but we do not know the starting or ending values of net income. Thus, we cannot calculate the change or original. To prove insufficiency, develop contradictory examples. Example 1: Gross Income = 100 -- 104 Deductions = 10 -- 11.5 Net Income = 90 -- 92.5 % Change = 2.5 / 90 = 2.78% ``` Example 2: Gross Income = 50 -- 52 Deductions = 30 – 34.50 Net Income = 20 – 17.5 % Change = -2.5 / 20 = -12.5% ```

Answer 24

OG13, HW2, Q92. We can rewrite the question to be ad < bc? We usually cannot cross multiply variables across an inequality because these variables could be negative. However, we are told that they are positive. (2) SUFFICIENT. (ad/bc)^2 < ad/bc is true of positive fractions. Thus, 0 < ad/bc < 1 and 0 < ad < bc (1) INSUFFICIENT. 0 < (c – a) / (d – b). In order for this to be true, both the numerator and denominator must be negative or positive. - If both are positive, then c > a and d > b - If both are negative, then a > c and b > d In order to prove insufficiency, use real numbers. (4) (2) < (1)(1) NO (1) (1) < (4)(2) YES

Answer 25

OG13, HW2, Q114. All of the numbers have to be integers and the number of employees must evenly divide each of x, y and z. (1) INSUFFICIENT. Even with w as an unknown multiplier, we do not know the number of employees x = 2w y = 3w z = 4w (2) INSUFFICIENT. 18 pens, 27 pencils and 36 pads. All of these numbers are divisible by 9. Assuming the items are distributed evenly, there could be 9, 3, or 1 employees. (C) INSUFFICIENT. 2:3:4 is the same as 6:9:12, which is the same as 18:27:36. Statement 1 provides no new information beyond what statement 2 has already given us

Answer 26

``` OG13, Lecture # 3, Q 156. Linear Equations. Answer A. Rephrasing the equation yield the question: “what is x/n?” S = (2/n) / [(1/x) + (2/3x)] S = (2/n) / [(3/3x) + (2/3x)] S = (2/n) / (5/3x) S = 6x / 5n ? ``` (1) SUFFICIENT. x = 2n can be substituted into the equation to get 6(2n) / 5n = 12/5 (2) INSUFFICIENT. n = 1/2 can be substituted into the equation but we still do not know the value of x

Answer 27

Quant2, Lecture #3, Q 46. Quadratic Equations. Answer B. We can rephrase the question to: x^2 – y^2 = (x – y)(x + y) = ? (2) SUFFICIENT. This can easily be rephrased to give us one value for the expression x – y = 1/(x + y) (x – y)(x + y) = 1 (1) INSUFFICIENT. x – y = y + 2. We can substitute this into the question but this will not help us.

Answer 28

``` Quant2, Lecture # 3, Q 73. Quadratic Equations. Answer A. We can rephrase the question using the equation given in the question stem: x^2 – 2xy + y^2 = ? x^2 + y^2 – 2xy = ? 29 – 2xy = ? What is the value of xy? ``` (1) SUFFICIENT. This gives us exactly what we need (2) INSUFFICIENT. x = 5 does not tell us the value of xy Takeaway: immediately use the information that is given to you in the question.

Answer 29

OG13, HW 3, Q 98. Linear Equations. Answer E. There is no rephrasing to be done for this linear equations question. (1) INSUFFICIENT. If rs = 1, then r could be anything (2) INSUFFICIENT. If st = 1, then r could be anything (C) INSUFFICIENT. We can combine the equations, trying to put rst on one side of the resulting equation. We cannot isolate rst on one side of the equation and have a number by itself on the other side. Therefore, we cannot find a specific value for rst, and we cannot confirm or deny that rst = 1 (rs)(st) = (1)(1) r * t * s^2 = 1 rst = 1/2

Answer 30

``` OG13, HW 3, Q 67. (1) INSUFFICIENT. We can solve the factored quadratic equation, which has two solutions. n(n + 1) = 6 n^2 + n – 6 = 0 (n + 3)(n – 2) = 0 n = -3 OR n = 2 (2) SUFFICIENT. We can simplify the equation and solve for one value of n 2^2n = 16 (2^2)^n = 16 4^n = 4^2 n = 2 ```

Answer 31

OG13, HW 3, Q 1. Linear Equations. Answer B. No need to rephrase the question (2) SUFFICIENT. If x^2 =4, then x must equal either 2 or – 2. Since ǀ2ǀ = 2 and ǀ-2ǀ = 2, it must be true that ǀxǀ = 2 (1) INSUFFICIENT. If x = -ǀxǀ, then we know that x must be negative or zero, since it equals the negation of an absolute value, which will always be positive or zero. However, this does NOT tell us anything about the specific numeric value of ǀxǀ. For example, ǀxǀ could equal 5 or 7, etc. Takeaways: - If you know a variable’s Square, you know its absolute value - the expression -ǀxǀ means that x must be negative or zero

Answer 32

OG13, HW 3, Q99. Quadratic Equations. Answer B. The roots of a quadratic equation are the solutions to that equation, so this equation tells us that: (x – r)(x – s) = x^2 + bx + c = 0 x^2 – sx – rx + rs = x^2 + bx + c = 0 x^2 – x(s + r) + rs = x^2 + bx + c = 0 So we know that: b = -(s + r) c = rs We can rephrase the question to ask: is c < 0? (2) SUFFICIENT. We know that either r is negative and s is positive or vice versa. Either way, rs would have to be less than 0 (1) INSUFFICIENT. If b is negative, we know that –(r + s) is negative, so r + s must be positive. Test numbers to prove insufficiency to the yes/no question. -(r + s) < 0 r + s > 0

Answer 33

OG 13, HW 3, Q 160. Exponents & Roots. Answer A. We can rephrase the question, translating from words to symbols. Is b – a ≥ 2(3^n – 2^n)? Is b – a ≥ 2(3^n) – 2(2^n) (2) INSUFFICIENT. This statement does not tell us anything about a or b. (1) SUFFICIENT. We can manipulate the information in this statement to get b – a on one side of an equation. To do so, subtract equations. The tricky step is recognizing that we have to rewrite 3^(n+1) as (3^n)(3), which applies a subtle exponents rule. Is b – a ≥ 2(3^n) – 2(2^n)? Is 3(3^n) – 2(2^n) ≥ 2(3^n) – 2(2^n)? Is 3(3^n) ≥ 2(3^n)? Is 3 ≥ 2? The answer to this question will always be Yes. Thus, this statement is sufficient.

Answer 34

OG13, HW 3, Q 163. Quadratic Equations. Answer D. Let M = # of members and C = dollar contribution per person. Thus, 60 = M * C or M = 60/C. The question can be rephrased to “what is 60/C?” or simply “what is C?” (1) SUFFICIENT. If C = 4, then M = 60/4 = 15 (2) SUFFICIENT. We have two equations and two variables, but we still need to solve to make sure there we can come to one value. From the question stem, we know that C = 60/M, which we can substitute into our new equation from statement 2. 60 = (m – 5)(c + 2) 60 = (m – 5)(60/m + 2) 60 = 60 + 2m – 300/m – 10 0 = -10 + 2m – 300/m 0 = 2m^2 – 10m – 300 0 = m^2 – 5m – 150 0 = (m – 15)(m + 10) m = 15 OR m = -10 Since the number of club members must be positive, the only valid solution is m = 15. BEWARE of assuming that a quadratic will always produce two solutions and thus fail to provide a sufficient answer for a value question. On difficult GMAT questions, one of the solutions may be either invalid or redundant, leaving only one (therefore sufficient) answer. The safest policy is to factor and solve any quadratic before answering the original question.

Answer 35

OG13, HW 3, Q 171. Linear Equations. Answer A. The common term of t – x in both the numerator and denominator allows us to do strategic direct algebra: (2t + t – x)/(t – x) = [(2t) + (t – x)] / (t – x) = 2t/(t – x) + (t – x)/(t – x) = 2t/(t – x) + 1 The problem hinted at this manipulation by writing the numerator as 2t + t – x, rather than as the more normal form 3t – x. The only value that is still unknown is 2t/(t – x). The question can be rephrased to ask, “what is the value of 2t/(t – x)”? (1) SUFFICIENT. This answers the rephrased question (2) INSUFFICIENT. We cannot manipulate t – x = 5 to determine the value of 2t/(t – x)

Answer 36

OG13, Lecture # 4, Q 118. Formulas. Answer D. This problem is asking us to figure out whether w or 10 is smaller. If w is greater than 10, then min(10, w) = 10. If w is less than 10, min(10, w) = w. (2) SUFFICIENT. w = max(10, w) means that w equals 10 or something greater than 10. In other words, w ≥ 10. Thus, we can substitute this value into the question. Min(10, 10 or something greater than 10) = 10 (1) SUFFICIENT. To deal with this statement, we need to think about possible values of z. Specifically, we need to consider values of z that are less than and greater than 20 (because that will change the value of max(20, z) If z = 15, then max(20, z) = 20 If z = 25, then max(20, z = 25 No matter what value we pick for z, the value of w will have to be at least 20 (i.e. 20 or something greater than 20), which means the value of min(10, w) = 10

Answer 37

OG13, Lecture # 5, Q 158. Algebraic Translations. Answer B. Assign C to the number of tapes that Carmen has and R to the number of tapes that Rafael has. The question stem states that C + 12 = 2R, which can be rewritten to say C = 2R – 12. We are asked if C < R, which can be rephrased to: 2R – 12 < R?; R < 12? Alternatively, we could simplify the expression in terms of C. If C = 2R – 12, then R = (C + 12)/2 and the inequality becomes: C < (C + 12)/2?; C < 12? We have two rephrasings for this question, either of which would be sufficient: “Is R < 12 OR is C < 12?” (1) INSUFFICIENT. R could be either greater than or less than 12 (2) SUFFICIENT. C is less than 12. This is a direct answer to the second rephrasing of the original question. ALTERNATIVE METHOD: Test numbers. With inequalities, we only care about the edges so test the limits! (1) INSUFFICIENT. R = 6+ Low end: R = 6, which means C = 0. Thus, C < R High end: R = 100, which means C = 188. Thus, C > R (2) SUFFICIENT. C = 11- High end: C = 10, which means R = 11. Thus, C < R Low end: C = 0, which means R = 6. Thus, C < R Takeaways: - For inequality questions, TEST THE LIMITS! - Beware of hidden constraints! (e.g. positive integer questions)

Answer 38

OG13, Lecture # 5, Q 110. Statistics. Answer B. We are told 120,000 = (T + J + S) / 3. Thus, the sum of T, J, and S is $360,000. We must find the median of T, J, and S. (1) INSUFFICIENT. Knowing that Tom’s house cost $110,000 tells us that Tom’s house is not the most expensive house because it is lower than the average and that J + S = $250. However, we do not know whether Tom’s house was the least expensive or the middle house. Test scenarios. If Tom’s house is the least expensive, then the middle house would cost more than $110,000, and the median would be some number greater than $110,000. If Tom’s house is the middle house, however, then the median would be $110,000 (2) SUFFICIENT. Test scenarios. Jane’s house cannot be less expensive than both of the others, or else the average would have been higher than $120,000. By the same token, we know that Jane’s house cannot be more expensive than bother of the others. Otherwise, the overall average would have to be less than $120,000. There are only two viable scenarios: (1) All three houses are worth different amounts and Jane’s house is in the middle (2) All three houses are worth $120,000. In either case, the median is $120,000 Takeaways: -Given average and one number equal to the average, we know that the median is equal to that number

Answer 39

OG13, HW 4, Q 48. Inequalities. Answer B. (1) If a < c, we know that the lower range of possibilities for x is smaller than the lower range for y. Using only this information, we are unable to say for certain that x itself is smaller than y. We can test numbers to verify our thinking. For example, suppose a = 0 and c = 2. If x = 2 and y = 3, then x < y If x = 4 and y = 3, then x > y (2) SUFFICIENT. If b < c, the upper limit for x is smaller than the lower limit for y. What this means is that the biggest x can be is smaller than the smallest y can be. Therefore, x must be smaller than y

Answer 40

OG13, HW 4, Q 52. Inequalities. Answer D. The question asks whether x < 10 < y (1) SUFFICIENT. Since xy = 100 and both x and y are positive, either x = y = 10, or one of the variables is greater than 10 and the other one is less than 10. Since this assessment also tells us that x < y, we know that x < 10 and y > 10 (2) SUFFICIENT. Since we know that x and y are both positive, we can square root this inequality, which tells us that x < 10 < y

Answer 41

OG13, HW 4, Q 96. Formulas. Answer C. The question asks whether 0 is the smallest integer that is greater than or equal to x. For this to be true, x would have to be less than or equal to 0 but greater than -1. The question can be rephrased as “Is -1 < x ≤ 0?” (1) INSUFFICIENT. If x = -0.5, then the smallest integer that is greater than x is 0. However, if x = 0.5, then the smallest integer that is greater than x is 1. (2) INSUFFICIENT. There is no way to solve for an exact x here, so a good approach is to try to prove insufficiency by testing numbers. We know that if x = -0.5, then the smallest integer that is greater than x is 0. However, if x = -3.5, then the smallest integer that is greater than x is -3. (C) SUFFICIENT. The combined statements tell us that -1 < x < 0. The smallest integer that is greater than every number that is greater than every number in that range is 0.

Answer 42

OG13, HW 5, Q 57. Answer E. Algebraic Translations. Let T equal the number of $10 gift certificates sold and F for the number of $50 gift certificates sold. The given information can be translated and the question can be rephrased as “T + F = ?” F > 5 T and F are both integers (1) INSUFFICIENT. We are told T 5, this is not enough information to tell use the value of T + F (2) INSUFFICIENT. We are told 10T + 50F = 460. If this were a simple two variable, linear equation, it would not be solvable. However, the integer constraints and F > 5 greatly reduces the number of possible solutions. We know that F must be at least 6, we know that 6 * 50 = $300 worth of $50 gift certificates were sold, leaving $160 worth of gift certificates to account for. However, $160 worth of gift certificates could be formed a few different ways, which means there is more than one possible answer to the question: 3(50) + 1(10) = $160 2(50) + 6(10) = $160 (C) INSUFFICIENT. Even if both statements are true, more than one example from our statement 2 analysis matches the information. There could be three $50 and one $10 or two $50 and six $10.

Answer 43

OG13, HW 5, Q 68. Answer A. Rates & Work. Set up a RTW chart for the three machines and the combination of all three machines working together. The question asks for the time it takes Machine K to complete the task. We would be able to solve for K’s time if we any of the following (a) K’s rate (2) M + P combined rate (b) M and P combined rate (c) M and P combined time. (1) SUFFICIENT. We are given the time it takes M and P to complete the task together (36 minutes). We know that k + m + p = Total Rate = 1/24. Thus, k + (1/36) = 1/24, which allows us to solve for K’s rate and K’s time. (2) INSUFFICIENT. This statement provides the time necessary for K and P to complete the task when working together. This time could be used to find the combined rate of K and P. However, we cannot determine the individual rate of K. Takeaways: -GMAT often gives you the relationship among three variables (k + m + p) and asks us for the value of one of the variables; the right combo of variables will do the trick!

Answer 44

OG13, HW 5, Q 105. Answer B. Statistics. We are told that X + Y = 10 and 3X + 5Y = Cost. We are asked if X > Y. Since X + Y = 10, we can rephrase the question is X > 10 – X OR is X > 5? Alternatively, the question can be phrased is Y 4, which means Y could equal 5, in which case X = Y. Or Y could equal 6, in which case Y > X. Or Y could equal 4.1, in which case Y 5 and X > Y Takeaways: -In weighted average problems, think about a see-saw

Answer 45

OG13, HW 5, Q 106. Answer D. Rates & Work. Draw a picture of the cars both NOW and LATER to ensure that you can visualize the problem. Use relative rates; T = Difference in Distance between NOW and LATER / Difference between the Rates. We need to find the difference in rates between X and Y. (1) SUFFICIENT. This statement gives us the speed of Cars X and Y directly, so we can compute the difference in speeds through subtraction. T = 1 / 10 = 6 minutes (2) SUFFICIENT. In three minutes, the distance between the two cars grew by 1/2 mile. R = 0.5 / 3 = 1mi / 6min. Thus, T = 6 minutes.

Answer 46

OG13, HW 5, Q 109. Answer C. Statistics. In order to answer this, we need to know what is required to calculate the standard deviation of a list of numbers. The standard deviation is just a measure of how spread out the numbers on a list are relative to the mean. The formula for standard deviation is: SD = √∑(x - µ)^2 / N. The important takeaway from this formula is that we need to know the values of all the numbers on the list in order to compute the list’s standard deviation. (1) INSUFFICIENT. Average = (Last + First)/2 = (22 + 4)/2 = 13. However, knowing that M’s average value of 13 is not enough. As long as we remove two integers whose average is 13 from the list (e.g. 4 + 22 or 12 + 14), the average value of the remaining numbers will also be 13, but the actual numbers of the list will be different, so the SD will be different. (2) INSUFFICIENT. We have no way of knowing what the second number removed is (C) SUFFICIENT. If 22 is not on list M and M’s average is 13, the removed number MUST be 4, because the only way to keep M’s average the same as the given list’s average is to remove two numbers with an average value of 13. Thus, the two statements together allow us to calculate the SD Takeaway: -Need to know the values of all the numbers on a list in order to compute a list’s standard deviation

Answer 47

OG13, HW 6, Q 21. Answer E. Overlapping Sets. There are several ways to determine the percent of the drama club members that are female (e.g. using ratios, percents, or fractions), so do not worry about finding specific values. However, we still need either a percent or ratio of female members of the drama club to total members of the drama club. (1) INSUFFICIENT. Knowing the percent of female students who are drama club members does not tell us the percent of drama club members who are female. For example, say there are 100 female students and 40%, or 40, of them are in the drama club. If the drama club has 200 members, then the percent would be 40/200 = 20%. But if the club has 400 members, then the percent would be 40/400 = 10% (2) INSUFFICIENT. Knowing the percent of male students who are drama club members does not tell us the percent of drama club members who are female (C) INSUFFICIENT. To demonstrate the information that we have been given, set up a double-set matrix. Percent of drama club members that are female students = 0.4F / (0.25M + 0.4F), which is unsolvable. Without knowing how many male and female students there are in the school, we have no way of knowing what percent of the drama club is female Takeaway: -You only get to choose one smart number (e.g. you cannot choose numbers for M and F above)

Answer 48

OG13, HW 6, Q 34. Answer E. Overlapping Sets. From the question stem, we know that each student attended a science fair for either 0, 1, 2, or 3 days. Name variables such that D0 is the number of students who attended for no days, D1 is the number of students that attended for 1 day, D2 is the number of students that attended for 2 days, and D3 is the number of students that attended for 3 days. Rephrase the question to: “what is D3?” D0 + D1 + D2 + D3 = 900 (1) INSUFFICIENT. 270 = D2 + D3. We cannot solve for D3 because we do not know D2 (2) INSUFFICIENT. We cannot solve for D3 because we do not know D1 or D2. We cannot solve by plugging this equation into the given equation either because we do not know the value of D0 0.1(D1 + D2 + D3) = D3 0.1D1 + 0.1D2 + 0.1D3 = D3 0.1D1 + 0.1D2 = 0.9D3 D1 + D2 = 9D3 (C) INSUFFICIENT. Even with all three equations, we cannot solve for D3 because we do not know the value of D0.

Answer 49

OG13, HW 6, Q 49. Answer C. Overlapping Sets. Set up a double-set matrix for each statement. (1) INSUFFICIENT. Note that 0 people are neither directors of R nor K. This is because we know that 17 represents the total number of directors. Thus, each of the 17 people is a member of at least 1 board. No one is on neither board. Nonetheless, we cannot arrive at a value for the number of directors of both R and K. (2) INSUFFICIENT. This is not enough information to solve for the missing value. (C) SUFFICIENT. Now place all of the same information into the same matrix. From here, we can easily fill in every other box. 3 people are directors of both Company R and Company K. Notice that without zero in neither, we would not have sufficiency at this point. R: K = ?, No K = 5, Total = 8 No R: K = 9, No K = 0, Total = 9 Total: K = 12, No K = 5, Total = 17

Answer 50

OG13, HW 6, Q 70. Answer C. Statistics. We know that a test was given to a group of men and women and that the average score for the group was 80. The question asks if the average score of the women was greater than 85. If we knew whether there were more women or more men, and we knew what the average of either the women or the men was, we would know whether the overall average of 80 was closer to the men’s average or to the women’s average. (1) INSUFFICIENT. Look at two scenarios: (1) Suppose there is only 1 man and 1 woman in the group. If the man has a score of 74, then the woman must have a score of 86 for the average to be 80 – YES, Women Average > 85 (2) Suppose there is still 1 man but there are 2 women in the group. If the man has a score of 74, the women must have an average score of 83 for the total average to be 80 – NO, Women Average

Answer 51

OG13, HW 6, Q 93. Answer C. Overlapping Sets. Set up a double-set matrix for each statement. Is X greater than Y? (1) INSUFFICIENT. 0.2X are members of both club X and club Y. But we do not know whether X > Y (2) INSUFFICIENT. 0.3Y are members of both club X and club Y. But we do not know whether X > Y (C) SUFFICIENT. Both statements gave us information for the Yes-Yes cell. This is the key step: we can set those two expressions equal to each other and solve for X in terms of Y. Since X is 1.5 times the value of Y, X is in fact greater than Y. Even though we do not know the value of either X or Y, we know that X is greater than Y. 0.2X = 0.3Y X = 1.5Y

Answer 52

OG13, HW 6, Q 112. Answer D. Consecutive Integers. By looking at the number line, we can write the following relationships. We can rephrase the question in any of several ways: “What is K?” or “what is X?” or “what is y – x?” X = 3K Y = 7K Y – X = 4K (1) SUFFICIENT. X = 1/2. If X = 1/2 = 3K, then K = 1/6 and Y = 7/6 (2) SUFFICIENT. If Y – X = 2/3 = 4K, then K = 1/6 and Y = 7/6 The tick marks represent an evenly spaced set, of which consecutive integer sets are a special example

Answer 53

CAT 2, Q 2. Answer A. Digits & Decimals. We are told that N is a positive three-digit number that is greater than 200 and that each digit of N is a factor of N itself. We must determine the value of N. (1) SUFFICIENT. We are told that the tens digit of N is 5, thus N = x5z. All multiples of 5 end with a units digit of either 5 or 0. However, the units digit of N cannot be 0, since 0 is not a factor of any number. Therefore, the units digit of N must also be 5. Therefore N has the form x55, with only the hundreds digit left to consider. After working through the possible values of x, we can conclude that N = 555 - X = 2, N = 255. Not possible – X cannot be an even number because N is odd and thus does not have any even factors - X = 3, N = 355. Not possible – N is not a multiple of 3 - X = 5, N = 555. This is a possible value for N - X = 7, N = 755. Not possible – N is not a multiple of 7. - X = 9, N = 955. Not possible – N is not a multiple of 9. (2) INSUFFICIENT. We are told the units digit of N is 5, thus N = xy5. If the units digit of N is 5, then 5 must be a factor of N. All integers ending in 5 are multiples of 5, though, so this fact doesn’t narrow the possibilities any further. N is a multiple of 5 and any number is a multiple of 1. Using only these digits, try to formulate two numbers that satisfy the statement. Both 515 and 555 satisfy statement (2) so it is not sufficient to answer the question.

Answer 54

CAT 2, Q 7. Answer C. Exponents & Roots. The question asks whether x^n is less than 1. In order to answer this, we need to know not only whether x is less than 1, but also whether n is positive or negative since it is the combination of the two conditions that determines whether x^n is less than 1. (1) INSUFFICIENT. If x = 2 and n = 2, x^n = 22 = 4. If x = 2 and n = -2, x^n = 2(-2) = 1/(22) = 1/4. (2) INSUFFICIENT. If x = 2 and n = 2, x^n = 22 = 4. If x = 1/2 and n = 2, x^n = (1/2)2 = 1/4. (C) SUFFICIENT. Taken together, the statements tell us that x is greater than 1 and n is positive. Therefore, for any value of x and for any value of n, x^n will be greater than 1 and we can answer definitively "no" to the question.

Answer 55

CAT 2, Q 8. Answer D. Polygons. Let the length (longer dimension) of each rectangular tile be called L, and the width (shorter dimension) of each tile W. Then each horizontal side of square ABCD has total length 2W + 3L, and each vertical side has total length 4L. (1) SUFFICIENT: Because ABCD is a square, these total lengths must be equal: 2W + 3L = 4L, which reduces to L = 2W. Therefore, each side of square ABCD is equal to 4L = 8W, and the total area of square ABCD is (8W)(8W) = 64W^2. The total area of the tiles is 14(L × W) = 14(2W × W) = 28W^2. The desired fraction is thus (28W^2)/(64W^2) = 28/64. (2) SUFFICIENT. Each horizontal side of square ABCD has total length 3L, and each vertical side has total length 4L – 2W. Because EFGH is a square, these total lengths must be equal: 3L = 4L – 2W, which reduces to L = 2W. Therefore, each side of square ABCD is equal to 3L = 6W. In turn, ABCD must also be a square, since each of its sides is 2W longer than the corresponding side of EFGH (i.e., longer by W on each side). Therefore, each side of ABCD is equal to 6W + 2W = 8W, and the total area of square ABCD is (8W)(8W) = 64W^2. The total area of the tiles is 14(L × W) = 14(2W × W) = 28W^2. The desired fraction is thus (28W^2)/(64W^2) = 28/64.

Answer 56

CAT 2, Q 10. Answer D. Divisibility & Primes. (1) SUFFICIENT. The factors of any number N can be sorted into pairs that multiply to give N. However, if N is a perfect square, one of these ‘pairs’ will consist of just one number: the square root of N. Since all of the other factors can be paired off, it follows that if N is a perfect square, then N has an odd number of factors. This statement then implies that N is not a perfect square. (e.g. perfect square 25 has 3 distinct factors – 1 and 25, 5 and 5) (2) SUFFICIENT. In order to test this statement, sum the number of distinct factors for several perfect squares. 4: factors are 1 and 4, 2 and 2. Sum of distinct factors = 7 (ODD) 16: factors are 1 and 16, 2 and 8, 4 and 4. Sum of distinct factors = 31 (ODD) 25: factors are 1 and 25, 5 and 5. Sum of distinct factors = 31 (ODD) 36: factors are 1 and 36, 2 and 18, 3 and 12, 4 and 9, 6 and 6. Sum of distinct factors = 91 (ODD) Following this pattern, if N is a perfect square, then the sum of its distinct factors should be odd. However, we are told that the sum of the distinct factors of N is even. Thus, N is not a perfect square. Takeaways: - If N is a perfect square, then it has an odd number of distinct factors - If N is a perfect square, then the sum of its distinct factors is odd

Answer 57

CAT 2, Q 14. Answer A. Circles & Cylinders. We are told that DAB is a right triangle. A right triangle inscribed in a circle will have a diameter of the circle as its hypotenuse. Therefore DB is a diameter. Denoting the center of the circle with O, and adding the auxiliary line segment (radius) OA, we get a clearer picture. We now see that both angle ADB and angle ACB are inscribed angles in the circle, and that they both span the same arc, ACB. Therefore, they must be equal in measure (and each equal to half the corresponding central angle, AOB. But we know how big angle ACB is—it’s 60 degrees, because ABC is an equilateral triangle. Thus, angle ADB is 60 degrees as well, and angle ABD must be 30 degrees. This is because the three internal angles of right triangle DAB add up to 180 degrees. At this point, we can recognize triangle DAB as a 30-60-90 triangle. (1) SUFFICIENT. Knowing that the short side of the 30-60-90 triangle is equal to 4, we can solve for the hypotenuse DB, which is 8. We know that DB is the diameter of the circle, so we can calculate the desired area. (2) INSUFFICIENT. This statement tells us nothing about the radius, diameter or circumference of the circle. In fact, in light of the advance work we did, we can see that it tells us nothing that we do not already know. Takeaway: -Two inscribed angles within a circle that span the same arc must be equal in measure (and each equal to half the corresponding central angle)

Answer 58

CAT 2, Q 17. Answer A. Linear Equations. Note that one need not determine the values of both x and y to solve this problem; the value of product xy will suffice. (1) SUFFICIENT. The statement can be rephrased to say x = -3y. If x and y are non-zero integers, we can deduce that they must have opposite signs: one positive, and the other negative. Therefore, this last equation could be rephrased as |x| = 3|y|. We don’t know whether x or y is negative, but we do know that they have the opposite signs. Converting both variables to absolute value cancels the negative sign in the expression x = -3y. We are left with two equations and two unknowns, where the unknowns are |x| and |y|: |x| + |y| = 32 |x| – 3|y| = 0 Subtracting the second equation from the first yields |y| = 8. Substituting 8 for |y| in the original equation, we can easily determine that |x| = 24. Because we know that one of either x or y is negative and the other positive, xy must be the negative product of |x| and |y|, or -8(24) = -192. (2) INSUFFICIENT: Provides two equations with two unknowns. Solving these equations allows us to determine |x| = 24 and |y| = 8. However, this gives no information about the sign of x or y. The product xy could either be -192 or 192. |x| + |y| = 32 |x| - |y| = 16

Answer 59

CAT 2, Q 24. Answer B. Formulas. The sequence described can be written as An+1 = kAn. Note that if the positive constant k is greater than 1, the sequence will be increasing, and if the positive constant k is less than 1, the sequence will be decreasing. Also note that k cannot equal 1, 0, or any negative number. (1) INSUFFICIENT. The ninth term is 81. Consider the possible cases for k: k > 1 and k 1 and k 1, then the sequence will be increasing, meaning the terms one through four must be less than 1 and the terms six through nine must be greater than 1. In this case, again, a total of 4 terms will be greater than 1. Because 4 out of the 9 terms are greater than 1 regardless of the value of k, this statement is sufficient.

Answer 60

CAT 2, Q 27. Answer D. Exponents & Roots. The best way to answer this question is to use the exponential rules to simplify the question stem, and then analyze each statement based on the simplified equation. (3^27)(35^10)(z) = (5^8)(7^10)(9^14)(x^y) (3^27)(5^10)(7^10)(z) = (5^8)(7^10)(3^28)(x^y) (3^27)(5^10)(7^10)(z) = (5^8)(7^10)(3^28)(x^y) 25z = 3x^y (1) SUFFICIENT. Analyzing the simplified equation above, we can conclude that z must have a factor of 3 to balance the 3 on the right side of the equation. Statement (1) says that z is prime, so z cannot have another factor besides the 3. Therefore z = 3. Since z = 3, the left side of the equation is 75, so x^y = 25. The only integers greater than 1 that satisfy this equation are x = 5 and y = 2, so statement (1) is sufficient. (2) SUFFICIENT. Analyzing the simplified equation above, we can conclude that x must have a factor of 5 to balance out the 5^2 on the left side. Since statement (2) says that x is prime, x cannot have any other factors, so x = 5. Therefore statement (2) is sufficient.

Answer 61

CAT 2, Q 29. Answer C. Positives & Negatives. (1) INSUFFICIENT. According to this statement, the magnitude of b is greater than 1; that is, either b > 1 or b < –1. On the other hand, a can have any value whatsoever (except zero). With that freedom, it’s possible to choose values of a and b that yield contrasting answers to the question. If a = 1 and b = 2, then the answer is YES (2 > 1/2) If a = –1 and b = 2, then the answer is NO (–2 < –1/2) (2) INSUFFICIENT. In interpreting this statement, note the influence of signs. The values ab and a/b have to have the same sign (either positive or negative). In order for their sum to be positive, they must both be positive. Therefore, the two variables a and b are either both positive or both negative. Within these constraints, it’s possible to choose values of a and b that yield contrasting answers to the question. If a = 1 and b = 2, then the answer is YES (2 > 1/2) If a = 1 and b = 1/2, then the answer is NO (1/2 < 2) (C) SUFFICIENT. Statement 1 indicates that b > 1 or b < –1. Statement 2 indicates that a and b are either both positive or both negative. First, consider the case in which both a and b are positive. In that case, b > 1. Consider a the “starting” number. Multiplying a by a number greater than 1 will make a larger. Dividing a by a number greater than 1 will make a smaller. In other words, ab must be greater than a/b. Now consider the remaining case, in which both a and b are negative. In this case, b < –1. Since both numbers are negative, the negative signs cancel when they are multiplied or divided—in other words, the outcome is exactly the same as in the case where both a and b are positive. Again, ab must be greater than a/b. Therefore, ab > a/b in all cases. Both statements together are sufficient.

Answer 62

CAT 2, Q 30. Answer B. Fractions. We are told that a = (b/100) * c, which can be rephrased to say c = 100a/b. We must find the value of c (1) INSUFFICIENT. a = (c/100) * b, which can be rephrased to say c = 100a/b. This statement is identical to the equation given in the question stem, so statement (1) does nothing to help determine the value of c. (2) SUFFICIENT. b = (c/100) * a, which can be rephrased to say c = 100b/a. Set the equation from the question stem and the equation from the statement equal to each other. We learn that a = b, which means that c = 100 100a/b = 100b/a 100a^2 = 100b^2 a^2 = b^2 (we are told that a, b and c are positive) a = b

Answer 63

Quant 2, Lecture # 7, Q 123. Answer A. Triangles & Diagonals. The question is asking: BC = BD = ? We can use all of the information in the question stem to determine that angle ABD = x. Thus, BD = AD. ``` DBC = 180 – 4x ABC = 180 – 3x ABC = ABD + DBC 180 – 3x = ABD + 180 – 4x x = ABD ``` (1) SUFFICIENT. We are given the length of AD, which means that AD = BD = BC = 6 (2) INSUFFICIENT. Knowing the degree of x does not help us determine the length of side BC

Answer 64

OG12, Lecture # 7, Q 173. Answer D. Triangles & Diagonals. We are told that arc PQR is a semicircle, which provides two key pieces of information (1) PR is the diameter, which equals a + b (2) angle PQR is a right triangle. Thus, the two triangles shown in the picture are similar triangles because we are given two inscribed right triangles. We can relate the length of b to the length of a accordingly. ``` 2/a = b/2 ab = 4 ``` (1) SUFFICIENT. We are told that a = 4. Thus, b = 1 and PR = 5 (2) SUFFICIENT. We are told that b = 1. Thus, a = 4 and PR = 5 Takeaway: - There are two common disguises for similar triangles (1) Two inscribed right triangles (as shown above) (2) Acorn shaped triangles

Answer 65

OG13, HW 7, Q 42. Answer E. Polygons. We can rephrase the question to say “Are the length and width of the glass rectangle greater than or equal to the length and width, respectively, of the tabletop rectangle?” (1) INSUFFICIENT. This statement tells us nothing about the size of the glass (2) INSUFFICIENT. This statement tells us nothing about the size of the tabletop (C) INSUFFICIENT. If the glass is 2,400 square inches, then it could be 40 inches wide and 60 inches long, since 40 * 60 = 2,400, so its dimension could be greater than or equal to those of the tabletop. However, the glass rectangle could also be 24 inches wide by 100 inches wide, since 24 * 100 = 2,400, which would mean that its width would be smaller than the tabletop’s width.

Answer 66

OG13, HW 7, Q 74. Answer A. Coordinate Plane. Draw a picture to help visualize the solution. (1) SUFFICIENT. If the x-coordinate of point R is -1, R is somewhere on the vertical line with the equation x = -1. Since the midpoint between the x-coordinates of the two points is -1, that line is the same distance from (-3, -3) and (1, -3), so any point on it is as well. (2) INSUFFICIENT. If point R is on the horizontal line y = -3, then it could be the point (1, -3) itself, in which case it is closer to this point than to (-3, -3). Alternatively, point R could be the point (-1, -3), in which case it is equidistant to (-3, -3 and (1, -3)

Answer 67

OG13, HW 7, Q 79. Answer C. Triangles & Diagonals. The question “Does Area(ABC) equal Area(DBA)?” can be rephrased to “Does ac = dx?” (1) INSUFFICIENT. a = d√2 is not enough information because we do not know anything about the values of c and x (2) INSUFFICIENT. a = c. Thus, we know that x = c√2, which is not enough because we do not know anything about the value of d. (C) SUFFICIENT. From the first statement, we know that a = d√2, and from the second statement, we know that x = c√2. Start by replacing a with d√2 ``` ac = (d√2)(c). Then we can rewrite the equation relating x and c in terms of c. x = c√2 c = x/√2. Replace x with x/√2 ac = (d√2)(x/√2) ac = dx ```

Answer 68

OG13, HW 7, Q 155. Answer A. Coordinate Plane. Redraw the picture exaggerating constraints (ie make sure that PQ is substantially longer than OP) so that our eyes will be working for us as we make sense of this diagram. We can rephrase the question as “Is (1/2)(b)(h) > 48 or is bh > 96?” (1) SUFFICIENT. Knowing the coordinates of point P tells us the value of h = 8, the y-coordinate of P. It also tells us that OX = 6, the x-coordinate of P. First consider what would happen if OP = PQ. If OP = PQ, then triangle OPQ would be isosceles, and the height would be the perpendicular bisector of the base OQ. Thus, we have that OX = XQ = 6. However, we know that OP OX. Thus, the area of the smaller triangle is A = (1/2)(6)(8) = 24. In turn, we know that the Area of OPQ = (1/2)(something greater than 12)(8) = greater than 48. (2) INSUFFICIENT. Knowing the coordinates of point Q tells us nothing about the height of triangle OPQ.

Answer 69

``` OG13, HW 7, Q 166. Answer C. Lines & Angles. The key is that the quadrilateral formed by the four lines has four interior angles that must sum to 360. Label interior angels p, q, r and s. With this information, we can rephrase the question to “what is the value of p + q?” p + q + r + s = 360 x + p = 180 y + q = 180 (x + p) + (y + q) = 360 x + y = 360 – p – q ``` (1) INSUFFICIENT. w = 95, which means that 95 + s = 180 or s = 85. Unfortunately, this does not allow us to find the value of p or q, or the sum of p and q. (2) INSUFFICIENT. z = 125, which means that 125 + r = 180 or r = 55. Unfortunately, this does not allow us to find the value of p or q, or the sum of p and q. (C) SUFFICIENT. Since we know that s + r = 85 + 55 = 140, we can determine that p + q must equal 360 – 140 = 220. We have answered our rephrased question. x + y = 360 – 220 = 140

Answer 70

OG12, Lecture # 8, Q 82. Answer B. Divisibility & Primes. Your first instinct in any divisibility problem should be to break down numbers into their prime factors. 105 = 3 * 5 * 7. Therefore, for xy to be divisible by 105, it needs to have at least one factor each of 3, 5 and 7 in its prime box. Because xy is the product of x and y, it must be divisible by the individual factors of x and y, namely 2 and 3 for x, and 2 and 7 for y. This assures us that the prime box of xy already contains at least one 3 and one 7 (as well as two 2’s, which are not relevant to the question at hand. The remaining question is whether xy also contains a 5. To answer this question with a Yes, we need to know that a 5 shows up in either x or y. Thus, we can rephrase the question as “is x or y divisible by 5?” (1) INSUFFICIENT. The fact that x is divisible by 9 does not tell us whether it is also divisible by 5. (2) SUFFICIENT. This statement tells us that y is divisible by 25. Any number that is divisible by 25 is also divisible by 5, since 5 is a factor of 25. Thus, y is divisible by 5, and we can answer the rephrased question

Answer 71

MGMAT, Lecture # 8. Answer C. Divisibility & Primes. Note that we are looking for the greatest POSSIBLE common factor rather than the GCF itself. The greatest possible common factor between two integers is the distance between them. (1) INSUFFICIENT. There are an infinite number of possibilities. (2) INSUFFICIENT. We do not know whether or not a is an integer. The greatest common factor only applies to integers (C) SUFFICIENT. If a is an integer, and b = a + 4, then b must also be an integer. The distance between a and b is 4 no matter what. The greatest POSSIBLE common factor between two integers is the distance between them. Thus, the greatest possible common factor is 4. Takeaway: - Pay close attention to every word in divisibility problems; in this problem, the inclusion of the word “possible” completes changes the answer - If you were to exclude the word “possible”, then you would not be able to determine the GCF and the answer would be E

Answer 72

OG13, Lecture # 135. Answer B. Divisibility & Primes. We want to know if n students can be divided evenly into m classrooms. We know that m and n must both be integers because they are quantities, so this must be a Divisibility problem! This question can be rephrased as, “is n divisible by m?” given that 3 13 (1) INSUFFICIENT. This statement tells us that 3n is divisible by m (3n/m = integer). Choose smart numbers. If we pick a number for n first, it seems more straightforward to figure out what m could be (because m can only be between 3 and 13) The question states that n > 13, so let’s try n = 14. The question states that 3n is divisible by m. If n is 14, then 3n = 3 * 14 = 42. If 3n is 42, what could m be? Of the numbers between 3 and 13, 42 is only divisible by 6 and 7. Now we know that, when n = 14, m can be 6 or 7. Now let’s go back to the original question. If n = 14 and m = 6, then n is not divisible by m (answer is NO). If n = 14 and m = 7, then n is divisible by m (answer is YES) Why is this statement insufficient? When n = 14, 3 * 14 = 3 * 2 * 7. For (3 * 2 * 7) to be divisible by m, it must have all the prime factors that m has. That is why the only two possible values for m are 6 (2 * 3) and 7. Now we can see why 3n being divisible by m did not guarantee that n would be divisible by m. m is able to have a 3 as a prime factor that did not originally come from n. (2) SUFFICIENT. Use the same methodology to see why statement 2 is sufficient. The statement tells us that 13n is divisible by m. The difference between the two statements is that there are possible values of m that are divisible by 3, but there are no possible values of m that are divisible by 13. Therefore, the only way that 13n can be divisible by m is if n contains all the prime factors of m. And that is just another way of saying that n must be divisible by m. A quick look at a few numbers confirms that this statement is sufficient. If n = 14, 13n = 13 * 2 * 7. Possible values of m include 7 (14 is divisible by 7) If n = 15, 13n = 13 * 3 * 5. Possible values of m include 5 (15 is divisible by 5)

Answer 73

OG13, HW 8, Q 69. Answer A. Positives & Negatives. Sometimes, it is good to exaggerate the picture somewhat to ensure that we do not find unjustified features, such as even spacing (1) SUFFICIENT. If q = -s, then q and s are equally spaced on opposite sides of zero. That is to say that zero is halfway between q and s (test numbers if unsure). If zero is halfway between q and s, r must be closer to zero than either one of them because it is definitely somewhere between q and s. If r is between 0 and s, then it is closer to 0 than s. But s and q are equidistant from 0, which means that r is also closer to 0 than q. The logic is the same if r is between q and 0. Finally, t is definitely farther from zero than s since t is to the right of s (and thus a larger positive number). Therefore, r is the closest to zero (2) INSUFFICIENT. If –t

Answer 74

OG13, HW 8, Q 83. Answer E. Divisibility & Primes. This remainders questions tells us that k is one of the numbers 57, 58, 59, 60, 61, 62, 63, 64, 65. (1) INSUFFICIENT. This statement can be rephrased “k is not a multiple of 2” because all multiples of 2, otherwise known as even numbers, have a remainder of 0 when divided by 2. Since any integer that is not an even number is an odd number, the simplest possible rephrase to this statement is “k is an odd number”. Thus, k = 57, 59, 61, 63, or 65 (2) INSUFFICIENT. This statement can be rephrased as “k + 1 is a multiple of 3”. We can prove insufficiency by looking at the possible values of k and seeing that there are multiple possible values of k 59 + 1 = 60 62 + 1 = 63 are both multiples of 3 (C) In order to combine both statements, we can make a table of the odd numbers in the range and determine whether k + 1 is divisible by 4 for more than one of them ``` k, k + 1, is k + 1 divisible by 3 57, 58, N 59, 60, Y 61, 62, N 63, 64, N 65, 66, Y ``` If both statements are true, k could be either 59 or 65, so we do not have enough information to determine the value of k

Answer 75

OG13, HW 8, Q 97. Answer C. Positives & Negatives. (1) INSUFFICIENT. There is no information about the relative values of x and y. They could be equal, y could be greater than x, or x could be greater than y and this could still be true (2) INSUFFICIENT. In order for y^x to be a negative number, y must be a negative integer and x must be an odd integer. This is not sufficient because x could be a negative odd integer, which does not have to be greater than y x = -1, y = -1, y^x = -1, x = y x = -1, y = -3, y^x = -1/3, x > y (C) SUFFICIENT. Statement 2 tells us that y is a negative number. If x + y > 0, and y is negative, then x must be positive

Answer 76

OG13, HW 8, Q 101. Answer A. Inequalities. The important constraint here is that x is a negative number. (1) SUFFICIENT. If x > 0, then x > 3. If x < 0, then x < -3. Since we are told that x is negative, we know that x < -3 (2) INSUFFICIENT. The tricky part of this statement is that the cube root of -9 is not an integer, so we cannot easily calculate the upper bound of x. However, we know that the upper bound of x must be between -2 and -3 because (-2)^3 = -8 and (-3)^3 = -27. This means that there are values greater than -3 that make the equation true. But it is also true that x can be less than -3. For instance, (-5)^3 = -125, which is less than -9. Therefore the statement is insufficient

Answer 77

OG13, HW 8, Q 173. Answer C. Odd & Evens. For the Combo x – y to be odd, one of these terms must be odd and the other even. Thus, we can rephrase the question somewhat as “are x and y opposites in terms of Odds & Evens?” (1) INSUFFICIENT. Create an odds/evens table. z = odd, z^2 = odd, x = odd z = even, z^2 = even, x = even (2) INSUFFICIENT. Create an odds/evens table. z = even, z – 1 = odd, x = odd z = odd, z – 1 = even, x = even (C) SUFFICIENT. Combine the tables. We can see that whenever x is even, y will be odd, and vice versa. Thus, x and y are opposite in terms of odds & evens, and x – y must be odd z = even, z – 1 = odd, y = odd, x = even z = odd, z – 1 = even, y = even, x = odd Alternatively, we could use direct algebra by subtracting the y equation from the x equation: x = z^2 y = z^2 – 2z + 1 x – y = z^2 – (z^2 – 2z + 1) x – y = 2z – 1 2z must be even; thus, 2z – 1 must be odd, and x – y is odd as well

Answer 78

OG13, HW 9, Q 122. Answer D. Circles & Cylinders. Let R be the radius of the larger circle and r be the radius of the smaller circle. We are asked for the difference in areas of these circles, i.e. piR^2 – pir^2. Dividing by pi, we can rephrase the question as “what is R^2 – r^2” (1) SUFFCIENT. AB is the radius of the smaller circle, so r = 3. Furthermore, AC is the radius of the larger circle, so R = 3 + 2 = 5. We thus have the values of both R and r directly (2) SUFFICIENT. CE is a radius of the larger circle, so R = CD + DE = 1 + 4 = 5. Using the fact that all radii of a circle are equal, we can also establish that AC = 5. Since CD = 1, we know that the diameter of the small circle, AD, equals AC + CD = 5 + 1 = 6. Therefore, r = 3, and again we have both R and r.

Answer 79

OG13, HW 9, Q 125. Answer B. Probability. Since there are only three chips, we can use the 1 – x principle to simplify the question. The only chips that are not white or blue are red, so we can rephrase as “what is the probability that the chip is red?” (1) INSUFFICIENT. P(B) = 1/5 and P(W or R) = 4/5. However, we do not have any way of determining the probability that the chip is red. All we know is that this probability is between 0 and 4/5 (2) SUFFICIENT. This statement provides the answer to our rephrased question above. If the probability that the chip will be red is 1/3, the probability that the chip will be white or blue is 2/3.

Answer 80

Quant2, HW 9, Q 122. Answer E. Probability. We know that m + w = 10. Thus, the probability that both representatives selected will be women is p = w(w – 1) / 90. w(w – 1) / 90 > 1/2? w(w – 1) > 45? The question can be rephrased as “is the number of female employees 8, 9 or 10?” (1) INSUFFICIENT. W > 5 is not enough to answer our rephrased question. W could be 6 or 7 (answers NO to the question) or w could be 8, 9, or 10 (answers YES to the question) (2) INSUFFICIENT. m(m – 1) / 90

Answer 81

CAT #3, Q 3. Answer B. Linear Equations. (1) INSUFFICIENT. Note that x and y do not have to be integers. Start by manipulating the equation by multiplying both sides by xy. The value of xy changes according to the values of x and y xy[(2/x) + (2/y)] = 3xy 2y + 2x = 3xy xy = (2y + 2x)/3 ``` (2) SUFFICIENT. If we manipulate this equation and solve for xy, we come up with a distinct value for xy. x^3 – (2/y)^3 = 0 x^3 = (2/y)^3 x^3 * y^3 = 8 (xy)^3 = 8 xy = 2 ```

Answer 82

CAT #3, Q 7. Answer A. Positives & Negatives. Because list S has nine integers (an odd number), the median must be the middle integer. (1) SUFFICIENT: The median is an integer. According to this statement, the nine numbers must collectively multiply to this same integer. None of the numbers is a fraction, and all nine are different numbers. The only way to multiply a bunch of integers together and arrive at the same starting point is to multiply by 0 or 1. In order to use 1, every number on the list except one would have to equal 1; for example, 1 × 1 × 1 × 3 = 3. In order to use 0, though, only one number has to equal zero; for example, 1 × 18 × -3 × 0 = 0. Therefore, the only way in which the product of nine different integers can equal just one of the integers on that list is if the product of the nine integers is zero. (Not convinced? Try to disprove that rule using some real numbers. If the list is -4, -3, -2, -1, 1, 2, 3, 4, 5 then the median is 1 but the product is definitely not 1. If the list is -10, -9, -8, -7, -6, -5, -4, -3, -2, the median is -6 but the product is not -6. If, on the other hand, the list is -4, -3, -2, -1, 0, 1, 2, 3, 4, then the median is 0 and the product is also 0.) Further, since the product of the nine integers equals the median, the median itself must also be zero. The median, then, is definitely not positive; the statement is sufficient to answer the question. (2) INSUFFICIENT: The median of list S is one of the nine integers in list S. Therefore, if the sum of all nine integers equals the median, it follows that the sum of the eight integers on either side of the median must be zero: Sum of all nine integers = (Median) + (Sum of other eight integers) Statement 2 says that the Sum of all nine integers equals the Median. Substitute that info into the above equation: Median = (Median) + (Sum of other eight integers) 0 = Sum of other eight integers This isn’t enough, though, to determine the median; in fact, any number can still be the median of the list, as long as the surrounding numbers sum to zero. For example, the list –10, –9, –8, –7, 5, 7, 8, 9, 10 satisfies the criterion and has a median value of 5; the list –10, –9, –8, –7, -1, 6, 8, 9, 11 also satisfies the criterion and has a median value of -1.

Answer 83

CAT #3, Q 8. Answer C. Statistics. Before looking at the statements, it is important to remember that nothing in the question assumes that the numbers in either list must all be different. For instance, if list P contains 3 numbers, {1, 2, 3} is one possible set, as is {1, 1, 1}. (1) INSUFFICIENT. Under this condition, it is possible for the median of the combined list R to be greater than the median of set P. For instance, if list P contains the numbers 1, 2, and 3, and list Q contains the numbers 4, 5, and 6, then the median of list R (which contains all six numbers) is 3.5, which is greater than 2 (the median of list P). It is also possible for the median of the combined set to equal the median of list P. For instance, if list P contains the numbers 1, 1, 1, and 1, and list Q contains the numbers 2, 2, and 2, then the median of list P is 1, and the median of list R is also 1. (2) INSUFFICIENT. This information does not tell us whether the median of list R is greater than the median of list P. For instance, if list P contains the numbers 1, 2, and 3, and list Q contains the numbers 4, 5, and 6, then the median of list R (which contains all six numbers) is 3.5, which is greater than 2 (the median of list P). It is also possible for the median of the combined set to equal the median of list P, as is perhaps most easily seen in cases in which all of the numbers are the same: e.g., if list P contains 1, 1, and 1, as does list Q, then the median of all three lists P, Q, and R is 1. (C) SUFFICIENT. If both conditions are true, then the set will have two middle values: the largest value in list P and the smallest value in list Q. According to statement (1), these values are different, so the median, found by averaging them, will be distinct from both; the median will be greater than the value from list P, and less than the one from list Q. Since the median is greater than the largest value in list P, it must be greater than every value in list P, and so greater than the median of P; therefore, this statement is sufficient.

Answer 84

CAT #3, Q 18. Answer E. Percents. Perhaps the most straightforward way to investigate this problem is by testing values. Since the problem deals with inequalities – the prompt question is an inequality, and there is also an inequality in one of the statements – we should pay particular attention to extreme values as test cases. In the following discussion, we will set the starting price to $100. (1) INSUFFICIENT: Considering extreme cases, if m = 99 and d = 1, that’s a 99% markup followed by a 1% markdown. If the original price is $100, the price after the markup is $199, and the value of the discount is only $1.99. In this case, the final price is much greater than the original, so this is a YES answer to the question. Can we find a NO answer? Considering the other extreme of d (large values), let m = 99 and d = 90. (We could investigate d up to 98 if necessary, but it’s best to stick to numbers that provide easy calculation.) With these values, the price after the markup is $199. The discount is 90% of $199, which is very close to 90% of 200 (= $180). Therefore, the price will be marked down by almost $180, substantially less than the original $100 price. This is a NO answer to the question. (2) INSUFFICIENT: Again, consider extreme cases. The smallest possible values of m and d are m = 3 and d = 2. (We can’t use d = 1 because then m will not be an integer.) In this case, the price after the 3% markup is $103; the discount is 2% of $103, or $2.06. If $2.06 is taken away from $103, the result is greater than the original $100, so this is a YES answer to the question. Can we find a NO answer? The largest possible values of m and d are m = 99 and d = 66. (To find these values, set m to the largest possible integer, 99. Then solve for d, given the equation m = 1.5d. If you do not receive an integer value for d, then try m = 98 and so on.) In this case, the price after the 99% markup is $199. The subsequent discount is 66% of $199, which is very close to 66% of 200 (= $132); this discount is large enough to bring the final price far below the original price of $100. This is a NO answer to the question. (C) INSUFFICIENT. Review the work already done, so as not to duplicate effort needlessly. The case m = 99 and d = 66 satisfies both statements 1 and 2, so that’s a NO answer to the question. Can we find a YES answer? Consider smaller values that satisfy both statements, say m = 45 and d = 30. In this case, the markup is 45%, taking the price from $100 to $145. The discount is 30% of $145, or $43.50. The final price is $145 - $43.50 = $101.50, which is greater than the original price; this is a YES answer to the question.

Answer 85

CAT #3, Q 19. Answer E. Ratios. The question asks us to find the ratio of gross revenue from computers to that from printers, given that the price of a computer is five times the price of a printer. (1) INSUFFICIENT: Statement (1) says that the ratio of computers to printers sold in the first half of 2003 was 3 to 2; one set of values satisfying this statement is 3 computers and 2 printers. Using example prices of $5 per computer and $1 per printer, we arrive at a gross revenue of $15 from computers and $2 from printers. During the second half of 2003, the ratio of computers to printers sold was 2 to 1; one set of values satisfying this statement is 2 computers and 1 printer, grossing $10 and $1 respectively if we use the sample prices chosen above. With these numbers, then, the full-year gross revenues are $25 from computers and $3 from printers. Alternatively, for the second half of 2003 Acme may have sold 4 computers and 2 printers, still in the required ratio of 2 to 1. In this case, Acme would have grossed $20 and $2 from computers and printers, respectively; with these numbers, then, the full-year gross revenues are $35 from computers and $4 from printers, yielding a different overall ratio. Therefore, statement (1) is insufficient to give us a definitive answer. (2) INSUFFICIENT: Statement (2) tells us that one of Acme’s computers cost $1,000 in 2003, but it tells us nothing about the ratio or numbers of computers and printers sold. (C) INSUFFICIENT: Statement (2) fixes the price of a computer at $1,000, but, if we inflate the sample prices from $5 and $1 to $1,000 and $200, both of the solutions given in the explanation of statement (1) will still be possible. (The arithmetic will be identical; the prices will be 200 times as great.) Therefore, statements (1) and (2) together are still insufficient.

Answer 86

CAT #3, Q 26. Answer B. Digits & Decimals. For fraction p/q to be a terminating decimal, the numerator must be an integer and the denominator must be an integer that can be expressed in the form of (2^x)(5^y) where x and y are integers. (Any integer divided by a power of 2 or 5 will result in a terminating decimal.) The numerator p, (2^a)(3^b), is definitely an integer since a and b are defined as integers in the question. The denominator q, (2^c)(3^d)(5^e), could be rewritten in the form of (2^x)(5^y) if we could somehow eliminate the expression 3^d. This could happen if the power of 3 in the numerator (b) is greater than the power of 3 in the denominator (d), thereby canceling out the expression 3^d. Thus, we could rephrase this question as, is b > d? (1) INSUFFICIENT. This does not answer the rephrased question "is b > d"? The denominator q is not in the form of (2^x)(5^y) so we cannot determine whether or not p/q will be a terminating decimal. (2) SUFFICIENT. This answers the question "is b > d?"

Answer 87

CAT #3, Q 30. Answer A. Consecutive Integers. Note that the daily increase is consistent: he always earns $10 more each day. This is essentially a consecutive integer problem (although, in this case, there are increments of 10, not 1). Use f to represent Dan’s earnings on the first day. Express his earnings on subsequent days as f + 10, f + 20, f + 30, … f + 140. (1) SUFFICIENT: This statement provides the sum of the most recent three days: (f + 120) + (f + 130) + (f + 140) = $690 3f + 390 = $690 3f = $300 f = $100 Dan earned $100 on his first day. (2) INSUFFICIENT: This statement provides the difference between the totals for the first three days and the last three days: First three days = (f) + (f + 10) + (f + 20) = 3f + 30 Last three days = (f + 120) + (f + 130) + (f + 140) = 3f + 390 Last three days − First three days = (3f + 390) − (3f + 30) = $360 This statement is always true, regardless of how much Dan made on the first day. (This information can be determined from the question stem alone!)

Answer 88

CAT #3, Q 33. Answer C. Odds & Evens. Divisibility has to do with the products of numbers, so it is likely going to be better to use the (a – b)(a + b) form. For a product to be divisible by 4, either one of the values (a – b) or (a + b) must be divisible by 4 or each value (a – b) and (a + b) must be divisible by 2 (that is, even). For both (a – b) and (a + b) to be even, either both a and b must be even or both must be odd. We can use the properties of odds and evens to see this. If we know that a and b have the same parity (that is, both are odd or both are even), then we know that the quantities (a – b) and (a + b) are even; the reverse is also true: if we know that either (a – b) or (a + b) is even, then we know that a and b have the same parity. Thus, we can prove sufficiency through any combination of these findings: (1) a and b have the same parity; (2) (a – b) is even OR (3) (a + b) is even. (1) INSUFFICIENT: a = (c – 1)^2. Here we are given an algebraic expression for a, but no information about b. There is no way to prove whether (a – b) or (a + b) is even. Furthermore, the statement doesn’t tell us whether a is odd or even. To see this, we can pick numbers for c. If c were even (e.g. 2) then a would be odd (i.e. 1), or if c were odd (e.g. 3), then a would be even (i.e. 4). Thus, we can see that a will be odd if c is even and a will be even if c is odd. (2) INSUFFICIENT: b = c^2 – 1. Here we are given an algebraic expression for b, but no information about a. There is no way to prove whether (a – b) or (a + b) is even. Furthermore, the statement doesn’t tell us whether b is odd or even. To see this, we can pick numbers for c. If c were even (e.g. 2) then b would be odd (i.e. 3), or if c were odd (e.g. 3), then b would be even (i.e. 8). Thus, we can see that b will be odd if c is even and b will be even if c is odd. (C) SUFFICIENT: There are two ways to combine the information in the statements and prove sufficiency: recognizing a pattern involving odds and evens in the statements or manipulating the algebraic expressions. In our individual examinations of statements 1 and 2, above, we determined that, if c is odd, then a must be even and b must be even. We also determined that, if c is even, then a must be odd and b must be odd. Either way, a and b must have the same sign, which is sufficient to answer the question. We can test either (a + b) or (a – b), but note that we don’t need to test both: (a + b) = 2c2 – 2c = 2(c2 – c) = even, because anything multiplied by an even number (in this case, 2) is even. We can stop here because we only need to show whether (a + b) or (a – b) is even. (a – b) = –2c + 2 = 2(–c + 1) = even, because anything multiplied by an even number (in this case, 2) is even. We can stop here because we only need to show whether (a + b) or (a – b) is even.

Answer 89

CAT #3, Q 36. Answer E. FDPs. Rephrase this question by solving the inequality for x: “is x > 1/16 + 5/8?” or “is x > 11/16?” (1) INSUFFICIENT: Translate this statement into math. 4x < 3 or x < 12/16 Some potential x values that are less than 12/16 are greater than 11/16 (x = 11.5/16 = 23/32, for example), but others are less than 11/16 (x = 0, for example). Therefore, it is uncertain whether x > 11/16. (2) INSUFFICIENT: Translate this statement into math (1/3)(2) < x or x > 2/3 First, how do 2/3 and 11/16 compare? Which one is bigger? Compare the two fractions by cross-multiplying, giving 3 * 11 = 33 for the left-hand fraction and 2 * 16 = 32 for the right-hand fraction. Because 33 is bigger than 32, 11/16 is bigger than 2/3. Draw a number line and place both 11/16 and 2/3 on that number line. If x > 2/3, there are some values between 2/3 and 11/16 (that is, some values of x are less than 11/16). There are also some values greater than 11/16. Therefore, it is uncertain whether x > 11/16. (C) INSUFFICIENT: Combining the information from the two statements gives a range for x: 2/3 < x < 12/16 Place these values on a number line: 2/3, 11/16, 12/16. Also draw in the range of possible values for x. This range still allows for values of x that are less than 11/16 and for values of x that are greater than 11/16. It is still uncertain whether x > 11/16.

Data Sufficiency Flashcards

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