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# GMAT Descriptive Statistics : Mean Median Range

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This difficulty GMAT quant question tests concepts in number properties and statistics including elementary properties of prime numbers, odd and even numbers, mean, median and range.

Consider a set S = {2, 4, 6, 8, x, y} with distinct elements. If x and y are both prime numbers and 0 < x < 40 and 0 < y < 40, which of the following MUST be true?

I. The maximum possible range of the set is greater than 33.
II. The median can never be an even number.
III. If y = 37, the average of the set will be greater than the median.

A. I only
B. I and II only
C. I and III only
D. III only
E. I, II, and III

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### GMAT Descriptive Statistics : Mean Median Range

1. 1. Descriptive statistics Mean, Median & Range Problem solving GMAT QUANTITATIVE REASONINGQ-51 Series
2. 2. Question Consider a set S = {2, 4, 6, 8, x, y} with distinct elements. If x and y are both prime numbers and 0 < x < 40 and 0 < y < 40, which of the following MUST be true? I. The maximum possible range of the set is greater than 33. II. The median can never be an even number. III. If y = 37, the average of the set will be greater than the median. A. I only B. I and II only C. I and III only D. III only E. I, II, and III
3. 3. S = {2, 4, 6, 8, x, y} with distinct elements. x and y are both prime. 0 < x < 40 and 0 < y < 40 Which of the following MUST be true? 01 Key data · Set S has 6 elements. · The elements of set S are distinct. · x and y are prime numbers. · 0 < x < 40 and 0 < y < 40 Because 2 is already an element in S, both x and y have to be odd.
4. 4. S = {2, 4, 6, 8, x, y} with distinct elements. x and y are both prime. 0 < x < 40 and 0 < y < 40 Which of the following MUST be true? I The maximum possible range of the set is greater than 33 · Unknowns: x and y. Both are prime and are greater than 2 and less than 40. · So, smallest number in the set is 2.
5. 5. S = {2, 4, 6, 8, x, y} with distinct elements. x and y are both prime. 0 < x < 40 and 0 < y < 40 Which of the following MUST be true? I The maximum possible range of the set is greater than 33 · Unknowns: x and y. Both are prime and are greater than 2 and less than 40. · So, smallest number in the set is 2. · Maximum value x or y can take is 37, the largest prime less than 40. · Maximum possible range: 37 – 2 = 35 > 33. Statement I is true.
6. 6. S = {2, 4, 6, 8, x, y} with distinct elements. x and y are both prime. 0 < x < 40 and 0 < y < 40 Which of the following MUST be true? II The median can never be an even number · Set has 6 numbers. The median is the average of the 3rd and 4th number. Approach Let us compute the median for all possible values that x and y can take and check whether the median is an even number
7. 7. S = {2, 4, 6, 8, x, y} with distinct elements. x and y are both prime. 0 < x < 40 and 0 < y < 40 Which of the following MUST be true? II The median can never be an even number · Set has 6 numbers. The median is the average of the 3rd and 4th number. · Let x = 3, y = 5. S = {2, 3, 4, 5, 6, 8}. Median 4.5. Not even. Let x = 3, y > 7. S = {2, 3, 4, 6, 7, 8}. Median 5. Not even. Let x = 5, y > 7. S = {2, 4, 5, 6, 7, 8}. Median 5.5. Not even. Let x = 7, y > 11. S = {2, 4, 6, 7, 8, 11}. Median 6.5. Not even. Let x > 11, y > 13. S = {2, 4, 6, 8, 11, 13}. Median 7. Not even. Statement II is true.
8. 8. S = {2, 4, 6, 8, x, y} with distinct elements. x and y are both prime. 0 < x < 40 and 0 < y < 40 Which of the following MUST be true? III If y = 37, the average of the set will be greater than the median · S = {2, 4, 6, 8, 37, x}, where x is a prime greater than 2 and less than 40. Approach Let us compute the average and the median for all possible values that x can take and check whether the average is greater than the median.
9. 9. S = {2, 4, 6, 8, x, y} with distinct elements. x and y are both prime. 0 < x < 40 and 0 < y < 40 Which of the following MUST be true? III If y = 37, the average of the set will be greater than the median · S = {2, 4, 6, 8, 37, x}, where x is a prime greater than 2 and less than 40. · Let x = 3. Average = 2+3+4+6+8+37 6 = 10. Median 5. Average > Median. Let x = 5. Average = 2+4+5+6+8+37 6 = 10.33. Median 5.5. Average > Median. Let x = 7. Average = 2+4+6+7+8+37 6 = 10.66. Median 6.5. Average > Median. Let x > 11. Average = 2+4+6+8+11+37 6 > 11.33 Median 7. Average > Median. Statement III is true. Statements I, II and III true. Choice E.
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