Statistics and Data Analysis Practice Questions — GMAT — Page 3

Set $S$ consists of five distinct integers: $14, 22, 18, 9,$ and $31$ . What is the range of the numbers in Set $S$ ?

Show answer & explanation

Answer: 22

Answer

The range of the numbers in Set

S

is

22

.

To find the range of a set of numbers, subtract the minimum value from the maximum value in the set. For Set

S = \{14, 22, 18, 9, 31\}

, the maximum value is

31

and the minimum value is

9

. Subtracting the minimum from the maximum gives

31 - 9 = 22

.

Step-by-Step Solution

1

Identify the maximum and minimum elements in the given set.

The maximum element is

31

and the minimum element is

9

.

The range of a dataset is defined as the difference between its greatest and least values.

2

Subtract the minimum value from the maximum value.

31 - 9 = 22

.

Applying the formula

\text{Range} = \text{Maximum} - \text{Minimum}

yields the range of the set.

Key Concept

Range of a Numerical Data Set

Question 42Question →

In a corporate firm of $N$ employees, each employee earns a distinct annual salary. An employee earning a salary of $\$78,000$ is at the $60\text{th}$ percentile of all salaries in the firm. The firm subsequently hires $10$ new employees, each of whom earns an annual salary strictly less than $\$78,000$ . If $\$78,000$ is now at the $70\text{th}$ percentile of all salaries in the expanded firm, how many employees were originally in the firm?

30

40

50

70

100

Show answer & explanation

Answer: 30

Answer

30

The percentile rank of a score indicates the percentage of values in the set that are less than or equal to that score. Originally,

60\%

of

N

employees earned

\le \$78,000

, giving

0.60N

employees. Adding

10

employees who all earn under

\$78,000

brings the count of employees earning

\le \$78,000

to

0.60N + 10

, while the total workforce becomes

N + 10

. Since

\$78,000

is at the 70th percentile of the new distribution,

0.60N + 10 = 0.70(N + 10)

. Expanding and solving yields

0.60N + 10 = 0.70N + 7

, so

0.10N = 3

, which gives

N = 30

.

Step-by-Step Solution

1

Define the initial number of employees earning at or below $78,000.

Initially,

0.60N

employees earn

\le \$78,000

.

By definition of percentile rank with distinct values, being at the 60th percentile of

N

scores means

60\%

of the total

N

salaries are less than or equal to

\$78,000

.

2

Determine the new count of employees and the updated count of employees earning at or below $78,000.

New total employees

= N + 10

; new count earning

\le \$78,000

is

0.60N + 10

.

All 10 newly hired employees earn salaries strictly less than

\$78,000

, increasing the count of salaries

\le \$78,000

by 10.

3

Set up an equation using the new percentile rank.

0.60N + 10 = 0.70(N + 10)

The salary

\$78,000

is now at the 70th percentile of the updated total group size of

(N + 10)

.

4

Solve the algebraic equation for

N

.

0.60N + 10 = 0.70N + 7 \implies 3 = 0.10N \implies N = 30

.

Subtracting

0.60N

and

7

from both sides gives

0.10N = 3

, which yields

N = 30

.

Key Concept

Percentile Rank in Expanding Data Sets

Question 43Question →

A dataset contains $20$ distinct test scores arranged in ascending order:

$10, 12, 15, 17, 20, 22, 25, 28, 30, 32, 35, 38, 40, 43, 45, 48, 50, 52, 55, 60$

If the $p\text{th}$ percentile of a dataset of $N$ values is defined as the value at position $k = \frac{p}{100} \times N$ when ordered from least to greatest, what is the $75\text{th}$ percentile of these test scores?

Show answer & explanation

Answer: 45

Answer

The 75th percentile of the given test scores is 45.

The 75th percentile corresponds to position k = (75/100) * 20 = 15 in the ordered list of 20 scores, which is equal to 45.

Step-by-Step Solution

1

Calculate the position index

k

for the

75\text{th}

percentile.

k = \frac{75}{100} \times 20 = 15

The rank formula determines the 1-based index of the target percentile value in an ordered set of size

N = 20

.

2

Locate the

15\text{th}

score in the ordered dataset.

The

15\text{th}

score is

45

.

Counting from the lowest score (

10

at position 1), the

15\text{th}

score in the sequence is

45

.

Key Concept

Percentiles and Quartiles

Question 44Question →

An investment portfolio consists of three asset classes: Equities, Bonds, and Real Estate. The total monetary value of Equities to Bonds is in the ratio $3 : 2$ , and the total value of Bonds to Real Estate is in the ratio $4 : 5$ . Over the past year, Equities yielded an average return of $12\%$ . If the overall weighted average return of the entire portfolio was $10.4\%$ , and the average percentage return of Bonds was $25\%$ higher than the average percentage return of Real Estate, what was the average percentage return of Bonds?

Show answer & explanation

Answer: 10.5

Answer

The average percentage return of Bonds was 10.5%.

The correct return of Bonds is obtained by first unifying the given ratios into a continuous ratio of weights

6 : 4 : 5

for Equities, Bonds, and Real Estate respectively. Substituting these weights and the algebraic relation between Real Estate return (

r

) and Bond return (

1.25r

) into the weighted average expression yields

r = 8.4\%

. Multiplying this by

1.25

gives the bond return of

10.5\%

.

Step-by-Step Solution

1

Determine the combined ratio of portfolio weights for all three asset classes

The ratio Equities : Bonds : Real Estate is 6 : 4 : 5, giving a total weight sum of 15 units.

Since Equities to Bonds is 3:2 (or 6:4) and Bonds to Real Estate is 4:5, the common term Bonds allows combining the ratios into 6 : 4 : 5.

2

Express the unknown asset returns in terms of a single variable

Return of Real Estate =

r\%

, Return of Bonds =

1.25r\%

.

The problem states the return of Bonds is 25% higher than that of Real Estate (

r + 0.25r = 1.25r

).

3

Set up and solve the weighted average formula for the portfolio

r = 8.4\%

.

Using

\frac{6(12) + 4(1.25r) + 5r}{15} = 10.4

, we get

72 + 10r = 156

, leading to

10r = 84

.

4

Calculate the targeted return for Bonds

Return of Bonds =

10.5\%

.

Multiplying

1.25

by

8.4

yields

10.5

.

Key Concept

Combining multiple ratios to establish weighted average component weights and solving multi-variable weighted average equations

Estimated Time:2m 30s

Question 45Question →

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 46Question →

A corporation has three regional branches: Branch X, Branch Y, and Branch Z. The ratio of the number of employees in Branch X to Branch Y is $1 : 3$ . The average monthly salary of employees in Branch Y is $40\%$ higher than the average monthly salary of employees in Branch X. Branch Z has twice as many employees as Branch Y, and the average monthly salary of employees in Branch Z is $x\%$ lower than the combined average monthly salary of employees in Branches X and Y. If the overall average monthly salary across all three branches combined is $\$910$ , and the average monthly salary in Branch X is $\$1{}000$ , what is the value of $x$ ?

35

40

50

70

100

Show answer & explanation

Answer: 50

Answer

50

The combined average salary of Branches X and Y is calculated by weighting Branch X's salary (

\$1{}000

) with 1 unit of weight and Branch Y's salary (

\$1{}400

) with 3 units of weight, giving

\frac{1{}000 + 4{}200}{4} = \$1{}300

. For the entire company of

10

units of employees with an average salary of

\$910

, the total payroll is

10 \times 910 = \$9{}100

. Subtracting the combined payroll of Branches X and Y (

\$5{}200

) leaves

\$3{}900

for Branch Z's

6

units of employees. Thus, Branch Z's average salary is

\frac{3{}900}{6} = \$650

. Comparing

\$650

to

\$1{}300

gives a percentage decrease of

\frac{1{}300 - 650}{1{}300} \times 100\% = 50\%

.

Step-by-Step Solution

1

Determine the average monthly salary of Branch Y.

Branch Y average salary =

1{}000 \times 1.40 = \$1{}400

.

Branch Y's average salary is given as

40\%

higher than Branch X's average salary of

\$1{}000

.

2

Express employee counts in terms of a single variable

n

.

Branch X has

n

employees, Branch Y has

3n

employees, and Branch Z has

2 \times 3n = 6n

employees. Total employees =

n + 3n + 6n = 10n

.

The ratio of Branch X to Branch Y employees is

1:3

, and Branch Z has twice as many employees as Branch Y.

3

Calculate the combined average salary of Branches X and Y.

Combined average salary

S_{XY} = \frac{n(1{}000) + 3n(1{}400)}{n + 3n} = \frac{5{}200n}{4n} = \$1{}300

.

The combined average of two sets is total combined earnings divided by total combined size.

4

Calculate total company payroll and determine Branch Z's average salary.

Total payroll =

10n \times 910 = 9{}100n

. Branch Z payroll =

9{}100n - 5{}200n = 3{}900n

. Average salary for Branch Z

S_Z = \frac{3{}900n}{6n} = \$650

.

Total payroll is total employees times overall mean. Subtracting the combined payroll of X and Y yields Branch Z's payroll.

5

Compute the percentage by which Branch Z's average salary is lower than the combined average salary of Branches X and Y.

x = \frac{1{}300 - 650}{1{}300} \times 100\% = 50\%.

Percentage decrease is calculated as (Base Value - New Value) / Base Value.

Key Concept

Weighted Averages of Combined Sets

Estimated Time:2m 0s

Question 47Question →

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 48Question →

The table below shows the distribution of scores achieved by $50$ candidates on a professional certification assessment:

Score	Number of Candidates
$60$	$4$
$70$	$11$
$80$	$20$
$90$	$10$
$100$	$5$

What is the interquartile range (IQR) of the assessment scores?

10

20

30

35

40

Show answer & explanation

Answer: $20$

Answer

The interquartile range of the assessment scores is

20

.

To find the interquartile range (IQR), first find

Q_1

(

25\text{th}

percentile) and

Q_3

(

75\text{th}

percentile). With

50

total candidates,

Q_1

is situated around position

12.5

, which falls into the score of

70

.

Q_3

is situated around position

37.5

, which falls into the score of

90

. Subtracting

Q_1

from

Q_3

yields

90 - 70 = 20

.

Step-by-Step Solution

1

Calculate the cumulative frequency distribution to find score positions.

Score

60

: candidates

1

to

4

; Score

70

: candidates

5

to

15

; Score

80

: candidates

16

to

35

; Score

90

: candidates

36

to

45

; Score

100

: candidates

46

to

50

. Total

N = 50

.

Cumulative frequencies identify the precise position of ranked scores.

2

Determine the first quartile (

Q_1

), which represents the

25\text{th}

percentile.

The

25\text{th}

percentile corresponds to the

0.25 \times 50 = 12.5\text{th}

position. Looking at the cumulative frequencies, candidate positions

5

through

15

all scored

70

, so

Q_1 = 70

.

The first quartile marks the score boundary below which

25\%

of the dataset falls.

3

Determine the third quartile (

Q_3

), which represents the

75\text{th}

percentile.

The

75\text{th}

percentile corresponds to the

0.75 \times 50 = 37.5\text{th}

position. Candidate positions

36

through

45

all scored

90

, so

Q_3 = 90

.

The third quartile marks the score boundary below which

75\%

of the dataset falls.

4

Compute the interquartile range

(\text{IQR} = Q_3 - Q_1)

.

\text{IQR} = 90 - 70 = 20.

The interquartile range measures the spread of the middle

50\%

of the distribution.

Key Concept

Interquartile Range (IQR) and Quartile Positions in Frequency Distributions

Question 49Question →

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 50Question →

Data Set $A$ consists of $n$ numbers with mean $\mu$ and standard deviation $\sigma$ , where $\sigma > 0$ . Data Set $B$ is created by transforming each element $x$ in Data Set $A$ into $y = 5 - 2x$ . If the variance of Data Set $B$ is equal to $k$ , which of the following expressions represents the standard deviation of Data Set $A$ in terms of $k$ ?

\frac{\sqrt{k}}{2}

-\frac{\sqrt{k}}{2}

\frac{k}{4}

\frac{\sqrt{k - 5}}{2}

\sqrt{\frac{k}{2}}

Show answer & explanation

Answer: $\frac{\sqrt{k}}{2}$

Answer

\frac{\sqrt{k}}{2}

Under a linear transformation

y = ax + b

, constant shifts do not change data dispersion, while multiplying elements by

a

scales the standard deviation by

|a|

and the variance by

a^2

. Here,

y = -2x + 5

, so

a = -2

. The variance of Data Set

B

is

\text{Var}(B) = (-2)^2 \cdot \text{Var}(A) = 4\sigma^2

. Given

\text{Var}(B) = k

, we get

4\sigma^2 = k

, which leads to

\sigma^2 = \frac{k}{4}

. Taking the positive square root yields

\sigma = \frac{\sqrt{k}}{2}

.

Step-by-Step Solution

1

Analyze the linear transformation rule for standard deviation and variance.

For

y = ax + b

,

\text{SD}(y) = |a| \cdot \text{SD}(x)

and

\text{Var}(y) = a^2 \cdot \text{Var}(x)

.

Adding a constant

b

shifts all data points equally without altering dispersion, while multiplying by

a

scales distance from the mean by

|a|

.

2

Apply the transformation

y = -2x + 5

to find the variance of Data Set

B

in terms of

\sigma

.

\text{Var}(B) = (-2)^2 \cdot \text{Var}(A) = 4\sigma^2.

The coefficient of

x

is

a = -2

, so variance scales by

a^2 = (-2)^2 = 4

.

3

Set the variance equal to

k

and solve for the standard deviation

\sigma

of Data Set

A

.

4\sigma^2 = k \implies \sigma^2 = \frac{k}{4} \implies \sigma = \frac{\sqrt{k}}{2}.

Standard deviation is strictly non-negative (

\sigma > 0

), requiring the positive square root.

Key Concept

Linear Transformation of Dispersion Measures

Estimated Time:2m 0s

Question 51Question →

In a dataset of $N$ distinct values arranged in ascending order, the $p\text{th}$ percentile is defined as the value at position $\left\lceil \frac{p}{100} \times N \right\rceil$ .

Initially, a dataset contains $100$ distinct test scores. The $40\text{th}$ percentile score is $S_1$ and the $70\text{th}$ percentile score is $S_2$ . A researcher adds $x$ new scores that are strictly less than $S_1$ , and $y$ new scores that are strictly between $S_1$ and $S_2$ , where $x$ and $y$ are positive integers.

If the $40\text{th}$ percentile score of the expanded dataset remains $S_1$ and the $70\text{th}$ percentile score of the expanded dataset remains $S_2$ , what is the maximum possible value of $x + y$ ?

2

3

4

5

6

Show answer & explanation

Answer: $3$

Answer

The maximum possible value of

x + y

is

3

.

The correct choice is

3

. By setting up the percentile rank equations using the ceiling function, the position of

S_1

in the expanded dataset requires

\lceil 0.4(x + y) \rceil = x

, while the position of

S_2

requires

\lceil 0.7(x + y) \rceil = x + y

. Letting

k = x + y

, the ceiling condition

\lceil 0.7k \rceil = k

implies

k - 1 < 0.7k \le k

, which simplifies to

0.3k < 1

or

k < 3.33

. Since

x

and

y

are positive integers,

k

can take integer values of

2

or

3

. Thus, the maximum possible value of

x + y

is

3

.

Step-by-Step Solution

1

Determine the initial rank positions of

S_1

and

S_2

in the 100-element dataset.

Position of

S_1 = \lceil 0.40 \times 100 \rceil = 40

; Position of

S_2 = \lceil 0.70 \times 100 \rceil = 70

.

Applying the given percentile formula to

N = 100

gives the exact index of each score in ascending order.

2

Express the new rank positions of

S_1

and

S_2

after inserting

x

scores below

S_1

and

y

scores between

S_1

and

S_2

.

In the new dataset of size

100 + x + y

,

S_1

is at position

40 + x

and

S_2

is at position

70 + x + y

.

Adding

x

scores less than

S_1

shifts

S_1

's rank by

x

. Adding

y

scores between

S_1

and

S_2

further shifts

S_2

's rank by

x + y

relative to its original position.

3

Formulate the percentile conditions for the expanded dataset of size

M = 100 + x + y

.

For

S_1

:

\lceil 0.40(100 + x + y) \rceil = 40 + x \implies \lceil 0.4(x + y) \rceil = x

.
For

S_2

:

\lceil 0.70(100 + x + y) \rceil = 70 + x + y \implies \lceil 0.7(x + y) \rceil = x + y

.

The percentile rank of

S_1

must equal its new position

40 + x

, and the percentile rank of

S_2

must equal its new position

70 + x + y

.

4

Solve for the total added elements

k = x + y

.

From

\lceil 0.7k \rceil = k

, we get

k - 1 < 0.7k \le k

, which simplifies to

0.3k < 1 \implies k < \frac{10}{3} \approx 3.33

.

By definition of the ceiling function,

\lceil z \rceil = k \iff k - 1 < z \le k

for integer

k

.

5

Evaluate integer solutions for

k = x + y

given

x, y \ge 1

.

For

k = 2

:

x = \lceil 0.4(2) \rceil = 1

,

y = 1

(valid).
For

k = 3

:

x = \lceil 0.4(3) \rceil = 2

,

y = 1

(valid).
The maximum integer value for

k = x + y

is

3

.

Since

x

and

y

are positive integers,

k \ge 2

. The maximum integer

k < 3.33

is

3

.

Key Concept

Percentile Rank and Indexing with Ceiling Functions

Estimated Time:2m 30s

Question 52Question →

A market research study examined the streaming service subscriptions of $250$ households. Every surveyed household subscribes to at least one of three streaming services: Service X, Service Y, or Service Z.

- $130$ households subscribe to Service X.
- $110$ households subscribe to Service Y.
- $120$ households subscribe to Service Z.
- $40$ households subscribe to both Service X and Service Y.
- $35$ households subscribe to both Service Y and Service Z.
- $45$ households subscribe to both Service X and Service Z.

How many households subscribe to all three streaming services?

Show answer & explanation

Answer: 10

Answer

10 households subscribe to all three streaming services.

According to the Principle of Inclusion-Exclusion for three sets,

|X \cup Y \cup Z| = |X| + |Y| + |Z| - (|X \cap Y| + |Y \cap Z| + |X \cap Z|) + |X \cap Y \cap Z|

. Since every household subscribes to at least one service,

|X \cup Y \cup Z| = 250

. Substituting the values yields

250 = 130 + 110 + 120 - (40 + 35 + 45) + |X \cap Y \cap Z|

, which simplifies to

250 = 240 + |X \cap Y \cap Z|

. Solving for

|X \cap Y \cap Z|

gives

10

.

Step-by-Step Solution

1

Set up the Inclusion-Exclusion formula for three overlapping sets

Total = |X| + |Y| + |Z| - (|X ∩ Y| + |Y ∩ Z| + |X ∩ Z|) + |X ∩ Y ∩ Z|

Because every household subscribes to at least one service, the total number of households equals the union of all three sets.

2

Sum the individual set sizes and the pairwise intersection sizes

Sum of individual sets = 130 + 110 + 120 = 360; Sum of pairwise intersections = 40 + 35 + 45 = 120

Aggregating individual and overlapping counts simplifies substitution.

3

Substitute the values into the formula and solve for the intersection of all three sets

250 = 360 - 120 + |X ∩ Y ∩ Z| => 250 = 240 + |X ∩ Y ∩ Z| => |X ∩ Y ∩ Z| = 10

Subtracting 240 from 250 yields the exact count of households subscribing to all three services.

Key Concept

Inclusion-Exclusion Principle for Three Sets

Estimated Time:1m 30s

Question 53Question →

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 54Question →

A panel of $150$ sommeliers evaluated three vintages of wine: Cabernet, Pinot Noir, and Syrah. Every sommelier rated at least one vintage as Exceptional. Overall, $79$ sommeliers rated Cabernet as Exceptional, $71$ rated Pinot Noir as Exceptional, and $84$ rated Syrah as Exceptional. Exactly $18$ sommeliers rated all three vintages as Exceptional, and exactly $27$ rated ONLY Cabernet as Exceptional. If the total number of sommeliers who rated BOTH Cabernet and Pinot Noir as Exceptional is twice the number of sommeliers who rated ONLY Pinot Noir and Syrah as Exceptional, how many sommeliers rated ONLY Cabernet and Syrah as Exceptional?

Show answer & explanation

Answer: 24

Answer

24

Applying the 3-set Inclusion-Exclusion Principle determines that

48

sommeliers rated exactly two vintages. Combining the Cabernet set total (

79

) with the given ONLY Cabernet count (

27

) establishes that

cp + cs = 34

. Substituting

cp = 2ps - 18

into these two relationships forms a system of linear equations (

3ps + cs = 66

and

2ps + cs = 52

). Solving this system yields

ps = 14

and

cs = 24

sommeliers who rated ONLY Cabernet and Syrah as Exceptional.

Step-by-Step Solution

1

Calculate the sum of all pairwise set overlaps using the Inclusion-Exclusion Principle.

|C \cap P| + |P \cap S| + |C \cap S| = 102

The total union is equal to the sum of individual set sizes minus the sum of pairwise intersections plus the three-set intersection.

2

Determine the sum of regions representing sommeliers who rated exactly two vintages.

cp + ps + cs = 48

Each pairwise intersection consists of an 'exactly two' region plus the 'all three' region (

18

). Subtracting

3 \times 18 = 54

from

102

leaves

48

.

3

Set up linear equations using the given ratio and Cabernet set total.

3ps + cs = 66

and

2ps + cs = 52

Expressing

cp = 2ps - 18

and substituting it into

cp + ps + cs = 48

gives the first equation; substituting into the Cabernet total

27 + cp + cs + 18 = 79

gives the second equation.

4

Solve the system of equations for the target region

cs

.

ps = 14

and

cs = 24

Subtracting the two equations yields

ps = 14

, and substituting back gives

cs = 24

.

Key Concept

Three-set Venn diagram algebraic modeling and inclusion-exclusion principle

Question 55Question →

In a university computer science department, students belong to one of three groups: Undergraduates, Master's students, or Doctoral candidates. The ratio of the number of Undergraduates to Master's students is $3 : 2$ , and the ratio of the number of Master's students to Doctoral candidates is $4 : 1$ . The average number of weekly research hours is $6$ hours for Undergraduates and $14$ hours for Master's students. If the combined average number of weekly research hours across all three groups is $11$ hours, what is the average number of weekly research hours for a Doctoral candidate?

13

18

20

21

29

Show answer & explanation

Answer: 29

Answer

The average number of weekly research hours for a Doctoral candidate is 29 hours.

To find the average weekly research hours for Doctoral candidates, first combine the group ratios. Given

U : M = 3 : 2

and

M : D = 4 : 1

, scale

U : M

to

6 : 4

so that the

M

term is identical in both ratios. This yields a single combined ratio

U : M : D = 6 : 4 : 1

, giving a total of

6 + 4 + 1 = 11

equal parts.

Next, calculate the total research hours across all parts:

6(6) + 4(14) + 1(x) = 36 + 56 + x = 92 + x

. Setting the overall weighted average to

11

yields

\frac{92 + x}{11} = 11

, which simplifies to

92 + x = 121

, so

x = 29

.

Step-by-Step Solution

1

Determine the combined ratio of students across the three groups.

Undergraduates : Master's : Doctoral =

6 : 4 : 1

.

We are given

U : M = 3 : 2

and

M : D = 4 : 1

. To combine these into a single ratio

U : M : D

, scale

U : M

by multiplying by

2

so that the

M

term matches:

U : M = 6 : 4

. Thus,

U : M : D = 6 : 4 : 1

.

2

Express total students and total research hours in terms of a multiplier

k

.

Total students =

11k

; Total hours =

36k + 56k + kx = 92k + kx

.

Let the number of students be

U = 6k

,

M = 4k

, and

D = 1k

. The total research hours contributed by Undergraduates is

6k \times 6 = 36k

, by Master's students is

4k \times 14 = 56k

, and by Doctoral candidates is

1k \times x = kx

.

3

Set up the weighted average formula and solve for

x

.

x = 29

.

The combined average is given by

\frac{\text{Total Hours}}{\text{Total Students}} = 11

. Substituting our values gives

\frac{92k + kx}{11k} = 11 \implies \frac{92 + x}{11} = 11 \implies 92 + x = 121 \implies x = 29

.

Key Concept

Weighted Average and Combined Sets with Compound Ratios

Estimated Time:2m 0s

Question 56Question →

A commercial airline conducted a fleet audit of $180$ transoceanic aircraft to evaluate their emergency navigation systems: Satellite Positioning ( $S$ ), Inertial Guidance ( $I$ ), and Terrestrial Radio Navigation ( $T$ ). The audit revealed the following data:
- $95$ aircraft are equipped with Satellite Positioning.
- $80$ aircraft are equipped with Inertial Guidance.
- $75$ aircraft are equipped with Terrestrial Radio Navigation.
- $30$ aircraft are equipped with Satellite Positioning and Inertial Guidance, but not Terrestrial Radio Navigation.
- $20$ aircraft are equipped with Inertial Guidance and Terrestrial Radio Navigation, but not Satellite Positioning.
- $25$ aircraft are equipped with Satellite Positioning and Terrestrial Radio Navigation, but not Inertial Guidance.
- $15$ aircraft are not equipped with any of these three navigation systems.

How many aircraft in the fleet are equipped with exactly one of these three navigation systems?

70

75

85

95

100

Show answer & explanation

Answer: $85$

Answer

The number of aircraft equipped with exactly one navigation system is

85

.

Subtracting the

15

unequipped aircraft from the total fleet of

180

gives

165

aircraft equipped with at least one navigation system. Letting

x

represent the number of aircraft with all three systems, the single-system counts are

(95 - 30 - 25 - x) = 40 - x

for Satellite Positioning only,

(80 - 30 - 20 - x) = 30 - x

for Inertial Guidance only, and

(75 - 25 - 20 - x) = 30 - x

for Terrestrial Radio Navigation only. Summing all seven mutually exclusive regions yields

(40 - x) + (30 - x) + (30 - x) + 30 + 20 + 25 + x = 165

, which simplifies to

175 - 2x = 165

, giving

x = 5

. Substituting

x = 5

into the single-system expressions gives

35 + 25 + 25 = 85

.

Step-by-Step Solution

1

Calculate the total number of aircraft equipped with at least one system.

180 - 15 = 165

aircraft.

Subtracting the

15

aircraft with no systems from the fleet total gives the union of sets

S

,

I

, and

T

.

2

Express the regions of the 3-set Venn diagram in terms of

x

, where

x

is the number of aircraft equipped with all three systems.

Only

S = 40 - x

, Only

I = 30 - x

, Only

T = 30 - x

.

For set

S

:

95 - (30 + 25 + x) = 40 - x

. For set

I

:

80 - (30 + 20 + x) = 30 - x

. For set

T

:

75 - (25 + 20 + x) = 30 - x

.

3

Set up an equation for the total number of aircraft with at least one system to solve for

x

.

(40 - x) + (30 - x) + (30 - x) + 30 + 20 + 25 + x = 165 \implies 175 - 2x = 165 \implies x = 5

.

Summing all

7

non-overlapping regions inside the Venn diagram equals the total union of

165

.

4

Compute the sum of aircraft equipped with exactly one system.

(Only

S

) + (Only

I

) + (Only

T

) =

(40 - 5) + (30 - 5) + (30 - 5) = 35 + 25 + 25 = 85

.

Substitute

x = 5

into each of the single-system region expressions.

Key Concept

Three-Set Inclusion-Exclusion Principle and Venn Diagram Region Decomposition

Estimated Time:2m 0s

Question 57Question →

A financial advisory board of $6$ members is to be selected from a pool of $6$ senior analysts and $5$ junior analysts. The board must include at least $2$ senior analysts and at least $2$ junior analysts. However, two specific senior analysts, $A$ and $B$ , cannot both serve on the board together. How many different $6$ -member boards can be formed under these conditions?

300

320

325

336

425

Show answer & explanation

Answer: 320

Answer

The total number of valid 6-member boards that can be formed is 320.

To find the number of valid boards, we first calculate all possible 6-member boards satisfying the minimum criteria of having at least 2 senior analysts and at least 2 junior analysts. The possible (senior, junior) distributions are (4,2), (3,3), and (2,4). Calculating each case gives

\binom{6}{4}\binom{5}{2} = 150

,

\binom{6}{3}\binom{5}{3} = 200

, and

\binom{6}{2}\binom{5}{4} = 75

, for a total of

150 + 200 + 75 = 425

boards. Next, we determine how many of these boards contain both senior analysts A and B. Fixing A and B requires choosing 4 more members from the remaining 4 seniors and 5 juniors such that the total junior count is at least 2. The valid remaining senior choices

s'

are 0, 1, or 2, yielding

\binom{4}{0}\binom{5}{4} = 5

,

\binom{4}{1}\binom{5}{3} = 40

, and

\binom{4}{2}\binom{5}{2} = 60

, totaling

5 + 40 + 60 = 105

boards. Subtracting these forbidden boards from 425 yields

425 - 105 = 320

.

Step-by-Step Solution

1

Calculate total valid boards with at least 2 senior analysts and at least 2 junior analysts without restriction on A and B.

Total valid boards without conflict constraint = 425

The valid (Senior, Junior) distributions for a 6-member board are (4,2), (3,3), and (2,4). Their combinations are

\binom{6}{4}\binom{5}{2} = 150

,

\binom{6}{3}\binom{5}{3} = 200

, and

\binom{6}{2}\binom{5}{4} = 75

, giving

150 + 200 + 75 = 425

.

2

Calculate the number of valid boards that contain BOTH senior analysts A and B.

Number of invalid boards containing both A and B = 105

If A and B are both selected, 2 senior spots are fixed. We need 4 remaining members from the remaining 4 seniors and 5 juniors. To satisfy the junior constraint (at least 2 juniors), the additional senior count

s'

can be 0, 1, or 2:

s'=0 \implies \binom{4}{0}\binom{5}{4}=5

,

s'=1 \implies \binom{4}{1}\binom{5}{3}=40

,

s'=2 \implies \binom{4}{2}\binom{5}{2}=60

. Summing gives

5 + 40 + 60 = 105

.

3

Subtract the conflicting boards containing both A and B from the total valid boards.

Valid boards = 425 - 105 = 320

Complementary counting gives the exact number of boards satisfying all conditions.

Key Concept

Combinations with Subgroup Constraints and Complementary Counting

Estimated Time:2m 0s

Question 58Question →

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 59Question →

A 5-digit passcode is to be created using five distinct digits chosen from the set $\{1, 2, 3, 4, 5, 6, 7\}$ . The passcode must include both the digit $3$ and the digit $5$ , and the digit $3$ must appear somewhere to the left of the digit $5$ in the passcode. How many such 5-digit passcodes can be formed?

Show answer & explanation

Answer: 600

Answer

600

To form a valid passcode, first select 3 digits from the 5 available digits

\{1, 2, 4, 6, 7\}

, which can be done in

\binom{5}{3} = 10

ways. Each chosen set of 5 distinct digits (including 3 and 5) can be arranged in

5! = 120

ways. Because the digits 3 and 5 are distinct, the digit 3 appears before the digit 5 in exactly half of these arrangements (

120 / 2 = 60

). Multiplying the 10 combinations of digits by the 60 valid arrangements gives a total of

600

valid passcodes.

Step-by-Step Solution

1

Determine the number of ways to choose the remaining digits.

\binom{5}{3} = 10

ways to choose 3 additional digits from

\{1, 2, 4, 6, 7\}

.

Since the passcode must contain both 3 and 5, 3 additional distinct digits must be selected from the 5 available remaining digits.

2

Calculate the total permutations of the 5 chosen digits.

5! = 120

total permutations.

Any set of 5 distinct digits can be arranged into a 5-digit sequence in

5!

ways.

3

Apply the positional restriction using symmetry.

\frac{120}{2} = 60

valid arrangements per set of digits.

In exactly half of all permutations containing both 3 and 5, the digit 3 appears before the digit 5.

4

Compute the total number of valid passcodes.

10 \times 60 = 600

passcodes.

Multiply the number of digit selections by the number of valid orderings per selection.

Key Concept

Combining selection (combinations) with symmetry-restricted arrangements (permutations)

Estimated Time:2m 0s

Question 60Question →

A security system generates 4-digit passcodes using distinct digits selected from the set $\{1, 2, 3, 4, 5, 6, 7\}$ . How many different 4-digit passcodes can be formed if the first digit must be odd and the last digit must be even?

Show answer & explanation

Answer: 240

Answer

240

To form a valid passcode under the given constraints, we analyze each position step-by-step. The first digit has 4 possible choices (odd numbers: 1, 3, 5, 7), and the fourth digit has 3 possible choices (even numbers: 2, 4, 6). Since the problem specifies that the digits in the passcode must be distinct, choosing the first and last digits consumes 2 of the 7 available digits, leaving 5 digits for the second position and 4 digits for the third position. Applying the Fundamental Counting Principle gives a total of

4 \times 5 \times 4 \times 3 = 240

unique passcodes.

Step-by-Step Solution

1

Determine choices for the first digit

4 choices

The set contains 4 odd digits: 1, 3, 5, and 7.

2

Determine choices for the fourth (last) digit

3 choices

The set contains 3 even digits: 2, 4, and 6.

3

Determine choices for the second and third digits

5 choices for the second digit, 4 choices for the third digit

All digits in the passcode must be distinct. Having used 2 digits for the first and last positions, 5 digits remain out of 7 for the second slot, and 4 digits remain for the third slot.

4

Apply the Fundamental Counting Principle

4 × 5 × 4 × 3 = 240

Multiply the number of available options for each position to find the total number of distinct passcodes.

Key Concept

Permutations with Position Restrictions

Statistics and Data Analysis

Answer

Step-by-Step Solution

Key Concept

Common Mistakes

Answer

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Common Mistakes

Answer

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Common Mistakes

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Common Mistakes

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Common Mistakes

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Common Mistakes

Answer

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Common Mistakes

Answer

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Common Mistakes

Answer

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Common Mistakes

Answer

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Common Mistakes

Answer

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Common Mistakes

Answer

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Common Mistakes

Answer

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Common Mistakes

Answer

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Common Mistakes

Answer

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Common Mistakes

Answer

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Common Mistakes

Answer

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Common Mistakes

Answer

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Common Mistakes

Answer

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Common Mistakes

Answer

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