Descriptive statistics

Mean, Median & Range

Problem solving

GMAT QUANTITATIVE REASONINGQ-51 Series

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

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.

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.

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.

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.

ApproachLet us compute the median for all possible

values that x and y can take and check

whether the median is an even number

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.

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.

ApproachLet 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.

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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