Word Problem Q Bank

During 2005, a company produced an average of 2,000 products per month. How many products will the company need to produce from 2006 through 2008 in order to increase its monthly average for the period from 2005 through 2008 by 200% over its 2005 average? - 148k - 172k - 200k - 264k - 288k

A 200% increase over 2,000 products per month would be 6,000 products per month. (Recall that 100% = 2,000, 200% = 4,000, and "200% over" means 4,000 + 2,000 = 6,000.) In order to average 6,000 products per month over the 4 year period from 2005 through 2008, the company would need to produce 6,000 products per month × 12 months × 4 years = 288,000 total products during that period. We are told that during 2005 the company averaged 2,000 products per month. Thus, it produced 2,000 × 12 = 24,000 products during 2005. This means that from 2006 to 2008, the company will need to produce an additional 264,000 products (288,000 - 24,000). The correct answer is D.

8, 5, x, 6 The median of the list of positive integers above is 5.5. Which of the following could be the average (arithmetic mean) of the list? - 3 - 5.5 - 6.25 - 7 - 7.5

In a list that contains an even number of values, the median is the average of the two middle values when the values are arranged in increasing order. Therefore, we can arrange 5, 6, and 8 and, using the given value for the median, determine possible values for x.If x is the smallest number in the list, { x, 5, 6, 8}, then the median would be . This conforms to the median value given in the question, so it is one possible arrangement of the values.If we consider the cases where x is not the smallest number, the ordered list now becomes one of the following (note that x cannot fall between 5 and 6 because the set must contain four integers):5, 6, x, 85, 6, 8, xIn both cases, the two middle values are 6 and some integer greater than or equal to 6 (The median will equal 6 if x = 6). The medians are the averages of these two values; in each case, the median will necessarily be greater than 6 and therefore cannot be 5.5. Therefore x must be the smallest number in the list; the only possible arrangement is { x, 5, 6, 8}Notice that the question asks for a value that could be the average (or arithmetic mean) of the list. The use of the word "could" in a question indicates that there are multiple potential values and we should therefore look for a pattern. Because x is a positive integer less than or equal to 5, the possible values of x are 1, 2, 3, 4, and 5. The average of the list will be lowest when x is at its minimum and highest when x is at its maximum. Thus we can calculate a range of possible averages and eliminate any answer choices that fall outside of that range.When x = 1, the average of the list is . The smallest possible value for the mean of the set is 5, eliminating answer choice A.When x = 5, the average of the list is . The largest possible value for the mean of the set is 6, eliminating answer choices C, D, and E.

What is the average of x and |y| ? (1) x + y = 20 (2) |x + y| = 20 12TEN

N The question asks for the average of x and |y|. Taking the absolute value of a number has no effect if that number is positive; on the other hand, taking the absolute value of a negative number changes the sign to positive. The most straightforward way to approach this question is to test positive and negative values for y. (1) INSUFFICIENT: We know that the sum of x and y is 20. Here are two possible scenarios, yielding different answers to the question: (2) INSUFFICIENT: We know that |x + y| = 20. The same scenarios listed for statement (1) still apply here. There is more than one possible value for the average of x and |y|, (1) AND (2) INSUFFICIENT: We still have the same scenarios listed above. Since there is more than one possible value for the average of x and |y|, both statements taken together are NOT sufficient. The correct answer is E.

If the standard deviation of Set Y is 4, what are the greatest and least values that are within one standard deviation of the mean? (1) The median of Set Y is 5. (2) The mean of Set Y is 6. 12TEN

Standard deviation is a measure of how far data points in a distribution fall from the mean. The problem tells us the standard deviation. In order to figure out the values of the greatest and least values that fall within one deviation, then, the only information we need is the mean. We can rephrase the question, "What is the mean of Set Y?"(1) INSUFFICIENT: This tells us nothing about the mean.(2) SUFFICIENT: If we know that the mean is 6, the greatest value that is one standard deviation from the mean is 10 and the least value that is one standard deviation from the mean is 2. The correct answer is B.

Trains A and B travel at the same constant rate in opposite directions along the same route between Town G and Town H. If Train A travels for exactly two hours before passing Train B, how long does it take Train B to travel the entire distance between Town G and Town H? (1) Train B started traveling from Town G toward Town H one hour after Train A started traveling from Town H toward Town G. (2) Train B travels at a rate of 150 miles per hour. 12TEN

The problem never specifies how far apart the two towns are so, at first glance, it appears to be impossible to tell how long it takes Train B to travel between the two towns. But the GMAT likes to set traps. Investigate this a little further. The question stem specifies that the two trains travel at the same constant rate. That's pretty unusual for a GMAT problem...note that down. The problem also doesn't actually say that one starts at one town and the other starts at the other town. It does say that Train A goes for exactly 2 hours before it passes Train B coming in the other direction. (But no info yet on how fast they're going or when or where each one started.) Finally, it asks how long it would take Train B to travel between the two towns. Since Train A travels at the same rate, it would be enough to know how long it takes either train to travel the full distance between the towns. Glance at the statements. The first one is much more complicated, so start with statement (2). (2) INSUFFICIENT: This gives you Train B's rate, so what you really need to find now is the distance between the two towns; if you know that, then you can calculate how long it would take Train B to travel that distance. Can you determine the distance between Towns G and H? If Train B travels at 150 miles per hour, so does Train A. The question stem states that Train A runs for 2 hours before it passes Train B, so Train A goes 300 miles before it passes Train B. But when does Train B start? If it starts at the same time as Train A, then it also travels 300 miles before they pass each other...but Train B might have started 3 hours earlier or an hour later. It's not possible to tell. This statement also doesn't specify that they each start in one of the towns; they could start somewhere in the middle. (1) SUFFICIENT: This statement is so tricky. First, the statement establishes that Train B starts in Town G and Train A starts in Town H. It also indicates when Train B started relative to Train A: Train B started one hour later. But there's still no information about how far apart the two towns are or how fast the trains are traveling. So how could this possibly be enough? When the GMAT provides information about relationships in a problem pertaining to rates, be cautious. The information may not seem sufficient at first glance, but could contain an unexpected path to solving. Consider drawing out what's happening, so you can visualize it. First, Train A starts at Town H and travels alone for an hour towards Town G. Let's say Train A starts at noon. Then, an hour later at 1pm, Train A keeps going and Train B starts out from Town G towards Town H. They each travel for one hour before they meet (since Train A travels a total of 2 hours before they meet). Distance-wise, Train A covers the same distance in the second hour as in the first hour, since it's traveling at a constant rate. And here's the key: Train B is traveling at that same rate, too, so it also covers the same distance in one hour that Train A covers in one hour. When they meet, collectively, they've covered the exact distance between Towns G and H. Train A has traveled for 2 hours and Train B has traveled for 1 hour. And since the trains are traveling at the exact same speed, Train B will continue on its path to Town H, taking the same time that A needed to reach their meeting point: 2 hours. You can essentially think of the three 1-hour increments as representing just one train, traveling at the same speed all the way from Town G to Town H. A train traveling at that exact speed will always take 3 hours to make the trip. In other words, it would take Train B (which is traveling at that speed) exactly 3 hours to make the trip from one town to the other.

What is the sum of all of the integers in the chart above? look image - 0 - 300 - 500 - 1500 - 6500

1500, The best approach to this problem is to attempt to find a pattern among the numbers. If we scan the table, we see that there are five sets of consecutive integers represented in the five columns: 98, 99, 100, 101, 102 -196, -198, -200, -202, -204 290, 295, 300, 305, 310 -396, -398, -400, -402, -404 498, 499, 500, 501, 502 To find the sum of a set of consecutive integers we can use the formula: Sum of consecutive set = (number of terms in the set) × (mean of the set). Each group contains 5 consecutive integers and the mean of a consecutive set is always equal to the median (or the middle term if there is an odd number of terms). In this way we can find the sum of the five sets: 5(100) = 500 5(-200) = -1,000 5(300) = 1,500 5(-400) = -2,000 5(500) = 2,500 Therefore the sum of all the integers is:500 + (-1,000) + 1,500 + (-2,000) + 2,500 = 1,500.The correct answer is D.

A certain bank has ten branches. What is the total amount of assets under management at the bank? (1) There is an average (arithmetic mean) of 400 customers per branch. When each branch's average (arithmetic mean) assets under management per customer is computed, these values are added together and this sum is divided by 10. The result is $400,000 per customer. (2) When the total assets per branch are added up, each branch is found to manage an average (arithmetic mean) of 160 million dollars in assets. 12TEN

2 Since Statement 2 is less complex than Statement 1, begin with Statement 2 and a BD/ACE grid.(1) INSUFFICIENT: When the average assets under management (AUM) per customer of each of the 10 branches are added up and the result is divided by 10, the value that is obtained is the simple average of the 10 branches' average AUM per customer. Multiplying this number by the total number of customers will not give us the total amount of assets under management. The reason is that what is needed here is a weighted average of the average AUM per customer for the 10 branches. Each branch's average AUM per customer needs to be weighted according to the number of customers at that branch when computing the overall average AUM per customer for the whole bank. Let's look at a simple example to illustrate: LOOK IMAGE If we take a simple average of the average number of apples per person from the two rooms, we will come up with (2 + 3) / 2 = 2.5 apples/person. This value has no relationship to the actual total average of the two rooms, which in this case is 2.6 apples. If we took the simple average (2.5) and multiplied it by the number of people in the room (10) we would NOT come up with the number of apples in the two rooms. The only way to calculate the actual total average (short of knowing the total number of apples and people) is to weight the two averages by the number of people in each room, in the following manner: (4*2 + 6*3) / 10. (2) SUFFICIENT: The average of $160 million in assets under management per branch spoken about here was NOT calculated as a simple average of the 10 branches' average AUM per customer as in statement 1. This average was found by adding up the assets in each bank and dividing by 10, the number of branches ("the total assets per branch were added up..."). To regenerate that original total, we simply need to multiply the $160 million by the number of branches, 10. (This is according to the simple average formula: average = sum / number of terms)

T is a set of y integers, where 0 < y < 7. If the average of Set T is the positive integer x, which of the following could NOT be the median of Set T ? - 0 - X - -X - 1/3y - 2/7y

2/7y, There is no possible way to create Set T with a median of 2/7y. Why? We know that y is either 1, 2, 3, 4, 5, or 6. Thus, 2/7y will yield a value that is some fraction with denominator of 7. The possible values of are as follows: 2/7, 4/7, 6/7, 1 1/7, 1 3/7, 1 5/7 However, the median of a set of integers must always be either an integer or a fraction with a denominator of 2 (e.g. 2.5, or 5/2). So (2/7) y cannot be the median of Set T.

A copy machine, working at a constant rate, makes 35 copies per minute. A second copy machine, working at a constant rate, makes 55 copies per minute. Working together at their respective rates, how many copies do the two machines make in half an hour? - 90 - 2700 - 4500 - 5400 - 324000

2700, To find the combined rate of the two machines, add their individual rates: 35 copies/minute + 55 copies/minute = 90 copies/minute. The question asks how many copies the machines make in half an hour, or 30 minutes. 90 copies/minute × 30 minutes = 2,700 copies.The correct answer is B.

30% of major airline companies equip their planes with wireless internet access. 70% of major airlines offer passengers free on-board snacks. What is the greatest possible percentage of major airline companies that offer both wireless internet and free on-board snacks? - 21% - 30% - 40% - 50% - 70%

30%, 30% of airlines have WiFi and 70% have free snacks. The question asks for the greatest possible percentage of airlines that have both WiFi and free snacks. Glance at the answers. They're really far apart, so estimate and logic this out. Imagine that you have 100 airlines. 30 of them offer WiFi and 70 of them offer snacks, but you don't know the overlap between who offers WiFi and who offers snacks. However, it's true that 30 out of the 100 offer WiFi, so at most, those 30 airlines offer both WiFi and snacks. It would be impossible for 40 airlines to offer both, since exactly and only 30 of them offer WiFi in the first place. Eliminate answers (C), (D), and (E). The two remaining answers are 21% and 30%. Is it possible that all of the companies who offer WiFi also offer snacks? Since the number of airlines that offer WiFi is less than the number of airlines that offer snacks, it is possible that all of the companies that offer WiFi also offer snacks. The problem offers no other constraints that limit this possibility. If all 30 airlines that offer WiFi also offer snacks, then 30 out of 100 airlines is the greatest number of airlines that offer both. So the greatest possible percentage is 30 out of 100 or 30%.

Billy has an unlimited supply of the following coins: pennies (1¢), nickels (5¢), dimes (10¢), quarters (25¢), and half-dollars (50¢). On Monday, Billy bought one candy for less than a dollar and paid for it with exactly four coins (i.e., he received no change). On Tuesday, he bought two of the same candy and again paid with exactly four coins. On Wednesday, he bought three of the candies, on Thursday four of the candies, and on Friday five of the candies; each day he was able to pay with exactly four coins. Which of the following could be the price of one candy in cents? - 8c - 13c - 40c - 53c - 66c

40c We can back solve this question by using the answer choices. Let's first check to make sure that each of the 5 possible prices for one candy can be paid using exactly 4 coins: (A) 8 = 5+1+1+1 (B) 13 = 10+1+1+1 (C) 40 = 10+10+10+10 (D) 53 = 50+1+1+1 (E) 66 = 50+10+5+1 So far we can't make any eliminations. Now let's check two pieces of candy: (A) 16 = 5 + 5 + 5 + 1 (B) 26 = 10 + 10 + 5 + 1 (C) 80 = 25 + 25 + 25 + 5 (D) 106 = 50 + 50 + 5 + 1 (E) 132 = 50 + 50 + 25 + 5 + 1 + 1 We can eliminate answer choice E here. Now three pieces of candy: (A) 24 = 10 + 10 + 1 + 1 + 1 + 1 (B) 39 = 25 + 10 + 1 + 1 + 1 + 1 (C) 120 = 50 + 50 + 10 + 10 (D) 159 = 50 + 50 + 50 + 5 + 1 + 1 + 1 + 1. We can eliminate answer choices A, B and D. Notice that at a price of 40¢, Billy can buy four and five candies with exactly 4 coins as well: 160 = 50 + 50 + 50 + 10 200 = 50 + 50 + 50 + 50 This problem could also have been solved using divisibility and remainders. Notice that all of the coins are multiples of 5 except pennies. In order to be able to pay for a certain number of candies with exactly four coins, the total price of the candies cannot be a value that can be expressed as 5x + 4, where x is a positive integer. In other words, the total price cannot be a number that has a remainder of 4 when divided by 5. Why? The remainder of 4 would alone require 4 pennies. We can look at the answer choices now just focusing on the remainder when each price and its multiples are divided by 5: The only price for which none of its multiples have a remainder of 4 when divided by 5 is 40¢. Notice that not having a remainder of 4 does not guarantee that exactly four coins can be used; however, having a remainder of 4 does guarantee that exactly for coins cannot be used!

Aki and Brune type at constant rates of 80 words per minute and 60 words per minute, respectively. Brune begins typing before Aki and has typed 600 words when Aki begins typing at 1:30 pm. If they continue typing at their respective rates, at what time will Aki have typed exactly 200 more words than Brune? - 1:40PM - 1:50PM - 2PM - 2:10PM - 2:20PM

Aki: 80 wpm Brune: 60 wpm Brune starts first and types 600 words before Aki starts at 1:30 pm. The question asks how long it will take before Aki will have typed exactly 200 more words than Brune. It's possible to write equations to solve this kind of problem...but why bother? Just sketch it out, step by step. The answer choices are in 10 min increments, so check each answer, in order, till you find the one that works. Every 10 minutes, Brune types 600 words and Aki types 800 words. Find the time at which Aki is 200 more than Brune. At 1:30 pm: B = 600 words, A = 0 words (A) 1:40 pm: B = 1,200 words, A = 800 words...Brune is ahead. Eliminate. (B) 1:50 pm: B = 1,800 words, A = 1,600 words...Brune is still ahead. Eliminate. (C) 2:00 pm: B = 2,400 words, A = 2,400 words...They're equal. Eliminate. Glance at the pattern. Notice anything? Aki is catching up by 200 words every 10 minutes. At 2 pm, they're even, so 10 minutes later, Aki should be 200 words ahead of Brune. If you're not sure, check the math. (D) 2:10 pm: B = 3,000 words, A = 3,200 words...Yep, Aki is 200 ahead of Brune. The correct answer is (D). Tip: Whenever you're sketching out a rate problem in the way shown above, if the people (or trains or whatever) are traveling at constant rates, then there will always be a pattern. It's very rare for the GMAT to have a rate problem in which the rate is not constant. So use that knowledge to your advantage when you're solving: Look for a pattern. In the above case, you can find the pattern after calculating answers (A) and (B) and then solve more quickly from there: (A) 1:40 pm: B = 1,200 words, A = 800 words...Brune is 400 ahead. Eliminate. (B) 1:50 pm: B = 1,800 words, A = 1,600 words...Brune is 200 ahead. Eliminate. (C) 2:00 pm: Aki is catching up by 200 words every 10 minutes, so they'll be even here. (D) 2:10 pm: Aki will be 200 ahead of Brune here.

What percentage of the current fourth graders at Liberation Elementary School dressed in costume for Halloween for the past two years in a row (both this year and last year)? (A) 60% of the current fourth graders at Liberation Elementary School dressed in costume for Halloween this year. (B) Of the current fourth graders at Liberation Elementary School who did not dress in costume for Halloween this year, 80% did not dress in costume last year. 12TEN

N, We can divide the current fourth graders into 4 categories: (1) The percentage that dressed in costume this year ONLY.(2) The percentage that dressed in costume last year ONLY.(3) The percentage that did NOT dress in costume either this year or last year.(4) The percentage that dressed in costume BOTH years.We need to determine the last category (category 4) in order to answer the question. (1) INSUFFICIENT: Let's assume there are 100 current fourth graders (this simply helps to make this percentage question more concrete). 60 of them dressed in costume this year, while 40 did not. However, we don't know how many of these 60 dressed in costume last year, so we can't divide this 60 up into categories 1 and 2. (2) INSUFFICIENT: This provides little relevant information on its own because we don't know how many of the students didn't dress up in costumes this year and the statement references that value.(1) AND (2) INSUFFICIENT: From statement 1 we know that 60 dressed up in costumes this year, but 40 did not. Statement 2 tells us that 80% of these 40, or 32, didn't dress up in costumes this year either. This provides us with a value for category 3, from which we can derive a value for category 2 (8). However, we still don't know how many of the 60 costume bearers from this year wore costumes last year.Since this is an overlapping set problem, we could also have used a double-set matrix to organize our information and solve. Even with both statements together, we can not find the value for the Costume Last Year / Costume This Year cell.

The table below represents three sets of numbers with their respective medians, means and standard deviations. The third set, Set [A+B], denotes the set that is formed by combining Set A and Set B. LOOK IMAGE If X - Y > 0 and L - M = 0, then which of the following must be true? I. Z > N II. R > M III. Q > R - 1 only - 2 only - 3 only - 1 and 2 only - none

None, If X - Y > 0, then X > Y and the median of A is greater than the mean of set A. If L - M = 0, then L = M and the median of set B is equal to the mean of set B. I. NOT NECESSARILY: According to the table, Z > N means that the standard deviation of set A is greater than that of set B. Standard deviation is a measure of how close the terms of a given set are to the mean of the set. If a set has a high standard deviation, its terms are relatively far from the mean. If a set has a low standard deviation, its terms are relatively close to the mean. Recall that a median separates the set into two as far as the number of terms. There is an equal number of terms both above and below the median. If the median of a set is greater than the mean, however, the terms below the median must collectively be farther from the median than the terms above the median. For example, in the set {1, 89, 90}, the median is 89 and the mean is 60. The median is much greater than the mean because 1 is much farther from 89 than 90 is. Knowing that the median of set A is greater than the mean of set A just tells us that the terms below set A's median are further from the median than the terms above set A's median. This does not necessarily imply that the terms, overall, are further away from the mean than in set B, where the terms below the median are the same distance from the median as the terms above it. In fact, a set in which the mean and median are equal can have a very high standard deviation if the terms are both far below the mean and far above it. II. NOT NECESSARILY: According to the table, R > M implies that the mean of set [A + B] is greater than the mean of set B. This is not necessarily true. When two sets are combined to form a composite set, the mean of the composite set must either be between the means of the individual sets or be equal to the mean of both of the individual sets. To prove this, consider the simple example of one member sets: A = [3], B = [5], A + B = [3, 5]. In this case the mean of A + B is greater than the mean of A and less than the mean of B. We could easily have reversed this result by reversing the members of sets A and B. III. NOT NECESSARILY: According to the table, Q > R implies that the median of the set [A + B] is greater than the mean of set [A + B]. We can extend the rule given in statement II to medians as well: when two sets are combined to form a composite set, the median of the composite set must either be between the medians of the individual sets or be equal to the median of one or both of the individual sets. While the median of set A is greater than the mean of set A and the median of set B is equal to the mean of set B, set [A + B] might have a median that is greater or less than the mean of set [A + B]. See the two tables for illustration:

Sequence A is defined by the equation An = 3n + 7, where n is an integer greater than or equal to 1. If set B is comprised of the first x terms of sequence A, what is the median of set B ? (1) The sum of the terms in set B is 275. (2) The range of the terms in set B is 30. 12TEN

Sequence A defines the infinite set: 10, 13, 16 ... 3n + 7. Set B is a finite set that contains the first x members of sequence A. Set B is based on an evenly spaced sequence, so its members are also evenly spaced. All evenly spaced sets share the following property: the mean of an evenly spaced set is equal to the median. The most common application of this is in consecutive sets, a type of evenly spaced set. Since the median and mean are the same, we can rephrase this question as: "What is either the median or the mean of set B?" (1) SUFFICIENT: Only one set of numbers with the pattern 10, 13, 16 ... will add to 275, which means only one value for n (the number of terms) will produce a sum of 275. For example,If n = 2, then 10 + 13 = 23If n = 3, then 10 + 13 + 16 = 39We can continue these calculations until we reach a sum of 275, at which point we know the value of n. If we know the value for n, then we can write out all of the terms, allowing us to find the median (or the mean). In this case, when n = 11, 10 + 13 + 16 + 19 + 22 + 25 + 28 + 31 + 34 + 37 + 40 = 275. The median is 25. (Though we don't have to do these calculations to see that this statement is sufficient.) Alternatively, the sum of a set = (the number of terms in the set) × (the mean of the set). The number of the terms of set B is n. The first term is 10 and the last (or nth) term in the set will have a value of 3n + 7, so the mean of the set = (10 + 3n + 7)/2.Therefore, we can set up the following equation: 275 = n(10 + 3n +7)/2 Simplifying, we get the quadratic: 3n2 + 17n - 550 = 0. This quadratic factors to (3n + 50)(n - 11) = 0, which only has one positive integer root, 11.Note that we can see that this is sufficient without actually solving this quadratic, however. The -550 implies that if there are two solutions (not all quadratics have two solutions) there must be a positive and a negative solution. Only the positive solution makes sense for the number of terms in a set, so we know we will have only one positive solution.Once we have the number of terms in the set, we can use this to calculate the mean (though, again, because this is data sufficiency, we can stop our calculations prior to this point):= (10 + 3n + 7)/2 = (10 + 3(11) + 7)/2 = 50/2 = 25.(2) SUFFICIENT: The first term of set B is 10. If the range is 30, the last term must be10 + 30 = 40. The mean of the set then must be (10 + 40)/2 = 25. This is sufficient.

A new electric car company holds a limited-time sales event. On the first day, 3 cars are sold. On each subsequent day of the event, 3 more cars are sold than on the previous day. For how many days does the event last? (1) If the event had lasted 2 more days, there would have been 84 more cars sold on the last day of the event than on the first. (2) On exactly 9 days during the event, the number of cars for the day is a multiple of 9. 12TEN

The daily sales figures for this event are consecutive multiples of 3. There are many ways the statements could give you enough information, so it's best just to move on to the statements without rephrasing.(1): SUFFICIENT. To figure out whether this statement helps, list out the sales for the first few days: 3, 6, 9, 12, 15... As you can see, there is a fixed difference between the first day and any potential last day, i.e. if 15 is the last day, the difference is 12, if 12 is the last day, the difference is 9. The statement tells you that the difference would be 84 if the event lasted two more days so the actual difference is 78 (6 less since it is 2 days fewer). If you continued this sequence out far enough, you could find the day that would have a difference of 78 from 3, i.e. that would have 81 as the number of cars sold. Algebraically, you could also represent the number of cars sold on the last day as 3n, since from the pattern you can see that on the third day 3(3) were sold, on the fourth days 3(4) were sold, etc. Two days after the last day, 3(n + 2) would be sold so the difference between that day and the first day could be set to equal 84, i.e. 3(n + 2) - 3 = 84, so n = 27. (2): INSUFFICIENT. Looking at the first few positive multiples of 3, you can see that every third number in the set is a multiple of 9:3, 6, 9, 12, 15, 18 . . .Therefore, the company will sell a multiple of 9 cars every third day of the event. If this happens exactly 9 times, you know the event must last at least 9 × 3 = 27 days, but fewer than 10 × 3 = 30 days. The event could last 27, 28, or 29 days.