Question 1
Statement 1: No. The only trick to this question is to recognize that it’s asking about outbreaks, not individual illnesses. That means we can find the answer without looking beyond the first column. There were 5 outbreaks in the first half of 2005 (any month between January and June would count as the first half of the year) and 8 outbreaks in the second half. There were only 3 outbreaks in the first half of 2006, and and 8 outbreaks in the second half. That makes for 8 outbreaks in the first half of both years (5 + 3 = 8) and 16 outbreaks in the second half of both years. This is exactly twice as many. Statement 2: No. Sort by the illness column to quickly find the five highest totals. One of them caused 5 deaths, three of them caused 0 deaths, and one of them is unknown. Because of the unknown, we cannot know if there were more deaths or not. Statement 3: Yes. There were three total outbreaks in June of 2005 and 2006 (1 in 2005, and 2 in 2006). The number of illnesses caused by these three outbreaks were 41, 115, and 13, making for 169 total illnesses. The hospitalizations were 13, 7, and 8 in number, making for a total of 28. To find the percent, we simply divide: 28/169 = approximately 16.5%, which is greater than 15%. Question 2
Statement 1: $15,500. This problem asks us about information i nformation pertaining only to a specific subset of the branches, those that saw an increase in in costs during the period from 7/2/2011 to 12/31/2011. This would be any branch with a positive Y coordinate, coordinate, so we will only be looking at the nine branches above the X axis. From this group of 9, we need to find the one with the median change in cost during the period beginning on 1/1/2011 (the X -axis). -axis). With 9 branches, the median will be the 5th (the physical middle). If we count over 5 from the left or 5 from the right, we see that the median branch appears to be the point at approximately (-3.5, 19). Now, we are asked to determine the net Variable Cost for this branch over the entire year. This is simply the sum of the costs for each period (in thousands), so -$3,500 + $19,000 = $15,500. Statement 2: 47%. To find branches with a net decrease in costs it will help to understand what is represented by each quadrant. Quadrant I: these branches had an increase over both periods (3 branches) Quadrant II: these branches had a decrease during the first period but an increase during the second (6 branches) Quadrant III: these branches had a decrease during both periods (1 branch) Quadrant IV: these branches had an increase during the first period but a decrease during the second (5 branches)
None of the branches that have an increase during both periods will have a net overall decrease, so we can ignore the branches in Quadrant I (3 branches). All branches that have a decrease during both periods will have a net overall decrease, so we count all branched in Quadrant III (1 branch). Now we must look at the more complicated situations – an increase in one period but a decrease in the other. To get a net overall decrease, we need a larger dollar decrease than increase. For Quadrant II (decrease in first period but increase in second), we need the X- coordinate coordinate (period 1) to be larger than the Y -coordinate -coordinate (period). This appears to be the case for 3 points: (-7, 5.5), (-15.5, 5.5), and (-16, 0.5). Finally, for Quadrant IV (decrease in the second period but increase in the first), we need the Y coordinate (period 2) to have larger magnitude than the X -coordinate -coordinate (period 1). This appears to be the case for 3 points (2.5, -4.5), (0.5, -7.5), and (11, -17). For all these points, the decrease outweighs the increase. So in total, we have 1+3+3=7 branches out of the 15, a number slightly less than ½ or 50%. Plug it into the calculator to get 7/15 = 0.4667 = 47%. As an alternative, you could find all the points satisfying satisfying the expression x +y <0, <0, or y <<-x . Mentally sketch the downward-sloping line y ==-x , which runs through the origin. origi n. What points fall below the line? You’ll find 7 points there. From here, the final calculation is the same.
Question 3
Statement 1: 50. In order to compare the differences dif ferences between pairs of countries, we begin by finding the approximate difference in 2008 cellular telephone subscribers between China and the US, and between the US and Italy. Using our eye as a guide for 2008 (the data point farthest to the right), we see that: China – US = 650 – 275 = 375 US – Italy = 275 – 90 = 185 The question is asking us to find approximately what percent of the difference between China and the United States is made up by the difference between the United States and Italy. This translates to, “185 is approximately what percent of 375.” We solve this by simply doing (185 / 375) × 100 = 49.3%, which is approximately 50%. Statement 2: 700. The formula for percent increase is given by [(final – initial) / initial] × 100. Here, the final value is the number United States cellular telephone subscriptions in 2008 and the initial value is the number of cellular telephone subscriptions in 1995. Using our eye to estimate, we have [(275 – 35) / 35] × 100 = (240 / 35) × 100 ≈ 686, which is approximately equal to 700.
None of the branches that have an increase during both periods will have a net overall decrease, so we can ignore the branches in Quadrant I (3 branches). All branches that have a decrease during both periods will have a net overall decrease, so we count all branched in Quadrant III (1 branch). Now we must look at the more complicated situations – an increase in one period but a decrease in the other. To get a net overall decrease, we need a larger dollar decrease than increase. For Quadrant II (decrease in first period but increase in second), we need the X- coordinate coordinate (period 1) to be larger than the Y -coordinate -coordinate (period). This appears to be the case for 3 points: (-7, 5.5), (-15.5, 5.5), and (-16, 0.5). Finally, for Quadrant IV (decrease in the second period but increase in the first), we need the Y coordinate (period 2) to have larger magnitude than the X -coordinate -coordinate (period 1). This appears to be the case for 3 points (2.5, -4.5), (0.5, -7.5), and (11, -17). For all these points, the decrease outweighs the increase. So in total, we have 1+3+3=7 branches out of the 15, a number slightly less than ½ or 50%. Plug it into the calculator to get 7/15 = 0.4667 = 47%. As an alternative, you could find all the points satisfying satisfying the expression x +y <0, <0, or y <<-x . Mentally sketch the downward-sloping line y ==-x , which runs through the origin. origi n. What points fall below the line? You’ll find 7 points there. From here, the final calculation is the same.
Question 3
Statement 1: 50. In order to compare the differences dif ferences between pairs of countries, we begin by finding the approximate difference in 2008 cellular telephone subscribers between China and the US, and between the US and Italy. Using our eye as a guide for 2008 (the data point farthest to the right), we see that: China – US = 650 – 275 = 375 US – Italy = 275 – 90 = 185 The question is asking us to find approximately what percent of the difference between China and the United States is made up by the difference between the United States and Italy. This translates to, “185 is approximately what percent of 375.” We solve this by simply doing (185 / 375) × 100 = 49.3%, which is approximately 50%. Statement 2: 700. The formula for percent increase is given by [(final – initial) / initial] × 100. Here, the final value is the number United States cellular telephone subscriptions in 2008 and the initial value is the number of cellular telephone subscriptions in 1995. Using our eye to estimate, we have [(275 – 35) / 35] × 100 = (240 / 35) × 100 ≈ 686, which is approximately equal to 700.
Question 4
Statement 1: No. To find the smallest ratio of "other" sources to total production, you should first sort by the Other column. (Of course, this doesn't guarantee that the lowest ratio will be with the lowest Other amount, but it gives you someplace to start; after all, if the total production were held constant, then that sort would be perfect.) Look at the first row, which now contains 2001. Total production is 71.89 that year, but Other is only 5.32. Punch this into the calculator to get 5.32/71.89 = approximately 0.074 = 7.4%, which is less than 8%. Statement 2: No. Sort by Year. Now look down the Nuclear column. You can see the numbers grow, even if irregularly, from 6's through 7's to 8's. This represents positive correlation. (If you were able to calculate the actual correlation coefficient, you'd find out that it is 0.946, which is very close to the maximum of 1.) Statement 3: No. Keep the sort on Year, for easier lookup. Now look up the Nuclear numbers for 2000 and for 2007. You get 7.86 and 8.46, respectively. You might be able already to see that the growth is less than 10% (which would be 0.786 in absolute numbers), or you can use the calculator quickly to find that the percent growth is approximately 7.6%, which is not more t han 10%. Question 5
In order to answer this logic problem, it might be best to s tart with a clear list of the constraints: • • • • • •
12 hours = total hours available 4 hours = total hours for walking wal king Minimum 4 art or architecture activities during the 2 days Maximum 1 art museum per day Minimum of 1 beach activity during the trip Minimum of 1 shopping activity each day Now we can start with the activities the family has already planned to see what has been accomplished from the list and what has not. On Day 1, the family has already planned 11 hours of activities, and only 1 hour of walking. Therefore there is only 1 hour left l eft to plan, and they could choose a walking activity if they would like. On Day D ay 2, the family has planned 9 hours of activities with 4 hours of walking. This means that there are 3 hours they can use for activities but they cannot include walking.
Using these 2 constraints, we see that the only activities possible for each day are the following: Day 1: Mirador De Colon, Montserrat, or La Pedrera Day 2: Montserrat Therefore, the family must choose the sightseeing trip to Montserrat on Day 2, leaving only Mirador De Colon and La Pedrera for Day 1. If we now look at the family preferences, we see that Mom already has her shopping on each day (Las Ramblas and Barri Gotico) and Little Brother has his beach activity on Day 2 (Nova Icària), but Big Sister only has 3 of her 4 art or architecture activities. Both Mirador De Colon and La Pedrera fit this category, but Dad will not go to more than 1 art exhibit on a single day, and the family will already visit Park Güell on Day 1. This means that they cannot visit La Pedrera as well. Therefore, the family will visit Mirador De Colon on Day 1. Column 1: The correct answer is A. Column 2: The correct answer is C. Question 6
Statement 1: Otherwise. In the last sentence of the Residency Residency Association statement, the president says “Such working conditions are dangerous for both patients and residents; some serious mistakes have already been made and were caught only at the last minute by senior staff.” The president, then, believes that the situation is dangerous and that there is at least the possibility of serious consequences from serious mistakes made by residents. In the last sentence of the Hospital Board statement, the spokesperson says “there is always an experienced physician on hand to verify the diagnosis, take over the procedure, or otherwise correct any potential errors” made by residents. The spokesperson, then, believes that there is an oversight system in place that will “correct any potential errors” – in other words, there is not the danger of lasting harm due to a resident’s mistake. (Note that the spokesperson doesn’t claim that the licensed physician won’t make mistakes, but the question asks only about lasting harm caused by residents .) .) Statement 2: Both Agree. The Residency Association president explicitly states that residents have averaged 17.5 hours during the past month, so the president does believe that residents are working more than 8 hours during a 24-hour shift. The Hospital Board spokesperson says that “residents are not expected to work 24 hours in a 24-hour shift, nor anywhere close to that” but also that “if residents choose to participate in many routine activities that can ably be handled by the nursing or medical staff, that is the residents' choice.” In other words, the spokesperson is acknowledging that it is possible for a resident to work long hours… if he or she chooses to do so. The two parties, then, both agree that residents may work longer than 8 hours during a 24-hour shift. Statement 3: Otherwise. The Hospital Board spokesperson states clearly that “until someone has finished the residency, he or she is not, and cannot reasonably be considered, a licensed physician.” The Residency Association president provides only indirect evidence that “Residents are responsible for all of the licensed physician tasks, including intake, the ordering of tests and labs, diagnosis and treatment, including surgery” (emphasis added). The spokesperson clearly states that no one can be considered a
licensed physician until the residency is complete. We might speculate either that the president disagrees or that there is not enough evidence to decide whether the president disagrees. Either way, the correct response is “otherwise.” Question 7
Statement 1: Cannot Infer. The Residency Association indicates that the residents are assumed to work 1/3 of the 24-hour shift, or 8 hours, but the president does not claim that residents should not work more than that amount. They claim only that they work too much. We can infer that they think they should work less than 17.5 hours, but we do not know what they think of an 8-hour cutoff. Likewise, the Hospital Board never indicates a specific time beyond which residents should not be able t o work. Statement 2: Both Accept. As part of their argument, the residents complain that their working conditions are dangerous and that, as a result, serious mistakes have been made. We can infer, then, that the residents believe the hospital is responsible for setting reasonable working hours in order to minimize or prevent errors made by residents (among other goals). The Hospital Board indicates that “when a resident” is working, “there is always an experienced physician on hand” in order to “correct any potential errors.” The board, then, also believes that the hospital is responsible for minimizing or preventing errors made by medical residents. Statement 3: Cannot Infer. The residents state that they “are taken advantage of terribly” and cite the fact that they are “undermined or treated poorly by senior doctors – sometimes when patients are present” as one piece of evidence. The residents, then, agree with this statement. We might surmise that any ethical hospital board also would not want doctors to treat medical residents poorly when patients are present (or at any time!), but the Hospital Board’s statement never addresses this aspect of the resident’s complaint. We do not know how the board feels about this i ssue. Question 8 According to the first tab, the residents are assumed to work “one-third of the hours” in an assigned shift. If the shift is 24 hours long, then, a resident is assumed to work for 8 hours. In that case, the hourly wage would be $105 / 8 hours = $13.125. Rounded to the hundredths place, this is $13.13. The resident indicates that the average time worked over the past month was actually 17.5 hours. The hourly wage in this case would be $105 / 17.5 hours = $6.00. The difference between the two wages is $13.13 - $6.00 = $7.13. The correct answer is E.
Question 9
To solve this fractions problem, we must find the ratio of the number of total boxed packed by the firstshift to the total number of boxes packed by both shifts together. From the information given, we know that there were 2/3 as many first-shift workers as second-shift workers and, inverting the second fraction, we know that each first-shift worker packed 3/4 as many boxes as each individual second-shift worker. From here, multiplying the ratio of workers by the ratio of work per individual gives the fraction of total first-shift boxes relative to the second shift. This is done as (2/3) × (3/4) = 6/12 = 1/2. Thus, the first shift packs half as many boxes as the second shift. We can compute the first-shift boxes relative to the total by: (first-shift fraction) / (first-shift fraction + second-shift fraction) = 1 / (1+2) = 1/3. Thus the first shift does 1/3 of the total work. We must look for two numbers in the table that are related by a factor of 3. The only two numbers are 12 and 36, meaning that the first shift packed 12 boxes, and the total number of boxes packed by both shifts was 3 × 12 = 36.
Alternatively, one could use numbers to establish the relationship between the number of total boxes packed by the first shift and the number of total boxes packed by the two shifts together. We use our fractional ratios to choose smart numbers and assign 2 workers to the first shift, 3 workers to the second shift, 3 boxes per individual on the first shift, and 4 boxes per individual on the second shift. This gives: Total First-shift Boxes = (2 workers) × (3 boxes per worker) = 6 boxes Total Second-shift Boxes = (3 workers) × (4 boxes per worker) = 12 boxes Total Boxes Overall = 12 boxes + 6 boxes = 18 boxes. Again, from this we can derive that the ratio of first-shift boxes to the total boxes packed is 6/18 = 1/3. Column 1: The correct answer is C. Column 2: The correct answer is E. Question 10
Statement 1: 60% and 65%. A person who was 30 in 1999 was 39 in 2008. Therefore, to answer this question, we need to multiply the probability that a person who was 30 in 1999 had health insurance by the probability that a person who was 39 in 2008 had health insurance. (The assumption of independence allows us to multiply these probabilities; in reality, these conditions are almost certainly not independent.) 79% of people aged 25-34 had health insurance in 1999, so a person who was 30 in 1999 had a 79% chance of having health insurance. Similarly, a person who was 39 in 2008 had an 80% chance of having health insurance. The probability we want is 79% × 80%. 0.79 × 0.80 = 0.632 = 63.2%, between 60% and 65%. Statement 2: 1,200,000 less. The easiest way to answer this question is to calculate the number of 3544 year olds who had health insurance in 1999 and 2008. The total US population is given in the information under the graph. The population in 1999 was 276,804,000; the population in 2008 was 301,483,000. If 35-44 year olds were 20% of the population in 1999, and 85% of 35-44 year olds had health insurance, then the total number of 35-44 year olds who had health insurance in 1999 was: 276,804,000 × 0.85 × 0.20 = 47,056,680 ≈ 47 million Similarly, 80% of 35-44 year olds had health insurance in 2008, and 35-44 year olds represented 20% of the population: 301,483,000 × 0.80 × 0.20 = 48.237,280 ≈ 48.2 million The question said approximately , so we don’t have to find the exact difference. 48.2 – 47 = 1.2. We’re talking about millions of people, so there were approximately 1,200,000 fewer 35-44 year olds insured in 1999 than in 2008. Question 11
The conclusion of the argument is that the Dante 5000 is more reliable than the company’s other stonecutting machines and the premise is that there have been fewer customer complaints about the Dante 500 than about the company’s other industrial stone-cutting machines over the l ast six months. This argument assumes a number of things! Do breakdowns in unreliable machines typically occur within the first six months? Have enough Dante 5000’s been sold to make a reasonable comparison with the other stone-cutting machines? Option A: The Dante 5000 is the most expensive stone-cutting machine produced by the company. The argument is not about cost. This is out of scope. Option B: There are other stone-cutting machines that are considered more reliable than the Dante 5000. The argument is only about this company’s stone-cutting machines. Other machines may be from other companies, so this is out of scope.
Option C: The Dante 5000 performed very well in initial testing.
The argument is reliability measured in terms of breakdowns, not about performance. This is out of scope. Option D: Stone-cutting machines usually break down very quickly under industrial use if they are not reliable. - Strengthens This option directly strengthens the conclusion that the Dante 5000 is reliable than the company’s other machines because it makes it more likely that six months would be enough time to see breakdowns if the machine were unreliable. Note that this statement does not plug all of the holes in the argument! It just makes this weak argument slightly stronger. Option E: Very few customers have purchased a Dante 5000. - Weakens This option directly weakens the conclusion by providing a good alternative interpretation of the low number of customer complaints. If very few customers have bought the machine, then a small number of complaints could actually represent a high rate (percent) of complaints. This choice points up the classic difference between absolute numbers and percents. Option F: The Dante 5000 employs a new technology that is more precise than that used by our previous stone-cutting machines. The argument is concerned with reliability measured in terms of breakdowns , not about precision. This choice is out of scope. Column 1: The correct answer is D. Column 2: The correct answer is E.
Question 12
Remember that R × T = D. We are given that the distance = 675, but only that the time for train A is t and that the time for train B is t – 3. So we know that R A × t = 675 and R B × (t – 3) = 675. So R A = 675/t and R B = 675/(t – 3), but we don’t have enough information to solve for either rate, because there are only two equations and three unknowns. What we do know though, is that the rate, or speed, of train A must be less than the speed of train B and that both rates must be in the chart. So we can test the smaller numbers as potential speeds of A. Solve for potential t ’s and (t – 3)’s and then test the (t – 3)’s until we find one that yields a speed that is in the chart. Use the calculator, of course, to move quickly through these computations. Potential value for A’s speed in mph
45.75
Implied value of t t =
675/45.75
14.754
=
Implied value of t – 3
Implied value of B’s speed in mph
t – 3 = 11.754
= 675/11.754 = 57.427 not on list
50
t = 675/50 = 13.5
t – 3 = 10.5
= 675/10.5 = 64.286 not on list
56.25
t = 675/56.25 = 12
t – 3 = 9
= 675/9 yes on list! Stop!
63 67.5 75 Since there can be only one set of answers, we can stop here.
= 75
Column 1: The correct answer is C. Column 2: The correct answer is F. Question 13
The passage as a whole strongly suggests that few cartels last more than 4 or 5 years because of the economic incentive that members have to cheat. Option A - No cartel will last more than 100 years.
This is much too strong of a statement to be logically inferred from “very few known cartels have lasted for more than 4 or 5 years.” It could be that one or more known (or unknown) cartels has indeed lasted 100 years, even though only 4 to 5 years is typical. Option B - As long as the members do not cheat, a cartel cannot be broken.
This is too strong of a statement to be inferred from “The main issue is that the members of a cartel all have an incentive to cheat and cut prices just a little in order to maximize their individual profits at the expense of profits of the cartel as a whole.” There could be other issues. Perhaps new suppliers that are not part of the market enter the cartel, or perhaps the cartel’s product is made technologically irrelevant. Option C - An effective system for preventing members from cheating would not increase the likelihood that a cartel would survive in the long-term. – Correct FALSE
The passage says that the main issue is that the members of the cartel have an incentive to cheat, so an effective system for preventing cheating WOULD likely increase the likelihood that a cartel would survive in the long run. Option D - Private cartels are not legal in most countries.
The passage does not discuss legality. Although this statement is likely true, it is not something that can be inferred from the passage. Option E – Cartels are inherently unstable and likely to fail in the long run. – Correct TRUE This can be directly inferred from “very few known cartels have lasted for more than 4 or 5 years” and “The main issue is that the members of a cartel all have an incentive to cheat and cut prices just a little in order to maximize their individual profits at the expense of profits of the cartel as a whole.” The economic incentive to cheat is inherent in the nature of a cartel and makes it inherently unstable. Option F – It is extremely difficult for competing firms to agree to fix prices, marketing, and production in the formation of a cartel. This choice is tempting, but the passage does not discuss the difficulty of forming a cartel agreement.
Although this statement is likely true in the real world, it is too extreme to infer language such as "extremely difficult" from the passage. Column 1: The correct answer is E. Column 2: The correct answer is C. It is extremely difficult for competing firms to agree to fix prices, marketing, and production in the formation of a cartel. Question 14
The key to answering this question is finding the relationship between X and Y . First, create an equation that matches the information provided. Revenue from widget sales will equal the number of widgets sold times the price per widget. Let w be the original number of widgets sold and let p be the original price per widget. “ X %” is the same as X/ 100; likewise, “Y %” is Y /100.
If the revenues are equal for the two months, then: wp = (Y /100) × (w ) × (X /100) × ( p )
Notice that both sides of the equation contain w and p . Cancel out w and p (which are non-zero, by the logic of the real world, so you’re allowed to divide them away). wp = (Y /100) × (w ) × (X /100) × ( p ) 1 = (Y /100) × (X /100)
You can go further (to prove that XY = 10,000), but let’s stay in “percent” land. What we have so far is that (Y /100) and (X /100) multiply together to 1. That is, they are reciprocals of each other. So let’s save time. Divide every answer choice by 100, to convert it to the decimal equivalent of a percent. We get 0.50 0.625 0.75 1.5 1.6 1.8 Now, to determine which pairs multiply together to 1, notice that you’ll need to pick one number smaller than 1 and the other number larger than 1. Which variable gets which? Since the factory “raised” the price to X %, we know that X % must be bigger than 100%, so X must be the one bigger than 1. Next, to check reciprocals quickly, see whether there’s a quick fraction equivalent of each of these decimals. Fortunately, there is! 0.50 = ½ 0.625 = 5/8 0.75 = ¾ 1.50 = 3/2 1.60 = 8/5 1.80 = 9/5 Now it’s easy to spot the reciprocals. Know your eighths! (1/8 = 0.125, etc.) 5/8 × 8/5 = 1, so the correct percents are 62.5% and 160%. Again, X must be 160 and Y must be 62.5. Column 1: The correct answer is E. Column 2: The correct answer is B. Question 15
Statement 1: Brazil. In order to solve this problem, you need to know that the definition of the median is the middle number in a set. Since there are 15 countries in the chart, the one with the 8 th largest national debt has the median national debt. To see this quickly, hold your finger or a piece of paper on the right edge of the chart, start moving it to the left, and count off countries as the piece of paper touches their bars until you get to the eighth longest bar, which is Brazil’s. Statement 2: between 50% and 70% . Use the lines on the graph to help with counting countries that fall into the ranges or “bins” given in the answer choices. Notice that the percent ranges are the midpoints between the lines shown. Since there are only two countries, Italy and Japan, with national debt as a percentage of GDP greater than 70%, we can quickly eliminate answers D and E. Five countries (United States, India, Germany, France, and Canada) have percentages between 50% and 70%. The United Kingdom, Spain, Brazil, and
possibly South Korea have values between 30% and 50%, for a total of 3 or possibly 4 countries in that range, but even 4 is less than the 5 countries between 50% and 70%. Australia, China, Mexico, and probably South Korea have percentages between 10% and 30%, for a total of 4 countries. Russia is the only country below 10%. So the answer must be between 50% and 70%. Question 16
Statement 1: No. First, compute the "turns" (as defined in the statement) of AV materials for 2004 (the date in the statement) by dividing "AV materials lent out to public" by "AV materials in collection" in the 2004 row. 6,683 / 2,456 = 2.72. Now the question is this: is there any year that has a smaller number? Sort by the numerator of the ratio you're examining, "Lent out to public AV materials," and look at the top row. The year in the top row is 2005, and the "Lent out to public AV materials" number is 6,059 (the smallest of any year). Does that generate a smaller ratio? Try it: 6,059 / 2,383 = 2.54, which is smaller than the 2004 ratio. You could tell in advance, if you looked closely, because 6,059 represents a bigger % decline from 6,683 (almost 10%) than 2,383 does from 2,456 (about 3%). So the AV turns in 2004 are not the lowest among years listed. Statement 2: No. First, re-sort by "Year" if necessary, so that your table is in chronological order. Measure the percent decline in "In collection Volumes": (37,464 - 42,760) / 42,760 ≈ -12%, or a 12% decline. Did any other characteristic listed in the table experience a greater percent decline? Scanning, we should see the first column ("Libraries Number") pop out at us: it fell from 508 to 351, a decline we should be able to estimate as 30% without using the calculator by rounding to 500 and 350 (the calculator puts the decline as approximately 31%). So the listed characteristic that experienced the greatest percent decline was not volumes in collection. Statement 3: No. For each year in question (2002 and 2003), compute the ratio of "In collection Books" to "Users Registered." 2002: 36,671 / 4,290 ≈ 8.55 2003: 35,371 / 4,206 ≈ 8.41 Notice that both raw numbers fell from 2002 to 2003 (books by about 3.5%, and users by about 2.0%), and the ratio only fell very slightly. So we should go ahead and use the calculator here, rather than risk error. The number of books in the collections per registered user did not rise from 2002 to 2003. Question 17
You are asked to figure out and apply these linguistic constraints to possible words. Let’s start with the classification of vowels as “brutish” ( a , o , and u ) and as “fragile” ( e and i ). Kurtish has stricter rules about the separation of these vowels—only one kind can appear within any word. Since this is a simple constraint, let’s apply it first. Which words fail this constraint? calzral – only a , brutish – passes fjp – NO vowels, but that’s okay (see note below) – passes aphueitse – both brutish ( a and u) and fragile ( e and i ) – fails, can’t be Kurtish brushmen – both brutish ( u ) and fragile (e ) – fails, can’t be Kurtish (don’t be fooled by the fact that brushmen sounds okay in English, though it’s not an English word) qudxatroua – only brutish ( a , o , and u ) – passes hzziigri – only fragile ( i ) – passes
So we know that our two answers must be in the middle, aphueitse and brushmen . One of these could be Laeglish; the other cannot. Now we have to apply the second constraint, the one about Laeglish: every consonant or group of consonants can only “directly touch” vowels of one type or the other. Let’s compare: aPHueiTSe – the PH touches a and u , which are both brutish – the TS touches i and e , which are both fragile
This passes the Laeglish test. Double-check the other word: BRuSHMeN – the BR only touches u , but the SHM touches u (brutish) and e (fragile). That breaks the Laeglish rule. So aphueitse could be Laeglish but not Kurtish; brushmen could not be either. Column 1: The correct answer is C. Column 2: The correct answer is D.
Note that fjp does not have any vowels as defined above, but no constraint demands that a word in either language contain vowels! In fact, a nod is given to the possibility of zero vowels within a word with the language “according to the vowels it [a word] contains, if any ” (emphasis added). Don’t apply outside knowledge inappropriately here (you expect words to have vowels). Question 18
Statement 1: Canada. This problem asks us to find the country with the highest standard deviation among all years. Standard deviation is a measure of the overall distance of the set from the mean value, or the “spread” of the set. Judging by eye from the plot, we see that Switzerland and the Netherlands are much closer to their mean values across all years than Sweden and the Canada, so we eliminate Switzerland and the Netherlands. Notice that Sweden and Canada start out at the same value of zero and then both trend upwards in the same way over the years. However, Canada is always higher than Sweden, meaning the overall spread of the set of values for Canada i s larger. Since they start at the same value and Canada always trends higher, the Canadian set is much more spread out, and therefore Canada has the larger standard deviation. Statement 2: 1,500. To solve this problem, we find the ratio of the two countries and then multiply by 100 to achieve a percent. This is given by: Percent = [(Sweden in 2000) / (Netherlands in 2000)] × 100 Finding the year 2000 along our x-axis, we use our eye to judge the approximate values of production for the Netherlands and Sweden in 2000. The Netherlands produced approximately 1 million toe of nuclear energy in 2000 and Sweden produced approximately 15 million toe of total nuclear energy in the same year. Plugging these values into the formula above: Percent = (15 million / 1 million) × 100 = 15 × 100 = 1,500. Question 19 According to the table on the Funding Sources tab, three teams have a Percent Funded figure exceeding 100%: Gymnastics, Soccer, and Track and Field. For each of these, we can calculate the amount of extra money generated by the respective team. The first tab indicates that there are four funding sources and that individual endorsements are not part of the overall team calculations, leaving three funding sources: TV rights, team sponsors, and government funding. These three sources are all detailed in the second tab. (Note that none of the three will receive government funding because all three already exceed 100% of their own expenses.) The gymnastics team has a total of 1,100 + 890 = 1,990 in funding and this represents 115% of the team’s monetary needs, or 1,990 = 1.15x. The actual amount needed, then, is 1,990 / 1.15 = 1,730, and
the surplus is 1,990 – 1,730 = 260. (Note that we can round our calculations because the problem asks for the approximate level of surplus funding.) The Soccer team has a total of 1,800 + 1,300 = 3,100 in funding and this represents 126% of the team’s monetary needs. The actual amount needed, then, is 3,100 / 1.26 = 2,460, and the surplus is 3,100 – 2,460 = 640. The Track and Field team has 1,000 + 345 = 1,345 in funding and this represents 103% of the team’s monetary needs. The actual amount needed, then, is 1,345 / 1.03 = 1,306, and the surplus is 1,345 – 1,306 = 39. The total surplus is 260 + 640 + 39 = 939, or approximately 940. The correct answer is A.
Question 20 Statement 1: No. The second paragraph of the Article tab indicates that “an individual team is considered underfunded if its funding (not including government sources) covers less than 95% of the team’s expected expenses.” There are 9 teams shown in the Funding Sources tab. Of these, only 4 are funded at less than 95% of expected expenses (basketball, boxing, equestrian, and volleyball). Statement 2: Yes. The Article tab tells us that surplus funding will be returned to the committee and reallocated among the teams who have not reached 100% funding. The government is willing to provide up to 5% of expenses to those teams who fall short, but it will not have to provide the full amount in all cases because there is a surplus among three teams that will be able to be distributed among the “short” teams. Therefore, the government will be paying some reduced amount. (In fact, the surplus of the three teams is larger than the shortfall of the other 6 teams, so the government will not be responsible for any funding at all.) Statement 3: No. It is true that the absolute amount of government funding is higher for the equestrian team (43) than for the basketball team (40). But the question asks about the funding as a percentage of total funds needed for that team . The passage tells us that the government will pay up to a maximum of 5% for any team that cannot meet its funding needs – but the government will only pay the full 5% if the team’s own funding is below 95% of its needed funding. Both the basketball and equestrian teams are below 95% of full funding, so both teams qualify for the maximum 5% in government funding. Question 21
To start, we need to calculate how much funding each team needs. If TV rights will increase by 15% for each sport, we need to recalculate the amount of funding from TV rights, and determine the new amount by which the team falls short of the goal (if any). Then we need to determine whether that is more or less than 5% of needed funding. Statement 1: Yes. Prior to the change, the basketball team had secured 400 + 250 = 650 in funding from TV rights and team sponsors, and this represented 82% of the team’s needed funding. Therefore, the team needed a total of 650 / (82/100) = 793 in funding. Now, the team will earn 400 × 1.15 = 460 in TV rights, or an increase of 60, so the team will have a total of 650 + 60 = 710 in funding. Calculate the percentage that this represents of total funding: 710 / 793 = 90%. The team is still more than 5% short, so it still qualifies for the maximum 5% government funding. Statement 2: No. Prior to the change, the boxing team had secured 130 + 185 = 315 in funding from TV rights and team sponsors, and this represented 92% of the team’s needed funding. Therefore, the team needed a total of 315 / (92/100) = 342 in funding. Now, the team will earn 130 × 1.15 = 149.5 in TV rights, or an increase of 19.5, so the team will have a total of 315 + 19.5 = 334.5 in funding. Calculate the percentage that this represents of total funding:
334.5 / 342 = 98%. The team is less than 5% short, so it no longer qualifies for the maximum 5% government funding. Statement 3: Yes. Prior to the change, the volleyball team had secured 100 + 130 = 230 in funding from TV rights and team sponsors, and this represented 86% of the team’s needed funding. Therefore, the team needed a total of 230 / (86/100) = 267 in funding. Now, the team will earn 100 × 1.15 = 115 in TV rights, or an increase of 15, so the team will have a total of 230 + 15 = 245 in funding. Calculate the percentage that this represents of total funding: 245 / 267 = 92%. The team is still more than 5% short, so it still qualifies for the maximum 5% government funding. Question 22
Statement 1: 1775. The graph gives you enough information to compute the number of characters in each document. Examine the labels of the columns: 1. Characters per word = Characters ÷ Word 2. Words per sentence = Words ÷ Sentence 3. Sentences per paragraph = Sentences ÷ Paragraph 4. Paragraphs If you multiply (1) and (2) together, the Words cancel, and you're left with "Characters per sentence." Keep multiplying all the way across, and you cancel out all the denominators! You're left with Characters. Now, one good and safe way forward is simply to estimate the numbers for each document (A through D), and multiply across, using the calculator. That might take a little time, but it's secure and not too time-consuming. Here's a set of sample estimations and results: (A) 5.22 × 45 × 1.5 × 26 ≈ 9,161 (B) 5.17 × 40 × 4.4 × 15 ≈ 13,648 (C) 5.06 × 38 × 1.1 × 32 ≈ 6,768 (D) 5.04 × 150 × 1 × 2 ≈ 1,500 So document B has the most characters. Checking against the text note, you can find that document B was published in 1775. Even if you estimate the numbers slightly differently, you'll still come out with document B as the clear winner. You can avoid a little computation by not computing out D's numbers, but if you notice, A and B are proportionally relatively close. A slight shift in a few points for each document, relatively unnoticeable, could cause the total characters of A to exceed those of B. The safe move is to estimate each number and multiply out. Statement 2: the ratio of words to sentences . "The ratio of characters to words" is another name for the "Characters per word" metric in the first column. Likewise, the other ratios in the answer choices are simply different names for the respective columns. So the question is this: Column 1 is most likely to be negatively correlated with which other column? Helpfully, column 1 is in ABCD order. What order are the other columns in? 2. DABC 3. BACD 4. CABD So they all preserve something of the original order, but notice that column 2 has the most drastic move (D moves to the very front). Moreover, in column 2 the letters besides D are all bunched up together, making their relative order less important; meanwhile, D has leapt out to a much greater number (150) than any of the others (which are somewhere between 35 and 45). This makes the effect of D's move much greater. (Yes, column 3 experiences something akin to this effect, but it's B that's out in front, and since B was already toward the beginning, it's less of a reversal.)
So the second column, with Words per sentence, that is most likely to be negatively correlated with the first column (Characters per word). You aren't responsible for computing the actual correlation coefficient until you get to b-school, but such computations bear out this result (the actual correlation coefficients between the first column and the others are -0.58, 0.43, and 0.33, respectively).
Question 23
Statement 1: No. Sort by column “1996” to find the range (maximum – minimum) for that year. The maximum value is 137.407 and the minimum value is 120.99, so, using the calculator, the range is 137.407 – 120.99 = 16.417. For the range to be greater than 15% of any single nation in 2000, the range must be greater than 15% of the maximum blood pressure posted in 2000. Sort by column “2000” to find the maximum value; 136.785. Use the calculator to compute 15% of this value; (0.15) (136.785) = 20.51775. Because the range in 1996 is less than 15% of the maximum value in 2000, (16.417 < 20.51775), this statement cannot be shown to be true. Statement 2: No. Sort each column independently, noting the country with the highest blood pressure for each year. The results are: 1980: Finland 1984: Finland 1988: Serbia 1992: Serbia 1996: Serbia 2000: Serbia 2004: Niger 2008: Niger So the nation that most frequently ranked highest in blood pressure was Serbia, which means that this statement cannot be shown to be true. Note that if the sorting was performed chronologically, as shown above, then we could have stopped after 2004, since at this point it was certain that Serbia had the highest blood pressure most frequently and not Finland. Statement 3: Yes. This prompt is asking us to find all nations that had an average male blood pressure below 125 mmHg in both 1980 and 2008. We can begin by sorting either column and identifying the nations that fit the criteria. For example, sorting the table by column “1980” shows that Papua New Guinea, Cambodia, India, Bhutan, Thailand, Vietnam, Nepal, and Maldives all had blood pressures below 125 mmHg in 1980. Now, sorting on “2008” shows that of these nations, only Papua New Guinea, Cambodia, Turkey, and Thailand remained below 125 mmHg in 2008. Therefore four nations fell below 125 mmHg in both years; this statement can be shown to be true. Question 24
Although you could draw a Venn Diagram or a Double-Set Matrix to answer this question, there is a formula we can use to save time. Suppose there is an overlap between two groups, group X and group Y. The total number of items equals the number of items in group X, plus the number of items in group Y, plus the number of items in neither, minus the number of items in both – because you double-counted them by adding X and Y. So here’s the formula: X + Y + Neither – Both= Total We can say the same for this problem: Spanish + French + Neither – Both = Total We know the total number of students is 350 and the number of students currently enrolled in Spanish is 230. Furthermore, we actually know that none of those 350 students is taking neither French nor Spanish.
230 + French + 0 – Neither = 350 French – Neither = 120 We need to find 2 choices that differ by 120, with the larger number representing the total number of students currently enrolled in French, and the smaller number representing the number of student enrolled in both French and Spanish. 260 and 140 are the only numbers in the list that have a difference of 120. Column 1: The correct answer is E. Column 2: The correct answer is A. Question 25
1st part
To calculate the interest rate at which the two options have equal total interest payments, figure out how much interest would be paid on the $240,000 loan at 5% and on the $190,000 loan at 4%. 5% of $230,000 = 0.05 × $240,000 = $12,000 <- use the calculator to do this quickly 4% of $190,000 = 0.04 × $190,000 = $7,600 For the total interest payment on Option 1 to be equal to the total interest payment, x , on Option 2, the interest on the $50,000 loan plus the interest on the $190,000 loan must be equal to the interest payment on the $230,000 loan. So $12,000 = $7,600 + x x = $12,000 – $7,600 = $4,400 Since you know the total interest payment for the $50,000 loan, you can compute the interest rate r% . Interest rate × principal = interest payment Interest rate = interest payment / interest rate r% = $4,400 / $50,000 = 0.088 = 8.8% 2nd part
To calculate the interest rate at which the interest payment on the $190,000 loan at 4% would be equal to twice the interest payment on the $50,000 loan at r% , first compute the interest payment on the $190,000 loan: 4% of $190,000 = 0.04 × $190,000 = $7,600 If the interest payment the $50,000 loan is equal to half the interest payment on the $190,000 loan, then you have: Interest payment on $50,000 loan = $7,600 / 2 = $3,800 Since you know the total interest payment for the $50,000 loan, you can compute the interest rate r% . Interest rate × principal = interest payment Interest rate = interest payment / interest rate r% = $3,800 / $50,000 = 0.076 = 7.6% Column 1: The correct answer is E. Column 2: The correct answer is D.
Question 26
Statement 1: True. Sort by total ore tonnage. The two countries with the greatest ore tonnage are the United States (2,464.8mm) and Australia (1,939.5mm). In order to calculate the exact values for the zinc and lead, we can multiply the percentages for each by the millions of metric tons of ore. There is a shortcut though: the problem asks us specifically to determine whether the zinc deposits are more than
1.5 times greater than the lead deposits. In the United States, lead deposits make up 2.38% of the ore. Multiply this number by 1.5: 2.38 × 1.5 = 3.57. The percentage for zinc grade is 3.88, which is greater than 3.57, so for the United States, it is true that zinc deposits are more than 1.5 times lead deposits. For Australia, lead deposits make up 4.39 percent. Multiply 4.39 × 1.5 = 6.59, which is less than the actual zinc grade percentage of 6.97 in Australia. In this case, too, zinc deposits are more than 1.5 times lead deposits. Statement 2: True. Sort by the Zinc Grade column. There are 12 countries, so the median will be the average of countries 6 and 7 (when sorted by Zinc Grade). The United States is #6, with 3.88% zinc grade and India is #7, with 4.06% zinc grade. Note: we do NOT need to calculate the actual value of the median. Any countries at or below 3.88% zinc grade are below the median, and any countries at or above 4.06% zinc grade are above the median. There are no countries between these two numbers (logically, there will never be, because these two figures are side by side on the list with nothing in between, by definition). Next, sort by the Lead Grade column. There are 6 countries with a Lead Grade percentage of 2 or greater (Germany, United States, Canada, Mexico, Russia, and Australia). Of these 6, how many have a zinc grade value of 3.88% or less? Three: Germany, United States, and Russia. It is the case, then, that exactly half (3 of 6) of the countries with a lead grade percentage of 2 or greater also have zinc grade percentages below the median. Statement 3: True. The term “range” refers to the difference between the largest number and the smallest number in a data set. First, note that both Total Ore Tonnage and Average Age are given in millions of years, so we can ignore the “millions” portion of each number and simply use the actual numbers shown in the table. Sort the table by Total Ore Tonnage. The range is 2,464.8 – 336.2 = 2,128.6. Next, sort by Average Age. The range is 1,791 – 15 = 1,776 (or notice that the larger number here, 1,791, is already smaller than the range of Total Ore Tonnage, which is more than 2,000).
Question 27
You’re given a long paragraph of text followed by a question. In this circumstance, it’s usually beneficial to read the question stem first; then you’ll have a good idea of which information will be most important while reading the passage. The question indicates that there is at least one activity that non-prescription medication manufacturers in Country X are required to perform, and there is at least one that is not permitted. You need to figure out what those two things are. The initial sentence indicates that Country X has instituted some new regulatory requirements (in simpler speech: laws) for non-prescription medication manufacturers. If companies import any components from other countries, they have to disclose those components, or tell the government what those components are. The second sentence details two additional requirements: (1) a company must use at least two different suppliers for any component, and (2) for certain medications (not all, only some), a company can’t rely on just one country for more than 70% of any single component. Note something else important about this sentence: item (1) does not mention anything about where the supplier is located. In other words, a company could have two suppliers in Country X, or one in Country X and one in another country. The final sentence explains three reasons why these regulations were implemented: to minimize contamination risks and / or financial dependencies on other countries, and to maintain a steady supply of critical drugs. Because you know that the question asks what activities are or are not allowed, most of your attention should be focused on the first two sentences – but you do still want to read this final sentence, though you may choose not to take detailed notes on it. One version of the notes might look like this: X: laws for NPM coms (1) disclose comps
(2) 2+ suppl per comp (3) for some meds, 1 comp < 70% from 1 country Why? 3 reasons Next, you examine the answer choices. What must these manufacturers do and what must they NOT do? Buying more than 70% of a component from one country : This choice is very tempting; the passage did mention a 70% figure. Be sure to verify the details in your notes. That requirement was true for some medications, but not all. If you’re talking about a medication that doesn’t fall into this category, then the manufacturer is not required to follow the 70% rule. This may be done but does not always have to be done; it is not the correct answer for either part of the question. Importing components from other countries : The passage is concerned with what rules the manufacturers must follow when they do import from other countries, so companies are permitted to import, but nowhere does the passage indicate that manufacturers must import from other countries. This is not the correct answer for either part of the question. Reporting to authorities the foreign components used : The first sentence said that these manufacturers “must disclose all sources of components that are imported from other countries.” In other words, the manufacturers are required to report any of the foreign components they use. This is the answer to the first part. Using a component for which only one supplier exists outside of Country X : This choice is very tempting;
be careful not to make assumptions that do not have to be true. The second sentence does indicate that companies must have at least two suppliers for any individual component, but the regulations do not specify where those suppliers must be. One or both could be in Country X. Nor does the paragraph indicate that this 2-supplier requirement applies only to foreign-sourced components; the sentence says only “for any individual component.” Manufacturing a drug whose components are all from a single supplier : The second sentence does indicate that a manufacturer must have at least two suppliers for any individual component. Even if a drug is made from one single component, the manufacturer must have at least two suppliers of that component. This is the answer to the second part.3 Column 1: The correct answer is C. Column 2: The correct answer is E.
Question 28
Statement 1: Conceptual Entity. First, find the box labeled “Body Location or Region.” It’s on the right side. The note tells you that an “is a” relationship is indicated by a solid rectilinear connector between a higher (parent) node and a lower (child) node. So, “Body Location or Region” has an “is a” relationship with “Spatial Concept,” which is its parent. In other words, Body Location or Region is a specific kind of Spatial Concept. (Note that “Body Space or Junction” is a sibling (on the same level), so there is no “is a” relationship with “Body Location or Region.”) Unfortunately, “Spatial Concept” is not an answer choice. However, you are told that “is a” relationships are inherited by grandchildren, great-grandchildren, etc. In other words, you can keep traveling up the tree. Body Location or Region is a specific kind of Idea or Concept, a specific kind of Conceptual Entity, and a specific kind of Entity as well. Of these, the only choice among the answers is Conceptual Entity. Body Location or Region is associated with other nodes via curved dotted connectors, but those do not represent “is a” relationships. Statement 2: Body Substance. The key words to notice are “spatially and functionally related.” These are two types of curved dotted connections between nodes. So the fast way to narrow down choices is to isolate nodes that are shown with two curved connections, then check the connection type.
Body Substance: has a produces relationship and a location of relationship with two other nodes. From the second hierarchy, you can see that produces is a functional relationship, while location of is a spatial relationship. This is the answer. The only other nodes that have two curved relationships shown are these: 1) Body Part, Organ, or Organ Component 2) Body Location or Region. Body Part, Organ, or Organ Component: has a part of relationship and a conceptual part of relationship with two other nodes. Part of is a physical relationship, while conceptual part of is a conceptual relationship. No good. Body Location or Region: has an adjacent to relationship and a location of relationship. These are both spatial relationships. No good. Question 29
To solve this logic problem, piece the clues together to see what restrictions they put on the final positions of runners. Work from the most restrictive clues to the least restrictive. The easiest clue to use first is that runner D finishes in fourth place. Graphically, draw this as: ___ ___ ___ _D_ ___ ___ ___ You know that runner C finishes in the top 50% but not in the first position, which gives two possibilities: ___ _(C)_ _(C)_ _D_ ___ ___ ___ You also know that A and E finish consecutively, which means they will always finish back-to-back, though you do not know in which order (A then E or E then A). Therefore, runners A and E could finish 1/2, 5/6, or 6/7. However, runner B finishes before F but after A, meaning B must also finish behind E, since E is consecutive to A. From this you can deduce that runners A and E must finish in slots 1/2, or there would not be enough spots left for B and F. As runner A does not appear as an option in the table, you must select E as the runner that could have finished first. Finally, the above tells you that the last 3 slots as taken by B, F, and G. Knowing that B finishes before F leaves 3 distinct possibilities for this placement: ___ ___ ___ _D_ _B_ _F_ _G_ ___ ___ ___ _D_ _B_ _G_ _F_ ___ ___ ___ _D_ _G_ _B_ _F_ Therefore, the runners that could have finished fifth place are B and G. Since G is not in the chart, select B as your answer. Column 1: The correct answer is E (fourth entry). Column 2: The correct answer is B (first entry).
Question 30
Statement 1: No. We cannot show definitively that the customer will not order the syringes from this supplier. The customer does not state that she will not order the syringes if she does not receive a discount specifically on the syringes. In fact, she states that her boss “might be willing” to place the order if the company receives a volume discount (and she doesn’t specify that this must be for the syringes); the supplier did offer a volume discount on the pumps. It’s possible, then, that she will also order the syringes from this supplier.
Statement 2: No. The supplier does claim that his syringes will “typically” last longer than the guaranteed 20 injections. If a syringe lasts for 30 injections, that would be (30-20)/20 = 50% more injections than guaranteed. If a syringe lasts for 40 injections, that would be (40-20)/20 = 100% more injections than guaranteed. The supplier claims that his syringes typically allow for 50% to 100% more injections than the base level guaranteed. Statement 3: Yes. In the 2nd email, the customer indicates that she is unhappy about the pricing and then states “If we were able to get a volume discount, though, he [the boss] might still be willing to give you the order.” This last part implies that they are already thinking seriously about switching to the other supplier. Question 31
In this “Except” question, we will need to find the “odd one out” answer. Four of the answer choices describe negotiating tactics used by the customer. The fifth one (the correct answer) does not. (A) Incorrect. The customer attempts to indicate the value of her business by pointing out to the supplier that her company is “ordering more than 1,000 pumps.” (B) Incorrect. The customer specifically asks for a large-volume discount. (C) Correct. The supplier offers a concession, but the question asks us to find negotiating tactics used by the customer, not the supplier. The customer asks for a volume discount on the pumps; the supplier offers a volume discount on the entire order (if the customer orders more). (D) Incorrect. The customer both tells the supplier to “hold off on the syringe order” and mentions that a volume discount might make her boss “willing to give you the order.” This implies that she is willing to consider walking away from the deal if needed. (E) Incorrect. The customer states that her boss is “upset” about previous price increases. The correct answer is C. Question 32
Statement 1: No. Email #1 indicates that the 20-injection syringes sell for $20 per 50 syringes, or $20 / 50 = $0.40 per syringe. Each syringe is guaranteed for 20 uses, or $0.40 / 20 = $0.02 per guaranteed use. Email #2 indicates that the 30-injection syringes sell for $30 per 50 syringes, or $30 / 50 = $0.60 per syringe. Each syringe is guaranteed for 30 uses, or $0.60 / 30 = $0.02 per guaranteed use. Per guaranteed use, the two syringes cost the same: $0.02. Statement 2: Yes. If the customer sticks with the original order, the price for the 1,200 pumps will be (1,200/10) ($175) = $21,000 less a 5% discount: ($21,000) (0.95) = $19,950. The price for the 50 syringes will be $20 and the price for the 550 needles will be (550/50) ($25) = $275. The total order will cost $20,245. If the customer orders an additional 200 needles, she’ll earn a 10% discount on the entire order. Before the discount, her cost will be $21,000 for the pumps, $20 for the syringes, and (750/50) ($25) = $375 for the needles, or a total of $21,395. Subtract 10%: ($21,395) (0.9) = $19,255.50. Therefore, the order that includes the additional 200 needles will be the less expensive order. There is a neat shortcut for the above math. What is the difference between the two scenarios? Scenario 1 units
Scenario discount
1
Scenario units
2
Scenario discount
pumps
1,200
5%
1,200
10%
syringes
50
none
50
10%
2
needles 550 none 750 10% The customer would buy the same amount of pumps and syringes under either scenario. For both products, she receives a larger discount under scenario 2. For the needles, though, while the customer
receives a larger discount under scenario 2, she also has to pay for 200 additional needles. 200 additional needles will cost an additional (200/50) ($25) = $100. In order for the customer to save money on the second scenario, then, the 10% discount must save the customer more than $100 on the entire order. The pumps cost $175 for a box of 10 and the customer wants to order 1,200 of them. 120 boxes at $175 per box is $21,000, so the customer is saving about $2,000 on the pumps alone, compared to half that (5%, or $1,000) under the first scenario. In other words, the savings on the pumps alone more than makes up for the extra $100 spent on needles. Statement 3: Yes. The syringes are guaranteed for a minimum of 20 uses. The needles are “single-use.” If she orders 50 syringes, they will be able to give a minimum of (50) (20) = 1,000 injections. There are more than enough syringes, then, to use up the 550 needles she plans to order.
Question 33
Statement 1: Yes. Sort by the “In continuous operation since” column. The 4 breweries that have been in continuous operation the longest have annual capacities of 1,700, 2,200, 2,200, and 2,700 liters. The average (arithmetic mean) is the sum of these numbers divided by 4. 1,700 + 2,200 + 2,200 + 2,500 = 8,800 8,800 / 4 = 2,200 The average annual capacity of the other four breweries is the sum of their capacities divided by 4. (5,800 + 1,000 + 1,500 + 115) / 4 = 2,103.75 The statement is true. Statement 2: No. Sort by Peating level. The two unpeated whiskies are at the top, with peating levels of 1.5 and 3.5, leading to a median of 2.5. Meanwhile, the other 6 whiskies, which are all peated, are sorted nicely below. Since the two middle peating levels among those whiskies are both 35, the median is 35. 35 – 2.5 = 32.5, which is NOT more than 35 ppm. The statement is false. Statement 3: No. The trick to this question is to only pay attention to breweries that have been in continuous operation for a short time. Ardbeg, Bruichladdich and Kilchoman have been in continuous operation for the least amount of time. Kilchoman, however, was founded in 2005, so the ratio is 1:1. That’s out. Ardbeg was founded 197 years before 2012 (2012 – 1815 = 197), and was in continuous operation for 15 years (2012 – 1997 = 15). 197/15 = 13.13. Bruichladdich was founded 131 years before 2012 (2012 – 1881 = 131), and was in continuous operation for 11 years (2012 – 2001 = 11). 131/11 = 11.9. Ardbeg has a higher ratio, so the statement is false. (Even if you “add 1 before you’re done” to count the full year of founding and the full year of 2012, you end up with Ardbeg having a higher ratio.) Question 34
This short problem is quite difficult because of the subject matter. The first sentence describes a value that is increasing steadily “by an unknown multiplier” or growth factor. So there is some multiplier that could be used to calculate the size of the population at the end of every hour. For example, if you know that a population starts out with 2 bacteria and the multiplier is 5, then at the end of the first hour, there will be 2 * 5 = 10 bacteria; at the end of the second hour, there will be 10 * 5 = 50 bacteria, and so on.
The second sentence then provides information to calculate that unknown multiplier, or growth factor; let’s call that multiplier r . At t = 1, the population is p . At t = 5, the population is p 2. You are told that the growth factor corresponds to hourly growth, so there are 4 one-hour intervals in the timeframe described. So the question becomes this: what value of r , multiplied by itself 4 times (for the 4 one-hour intervals), would take you from p to p 2? Multiplying r by itself 4 times is the same thing as raising r to the 4th power: (r )(r )(r )(r ) = r 4. In the four hours described, the bacteria grew from p to p 2 (or p times p ) so the bacteria grew by a factor of p . Therefore, p = r 4, or p 1/4 = r . (“ p to the one-fourth power” is the same thing as the fourth root of p .) p. The hourly growth factor, r , is p 1/4, or 4√ Here’s a way to look at the growth, step by step: t =
1
given p
calculation
(given) 1/4
=
5/4
2
×
3
54
×
14
=
64
4
3/2
×
1/4
=
7/4
=
32
2 7/4 5 × 1/4 = 8/4 = 2 You can now use the growth factor, p 1/4, to calculate the population size at t = 0. If you were going from hour 1 to hour 2, you would multiply the hour 1 population by the growth factor in order to get to the hour 2 population (as in the above table). Because you are working backwards, from hour 1 to hour 0, you instead divide hour 1’s population by the growth factor: p / p 1/4: p / p 1/4 = p 1 – 1/4 = p 3/4 “p to the ¾ power” is the fourth root of p 3, or 4√( p 3).
Column 1: The correct answer is A. Column 2: The correct answer is D.
Question 35
Statement 1: February 17. Earthquakes of magnitude 3 or greater are coded with darker shades of gray, all the way to black, whereas magnitude 1 earthquakes are white (on the bottom of the stack) and magnitude 2 earthquakes are light gray (just on top of the whi te columns). So the question becomes this: which day (other than February 14) has the greatest proportion of dark grays & blacks? By inspecting the columns, February 17 has the greatest number of dark grays & blacks: at least 20, whereas none of the others has as many. Meanwhile, only February 19 has fewer earthquakes overall, in fact only slightly fewer (February 17 seems to have 115 earthquakes, while February 19 has at least 111). February 19 only has ~15 earthquakes of magnitude 3 or greater. So the proportion of ~20 magnitude 3+ to 115 earthquakes is the largest of any day other than February 14. Statement 2: 22%. You can solve this problem by summing up on the calculator all the daily totals, estimating the exact numbers as best you can: 177 + 133 + 142 + 115 + 115 + 111 = 793. Finally, 177 ÷ 793 = 22.3%. Alternatively, you could estimate more roughly and get to the right answer, since the 5% difference between successive answer choices is actually rather large. For instance, you could eliminate A quickly by noticing that 17% is just a shade greater than 1/6 (= approximately 16.7%); however, February 14 showed a markedly greater number than 1/6 of the total. It might be hard to eliminate 27%, but 32%
and especially 37% should seem outlandishly big – after all, 32% is just slightly less than 1/3, and February 14’s total is not large enough to cover two average days. Question 36
Statement 1: 274. This problem asks you to find the difference in the number of billionaires between the US and the total of Russia, the UK, and Germany combined. Because the table is given in percents, you can first find the total number of billionaires in these two groups. Using the labels on the chart, you can see that the United States had 44% of the worldwide billionaires, Russia had 6%, the United Kingdom had 3%, and Germany had 6%. This means that Germany, Russia, and the United Kingdom had 6 + 6 + 3 = 15% of the total worldwide billionaires. You now convert these from percents to actual numbers, knowing that there were 946 billionaires worldwide. Number of US Billionaires = (44%)(946) = (0.44)(946) = 416.24 ≈ 416 Number of Russia, UK, and Germany Billionaires = (15%)(946) = (0.15)(946) = 141.9 ≈ 142 Therefore, the difference between these two groups is given by 416 – 142 = 274. Alternatively, you could subtract the percents first, which would give a 44% - 15% = 29% change. This also gives 0.29(946) = 274.34 ≈ 274. Statement 2: 35. You are asked to find the number of billionaires for each country grouped in the “Other” category, assuming that there are 7 countries in this category and each has the same number of billionaires. First, you must find the total number of billionaires given by “Other”. Using the table, you see that “Other” makes up 26% of the total 946 billionaires. Therefore, the total number of billionaires in “Other” is given by (26%)(946) = (0.26)(946) = 245.96 ≈ 246. Because there are 7 nations contributing the same number, you divide the total by 7 to achieve the desired result: 246 / 7 = 35.14 ≈ 35. Alternatively here you could take 1/7th of the percent first, and then calculate that fraction of the total. This gives 26% / 7 = 3.71%, and so (0.0371)(946) ≈ 35. (By the way, in actuality, it would be unlikely that there would only be 7 countries in the “Other” category. Typically, in a pie chart such as this one, the “Other” category would contain countries whose wedges would all be smaller than that of the smallest country shown individually, which would be Canada or China at 2%. Since the “Other” wedge is 26%, you’d expect at least 26 / 2 = 13 countries within. However, you should take the given condition that there are 7 countries in the Other group at face value and proceed as directed. When you are given a condition, don’t challenge it!)
Question 37
Statement 1: No. Sort by “% wages” to put that column in ascending order. You can readily observe that the population is also generally increasing from top to bottom, indicating that the correlation between these two quantities is positive, not negative. Statement 2: Yes. Here, you need to deal with a derived number: the percent of income derived from sources other than those listed (wages, salaries, dividends, and interest). If you had Excel and could build another column, you would add the two percent columns that you do have and subtract them from 100%. Then you would sort by that new column. You want the lowest number in that column, so you want the highest sum of the two percents. Since the wage numbers seem to have more variability, sort by that column. Texas is at the bottom, with 76.8% wage income. Together with 4.4% dividends & interest, you get 81.2% (for 18.8% other income). Will any other states give you a higher sum? As you scan upwards, you should see that no other state will have as large a sum of % wages and % dividends & interest (of course, check any possibilities with your calculator). So Texas is the subject of the sentence.
What about the predicate? Does Texas have an income per capita (meaning per person ) of more than $20,000? Well, divide the income per tax household by the size of tax household: $51,747 / 2.58 ≈ $20,057, which is more than $20,000. The statement is true. Statement 3: No. To compute the household income in dollars from dividends and interest in any state, multiply the household income in dollars by the “% dividends & interest” number. Georgia: $50,320 × 4.5% = $2,264 New Mexico: $44,129 × 4.9% = $2,162 Even though New Mexico has a higher percent of this sort of income, the average income is lower—by enough to bring the dollar figure below that of Georgia. The statement is false. Question 38
The charity plans to auction off 100 lots this year. If they average $1,500 per lot, they will raise $150,000, leaving an additional $100,000 to be raised by pledges. The capacity of the room is 1,098, but not all of those people will be invited guests; the executive director indicates in the third email that approximately 10% will be servers and other workers. That leaves 1,098 × 0.9 = 988 invited guests. The average pledge per guest would need to be $100,000 / 988 = $101, or approximately $100 per guest. The correct answer is B.
Question 39
Statement 1: No. In the third email, the executive director does introduce the idea of changing to a larger venue, but says only that “maybe” they “should think about” doing so. The paragraph concludes with a statement that the current venue should be “fine” if the number projections are accurate, indicating that the director is willing to stick with the current venue. Statement 2: Yes. In the second email, the auction coordinator indicates that they raised a total of $131,000 + $73,000 = $204,000 last year, or $46,000 short of this year’s goal. The coordinator then indicates that the additional items for auction this year “should go for about the same price on average” and therefore “should be enough to raise the additional $50,000 needed .” Notice the final 4 words (emphasis added): the coordinator is inferring that the charity will raise the same amount of money as it did last year plus the additional $50,000 to meet the new goal. Statement 3: No. In the second email, the auction coordinator says that the charity “ could probably plan on similar response and attendance rates this year.” This language indicates a degree of confidence, but not certainty. Note that there is a “mix-up” trap here; later in the same paragraph, the auction coordinator says “I’m certain we can reach 100 items.” Question 40
Statement 1: No. For the previous year, 1,378 people attended and there were $131,000 in pledges. The pledge per person, then, was $131,000 / 1,378 = $95.07 per person. This year, the capacity is 1,098, but 10% of the people in the room will be workers, not invited guests. Therefore, there will be a maximum of 1,098 × 0.9 = 988.2 = 988 invited guests. If they pledge at the same rate, they will pledge a total of approximately $95 × 988 = $93,860. There were 58 auction lots last year, from which $73,000 was raised. So the per-item rate was $73,000 / 58 = $1,258 (approximately). This year, the charity plans to have 100 lots; at the same per-item rate, they can expect to raise $1,258 × 100 = $125,800.
$94,000 and $125,800 add up to less than $250,000. Statement 2: No. At first glance, this statement seems to ensure the $250,000+ goal: the charity will have more people attending than last year and will also offer more items for auction than last year. This statement, however, neglects to provide any information about the pledges made by the guests or the amount raised by the auction items (either averages or any other figures). We cannot say for sure, then, how much the charity would raise. Statement 3: Yes. We can re-use some of our calculations from the first statement. We know that there are 988 invited guests; if they pledge an average of $150 each, they will pledge $148,200 collectively. We also know from statement 1 that there were 100 auction lots at an average of $1,258 for a total of $125,800. That alone is enough to raise the charity above the $250,000 goal, without even accounting for the higher average ($1,300) given in this statement. This plan indicates that the charity will raise more than $250,000. Question 41
Statement 1: strong nuclear. You are told that bosons carry particular forces, and that the blue lines show what interactions are possible. You are asked to identify the force that affects quarks but not leptons. Since the bosons correspond to forces, you can rephrase the problem: what boson “touches” quarks but not leptons via a blue line? Following the blue lines, you can see that the only boson that touches quarks but not leptons is the gluon. The notes indicate that the gluon carries the strong nuclear force. Statement 2: Z boson. Some bosons have a loop back to themselves, indicating that they can interact with other particles of precisely the same type. Which bosons do not have a loop back to themselves? Only two: photons and Z bosons, either of which would be possible answers. Question 42
In the given argument, Colleague A indicates displeasure with one aspect of the current office policy for “personal event” parties given by employees: because the parties are limited to one per month, everyone celebrating something is “lumped together” for that month. Note that colleague A does not actually go so far as to call for individual parties, one for each person celebrating something. Colleague B provides 3 reasons why, in her opinion, offering individual parties would not be a good idea: costs might go up, productivity might go down, and someone might be overlooked. We’re asked to find two things. First, we need to find something to which Colleague A would likely object . The first two statements are “opposites”: there are either “significant disadvantages” (answer A) or “significant advantages” (answer B) to changing the current policy in this matter. Colleague A does dislike something about the current policy, so it’s unlikely that this colleague would object to the statement that there are advantages to changing the policy. Would Colleague A argue that there are no significant disadvantages to such a policy change? We can’t quite go as far as that – Colleague A mentioned one advantage but did not discuss disadvantages, nor did he indicate that there are not any disadvantages to changing the policy. Answer C says that the company should maintain its current policy; Colleague A is very clear that he dislikes at least one aspect of the current policy. Colleague A, then, would be likely to object to this statement. Answers D and E address who should pay for the parties; Colleague A does not address this aspect of things at all.
Therefore, Colleague A’s objection is that the company should not maintain its current policy. How would Colleague B be likely to respond? Colleague B listed three problematic consequences of changing to another policy, so Colleague B would respond that there are significant disadvantages involved with changing the current policy. Column 1: The correct answer is C. Column 2: The correct answer is A. Question 43
Statement 1: No. Sort the table by Act / Hyp; there are seven sequences in the Hyp category. Scan the Max Identity category for these seven sequences. Six of them are less than 50%, but Ovalbumin-related protein Y is 99%. Statement 2: No. Sort the table by Serpin Clade. The four sequences in serpin clade A all have lengths in the 400s. The three sequences in serpin clade B all have lengths in the 300s. The average length of sequences in serpin clade B, then, must be lower. Statement 3: Yes. Sort the table by Length. 5 of the 11 sequences have a length less than 415; therefore 6 of the 11 have length 415 or greater. If choosing at random from among the 11 sequences, you have a 6/11 chance to choose one with length 415 or greater. This is greater than ½ (or 50%). Question 44
The question is a classic overlapping sets problem with three groups so we use the Venn Diagram approach to solve. Construct the Venn Diagram as follows:
The region in the center is the overlap of all three groups, which we know from the problem to be 5. We are also told that the same number of students overlap between any two groups, which means that the three overlapping regions in the center (excluding the middle) must all have the same value; we can call this value x . Since every class must contain 20 students and all the circles are identical, we know that the same number of students is missing from each class; we can represent this with another variable, y . The resulting Venn Diagram looks like this:
The question wants us to find a possible number of students who are in only one class ( y + y + y = 3 y ), and the corresponding number of students who overlap between any pair of classes (5+ x ). To find out more information about y and x, we can use the fact that every class must contain 20 students. Therefore, the following relation must hold: 5 + 2x + y = 20 2x + y = 15 Using the Venn Diagram, we also find that the number of students in only one class is 3 y and that the number of students overlapping between any pair of classes is 5 + x . We proceed by testing the answer choices in order. Immediately we recognize that the number of students common to any two classes cannot be the first option, 2, since it must be at least 5. Next we test the overlap of 7, which means that x = 2. Substituting in we find that 2x + y = 15 2(2) + y = 15 y = 11 This means that the number of students in only one class is equal to 3(11) = 33. Both answers appear in the table, which means that this is the correct answer. No need to test further choices. As an alternate solution, we might notice that 2x + y = 15 only has 8 possible ( x ,y ) solutions because x and y must be integers. Testing systematically we start from x = 1 and find that 5 + x = 5 + 1 = 6. This answer is not in the table, so we rule out x = 1. Trying x = 2 we find that 5 + x = 5 + 2 = 7, which is in the table, so we proceed. From here we find that y = 13 and 3y = (3)(13) = 39. As this answer is also in the table, we must have found the correct combination. Column 1: The correct answer is E. Column 2: The correct answer is B. Question 45
Statement 1: 1/2. Looking quickly at the drop down menu choices, we see that we are going to have to find a fraction comparing the number of households in the Midwest that do not use electricity, to the number of households in the US that do. This means we will need to find the following: (Midwest households that do NOT use electricity) ÷ (US households that DO use electricity) We don’t know these actual numbers in advance, but we do know the total numbers of Midwest and US households and the percents that use various types of heating sources, so we can calculate. Let’s start with the number of Midwest households that DO NOT use electricity. First, we need to find the fraction of Midwest households that fit this qualification. Rather than add up all of the NON-electricity categories, we can simply take the electricity percent (18%) and subtract that from 100% to get 100% – 18% = 82%. Now, to move from proportion to an actual number of households, we need to find 82% of the total, which is 25.9 million. Using our calculator, we find that 0.82(25.9) = 21.238 million Midwest households. Now we can do the same thing for the number of US households that DO use electricity. First, we can read from the chart that 35% of US households fit this qualification. Now, to move from proportion to an actual number of households, we need to find 35% of the total (113.6 million) = 0.35(113.6) = 39.76 million US households. Finally, we can plug these values into the fraction we are trying to sol ve: 21.238/39.76 = 0.53 ≈ ½. Statement 2: 1,662. This question is asking about an average number of households in the Midwest region per state. That means that we will need to find the total number of households that use either propane or natural gas (8% and 69% respectively = 77%) and then divide that by the number of states in the region (12 states). So, to find the total number of states, we need to figure out 77% of the total (25.9) = 0.77(25.9) = 19.943 million households. To find the state average, we then divide by 12 to get 19.943/12 = 1.6619 million households. But the question asks us for the number in thousands of houses, so we need to convert millions to thousands. There are 1,000 thousands in a million, so multiply 1.662 by 1,000 thousands per million to get approximately 1,662 thousand. (1.662 million = 1,662 thousand.)
Question 46
A.
B. C. D. E. F.
This question is exactly like a Critical Reasoning Assumption question. To that end, it may be helpful to find the basic components of the argument. The conclusion is the last sentence: “Considering this initial success, DSSCs are going to be the most useful form of solar cell technology.” The premises are, more or less, everything else. In order to prove the conclusion, we need to make sure that the other forms of solar cell technologies (which are not discussed anywhere in the prompt) will be less useful. Possible fact. This statement of fact would provide the most support for our assumption (F). The two descriptions here (“least expensive and most efficient”) are both superlatives along important dimensions, expense and efficiency, that would both significantly affect the utility (usefulness) of DSSCs. Bringing the cost down is positive, but it’s not enough to make DSSCs the MOST useful form of solar cell technology. We don’t know, for instance, how those cost decreases compare to those possible with other technologies. Again, this is good news for DSSCs, but the “plausibility” of DSSCs does not make them the best solar cell technology available. The efficiency of the nanoparticles is irrelevant. It’s nice that this efficiency is “above average”; however, it provides at best only weak support for the assumption, since we need to assume that DSSCs are the most useful. Required Assumption. This is the assumption we need. If no other forms of solar cell technologies will be more useful than DSSCs, than DSSCs will have to be the most useful. Column 1: The correct answer is F. Column 2: The correct answer is A. Question 47
Statement 1: Slovenia. The average national fourth grade TIMSS scores for 2003 and 2007 are coded in red and green respectively. This question asks us to find a nation that has a lower red and a higher green TIMSS score than New Zealand. New Zealand’s TIMSS scores fall below more than half the countries on the graph, so it is advantageous to start by finding the countries that have a lower red score, as fewer options will remain. Of the countries shown, only Iran, Norway, and Slovenia meet this criterion. We now search among the remaining three countries to find one that also has a higher green TIMSS score. Of the three, only Slovenia matches. Statement 2: Hong Kong, China. You can solve this problem by calculating the percent change: (Final – Initial) / Initial Calculate this percent change from 1995 to 2007 for each of the five options in the dropdown list. Estimating by eye and using the calculator we find that: Australia: (515 – 495) / 495 = 4.04% Hong Kong, China: (605 – 555) / 555 = 9.09% Iran, Islamic Republic: (400 – 380) / 380 = 5.26% Norway: (470 – 475) / 475 = -1.05% United States: (525 – 515) / 515 = 1.94% Note that it is much easier to calculate using only the 5 answer choices than to calculate using all countries on the graph.
Question 48
To solve this Rate-Time-Distance problem, we first recognize that Car A starts off behind Car B but then gains enough to actually be ahead in 4 hours. This means that we have a situation where Car A is chasing Car B, telling us that the two cars are working against each other (Car B keeps driving, making it harder for Car A to actually catch-up ). When things work against one another, we can simplify by subtracting their rates to get the relative rate at which Car A is gaining on Car B. Let A denote the speed of Car A and B denote the speed of Car B, so their relative rate is given by A – B . Physically, this is the rate at which Car A is gaining on Car B. We are then told that Car A starts out 20 miles behind Car B but finishes 4 miles in front at a time 4 hours later. Thus Car A gains on Car B by a total of 20+4 = 24 miles in 4 hours. We now make use of the formula Rate × Time = Distance. Notice that the combined rate, A – B , multiplied by the time of 4 hours must give us a total distance of 24 miles. Solving for A – B we find that: (A – B ) × 4 = 24 A – B = 6. Therefore, the speed of Car A must be 6 mph more than the speed of Car B, so we look for answers in the table that can match this criteria. The only two values with a difference of 6 are 54 mph and 48 mph. As Car A is the faster car, we mark its speed as 54 mph and Car B’s speed as 48 mph. We could also solve this problem algebraically. Car B has a rate of B , travels for 4 hours and will go some unknown distance d . Car A has a rate of A , also travels for 4 hours, but will have to travel not only d miles but also the 20 miles it is behind at the start, as well as the 4 miles it goes beyond B. This gives us 2 equations: 4B = d 4 A = d + 20 +4 Substituting in for d to gives the following relationship for A and B: 4 A = 4B + 24 4 A - 4B = 24 4( A – B ) = 24 A – B = 6 Column 1: The correct answer is D. Column 2: The correct answer is B.
Question 49
Statement 1: 50. If the five smallest states are estimated (under the given simplification) to have the same population each, then they have about 600,000 people each (according to the text), or 3,000,000 altogether. From there, we have to approximate the number of farmers markets. The five smallest dots are all to the left on the graph. Two have about 35 farmers markets each; one has about 25 markets; one has about 10; and the last has about 55. Together, that adds up to 160 markets for 3,000,000 people. To change the denominator to 100,000 people, we divide by 30, so we must do the same to the top, yielding 5.33…, or approximately 5 farmers markets per 100,000 people. Notice that if you misidentify one of the states (say, picking one of the cluster around 70-80 markets each), it will not make a real difference, since the answer choices all differ by powers of ten. Statement 2: 1 : 13,000. The two largest states are clearly the two on the right. One looks slightly bigger than the other, so we can estimate their populations at 13,000,000 and 12,000,000, for a combined total of 25,000,000. From there, we need to estimate the number of obese people in each
state. The 13,000,000 state looks to be at about 28.75% (halfway between 27.5 and 30), while the 12,000,000 state is slightly higher, more like 29.5%. 13,000,000 × 0.2875 + 12,000,000 × 0.295 = 3,737,500 + 3,540,000 = 7,277,500, which is the number of obese people in the two largest states. Now, we simply find the number of farmers markets in those two states, which looks to be about 300 + 275, or 575. 7,277,400 ÷ 575 ≈ 13,000, or 13,000 : 1. That’s the ratio we’re looking for. Question 50
Statement 1: 2005. The table is already sorted by year, so there is no need to re-sort to answer this question. Of the 9 movies that were filmed in 2004, 4 had English as one of the spoken languages. 4/9 ≈ 44.4%. Of the 12 movies that were filmed in 2005, 6 had English as one of the spoken languages. 6/12 = 50%. A greater percentage of movies in 2005 had English as one of the spoken languages. Statement 2: 2004. Sort by the “B/W?” column, and look for the “No” entries. The movies that were shot entirely in color in 2004 had lengths in minutes of 106, 105, 85, 90, 88, and 76. The movies that were shot entirely in color in 2005 had lengths in minutes of 100, 80, 75, 96, 91, 92, 80, 95, and 90. The average length of the movies from 2004 = (106 + 105 + 85 + 90 + 88 + 76) / 6 = 91.7 minutes. The average length of the movies from 2005 = (100 + 80 + 75 + 96 + 91 + 92 + 80 + 95 + 90) / 9 = 88.8 minutes. Statement 3: 2005. Sort by the “Medium” column. 6 movies were filmed in 2004 using 35mm film. Of those, only 1 was also shot in black and white. There were also 6 films shot in 2005 using 35mm film. Of those, 2 were also shot in black and white. A higher percentage of films shot in 35mm film in 2005 were also shot in black and white. Question 51
Statement 1: No. The information to answer this statement is all contained in the third tab. The last four elements have the shortest half lives, and only three of the four allow EC as a decay mode (element #100, fermium, does not). Statement 2: Yes. First, check the possible decay modes of californium. They are listed as Alpha particle emission and Spontaneous Fission (SF). Now, we must turn back to the descriptions of the various decay modes provided in the first tab. SF has no effect on the mass number of an element, which leaves only Alpha particle emission. This decay mode is described as causing an element to drop “the atomic number by 2 and the mass number by 4.” The atomic number of californium is 2 greater than the atomic number of curium, and the mass number is 4 greater. This fits the description of the decay mode exactly. Statement 3: Yes. This statement revolves around the percent difference formula (difference/”original” * 100), and a basic understanding of statistics. Because there are 11 elements listed in this question, the median will typically be the 6 th element (counting from either the top or the bottom). All the relevant data for this statement is to be found in the second tab. However, because there are only six data points for density, the median density will actually be the average of the melting points of the third and fourth elements. The element with the median melting point is Californium, at 1,173 degrees. The highest melting point is 1,900 degrees. The percent difference is (1,900 – 1,173) / 1,900 =0.3826, which gives us a percent difference of 38.3%. You are told that the “original” (what you’re taking the percent of ) is the highest melting point, so that’s why we divide by 1,900.
The median for density is the average of the middle two densities, 15,100 and 14,780. (15,100 + 14,780) / 2 = 14,940. The highest density is 20,450. The percent difference will thus be (20,450 – 14,940) / 20,450. Just by looking, we can see that this will be about 25%, which is much smaller than the other percent difference. (The calculator tells you 0.269…)
Question 52
If an element has more than one possible decay mode, it is impossible to know its atomic number after one cycle of decay (because three of the four decay modes change the atomic number in three different ways, and the fourth, spontaneous fission or SF, breaks the nucleus apart entirely, changing the atomic number unpredictably). So, to know the result, we’d need to be sure that the element had only one decay mode (and that decay mode can’t be SF). Statement 1: Cannot Be Determined. Neptunium (#93) has two decay modes. Statement 2: Can Be Determined. Mendelevium (#101) has only one decay mode, which is not SF. Statement 3: Cannot Be Determined. Einsteinium (#99) has three decay modes. Question 53
To begin, we must isolate the elements that meet the initial criterion, ‘two possible decay modes.’ Looking at the information in the third tab, we see that six elements have two possible decay modes: 93, 94, 95, 98, 100, 103. In order to answer the question, we need the statistics from the second tab for the six elements with two decay modes: Mass # of Smallest Mass # Atomic Most Stable Element of an Number Isotope Attested Isotope (MSI) 93
Neptunium
237
225
94
Plutonium
244
228
95
Americium
243
231
98
Californium
251
237
100
Fermium
257
242
103 Lawrencium 262 251 At this point, the description given in the first tab comes in handy. “An element is defined by its atomic number, the number of protons in the atomic nucleus...The mass number of an isotope is the total number of protons and neutrons.” This tells us that the number of neutrons in a given isotope is simply: mass number minus atomic number. But the atomic number is the same for every isotope of a given element! That means that the difference between the mass number of the most stable isotope and the smallest mass number of an attested isotope IS the difference between the neutrons (as the proton number always remains the same).
From here, finding the answer is relatively straightforward. We just subtract the value in the last column from the value in the second to last column for each element (we can skip Californium, as it doesn’t show up in the answer choices). The largest difference is 16, for Plutonium. Plutonium has an atomic number of 94, which is also its number of protons. Question 54
The only way to answer this question is to quickly work out all the possible patterns of 6 rounds that would result in a score of 4 points. Basically, Player X could have won 1, 2, or 3 games (any more than that and the score is too big, while draws alone can’t bring the score up to 4 points in only 6 games). Let’s consider these three possibilities: 1 Win: With only 1 win, the only way to get up to 4 points would be to draw the next 4 games, and sit the last one out. This will turn out to be the situation that the answer choices fit. 2 Wins: Looking at the answer choices, we can see that 2 isn’t an option. However, with 2 wins, the only way to reach 4 points is to draw twice and then lose once (and sit the final round out). 3 Wins: With three wins (adding up to six points), the only way to get down to 4 points is to lose two matches in a row (and sit out the final round). While 3 is an option in the answer choices, we already know that 2 isn’t, so this can’t be the correct situation. Question 55
Since the “Butterflies of Asia” exhibit already has two Asian butterflies and one non-Asian butterfly (the Australian Painted Lady), any Asian butterfly can be added to it; however, no butterfly that is not found in Asia can be added. Similarly, since the “Blue Butterflies” exhibit already has two blue colored butterflies and one butterfly that is not either blue or purple (the Great Nawab), any blue or purple butterfly can be added to it, but no butterfly that is not at least partially blue or purple can be. So a butterfly that CAN be added to both groups must be blue and Asian. A butterfly that CANNOT be added to either group must be neither blue nor Asian. Option F – Could be added to either exhibit
The Emerald Swallowtail butterfly is both partially blue and lives in South Asia, so it could be added to either exhibit. Option D – Could NOT be added to either exhibit
The Monarch Butterfly contains no blue and its natural habitat is only North America, so it can’t be added to either exhibit. Options A, B, C, and E
All of these butterflies are either found in Asia or partially blue or purple, but not both, and so could be added to one exhibit but not the other. Column 1: The correct answer is F. Column 2: The correct answer is D. Question 56
The question is this: what number P must be in the pool, and what number T must be on the team, such that when you choose T from P , you have between 20 and 25 possibilities (not-inclusive)? We also need T to be at least half of P. This is a combinatorics problem. The formula we want to use for the selection of a team (for which order of selection doesn’t matter, only the composition of the team) is this:
The “Ins” represent the team T , and the “Outs” represent those not chosen, or P – T. To get a number as high as 21 to 24, we should start at the high end for the pool P , and then test possibilities for T. 8!/(8!0!) = 1 (always just 1 way to choose all the candidates from a pool!) 8!/(7!1!) = 8 no 8!/(6!2!) = (8)(7)/2 = 28 no 8!/(5!3!) = (8)(7)(6)/6 = 56 no 8!/(4!4!) = (8)(7)(6)(5)/24 = too big Notice that you don’t have to check T = 3 or lower, because T must be at least half of P . Now try P = 7, skipping right to T = 6 and going down. 7!/(6!1!) = 7 no 7!/(5!2!) = (7)(6)/2 = 21 yes! There can be only one set of answers, so we can stop here. Column 1: The correct answer is E. Column 2: The correct answer is C. Question 57
Statement 1: 1/3. In order to solve this problem, you need to remember that square footage is a measure of area, and area = length × width. By inspecting the diagram, you can see that the width of the apartment is the width of the two bedrooms taken together, or 19.5 feet. You can also see that the length of the apartment is 41 feet and the length of the two bedrooms is 13.5 feet. So the area of the two bedrooms taken together is 19.5 × 13.5, and the area of the entire apartment is 41 × 13.5. So the square footage of the bedrooms is (19.5 × 13.5)/(41 × 13.5) of the area of the apartment. This simplifies down to 19.5/41 = 0.3293, which is much closer to 1/3 = 0.3333 than to ¼ = 0.25, 3/8 = 0.375, 2/5 = 0.4, or ½ = 0.5. Statement 2: 45 feet. In order to solve this problem, you need to recognize that the apartment is a rectangle (you are told to ignore the outdoor patio). The distance from one corner of the apartment to the farthest corner from it is the length of the diagonal of the rectangular apartment. From the diagram, you can determine that length and width of the apartment are 41 feet and 19.5 feet, respectively. The Pythagorean theorem, which is a2 + b 2 = c 2, can then be used to calculate the length of the diagonal of the apartment. So the length of the diagonal is the square root of 19.5 2 + 41 2. Use the IR calculator to compute these messy numbers! 380.25 + 1681 = 2061.25, and the square root of 2061.25 can be calculated using the IR calculator’s square root key (sqrt). Doing so, you get 45.4010. The closest answer is 45 feet. Question 58
Option B - strengthen
Most of the growth in the new suburbs is on the north side of the city, and it is expected that most of the new residents will work at the large companies headquartered in that area, primarily using Austia’s freeways only to travel to central Austia during off-peak hours. This option tells us that the new residents will not use Austia’s freeways during high traffic commute times as much as the existing residents do because they will not be leaving the area to get to work. Option C - weaken
Since Austia has limited public transportation outside of the city core, most commuters who drive from the outlying neighborhoods to the city center have no practical alternative to driving available. This option tells us that the new residents cannot use public transit instead of driving to the city center will have to use Austia’s freeways to drive to the city center to work, a dding to the freeway congestion. Options A, D, E, and F
These options do not directly address the issue of the new population’s freeway utilization, and so do not significantly strengthen or weaken the conclusion. Column 1: The correct answer is B. Column 2: The correct answer is C.
Question 59
Statement 1: Yes. Sort by DWT. There are only 3 TC vessels capable of transporting over 50,000 metric tons: Amore Mio II, Ayrton II, and Agamemnon II. For each, we need to multiply the daily rate (net) by 30. Amore Mio II: 30 × 25,000 = 750,000 Ayrton II: 30 × 22,000 = 660,000 Agamemnon II: 30 × 22,000 = 660,000 In all three cases, the charterer is paying less than $800, 000 per month to the owner. Statement 2: Yes. Sort the table by Type. There are 8 BC vessels and 11 TC vessels. The median of the BC vessels will be between the 4th and 5th. Count from lowest to highest to find the 4 th and 5th, which are both $15,000. This means the median of the BC vessels is $15,000. Because there are 11 TC vessels, the median will be the 6 th highest or lowest value. The 6 th is only $12,838. Statement 3: Yes. Sort by Class. Though many ships can be categorized as Ice Class IA, we can ignore anything in the BC category, because the daily operating expenses are so low. This leaves 8 ships. From here, we can quickly eyeball our way to the ship with the highest Daily operating expense to Daily rate ratio: Alkiviadis. 7,000/12,838 = 54.5%, which is less than 55%. You can use 50% as a benchmark, noting that all the other ships in the Ice Class IA category have daily operating expenses that are less than half of the daily charter rate.
Question 60
Statement 1: 130. FHR (fetal heart rate) is tracked on the top graph. You are told that baseline FHR is the average FHR over a 10-minute window, excluding certain time periods. One of the excluded periods
is a deceleration, and you are told that between minutes 3 and 5, the ultrasound shows a deceleration. So you must exclude all the FHR data between minutes 3 and 5 (the long dip and recovery). The rest of the heart rate data is clearly centered on 130 bpm, with only slight variability. Statement 2: 70. In the last sentence of the note, you are told that “the duration of a uterine contraction is measured from the onset of a sharp increase in uterine pressure from a baseline (typically 0 mmHg as shown) to the return to the baseline.” The lower graph demonstrates just such a sharp uptick in pressure starting at 120 seconds (the start of minute 2). This elevated pressure lasts until 190 seconds (one tick past minute 3). 190 – 120 = 70. Question 61
Statement 1: B - Low-fat milk (1%) and skim milk . You are looking for the greatest magnitude percent change, so a large percent decrease could be the winner. Let’s investigate each case by comparing visually the size of the column segment in 2005 with the corresponding segment in 1980, looking for large proportional changes either up or down. a) Flavored milks: Very little proportional change in size from 1980 to 200 5. b) Low-fat milk (1%) and skim milk: Appears to almost double in size, which would represent nearly a 100% increase. c) Plain reduced-fat milk (2%): Very slight growth, proportionally. d) Plain whole milk: A big decrease in absolute terms, but percent-wise, only a little more than a 50% decline – certainly not close to 100% decline. So the clear winner in percent terms is low-fat, with its nearly 100% increase. Closer inspection would reveal an approximately 80% increase, whereas plain whole fell only about 60%. Statement 2: A - 1980. This answer should jump out: the absolute amount of plain reduced-fat milk consumed was lowest in 1980, by visual inspection. In that same year you had the highest overall milk consumption. So it must be that the ratio of (a) plain reduced-fat to (b) overall milk consumption , or quantity (a) divided by quantity (b), was lowest in 1980. Question 62
The passage as a whole strongly suggests that foreshock, main shock, and aftershock are part of the same causal chain of events because the foreshocks, main shock, and aftershocks all occur on the same system of faults and are temporally related. Option A - All main shocks are preceded by foreshocks.
Although we do know that at least some main shocks are preceded by foreshocks because the third sentence, which says that a foreshock is only a foreshock if it occurs before a larger quake on the same fault system, we do are not told whether all main shocks are preceded by foreshocks. Option B - All main shocks are preceded by foreshocks.
Although we do know that at least some main shocks are followed by aftershocks because of sentence three, which says that an aftershock is only an aftershock if it occurs after a larger quake on the same fault system, we do are not told whether all main shocks are followed by aftershocks. Option C - Aftershocks are more common that foreshocks.
The passage doesn’t give us any information about the frequency of aftershocks relative to foreshocks. Option D - Foreshocks are generally weaker than aftershocks.
Although the passage says that neither foreshocks nor aftershocks are as strong as main shocks, it does not give us any information about their strength relative to each other. Option E – Main shocks can be triggered by foreshocks. - TRUE Since a foreshock is a smaller earthquake that happens before the bigger main shock on the same fault system, and there is a probability that a small earthquake can trigger a larger earthquake on the same fault system that depends on the smaller earthquakes “location and interaction with the fault system,” it must be possible for some foreshocks to trigger main shocks.
Option F – An aftershock can be a bigger earthquake than the main shock that preceded it. -
FALSE The fourth sentence says that an aftershock occurs only after a bigger quake on the same fault system, which is the exact opposite of what this statement says so i t cannot be inferred from the passage. Column 1: The correct answer is E. Column 2: The correct answer is F. Question 63
Statement 1: D - $9,765,600,000 . This graph does not give the exact per person spending of Minnesota and Iowa. Instead, it gives ranges of possible values. To find the greatest possible difference between the total spending of Minnesota and Iowa, you need to assume that Minnesota spent the maximum amount per person ($2,712) in its category (darkest). Meanwhile, assume also that Iowa spent the minimum amount per person ($1,536) in its category (one step up from the lightest/bottom). Multiply those values by the populations of the respective states to calculate total public and elementary school expenditures for each state: Minnesota = 5.3 million × $2,712 = $14,373,600,000 Iowa = 3 million × $1,536 = $4,608,000,000 Now subtract to find the difference: $14,373,600,000 − $4,608,000,000 = $9,765,600,000 Statement 2: B - 3.8 million. The easiest way to start answering this question is to first find a range for the possible values of the total state expenditures of Missouri. Multiply Missouri’s population by both the maximum possible expenditure per person and by the minimum possible expenditure per person to find the range: Min = 5.9 million × $1,161 = $6,849,900,000 Max = 5.9 million × $1,535 = $9,056,500,000 This information gives us the range of Kansas’ total expenditures as well, because we know that the two states spent the same amount. Now we need to use these values to find the range of possible values for the population of Kansas. The formula for population that we can use is this: Population = Total Expenditures ÷ Expenditures Per Person To verify this conceptually, imagine a state that spent $1,000,000,000 on Total Expenditures, and imagine that you separately know that the state spent $1,000 per person. Then there must be 1,000,000 people in that state. Now, to find the minimum possible population of Kansas, we need to divide the smallest possible Total Expenditures by the largest possible Expenditures Per Person. Minimum Population = $6,849,900,000 ÷ $1,970 ≈ 3.5 million Similarly, to find the maximum possible population, we need to divide the largest possible Total Expenditures by the smallest possible Expenditures Per Person. Maximum Population = $9,056,500,000 ÷ $1,843 ≈ 4.9 million The only answer in that range is 3.8 million. Question 64
Statement 1: Yes. Sort by the Washer and Dryer column. Note that all three of the units with a washer and dryer also have a pool. Statement 2: No. Sort by the Square Feet column. Since there are 10 (an even number of) apartments on the list, the median square footage is the average of the square footage of the 5 th and 6th (middle
two) apartments, or 660 and 678. Since the average (arithmetic mean) apartment size is given as 692 square feet, and we know that the median must be less than 678, this statement cannot be true. Statement 3: Yes. In order to solve this problem, you need to know that a positive correlation between rent per square foot and the state of having either covered or garage parking means that apartments with higher rents per square foot are more likely to have either covered or garage parking. You can see that this is the case by sorting by rent per square foot. Note that the four apartments with the highest rents per square foot also have either covered or garage parking and that none of the apartments with lower rents per square foot have either covered or garage parking. Question 65
To solve this statistics question, we must first recall the A verage Formula, which states that the Average = (Sum) / (Number of Terms). Rearranging, we see that Sum = (Average) × (Number of Terms). Therefore, when the prompt tells us that the average of all 4 numbers is 18, it’s really telling us that the sum of all four numbers is equal to (18) × (4) = 72. As we know that one of the large marble was 48, we know that the sum of the 3 remaining numbers is given by 72 – 48 = 24. Therefore, the sum of the other two large marbles and the small marble must be 24. This tells us to look among the answer choices for pairs that add up to 24. Within the table there are two such pairs: (11, 13) and (5,19). Because the small marble is limited to the range 1-10, inclusive, and both numbers in the set (11, 13) are greater than 10, neither of these could be the value of the small marble. Therefore, we are left with just one set: (5,19). As 5 is less than 10, we choose this value for the small marble and leave 19 a s the sum of the other two large values. Question 66
To solve this logic problem, we place the parents and children in their respective groups working from the most to the least restrictive clues. The most restrictive clue states that X is i n group 2: __________ _X________ __________ From here, we know that A and B must be split among groups 1 and 3. This means that one place in groups 1 and 3 are occupied by A and B, leaving only one “child” space in each of these groups. As no group can have more than two children and E must be in the same group as F, this means that both E and F must be in group 2. __________ _X_E_F____ __________ Another clue tells us that if F is in group 2 then A must be in group 1. __A_______ _X_E_F____ __________ Consequently, since A and B must be spli t among groups 1 and 3, we know that B must be in group 3. __A_______ _X_E_F____ __B_______ Finally, the only clue we have no used is that Y must be in a different group that A and D, meaning Y cannot be in group 1. Scanning the answer choices, we see that there are three adults that could potentially be in group 1 – X, Y, and Z. From the above we have ruled out X and Y, and so only Z could still be an adult in group 1. The possibilities for children in group 3 are A, D, and E. From the above, we know that neither A nor E can be in group 3, so the only possibility is D. Column 1: The correct answer is the 6 th choice. Column 2: The correct answer is the 2 nd choice. Question 67
Statement 1: Yes. In tab 2, we learned that the doctoral candidate proposed a study on humans. An age range for the study subjects was not specified, so both committees 1 and 2 would have had to accept the proposal in order for this study to be approved. Statement 2: Yes. In tab 2, the botany professor’s proposal targets “all living organisms native to a nature preserve.” Committees 1 through 5 all deal with living organisms, but the nature preserve “usually does not permit access to people,” so no people are living in the preserve. The relevant committees, then, must be committees 3, 4, and 5. Committees 3 and 4 must have a minimum of 4 members and can pass a proposal with at most a single “no” vote, so the botany professor must have received a minimum of 6 “yes” votes from these two committees combined. Finally, committee 5 must have at least 5 people and a simple majority is needed to approve a proposal, so the professor must have received at least 3 “yes” votes from this committee, for a total of at least 9 “yes” votes. Statement 3: No. The first paragraph of the first tab told us that the task of the committees is to “ensure that the proposed research complies with all laws and meets the university’s standards for ethical experimentation.” If a committee rejected a particular proposal, then that proposal must have failed to meet at least one of these two criteria, but it is entirely possible that a proposal might pass an ethical test and yet still fail to comply with some law.
Question 68
To compute the minimum number of board members, we must determine how many people we need to staff each of the 6 committees at the minimum level. The second paragraph tells us that each committee must have a minimum of 4 members; there are 6 committees, so it is tempting to say that we must need a minimum of 24 board members. Each board member, however, is allowed to sit on up to 3 committees. In addition, committees 5 and 6 must have an odd number of members – and 4 is not an odd number. Let’s begin by assuming that, if we want to minimize the number of total board members, then we must allow each member to sit on the maximum 3 committees. Further, because there are 6 committees total, let’s group our board members into two large groups. The first group will sit on committees 1, 2, and 3. How many members do we need to fill those three committees? These three committees can have the minimum 4 members each, so we need 4 board members total in order to fully staff the first three committees. What about the second set of three committees? Committee 4 can have a minimum of 4 members, but committees 5 and 6 must have an odd number of members (and a minimum of 4). Committees 5 and 6, then, must each have a minimum of 5 members. The same 5 people can sit on both committees, and 4 of those 5 can also sit on the 4 th committee, so we need a minimum of 5 board members to fully staff committees 4, 5, and 6. Together, we need a minimum of 4 + 5 = 9 board members to fully staff all six committees. The correct answer is C.
Question 69
Statement 1: No. It is true that the undergraduate’s proposal had to be reviewed by both committees 1 and 2 (because no age range was specified for the human subjects of the study). It is also true that each individual committee must vote unanimously in favor of a proposal in order for it to be accepted. But it is possible that only one of the two committees rejected the undergraduate student’s proposal. It is possible, therefore, that only one person on one committee rejected this proposal.
Statement 2: No. Under such circumstances, her proposal might be accepted, but it does not absolutely have to be true that it will be accepted. Tab 2 indicates that the sociology professor’s proposal was rejected when a certain fact became apparent, but does not indicate whether there may have been additional factors that would also have resulted in rejection. It might be the case that, even if she fixed the one noted problem, another problem may exist or arise. The way in which she fixes the existing problem might itself be something that is cause for rejection. Statement 3: Yes. A study involving humans would require the approval of Committees 1 and 2. The pets would require the involvement of committees 3 and 4 (mammals and all other animals). All four committees must have a minimum of 4 members, and the researcher would have to gain unanimous approval from Committees 1 and 2 (4 votes each) and at least 3 “yes” votes each from Committees 3 and 4, for a minimum of 14 votes.
Question 70
We need to maximize the area of a cylinder by selecting a radius and a height given certain limiting parameters. First, examine the formula for the volume of a cylinder: V = πr 2h . The value for π is a constant – that is, it will never change – so we can ignore it for our purposes. We can control two variables, therefore: the radius and the height. In the formula, the radius will be squared, so to maximize the area, we should make the radius as large as possible. Once we do that, the height will be determined, because the radius and the height must add up, at most, to 16. We’re told that “the height of the cylinder cannot be more than twice the radius, nor can the reverse be true.” The second half of that sentence is the most important: the radius cannot be more than twice the height. If the two add up to 16, then what would the radius and height have to be in order for the radius to be exactly twice the height? Divide 16 by 3. The height would be 5 1 /3 and the radius would be twice that, 10 2 /3. The problem specified, though, that the values must be integers. Should we set the radius at 10 or 11? 11 is the bigger number… but then the height would have to be 5 and the radius would be more than twice as big as the height. 10, then, is the largest possible value for the radius, and if the radius is 10, then the maximum value for the height is 6. Column 1: The correct answer is D. Column 2: The correct answer is B. Question 71
Statement 1: No. First we should verify that Fairview had the lowest profits during the second quarter (sort by Q2 Profits). We see that they do, in fact, have the lowest profits, so now we must test the second part of the statement: “Fairview has the lowest profit margin during the same quarter.” To find the profit margin for Fairview, we will use the formula Profit Margin = Profits / Revenues. During the second quarter (Q2), Fairview had profits of $1,130,0 00 and revenues of $5,014,000. Thus, Fairview’s profit margin that quarter was $1,130,000/$5,014,000 ≈ 0.23, or 23%. Now, to determine whether Fairview has the lowest profit margin we could try to calculate this value for every city, but that would take far too much time. It would be better to try to estimate values and then test those that might seem likely. We know that Fairview has the lowest profits, so every other city will have a larger numerator (potentially making Profit Margin larger), but if we can find a city whose profits are similar but whose revenue is substantially greater, that would be worth testing.
Let’s look at the list of cities around Fairview in this sort, e.g. Mt Pleasant, Greenwood and Lexington. All of these cities have similar profits to one another. Of these cities, Greenwood has by far the largest revenue, a value almost double that of Fairview. This makes Greenwood a likely candidate to have a smaller Profit Margin; let’s test. Greenwood’s profit margin = $1,627,000/$9,982,000 ≈ 0.16, or 16%. We have found a city that has a smaller profit margin than Fairview does, so Fairview’s cannot be the lowest. Statement 2: Yes. To determine the cities with the fewest stores, we must sort by column Stores. From this list we see that the four stores with the fewest are York (4), Milton (5), Oak Grove (5), and Springfield (6). Now we need to scan across the rows to see what happened to profits across the 4 quarters for each city, and determine if only one city saw profits fall steadily across all 4 quarters. We must be careful not to only look at the difference between Q1 and Q4, because an overall decrease is not the same as a steady decrease (meaning a decrease from Q1 to Q2, then another decrease from Q2 to Q3, and then a final decrease from Q3 to Q4). York saw a decrease from Q1 to Q2 (from $3,908 to $1,945, in thousands), and from Q2 to Q3 (from $1,945 to $1,631), but an increase from Q3 to Q4 (from $1,631 to $2,169). Although York did see an overall decrease in sales from Q1 to Q4 (from $3,908 to $2,169), it did not see decreases between all quarters, so it did not have a steady decrease across the entire period. Milton saw an increase from Q1 to Q2 (from $3,879 to $11,474), so cannot have a decrease in all quarters (incidentally, Milton saw an increase over the entire period, so Milton could not have had a steady decrease across all periods). Oak Grove saw a decrease from Q1 to Q2 (from $2,848 to $2,094), from Q2 to Q3 (from $2,094 to $1,017), and from Q3 to Q4 (from $1,017 to $731), so Oak Grove saw a decrease across all quarters. Springfield saw a decrease from Q1 to Q2 (from $5,809 to $3,851) but an increase from Q2 to Q3 (from $3,851 to $8,834). Although Springfield did see an overall decrease in sales from Q1 to Q4 (from $5,809 to $970), it did not see decreases between all quarters, so it did not have a steady decrease across the entire period. Therefore, only Oak Grove saw a steady decrease in profits across all four quarters. Statement 3: Yes. We want to try to find the city with the largest positive change in revenues from the first to the fourth quarter = Q4 Revenue – Q1 Revenue. Since we are trying to make this number as large as possible, it is easier to sort by the first number (as this moves in the same direction as our goal number). Therefore we sort by column Q4 Revenue and look for the largest value. Portsmouth has Q4 Revenues of $83,796 (in thousands) and Q1 revenues of $20,681, for a difference of $63,115. This is a very large change, but is it the largest? We could calculate the value for every city, but notice that no other city has Q4 Revenues above $50,000. This means that they could NOT have increased revenues by more than their Q4 values, so $63,115 must be the largest dollar increase in revenues between the two quarters. Now we just need to determine if Portsmouth had an increase in profits that exceeded $25,000,000. Remember, the numbers shown are in thousands of dollars, so add three zeroes. Q4 Profits – Q1 Profits = $35,452,000 - $6,004,000 = $29,448,000 > $25,000,000. Question 72
Statement 1: D - 220% . First, be sure to read the dotted line (2000) on the left side (male). Next, be careful about the vertical divisions—the minor gridlines represent 4 (hundred thousand), because they separate a division of 20 into 5 smaller parts. Reading carefully, you can see that the population of 35-to-39-year-old men was about 113 (hundred thousand), whereas the population of 60-to-64-year-old men was about 52 (hundred thousand). Now, as a percent of 52, 113 is 113/52 × 100% ≈ 2.17 = 217%. The closest answer is 220%.
