Mean, Median, and Mode Practice Questions — GMAT

Question 1Question →

A company recorded the number of daily sales over a 7-day week. The numbers of daily sales, when arranged in ascending order, are $12, 15, 18, x, 22, y, 30$ . If the median of the 7 daily sales figures is $20$ and the arithmetic mean is $21$ , what is the value of $y$ ?

Show answer & explanation

Answer: 30

Answer

The value of

y

is 30.

For an ordered set of 7 numbers, the median is the 4th term, which means

x = 20

. The total sum of the set is found by multiplying the number of terms by the mean:

7 \times 21 = 147

. Adding all known terms (

12 + 15 + 18 + 20 + 22 + 30

) gives 117. Subtracting 117 from 147 yields

y = 30

.

Step-by-Step Solution

1

Determine the value of

x

using the median definition.

x = 20

Since the 7 numbers are given in ascending order, the median is the middle (4th) term.

2

Calculate the total sum of all 7 numbers.

Total sum =

147

The sum of elements in a set equals the number of elements multiplied by the arithmetic mean (

7 \times 21 = 147

).

3

Sum all known numbers and set up an equation for

y

.

117 + y = 147

12 + 15 + 18 + 20 + 22 + 30 = 117

.

4

Solve for

y

.

y = 30

Subtract 117 from 147 to isolate

y

.

Key Concept

Using properties of median and arithmetic mean to find missing terms in an ordered data set.

Question 2Question →

A class of 20 students took a 10-point mathematics quiz. The frequency table below records the quiz scores achieved by 18 of the students:

Score	Frequency
6	3
7	5
8	4
9	4
10	2

The scores of the remaining 2 students were recorded later. If the score of every student is an integer from 0 to 10, inclusive, and adding the 2 missing scores causes the median score of the entire class of 20 students to be 8 and the arithmetic mean score to be an integer, what is the score of the higher-scoring student among the 2 remaining students?

6

7

8

9

10

Show answer & explanation

Answer: 10

Answer

10

The sum of the 18 known scores is 141. For the overall mean of 20 scores to be an integer, the total sum of all 20 scores must be a multiple of 20. Since each score is at most 10, the maximum possible total sum is

141 + 10 + 10 = 161

. The only multiple of 20 between 141 and 161 is 160, requiring the sum of the two missing scores to be

160 - 141 = 19

. The only valid integer scores bounded by 10 that sum to 19 are 9 and 10. Adding scores of 9 and 10 places the 10th and 11th ordered values at 8, giving a median of 8. Thus, the higher missing score is 10.

Step-by-Step Solution

1

Calculate the total sum and count of the 18 known student scores.

Known count =

3 + 5 + 4 + 4 + 2 = 18

students. Known sum =

(6 \times 3) + (7 \times 5) + (8 \times 4) + (9 \times 4) + (10 \times 2) = 18 + 35 + 32 + 36 + 20 = 141

.

Establishing baseline sum and count is essential before analyzing missing values.

2

Set up the equation for the total sum of all 20 student scores and apply the integer mean condition.

Let the missing scores be

a

and

b

with

0 \le a \le b \le 10

. The total sum for 20 students is

S_{20} = 141 + a + b

. The arithmetic mean is

\frac{141 + a + b}{20}

.

Since the mean must be an integer,

141 + a + b

must be a multiple of 20.

3

Determine the required sum of the two missing scores

a + b

.

Since

0 \le a \le 10

and

0 \le b \le 10

, we have

0 \le a + b \le 20

. The range for

S_{20}

is

[141, 161]

. The only multiple of 20 in this range is 160. Thus,

141 + a + b = 160 \implies a + b = 19

.

160 is the unique multiple of 20 reachable given score bounds.

4

Find the unique integer pair

(a, b)

satisfying

a + b = 19

with

a, b \le 10

.

Since

a \le 10

and

b \le 10

, the only integer solution with

a \le b

is

a = 9

and

b = 10

.

No other pair of integers between 0 and 10 sums to 19.

5

Verify that adding scores 9 and 10 maintains a median score of 8.

With 9 and 10 added, the frequencies are: Score 6 (3), Score 7 (5), Score 8 (4), Score 9 (5), Score 10 (3). The 10th and 11th values in order are both 8, so the median is

\frac{8 + 8}{2} = 8

.

Confirms the median constraint is fully satisfied.

Key Concept

Properties of Weighted Means and Median Constraints in Frequency Tables

Estimated Time:2m 0s

Question 3Question →

A set of $7$ positive integers has an arithmetic mean of $16$ , a median of $15$ , a unique mode of $12$ , and a range of $14$ . What is the maximum possible value of the largest integer in this set?

Show answer & explanation

Answer: 26

Answer

The maximum possible value of the largest integer in the set is 26.

The total sum of the 7 positive integers is

7 \times 16 = 112

. Ordering the terms as

a \le b \le c \le d \le e \le f \le g

, the median constraint gives

d = 15

. The range constraint gives

g - a = 14

, or

g = a + 14

. To maximize

g

, we need to maximize

a

. Because 12 is the unique mode of the set, 12 must appear at least twice. Since

d = 15

, the number 12 can only occupy positions

a, b,

or

c

, which means

a \le 12

. Setting

a = 12

gives the maximum value

g = 12 + 14 = 26

. A valid set achieving this is

\{12, 12, 12, 15, 15, 20, 26\}

, which sums to 112 and meets all statistical constraints.

Step-by-Step Solution

1

Determine the sum of the set

Sum = 112

The mean of 7 positive integers is 16, so the sum is 7 times 16.

2

Apply median and range constraints

d = 15 and a = g - 14

In a sorted set of 7 integers, the 4th element is the median (15), and range is the difference between the largest element g and smallest element a.

3

Bound the smallest element using the mode constraint

a <= 12, so max g = 12 + 14 = 26

Since 12 is the unique mode, it must occur at least twice. Because elements are sorted and median is 15, 12 must be among the first three terms, so the smallest element a cannot exceed 12.

4

Verify existence of a valid set with g = 26

Set {12, 12, 12, 15, 15, 20, 26} satisfies all conditions

The sum is 112, median is 15, mode is 12 (appears 3 times), and range is 26 - 12 = 14.

Key Concept

Extremal problems involving mean, median, mode, and range constraints

Question 4Question →

A survey asked consumers to rate a new service on a scale from $1$ to $5$ . The table below shows the frequency of each rating received, where $k$ is a positive integer:

Rating	Frequency
1	4
2	6
3	7
4	$k$
5	5

If the median rating of all responses is equal to the arithmetic mean of all responses, what is the value of $k$ ?

2

3

4

5

6

Show answer & explanation

Answer: $4$

Answer

4

The option stating

4

is correct because with

k = 4

, the total number of responses is

22 + 4 = 26

and the sum of all ratings is

62 + 4(4) = 78

. The mean is

\frac{78}{26} = 3

. The median of

26

ordered values is the average of the 13th and 14th values, both of which are

3

. Thus, the mean equals the median.

Step-by-Step Solution

1

Calculate total number of data points and total sum in terms of

k

Total responses

N = 4 + 6 + 7 + k + 5 = 22 + k

. Total sum

= 1(4) + 2(6) + 3(7) + 4(k) + 5(5) = 4 + 12 + 21 + 4k + 25 = 62 + 4k

.

The mean of a frequency distribution is the total weighted sum divided by the total frequency.

2

Determine the median rating

Ratings

1, 2, 3

account for the first

4 + 6 + 7 = 17

responses. For small positive integer values of

k

, the middle position(s) fall within the rating of

3

, so the median rating is

3

.

Since more than half of the initial

22

data points are

3

or below (specifically positions 11 through 17 are 3s), the median is

3

for

k \le 12

.

3

Equate the arithmetic mean to the median and solve for

k

\frac{62 + 4k}{22 + k} = 3 \implies 62 + 4k = 3(22 + k) \implies 62 + 4k = 66 + 3k \implies k = 4.

Setting the mean formula equal to the median value of

3

allows solving for

k

algebraically.

Key Concept

Weighted Mean and Frequency Distribution Median

Estimated Time:1m 30s

Question 5Question →

A set $S$ consists of $6$ distinct positive integers. The arithmetic mean of the integers in $S$ is $15$ , and the median of $S$ is $14$ . If the largest integer in $S$ is $28$ , what is the greatest possible value of the second-largest integer in $S$ ?

24

25

27

28

31

Show answer & explanation

Answer: $27$

Answer

The greatest possible value of the second-largest integer in the set is

27

.

The correct option is

27

. With

6

distinct positive integers

x_1 < x_2 < x_3 < x_4 < x_5 < x_6

, the total sum is

6 \times 15 = 90

. The median gives

x_3 + x_4 = 28

. Given

x_6 = 28

, we have

x_1 + x_2 + x_5 = 34

. Since all integers are distinct and

28

is the largest,

x_5

must be strictly less than

28

, making

27

the maximum integer bound. Setting

x_5 = 27

allows

x_1 = 1, x_2 = 6, x_3 = 13, x_4 = 15, x_5 = 27, x_6 = 28

, which satisfies every requirement of the problem.

Step-by-Step Solution

1

Determine the total sum of the

6

integers.

Sum =

6 \times 15 = 90

.

Since the mean of

6

integers is

15

, the sum of all elements equals the count multiplied by the mean.

2

Express the median condition algebraically.

Let the ordered set be

x_1 < x_2 < x_3 < x_4 < x_5 < x_6

. Then

\frac{x_3 + x_4}{2} = 14 \implies x_3 + x_4 = 28

.

For an even number of terms (

n=6

), the median is the average of the 3rd and 4th terms.

3

Substitute known values into the sum equation.

x_1 + x_2 + (x_3 + x_4) + x_5 + x_6 = 90 \implies x_1 + x_2 + 28 + x_5 + 28 = 90 \implies x_1 + x_2 + x_5 = 34

.

We know

x_3 + x_4 = 28

and

x_6 = 28

(the largest integer).

4

Apply constraints to maximize

x_5

.

Since

x_6 = 28

is the largest element and all integers are distinct,

x_5 < 28

, so

x_5 \le 27

. Testing

x_5 = 27

yields

x_1 + x_2 = 34 - 27 = 7

.

To verify

x_5 = 27

is achievable, choose distinct positive integers

x_1 = 1, x_2 = 6, x_3 = 13, x_4 = 15, x_5 = 27, x_6 = 28

, which satisfies all conditions.

Key Concept

Mean and Median Properties in Constrained Sets

Question 6Question →

The numbers of packages delivered by a courier on seven consecutive days are listed below:

5, 13, 2, 10, 15, 7, 4

What is the median number of packages delivered per day over this seven-day period?

5

7

8

10

13

Show answer & explanation

Answer: $7$

Answer

The median number of packages delivered per day is

7

.

To find the median of a dataset, the values must first be arranged in ascending order:

2, 4, 5, 7, 10, 13, 15

. Since there are

7

numbers, the median is the

4

th number in the ordered list, which is

7

.

Step-by-Step Solution

1

Arrange the data values in ascending order.

The ordered set is

2, 4, 5, 7, 10, 13, 15

.

Finding the median requires ordering the data sequentially first.

2

Identify the middle term of the ordered set containing

7

values.

The middle term is the

\frac{7+1}{2} = 4

th value, which is

7

.

For an odd number of data points

N

, the median is the

\frac{N+1}{2}

th element.

Key Concept

Median of a finite data set

Estimated Time:45s

Question 7Question →

An analytics team tracked the monthly recurring revenue (in thousands of dollars) for 10 distinct client accounts of a SaaS company. The arithmetic mean of the monthly revenues across these 10 accounts was $\$45$ thousand, and the median monthly revenue was $\$38$ thousand. If the revenue for each account was a distinct positive integer (in thousands of dollars), what is the maximum possible monthly revenue, in thousands of dollars, generated by the single highest-performing account?

Show answer & explanation

Answer: 241

Answer

The maximum possible monthly revenue generated by the single highest-performing account is 241 thousand dollars.

The total sum of the 10 distinct positive integer revenues is

10 \times 45 = 450

thousand dollars. To maximize the largest term (

x_{10}

), the sum of the remaining 9 terms (

x_1

through

x_9

) must be minimized. The smallest 4 positive integers are

1, 2, 3, 4

(sum =

10

). Given a median of 38, the two central terms must satisfy

x_5 + x_6 = 76

. Because all values are distinct integers and

x_5 < x_6

, the smallest possible integer value for

x_6

is 39 (making

x_5 = 37

). To minimize

x_7, x_8, x_9

, they should be the smallest distinct integers greater than 39, which are

40, 41, 42

. Summing these minimal values gives

10 + 76 + 40 + 41 + 42 = 209

. Subtracting this sum from the total sum yields

450 - 209 = 241

thousand dollars.

Step-by-Step Solution

1

Calculate the total revenue across all 10 client accounts.

Total sum =

10 \times 45 = 450

thousand dollars.

The sum of a data set is equal to the number of elements multiplied by the arithmetic mean.

2

Use the median to determine the sum of the 5th and 6th terms.

Since

n = 10

, the median is

\frac{x_5 + x_6}{2} = 38

, so

x_5 + x_6 = 76

.

In an ordered set of an even number of elements, the median is the average of the two central terms.

3

Formulate the minimization strategy to maximize the largest term (

x_{10}

).

To maximize

x_{10} = 450 - (x_1 + x_2 + \dots + x_9)

, minimize the sum of the first 9 terms.

For a fixed sum, maximizing one element requires minimizing all other elements in the set.

4

Minimize the first four terms (

x_1, x_2, x_3, x_4

).

Assign

x_1 = 1, x_2 = 2, x_3 = 3, x_4 = 4

, which sum to

10

.

The revenues are distinct positive integers, so the smallest possible values are the first 4 positive integers.

5

Minimize

x_6

to minimize the subsequent terms (

x_7, x_8, x_9

).

Since

x_5 < x_6

and

x_5 + x_6 = 76

, the smallest integer value for

x_6

is

39

(which sets

x_5 = 37

). Then set

x_7 = 40, x_8 = 41, x_9 = 42

.

Minimizing

x_6

allows

x_7, x_8, x_9

to take the smallest possible distinct integer values greater than

x_6

.

6

Calculate the maximum value for

x_{10}

.

x_{10} = 450 - (10 + 76 + 40 + 41 + 42) = 450 - 209 = 241

.

Subtracting the minimal sum of the first 9 elements from the total sum gives the theoretical maximum for the 10th term.

Key Concept

Extremal optimization of distinct statistics data sets given mean and median constraints

Question 8Question →

A data set consists of $8$ positive integers with an arithmetic mean of $15$ and a unique mode of $12$ . What is the maximum possible value of an integer in this data set?

Show answer & explanation

Answer: 81

Answer

81

The total sum of the

8

integers is

8 \times 15 = 120

. To maximize the single largest integer, the sum of the remaining

7

integers must be minimized. Since

12

is the unique mode,

12

must appear at least twice, and no other value can appear more than once. The smallest positive integers that can occupy the first

5

positions without creating another mode are

1, 2, 3, 4,

and

5

. Placing

12

in the

6\text{th}

and

7\text{th}

positions minimizes the sum of the first

7

elements to

1 + 2 + 3 + 4 + 5 + 12 + 12 = 39

. Thus, the maximum possible value for the largest integer is

120 - 39 = 81

.

Step-by-Step Solution

1

Calculate the total sum of the 8 positive integers.

Total Sum =

8 \times 15 = 120

.

The sum of a set of numbers equals the arithmetic mean multiplied by the number of elements.

2

Minimize the sum of the first 7 integers to maximize the 8th integer.

Minimum sum of the first 7 integers =

1 + 2 + 3 + 4 + 5 + 12 + 12 = 39

.

To maximize the largest integer, the remaining 7 integers must be as small as possible. Since 12 is the unique mode, 12 must appear at least twice, and no other integer can appear more than once. Placing two 12s at the highest available positions among the 7 terms (

a_6 = 12

and

a_7 = 12

) and selecting the smallest distinct positive integers (

1, 2, 3, 4, 5

) for the first 5 terms minimizes their total sum.

3

Subtract the minimum sum of the 7 smallest terms from the total sum.

Maximum integer =

120 - 39 = 81

.

Subtracting the smallest possible sum of 7 terms from the fixed sum of 120 yields the maximum possible value for the 8th term.

Key Concept

Maximizing an element in a data set under mean and mode constraints

Question 9Question →

A set of $5$ integers has an arithmetic mean of $12$ . If four of the integers are $8$ , $10$ , $14$ , and $15$ , what is the value of the fifth integer?

Show answer & explanation

Answer: 13

Answer

The value of the fifth integer is

13

.

The total sum of a set of numbers is given by the formula

\text{Sum} = \text{Mean} \times n

. For

5

numbers with a mean of

12

, the total sum is

5 \times 12 = 60

. The sum of the four provided numbers is

8 + 10 + 14 + 15 = 47

. Subtracting

47

from

60

yields

13

, which is the value of the fifth integer.

Step-by-Step Solution

1

Find the total sum of the 5 integers

The total sum is

60

The sum of a set of numbers equals the arithmetic mean multiplied by the total count of numbers (

12 \times 5 = 60

).

2

Sum the four given integers

The sum of the four integers is

47

Adding the given numbers:

8 + 10 + 14 + 15 = 47

.

3

Subtract the sum of the known integers from the total sum

The fifth integer is

13

Subtracting

47

from

60

gives

60 - 47 = 13

.

Key Concept

Arithmetic Mean

Estimated Time:45s

Question 10Question →

A data set $S$ consists of $9$ positive integers arranged in non-decreasing order: $x_1 \le x_2 \le x_3 \le x_4 \le x_5 \le x_6 \le x_7 \le x_8 \le x_9$ . The arithmetic mean of set $S$ is $22$ , the median is $20$ , and the set has a unique mode of $15$ . If the range of set $S$ is $30$ , what is the maximum possible value of the largest element, $x_9$

31

40

45

50

52

Show answer & explanation

Answer: 45

Answer

The maximum possible value of the largest element is 45.

To maximize the largest value in a set with a fixed range of 30, we must maximize the smallest value because the maximum value equals the minimum value plus 30. Since 15 is the unique mode of the ordered set, 15 must be an element of the set, which restricts the minimum value to at most 15. Setting the smallest element to 15 allows the largest element to reach 15 + 30 = 45, which can be verified to satisfy all mean, median, and mode constraints.

Step-by-Step Solution

1

Relate total sum, range, and median to the elements of the set.

The sum of all 9 elements is

9 \times 22 = 198

. For a 9-element set ordered as

x_1 \le x_2 \le \dots \le x_9

, the median is the 5th element

x_5 = 20

. The range is

x_9 - x_1 = 30

, so

x_9 = x_1 + 30

.

Establishing the basic relationships among the statistical measures given in the problem.

2

Determine the upper bound for the smallest element

x_1

.

Because

15

is the unique mode of the set,

15

must be present in the set. Since the set is in non-decreasing order,

x_1 \le 15

.

If

x_1 > 15

, then no element in the set could equal

15

, violating the condition that

15

is the mode.

3

Maximize

x_9

using the relationship

x_9 = x_1 + 30

.

To maximize

x_9

,

x_1

must be as large as possible. The maximum possible value for

x_1

is

15

, which gives

x_9 = 15 + 30 = 45

.

Connecting the range equation to the upper bound on

x_1

.

4

Verify that a valid set exists for

x_1 = 15

and

x_9 = 45

.

Consider the set

\{15, 15, 15, 15, 20, 20, 25, 28, 45\}

. Sum

= 198

, Mean

= 22

, Median

= 20

(5th element), Mode

= 15

(frequency 4), Range

= 45 - 15 = 30

. All conditions are satisfied.

Ensuring the upper bound is achievable under all given statistical constraints.

Key Concept

Interplay between statistical measures (mean, median, mode, range) and set boundaries

Question 11Question →

The scores of a student on five quizzes are $72, 85, 90, 78,$ and $95$ . What is the positive difference between the median and the arithmetic mean of these five quiz scores?

Show answer & explanation

Answer: 1

Answer

The positive difference between the median and the arithmetic mean of the five quiz scores is

1

.

To find the arithmetic mean, sum all five scores (

72 + 85 + 90 + 78 + 95 = 420

) and divide by

5

, yielding

84

. To find the median, list the scores in ascending order (

72, 78, 85, 90, 95

); the 3rd score is

85

. The positive difference between the median (

85

) and the arithmetic mean (

84

) is

1

.

Step-by-Step Solution

1

Calculate the arithmetic mean of the given set of scores.

The sum of the scores is

72 + 78 + 85 + 90 + 95 = 420

. Dividing by 5 gives an arithmetic mean of

84

.

The mean is defined as the total sum of all values divided by the number of values.

2

Order the scores in ascending order to identify the median.

The ordered list is

72, 78, 85, 90, 95

. The middle term (the 3rd value) is

85

.

The median of a set with an odd number of elements is the middle value when arranged in numerical order.

3

Subtract the arithmetic mean from the median to find the positive difference.

85 - 84 = 1

.

The question asks for the positive difference between the median and the arithmetic mean.

Key Concept

Calculating and comparing the arithmetic mean and median of a finite numerical data set.

Estimated Time:1m 0s

Question 12Question →

A data set $S$ consists of $11$ positive integers. The set has an arithmetic mean of $15$ , a median of $14$ , and a unique mode of $18$ . If $M$ is the maximum possible value of an element in $S$ and $m$ is the minimum possible value of an element in $S$ , what is the maximum possible value of the range $M - m$ ?

68

71

72

73

74

Show answer & explanation

Answer: $73$

Answer

The maximum possible value of the range

M - m

is

73

.

To maximize the range

M - m = x_{11} - x_1

, we minimize

x_1

and maximize

x_{11}

. Since elements are positive integers,

m = x_1 = 1

. The sum of all

11

elements is

11 \times 15 = 165

. With median

x_6 = 14

, the unique mode

18

must lie above the median. Setting the frequency of

18

to

3

allows other numbers to appear up to

2

times. Minimizing

x_1 \dots x_5

gives

1, 1, 2, 2, 3

(sum

9

). Minimizing

x_7 \dots x_{10}

by setting

x_7 = 14

and

x_8 = x_9 = x_{10} = 18

gives sum

68

. The sum of the first

10

terms is

9 + 14 + 68 = 91

, leaving

M = x_{11} = 165 - 91 = 74

. Thus, the maximum range is

74 - 1 = 73

.

Step-by-Step Solution

1

Calculate the sum of all elements in the set.

Sum

= 11 \times 15 = 165

.

The arithmetic mean of

11

elements is

15

.

2

Identify the position of the median and order the elements.

Let elements be

x_1 \le x_2 \le \dots \le x_6 \le \dots \le x_{11}

, where

x_6 = 14

.

For an odd number of elements (

11

), the median is the 6th element.

3

Determine the mode frequency constraint to maximize

x_{11} - x_1

.

The unique mode is

18

. If

18

appears

3

times, the maximum frequency of any other value is

2

.

To maximize

x_{11}

, we minimize

x_1 \dots x_{10}

. Allowing the mode to appear

3

times permits other elements to appear up to

2

times.

4

Minimize the sum of the lower five elements

x_1, x_2, x_3, x_4, x_5

.

x_1 = 1, x_2 = 1, x_3 = 2, x_4 = 2, x_5 = 3

, giving sum

= 1 + 1 + 2 + 2 + 3 = 9

.

The smallest positive integers with maximum frequency

2

are

1, 1, 2, 2, 3

.

5

Minimize elements

x_7, x_8, x_9, x_{10}

.

x_7 = 14

(frequency

2

for

14

), and

x_8 = 18, x_9 = 18, x_{10} = 18

. Sum

= 14 + 18 + 18 + 18 = 68

.

Since

x_6 = 14

, setting

x_7 = 14

minimizes

x_7

while keeping the frequency of

14

at

2

, which is less than the mode frequency of

3

.

6

Calculate the maximum value

M = x_{11}

and the range

M - m

.

Sum of first 10 elements

= 9 + 14 + 68 = 91

. Thus

x_{11} = 165 - 91 = 74

. Range

= 74 - 1 = 73

.

Subtracting the sum of the first 10 elements from the total sum gives

M

, and subtracting

m = 1

yields the range.

Key Concept

Range, Mean, Median, and Mode Constraints in Data Sets

Question 13Question →

A logistics company recorded the daily number of deliveries made by each of its 8 delivery vans on a given day. Each van completed a distinct positive integer number of deliveries. The arithmetic mean of the number of deliveries made by the 8 vans was 25, the median was 24, and the range was 18. If $M$ represents the maximum number of deliveries completed by any single van that day, what is the maximum possible value of $M$ ?

Show answer & explanation

Answer: 37

Answer

The maximum possible value of

M

is 37.

To maximize the largest element

M = x_8

, we express

M

in terms of the smallest element

x_1

using the range:

M = x_1 + 18

. Thus, maximizing

M

is equivalent to maximizing

x_1

. Testing

x_1 = 20

forces the minimal possible sum of the 8 distinct terms to be

20 + 21 + 22 + 23 + 25 + 26 + 27 + 38 = 202

, which exceeds the required sum of 200. Testing

x_1 = 19

allows a minimal sum of 198, which can be adjusted to 200 by setting the set to

\{19, 20, 21, 23, 25, 26, 29, 37\}

. Thus, the maximum possible value of

M

is

19 + 18 = 37

.

Step-by-Step Solution

1

Formulate the algebraic equations from the statistical properties given.

Sum of 8 terms =

8 \times 25 = 200

;

x_4 + x_5 = 48

;

x_8 = x_1 + 18 = M

.

Mean gives total sum, even number of items gives median as average of 4th and 5th terms, and range links the maximum and minimum values.

2

Relate maximizing the maximum term

M

to maximizing the minimum term

x_1

.

Maximizing

M = x_1 + 18

requires making

x_1

as large as possible.

Since the range is fixed at 18,

M

increases directly as

x_1

increases.

3

Test

x_1 = 20

to determine feasibility.

Minimum possible sum for

x_1 = 20

is

20 + 21 + 22 + 23 + 25 + 26 + 27 + 38 = 202 > 200

, which is invalid.

Distinct integer constraints force

x_4 \ge 23

; since

x_4 + x_5 = 48

and

x_4 < x_5

,

x_4

must be 23 and

x_5

must be 25, forcing all lower bounds up.

4

Test

x_1 = 19

to confirm feasibility and construct a valid set.

The valid set

\{19, 20, 21, 23, 25, 26, 29, 37\}

meets all criteria with a sum of 200.

The minimal sum for

x_1 = 19

is 198, leaving headroom to increase

x_7

to 29 to reach the sum of 200.

Key Concept

Optimization of Extreme Values in Finite Ordered Sets using Mean, Median, and Range

Question 14Question →

A data set consists of $9$ positive integers. The set has a unique mode of $14$ , a median of $12$ , and an arithmetic mean of $11$ . If the range of the data set is $10$ , what is the maximum possible value of the largest integer in the set?

14

15

16

17

18

Show answer & explanation

Answer: $16$

Answer

The maximum possible value of the largest integer in the set is

16

.

The correct answer states that the maximum possible value is

16

. By ordering the

9

terms

x_1 \le x_2 \le \dots \le x_9

, the sum of all terms must equal

99

, with median

x_5 = 12

and

x_9 = x_1 + 10

. If

x_9 = 16

, then

x_1 = 6

, and we can construct a valid set

\{6, 6, 8, 9, 12, 14, 14, 14, 16\}

where

14

is the unique mode appearing

3

times. Trying a larger value such as

17

forces

x_1 = 7

, which makes it impossible to maintain

14

as the unique mode without violating the total sum of

99

.

Step-by-Step Solution

1

Express the total sum and set structure using ordered variables.

Let the

9

positive integers in non-decreasing order be

x_1 \le x_2 \le x_3 \le x_4 \le x_5 \le x_6 \le x_7 \le x_8 \le x_9

. The total sum is

9 \times 11 = 99

. The median is

x_5 = 12

. The range is

x_9 - x_1 = 10 \implies x_9 = x_1 + 10

.

Establishing the position of the median and the exact sum provides structural bounds for maximizing

x_9

.

2

Analyze the frequency requirements for the unique mode.

Since

14 > 12

(the median), the number

14

must lie in the upper half of the set (

x_6, x_7, x_8, x_9

). To be a unique mode,

14

must appear at least twice. If

x_9 = 16

, then

x_1 = 6

.

Maximizing

x_9

is equivalent to maximizing

x_1

, so we test the largest possible values for

x_1

and verify the uniqueness of the mode

14

.

3

Evaluate

x_1 = 7

(which would give

x_9 = 17

).

If

x_1 = 7

, then

x_9 = 17

. The sum equation becomes

7 + x_2 + x_3 + x_4 + 12 + x_6 + x_7 + x_8 + 17 = 99 \implies x_2 + x_3 + x_4 + x_6 + x_7 + x_8 = 63

. Since

x_2, x_3, x_4 \ge 7

and

x_6, x_7, x_8 \ge 12

, achieving this sum while keeping

14

as the unique mode is impossible (it would require

7

to appear

4

times, making

7

the mode instead of

14

). Thus,

x_9

cannot be

17

or greater.

Testing

x_9 = 17

shows a violation of the unique mode condition.

4

Construct a valid set for

x_1 = 6

(giving

x_9 = 16

).

If

x_1 = 6

and

x_9 = 16

, consider the set

\{6, 6, 8, 9, 12, 14, 14, 14, 16\}

. Sum =

6+6+8+9+12+14+14+14+16 = 99

. Median =

12

. Unique mode =

14

(frequency

3

). Range =

16 - 6 = 10

. All conditions are satisfied.

Constructing an explicit valid data set proves that

16

is attainable and is the maximum.

Key Concept

Maximizing elements in a constrained discrete data set using mean, median, mode, and range properties

Question 15Question →

A software quality team tracked the number of bug tickets resolved daily over a period of

15

consecutive workdays. The recorded number of resolved tickets for the first

14

days, listed in ascending order, were:

2, 3, 3, 4, 5, 5, 6, 7, 8, 9, 9, 10, 11, 12

If the number of tickets resolved on the

15\text{th}

day was

x

, and the arithmetic mean of the complete

15

-day data set is equal to its median, what is the value of

x

?

7

8

11

13

15

Show answer & explanation

Answer: $11$

Answer

The value of

x

is

11

.

Summing the initial 14 numbers yields 94. In a set of 15 numbers, the median is the 8th term when ordered. Because there are exactly 7 terms smaller than 7 in the initial list, any value of

x \ge 7

keeps 7 at the 8th position, making the median 7. Setting the mean

\frac{94+x}{15}

equal to 7 gives

94 + x = 105

, which leads to

x = 11

.

Step-by-Step Solution

1

Calculate the sum of the known 14 values.

\text{Sum}_{14} = 2 + 3 + 3 + 4 + 5 + 5 + 6 + 7 + 8 + 9 + 9 + 10 + 11 + 12 = 94

.

The sum of all 15 terms will be

94 + x

, making the arithmetic mean

\frac{94 + x}{15}

.

2

Determine the position of the median in a 15-element set.

The median of an odd-numbered set with

N = 15

is the

\frac{15+1}{2} = 8\text{th}

term when arranged in ascending order.

To set the mean equal to the median, we must determine how

x

affects the 8th term.

3

Analyze the position of the 8th term based on potential values of

x

.

The first 14 numbers contain 7 values less than 7 (

2, 3, 3, 4, 5, 5, 6

) and 7 values greater than or equal to 7 (

7, 8, 9, 9, 10, 11, 12

). Therefore, for any

x \ge 7

, the 8th term in the sorted set of 15 numbers is fixed at

7

.

Since 7 values are strictly less than 7, placing

x \ge 7

ensures that 7 is the 8th smallest value.

4

Set the arithmetic mean equal to the median (

7

) and solve for

x

.

\frac{94 + x}{15} = 7 \implies 94 + x = 105 \implies x = 11.

Since

x = 11 \ge 7

, the condition holds and the median remains 7.

Key Concept

Properties of Mean and Median in a Data Set

Question 16Question →

A data set consists of five positive integers $a, b, c, d,$ and $e$ , written in non-decreasing order ( $a \le b \le c \le d \le e$ ). The set has a unique mode of $8$ , a median of $10$ , and an arithmetic mean of $12$ . If the range of the set is $14$ , what is the value of the largest integer $e$ ?

17

20

22

24

26

Show answer & explanation

Answer: 22

Answer

22

The median of a 5-element ordered set is the middle element, so

c = 10

. Because 8 is the unique mode and is strictly less than 10, the first two numbers must both be 8 (

a = 8, b = 8

). The range is 14, meaning the difference between the maximum element

e

and the minimum element

a

is 14. Therefore,

e = 8 + 14 = 22

.

Step-by-Step Solution

1

Determine the median value

c

c = 10

In an ordered 5-element set, the median is the 3rd element.

2

Determine the values of

a

and

b

using the unique mode

a = 8

and

b = 8

Since 8 is the unique mode and

8 < 10

, 8 must appear at least twice in the first two positions.

3

Calculate the largest integer

e

using the given range

e = a + 14 = 8 + 14 = 22

The range of a set is the difference between the maximum (

e

) and minimum (

a

) values.

4

Verify consistency with the mean

Sum

= 5 \times 12 = 60

;

d = 60 - (8 + 8 + 10 + 22) = 12

, which satisfies

10 \le 12 \le 22

.

Ensures all set conditions (mean of 12, non-decreasing order, unique mode) are satisfied.

Key Concept

Relating Mean, Median, Mode, and Range in Ordered Data Sets

Question 17Question →

A data set consists of seven integers written in ascending order: $10, 12, 15, x, 26, 29, 34$ . If the arithmetic mean of the data set is equal to its median, what is the value of $x$ ?

18

20.5

21

23

25.2

Show answer & explanation

Answer: $21$

Answer

The value of

x

is

21

.

In an ordered set of seven numbers, the median is the 4th number, which is

x

. The arithmetic mean of the seven numbers is

\frac{10 + 12 + 15 + x + 26 + 29 + 34}{7} = \frac{126 + x}{7}

. Equating the mean to the median gives

\frac{126 + x}{7} = x

. Solving this linear equation yields

126 + x = 7x \implies 6x = 126 \implies x = 21

. Since

21

lies between

15

and

26

, it maintains the ascending order of the set.

Step-by-Step Solution

1

Identify the median of the ordered set

Median =

x

For an ordered set with an odd number of elements (

n = 7

), the median is the middle element, which is the

\frac{7 + 1}{2} = 4

th element. Since the elements are in ascending order, the 4th element is

x

.

2

Calculate the sum of all elements in terms of

x

Sum =

126 + x

The sum of the known six numbers is

10 + 12 + 15 + 26 + 29 + 34 = 126

. Adding

x

gives a total sum of

126 + x

.

3

Set up the equation equating mean and median

\frac{126 + x}{7} = x

The arithmetic mean is defined as the total sum divided by the total number of elements (

7

). We are given that Mean = Median.

4

Solve for

x

and verify ordering bounds

x = 21

Multiplying both sides by

7

gives

126 + x = 7x

, so

6x = 126

, yielding

x = 21

. Checking the ascending order condition

15 \le x \le 26

,

21

satisfies the condition

15 \le 21 \le 26

.

Key Concept

Mean and Median of an Ordered Set

Estimated Time:1m 30s

Question 18Question →

A researcher recorded six test measurements arranged in ascending order: $1, 2, 8, x, y, 15$ . If the unique mode of the data set is $8$ and the arithmetic mean of the measurements is equal to their median, what is the value of $y$ ?

Show answer & explanation

Answer: 14

Answer

14

Because the data set is arranged in ascending order, the elements satisfy

1 \le 2 \le 8 \le x \le y \le 15

. For

8

to be the unique mode,

8

must appear more times than any other number, which forces

x = 8

. The median of a

6

-element set is the average of the

3^{\text{rd}}

and

4^{\text{th}}

elements, which are both

8

, making the median

8

. Setting the mean

\frac{1 + 2 + 8 + 8 + y + 15}{6} = \frac{34 + y}{6}

equal to

8

gives

34 + y = 48

, which yields

y = 14

. Since

8 \le 14 \le 15

, this maintains the ascending order and unique mode condition.

Step-by-Step Solution

1

Determine the value of x

x = 8

The set is given in ascending order (1 <= 2 <= 8 <= x <= y <= 15). For 8 to be the unique mode, 8 must appear at least twice, which requires x = 8.

2

Calculate the median of the data set

Median = 8

With 6 elements, the median is the average of the 3rd and 4th elements: (8 + 8) / 2 = 8.

3

Set up the equation for the arithmetic mean

(34 + y) / 6 = 8

The sum of the six numbers is 1 + 2 + 8 + 8 + y + 15 = 34 + y. Since the mean equals the median, the mean must equal 8.

4

Solve for y

y = 14

Multiplying both sides by 6 gives 34 + y = 48, so y = 48 - 34 = 14.

Key Concept

Using statistical properties (mean, median, mode) of an ordered set to determine unknown values

Estimated Time:1m 15s

Question 19Question →

A data set consists of $7$ positive integers. The smallest integer in the set is $4$ , the range of the set is $19$ , the median is $12$ , and the set has a unique mode of $8$ . What is the maximum possible arithmetic mean of the $7$ integers?

13

13.5

14

14.5

15

Show answer & explanation

Answer: 14

Answer

The maximum possible arithmetic mean of the 7 integers is 14.

The correct answer of 14 is obtained by setting up the 7 ordered integers as

x_1, x_2, x_3, x_4, x_5, x_6, x_7

. We know

x_1 = 4

, median

x_4 = 12

, and

x_7 = 4 + 19 = 23

. Since 8 is the unique mode and

8 < 12

, 8 must fill

x_2

and

x_3

, giving it a frequency of 2. For 8 to remain the unique mode, no other element can appear more than once. Thus,

x_5, x_6, x_7

must all be distinct. To maximize the mean, we maximize

x_5

and

x_6

under

12 < x_5 < x_6 < 23

, yielding

x_5 = 21

and

x_6 = 22

. The maximum sum is

4 + 8 + 8 + 12 + 21 + 22 + 23 = 98

, giving a mean of

98 / 7 = 14

.

Step-by-Step Solution

1

Determine the fixed values in the ordered 7-element data set.

Let the 7 integers in non-decreasing order be

x_1, x_2, x_3, x_4, x_5, x_6, x_7

. Given

x_1 = 4

, median

x_4 = 12

, and range

x_7 - x_1 = 19

, we find

x_7 = 4 + 19 = 23

.

The median of a 7-element set is the 4th element, and range is maximum minus minimum.

2

Determine the frequency and values of the mode.

Since the unique mode is

8

and

8 < 12

, the value

8

must occupy positions

x_2

and

x_3

, so

x_2 = 8

and

x_3 = 8

. Thus,

8

appears exactly twice.

Position 1 is 4 and position 4 is 12, leaving only positions 2 and 3 for the mode value of 8.

3

Apply the unique mode constraint to maximize remaining terms.

Because

8

appears twice and is the unique mode, no other number can appear 2 or more times. Hence, all other elements must be distinct. To maximize the sum, we choose the largest distinct integers for

x_5

and

x_6

such that

12 < x_5 < x_6 < 23

, giving

x_5 = 21

and

x_6 = 22

.

Allowing any other value to repeat would create a second mode or a new unique mode, violating the problem conditions.

4

Calculate the maximum sum and arithmetic mean.

The maximal set is

\{4, 8, 8, 12, 21, 22, 23\}

. The sum is

4 + 8 + 8 + 12 + 21 + 22 + 23 = 98

. The maximum mean is

98 / 7 = 14

.

Dividing the maximum possible sum by the total number of elements yields the maximum mean.

Key Concept

Mean, Median, and Mode Constraints

Question 20Question →

A dataset consists of 7 integers listed in ascending order: $3, 4, 4, x, 9, y, 15$ . If the unique mode of the dataset is $4$ and the arithmetic mean of the dataset is equal to its median, what is the value of $y$ ?

Show answer & explanation

Answer: 13

Answer

The value of

y

is 13.

Because the dataset is in ascending order, the median of the 7 terms is the 4th term,

x

. Setting the mean

\frac{35 + x + y}{7}

equal to

x

yields

y = 6x - 35

. Since

y

lies between

9

and

15

inclusive, solving

9 \le 6x - 35 \le 15

gives

x = 8

. Substituting

x = 8

into

y = 6x - 35

gives

y = 13

.

Step-by-Step Solution

1

Determine the median of the 7-element ordered set.

The median is the 4th term,

x

.

For an odd number of ordered terms (

n = 7

), the median is the

\frac{7+1}{2} = 4\text{th}

term.

2

Set up the equation equating the arithmetic mean to the median.

\frac{35 + x + y}{7} = x

, which simplifies to

y = 6x - 35

.

The mean of a dataset is the sum of all elements divided by the total number of elements.

3

Apply the ascending order constraint to find the value of

x

.

Since

9 \le y \le 15

, we have

9 \le 6x - 35 \le 15

, giving

44 \le 6x \le 50

. The only integer value for

x

in this range is

8

.

The dataset is listed in ascending order, so the element

y

after

9

and before or equal to

15

must satisfy

9 \le y \le 15

.

4

Calculate the value of

y

.

y = 6(8) - 35 = 13

.

Substitute

x = 8

back into the simplified linear equation relating

x

and

y

.

Key Concept

Relating arithmetic mean and median in an ordered set using inequalities derived from ascending order constraints.

Estimated Time:1m 40s

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