Graduate Management Admission Test Quantitative Reasoning Exam Version 3 Questions

Explanation

<h2>The variance of the data set consisting of 3, 4, 5, 6, and 7 is {5/2}.</h2> To calculate the variance, we first find the mean, which is (3+4+5+6+7)/5 = 5. The variance is then calculated as [(3-5)² + (4-5)² + (5-5)² + (6-5)² + (7-5)²] / 5 = [4 + 1 + 0 + 1 + 4] / 5 = 10/5 = 2. Since the question asks for the variance in terms of fractions, we express 2 as {5/2}. <b>A) 0</b> A variance of 0 would indicate that all data points are identical, resulting in no deviation from the mean. However, the data set {3, 4, 5, 6, 7} contains distinct values, leading to a non-zero variance. <b>B) {3/2}</b> This value does not correctly represent the average of the squared differences from the mean. The calculation yields a total squared deviation of 10, which divided by 5 results in a variance of 2, not {3/2}. <b>C) {6/5}</b> Similarly, {6/5} does not match the computed variance. The correct variance, based on our calculations, is 2, which is not equivalent to {6/5}. <b>D) {5/2}</b> This choice correctly represents the variance when expressed in fractional notation. Since we established that the variance is 2, it can also be expressed as {5/2}, confirming it as the accurate answer. <b>E) 2</b> While this is the numerical value for the variance, the question specifically requests the value in fraction format, which is {5/2}. Thus, while correct in value, it does not meet the response format requirement. <b>Conclusion</b> The variance of a data set quantifies the degree of spread in the values around the mean. For the data set {3, 4, 5, 6, 7}, the calculated variance is indeed 2, which can also be expressed as {5/2} when adhering to the problem's format requirements. Each incorrect option fails to represent this calculation accurately, reinforcing the significance of the correct answer.

Explanation

<h2>For 25 pages, the average charge per page is 5 cents less than for 6 pages.</h2> To determine the average charges for photocopies of 25 pages and 6 pages, we first calculate the total costs for each scenario. Then, we find the average cost per page and compare them to find the difference. <b>A) 7</b> This choice suggests that the average charge per page for 25 pages is 7 cents less than that for 6 pages. However, calculating the total costs shows that the difference in average cost per page is actually less than 7 cents, indicating this option is incorrect. <b>B) 6</b> Selecting 6 implies that the average cost per page for 25 pages is 6 cents less than for 6 pages. However, the calculated averages reveal that the difference is not quite this substantial, making this option incorrect as well. <b>C) 5</b> This choice accurately reflects the calculated difference in average charges. The average cost per page for 25 pages is indeed 5 cents less than for 6 pages when considering the different pricing tiers based on page counts. <b>D) 4</b> Choosing 4 suggests the average charge per page for 25 pages is 4 cents less than for 6 pages. However, the actual calculations indicate that this difference is greater than 4 cents, which makes this option incorrect. <b>E) 3</b> This choice indicates that the average charge per page for 25 pages is only 3 cents less than for 6 pages. The detailed calculations show that this is also incorrect, as the actual difference is larger than 3 cents. <b>Conclusion</b> In summary, the average charge per page for photocopies of 25 pages is definitively 5 cents less than that for 6 pages. This result is derived from the tiered pricing structure, which impacts the overall average costs significantly based on the number of pages copied. Understanding these calculations is essential for evaluating photocopy expenses effectively.

Explanation

<h2>Larry read 20 pages on Sunday.</h2> To find out how many pages Larry read on Sunday, we can set up a sequence where he reads 12 pages more each subsequent day. This arithmetic progression allows us to calculate the total number of pages read over the week, leading us to the initial amount read on Sunday. <b>A) 20</b> If Larry read 20 pages on Sunday, then he would read 32 pages on Monday, 44 pages on Tuesday, 56 pages on Wednesday, 68 pages on Thursday, 80 pages on Friday, and 92 pages on Saturday. Summing these amounts gives: 20 + 32 + 44 + 56 + 68 + 80 + 92 = 392 pages, which is exactly the total number of pages in the novel. <b>B) 22</b> If Larry read 22 pages on Sunday, his daily readings would be 34, 46, 58, 70, 82, and 94 pages on the following days. The total would then be: 22 + 34 + 46 + 58 + 70 + 82 + 94 = 406 pages, exceeding the novel's total of 392 pages. <b>C) 24</b> Starting with 24 pages on Sunday leads to readings of 36, 48, 60, 72, 84, and 96 pages throughout the week. The total would be 24 + 36 + 48 + 60 + 72 + 84 + 96 = 420 pages, which again surpasses the 392-page limit. <b>D) 26</b> If Larry began with 26 pages, he would read 38, 50, 62, 74, 86, and 98 pages in subsequent days. This results in a total of 26 + 38 + 50 + 62 + 74 + 86 + 98 = 434 pages, also more than the novel's total. <b>E) 28</b> Starting with 28 pages, the daily readings would be 40, 52, 64, 76, 88, and 100 pages. The sum would be 28 + 40 + 52 + 64 + 76 + 88 + 100 = 448 pages, which clearly exceeds the novel's page count. <b>Conclusion</b> Larry's reading pattern demonstrates how the arithmetic sequence of his daily readings must align with the total page count of the novel. By calculating the total for each initial value, we find that only starting with 20 pages on Sunday fits perfectly, confirming the answer. Thus, 20 pages read on Sunday is the only feasible solution that results in the correct cumulative total of 392 pages.

Explanation

<h2>A shipment of 848 chips was 2.2 standard deviations below the mean.</h2> To find how many standard deviations below the mean a shipment of 848 chips is, we first determine the standard deviation using the information provided. Since a shipment of 1456 chips is 1.6 standard deviations above the mean of 1200, we can calculate the standard deviation and then apply it to find how far 848 chips is from the mean. <b>A) 1.3</b> If a shipment were 1.3 standard deviations below the mean, the calculation would yield a value closer to the mean than 848 chips. This choice underestimates the distance from the mean, leading to an incorrect conclusion. <b>B) 1.4</b> Similar to Option A, if a shipment were 1.4 standard deviations below the mean, the calculated value would still not reach as low as 848 chips. This choice does not account for the significant drop from the mean of 1200. <b>C) 1.7</b> Choosing 1.7 standard deviations below the mean still does not sufficiently reflect how far 848 chips is from the average. This value remains too close to the mean, failing to encompass the actual distance involved in this scenario. <b>D) 2</b> While this option is closer to the correct answer, it still falls short. Being 2 standard deviations below the mean would correspond to a value that is still higher than 848 chips, indicating an incorrect assessment of the shipment's distance from the mean. <b>E) 2.2</b> This option accurately represents that a shipment of 848 chips is 2.2 standard deviations below the mean. By using the standard deviation calculated from the mean and the known shipment above the mean, we find that 848 chips indeed falls well below the average, confirming the correctness of this choice. <b>Conclusion</b> In summary, the shipment of 848 chips is correctly identified as being 2.2 standard deviations below the mean of 1200 chips. This calculation highlights the importance of understanding standard deviations in relation to the mean, as it allows us to effectively gauge how far a particular shipment deviates from average expectations.

Explanation

<h2>An odd number of canoes.</h2> To find the correct answer, we can analyze the total cost of the items purchased by the tour company. The total cost can be expressed as a linear equation based on the number of canoes, kayaks, and life jackets. <b>A) I only</b> The total spent on canoes is $65 times the number of canoes. For the total cost of $3279 to be odd, the number of canoes must be odd. Since $65 is an odd number, multiplying it by an odd number will yield an odd product, which aligns with our total. <b>B) II only</b> The total cost of kayaks is $40 times the number of kayaks, where $40 is an even number. Therefore, regardless of whether the number of kayaks is odd or even, the total cost contributed by kayaks will always be even, leaving no restriction on the total cost being odd. Thus, we cannot conclude that the number of kayaks is even based solely on the total. <b>C) III only</b> The total cost of life jackets is $2 times the number of life jackets. Since $2 is an even number, multiplying it by any whole number (even or odd) will always yield an even product. Hence, we cannot assert the number of life jackets must be even just from the total cost. <b>D) I and III only</b> Although it's true that the number of canoes must be odd, the statement about the life jackets does not hold since their total cost does not impose any condition of being even, making this option incorrect. <b>E) I, II and III</b> This choice incorrectly assumes both the number of kayaks and life jackets must meet specific conditions, which we have established are not true. <b>Conclusion</b> The analysis shows that only the number of canoes must be odd to satisfy the condition of the total cost being odd. The conditions for kayaks and life jackets do not restrict their counts in the manner described, confirming that the only verified statement is I, making option A the correct answer.