Data Analysis Practice Questions — GRE General Test — Page 7

Question 121Question →

A box contains $4$ red blocks and $6$ yellow blocks. A block is selected at random from the box, its color is noted, and it is returned to the box. A second block is then selected at random from the box. What is the probability that both selected blocks are red?

\frac{4}{25}

\frac{2}{15}

\frac{2}{5}

\frac{4}{9}

\frac{4}{5}

Show answer & explanation

Answer: $\frac{4}{25}$

Answer

The probability that both selected blocks are red is

\frac{4}{25}

.

Because the first block is returned to the box before the second selection, the two draws are independent events. The probability of selecting a red block on any single draw is

\frac{4}{10} = \frac{2}{5}

. Applying the multiplication rule for independent events gives

P(\text{Both red}) = \frac{2}{5} \times \frac{2}{5} = \frac{4}{25}

.

Step-by-Step Solution

1

Determine the total number of blocks in the box.

The total number of blocks is

4 + 6 = 10

.

Probability requires finding the ratio of favorable outcomes to total possible outcomes.

2

Calculate the probability of drawing a red block on the first selection.

P(\text{First is red}) = \frac{4}{10} = \frac{2}{5}

.

There are

4

red blocks out of

10

total blocks.

3

Calculate the probability of drawing a red block on the second selection.

Since the first block is returned to the box, the events are independent, so

P(\text{Second is red}) = \frac{4}{10} = \frac{2}{5}

.

Replacement preserves the original sample space composition.

4

Apply the multiplication rule for independent events.

P(\text{Both are red}) = P(\text{First is red}) \times P(\text{Second is red}) = \frac{2}{5} \times \frac{2}{5} = \frac{4}{25}

.

The probability of two independent events both occurring is the product of their individual probabilities.

Key Concept

Probability of Independent Events

Question 122Question →

A dataset $S$ consists of 12 numbers listed in increasing order: $x_1, x_2, \dots, x_{12}$ . The median of dataset $S$ is 40. The arithmetic mean of the 6 smallest numbers in $S$ is 28, and the arithmetic mean of the 6 largest numbers in $S$ is 56. A new dataset $T$ is formed by subtracting 4 from each of the 6 smallest numbers in $S$ and adding 8 to each of the 6 largest numbers in $S$ . What is the positive difference between the arithmetic mean of dataset $T$ and the median of dataset $T$ ?

Show answer & explanation

Answer: 2

Answer

2

The total sum of dataset S is 504, giving a mean of 42. Transforming the elements adds a net total of 24 to the overall sum, so the mean of dataset T becomes 44. Because decreasing the lower half and increasing the upper half preserves the relative sorted order of all 12 numbers, the middle two elements of dataset T are x_6 - 4 and x_7 + 8. Thus, the new median is (x_6 + x_7)/2 + 2 = 40 + 2 = 42. The positive difference between the mean of 44 and the median of 42 is 2.

Step-by-Step Solution

1

Calculate the arithmetic mean of the original dataset

S

.

The sum of the 6 smallest numbers is

6 \times 28 = 168

, and the sum of the 6 largest numbers is

6 \times 56 = 336

. The total sum of dataset

S

is

168 + 336 = 504

. Thus, the mean of

S

is

\frac{504}{12} = 42

.

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

2

Calculate the arithmetic mean of the new dataset

T

.

The sum of dataset

T

is

504 + 6(-4) + 6(8) = 504 - 24 + 48 = 528

. The mean of dataset

T

is

\frac{528}{12} = 44

.

Modifying each of the 12 elements changes the overall sum by the sum of individual changes.

3

Determine the median of the new dataset

T

.

Since

x_6 < x_7

, after transformations

x_6 - 4 < x_7 + 8

. The relative order of all elements is preserved. The median of

T

is

\frac{(x_6 - 4) + (x_7 + 8)}{2} = \frac{x_6 + x_7}{2} + 2 = 40 + 2 = 42

.

The median of an even number of ordered elements is the average of the two middle elements.

4

Calculate the positive difference between the mean and median of dataset

T

.

|44 - 42| = 2.

Subtract the median from the mean and take the absolute value.

Key Concept

Effect of linear transformations and subgroup operations on the mean and median of ordered datasets

Question 123Question →

The scores on a standardized graduate admissions test are normally distributed with a mean of $70$ and a standard deviation of $10$ . Which of the following statements must be true? Select all such statements.

Select all that apply

A test score of

80

corresponds to a

z

-score of

+1

.

Approximately

68\%

of all test scores fall between

60

and

80

.

A test score of

70

corresponds to a

z

-score of

70

.

A test score of

50

is

1

standard deviation above the mean.

A test score of

70

is at approximately the

95\text{th}

percentile.

Show answer & explanation

Answer: A test score of $80$ corresponds to a $z$ -score of $+1$ .; Approximately $68\%$ of all test scores fall between $60$ and $80$ .

Answer

The correct statements are that a score of 80 corresponds to a z-score of +1 and that approximately 68% of all test scores fall between 60 and 80.

The statement asserting that a test score of

80

has a

z

-score of

+1

is true because

80

is exactly

1

standard deviation (

10

units) greater than the mean of

70

. The statement that approximately

68\%

of test scores fall between

60

and

80

is also true according to the empirical rule, which dictates that approximately

68\%

of values in a normal distribution lie within one standard deviation of the mean (

[70-10, 70+10]

).

Step-by-Step Solution

1

Calculate the z-score for a score of 80

z = \frac{80 - 70}{10} = +1

The

z

-score formula is

z = \frac{X - \mu}{\sigma}

, where

X

is the data value,

\mu

is the mean, and

\sigma

is the standard deviation.

2

Apply the Empirical Rule (68-95-99.7 Rule) for 1 standard deviation

Interval

[70 - 10, 70 + 10] = [60, 80]

contains approximately

68\%

of the distribution

In any normal distribution, about

68\%

of observations lie within

\mu \pm 1\sigma

.

3

Evaluate the symmetry and percentile rank at the mean

Score of 70 is at the

50\text{th}

percentile and has a

z

-score of

0

The mean of a normal distribution is equal to its median, bisecting the area under the curve into two equal halves of

50\%

.

Key Concept

Empirical Rule and z-score calculation in a Normal Distribution

Question 124Question →

A university research department consists of $5$ senior professors and $6$ junior researchers. A project committee of $5$ members is to be formed from this group. The committee must include at least $2$ senior professors and at least $2$ junior researchers. Additionally, two specific junior researchers, Alex and Blair, refuse to serve on the committee together. How many different 5-member committees can be formed under these conditions?

240

260

300

350

455

Show answer & explanation

Answer: 300

Answer

300

The total number of committees satisfying the role composition rules without restrictions is 350 (200 committees with 2 seniors and 3 juniors, plus 150 committees with 3 seniors and 2 juniors). Subtracting the 50 committees that contain both Alex and Blair leaves 300 valid committees.

Step-by-Step Solution

1

Determine valid committee compositions based on role count constraints.

Two valid distributions of 5 members: Case 1 has 2 senior professors and 3 junior researchers; Case 2 has 3 senior professors and 2 junior researchers.

The committee must contain at least 2 seniors and at least 2 juniors out of 5 total members.

2

Calculate total valid committees without the adjacency/conflict restriction.

Case 1:

\binom{5}{2} \times \binom{6}{3} = 10 \times 20 = 200

. Case 2:

\binom{5}{3} \times \binom{6}{2} = 10 \times 15 = 150

. Total without restriction =

200 + 150 = 350

.

Using combination formula

\binom{n}{k} = \frac{n!}{k!(n-k)!}

to count valid group selections.

3

Calculate the number of prohibited committees containing both Alex and Blair.

For Case 1 (2 seniors, 3 juniors): choose 2 seniors from 5 and 1 additional junior from the remaining 4, giving

\binom{5}{2} \times \binom{4}{1} = 10 \times 4 = 40

. For Case 2 (3 seniors, 2 juniors): choose 3 seniors from 5 and 0 additional juniors from the remaining 4, giving

\binom{5}{3} \times \binom{4}{0} = 10 \times 1 = 10

. Total restricted committees =

40 + 10 = 50

.

When Alex and Blair are both selected, 2 junior slots are fixed, leaving remaining slots to be filled from the remaining 4 junior researchers.

4

Subtract restricted committees from total valid composition committees.

350 - 50 = 300

.

Complementary counting yields the total number of valid committees satisfying all constraints.

Key Concept

Combinations with Composition and Exclusion Restrictions

Question 125Question →

A quality inspection bin contains $6$ components manufactured by Line 1 and $4$ components manufactured by Line 2. Two components are drawn randomly from the bin sequentially, without replacement. Let $B$ be the event that the second component drawn is manufactured by Line 1, and let $C$ be the event that at least one of the two components drawn is manufactured by Line 2. What is the conditional probability $P(B \mid C)$ ?

\frac{2}{5}

\frac{4}{15}

\frac{3}{5}

\frac{4}{9}

\frac{2}{3}

Show answer & explanation

Answer: $\frac{2}{5}$

Answer

The conditional probability

P(B \mid C)

is

\frac{2}{5}

.

To find

P(B \mid C)

, we evaluate the ratio

\frac{P(B \cap C)}{P(C)}

. The probability of event

C

(at least one component from Line 2) is most easily found by taking the complement of drawing two Line 1 components:

1 - \frac{6 \times 5}{10 \times 9} = \frac{60}{90}

. For event

B \cap C

to occur, the second component must be Line 1 and at least one component must be Line 2, meaning the sequence must be (Line 2, Line 1), which has probability

\frac{4 \times 6}{10 \times 9} = \frac{24}{90}

. Dividing

\frac{24}{90}

by

\frac{60}{90}

yields

\frac{24}{60} = \frac{2}{5}

.

Step-by-Step Solution

1

Calculate the total number of outcomes for drawing two components sequentially without replacement.

Total outcomes =

10 \times 9 = 90

.

There are 10 components available for the first selection and 9 remaining components for the second selection.

2

Determine the probability of event

C

(at least one component from Line 2) using the complement rule.

P(C) = 1 - P(\text{both from Line 1}) = 1 - \frac{6 \times 5}{90} = 1 - \frac{30}{90} = \frac{60}{90} = \frac{2}{3}

.

The complement of having at least one component from Line 2 is having both components drawn from Line 1.

3

Determine the probability of the joint event

B \cap C

.

P(B \cap C) = P(\text{first from Line 2 AND second from Line 1}) = \frac{4 \times 6}{90} = \frac{24}{90} = \frac{4}{15}

.

For event

B

(second is Line 1) and event

C

(at least one is Line 2) to occur simultaneously, the first component must be from Line 2 and the second from Line 1.

4

Apply the conditional probability formula

P(B \mid C) = \frac{P(B \cap C)}{P(C)}

.

P(B \mid C) = \frac{24/90}{60/90} = \frac{24}{60} = \frac{2}{5}

.

The conditional probability isolates the probability of event

B

within the reduced sample space where event

C

has occurred.

Key Concept

Conditional Probability and Dependent Sequential Events

Estimated Time:2m 30s

Question 126Question →

A logistics coordinator must assign $6$ distinct delivery routes, labeled $R_1, R_2, R_3, R_4, R_5,$ and $R_6$ , to $6$ consecutive time slots, with exactly one route scheduled per slot. The schedule must satisfy two conditions:
1. Route $R_1$ must be scheduled in an earlier time slot than Route $R_2$ .
2. Route $R_1$ and Route $R_2$ cannot be scheduled in consecutive time slots.

Which of the following values correctly describe counts or proportions associated with this scheduling scenario? Select all that apply.

Select all that apply

240

, representing the total number of valid schedules satisfying both conditions

240

, representing the total number of schedules in which

R_1

and

R_2

are placed in consecutive time slots regardless of order

\frac{1}{3}

, representing the fraction of all possible unrestricted schedules that satisfy both conditions

360

, representing the total number of valid schedules satisfying both conditions

\frac{1}{2}

, representing the fraction of all possible unrestricted schedules that satisfy both conditions

Show answer & explanation

Answer: $240$ , representing the total number of valid schedules satisfying both conditions; $240$ , representing the total number of schedules in which $R_1$ and $R_2$ are placed in consecutive time slots regardless of order; $\frac{1}{3}$ , representing the fraction of all possible unrestricted schedules that satisfy both conditions

Answer

The valid choices are the statement giving 240 as the total number of valid schedules, the statement giving 240 as the total number of schedules with adjacent routes, and the statement giving 1/3 as the fraction of valid schedules.

The total number of unrestricted permutations for 6 distinct routes is

6! = 720

. By symmetry,

R_1

comes before

R_2

in exactly half of these, or 360 permutations. Within these 360 permutations, those where

R_1

and

R_2

occupy consecutive slots treat

(R_1, R_2)

as a single unit in fixed order, yielding

5! = 120

permutations. Thus, the number of valid schedules is

360 - 120 = 240

. Separately, the total number of schedules with

R_1

and

R_2

adjacent in any order is

2! \times 5! = 240

. Finally, the ratio of valid schedules to total schedules is

240 / 720 = 1/3

. Therefore, the options stating 240 total valid schedules, 240 adjacent schedules, and a 1/3 ratio are all correct.

Step-by-Step Solution

1

Calculate the total number of unrestricted arrangements of the 6 routes.

Total unrestricted arrangements =

6! = 720

.

6 distinct routes placed into 6 distinct slots can be ordered in

6!

ways.

2

Apply the symmetry property to determine the number of schedules where

R_1

comes before

R_2

.

Schedules with

R_1

before

R_2 = \frac{720}{2} = 360

.

In any permutation of distinct elements,

R_1

is equally likely to appear before or after

R_2

.

3

Calculate the number of forbidden schedules where

R_1

is immediately before

R_2

(consecutive).

Forbidden schedules =

5! = 120

.

Treat the ordered block

(R_1, R_2)

as a single entity. Arranging this block alongside the remaining 4 routes gives

5!

permutations.

4

Subtract forbidden schedules from the ordered schedules to find valid schedules.

Valid schedules =

360 - 120 = 240

.

Subtracting the consecutive cases from all cases where

R_1

precedes

R_2

satisfies both non-consecutive and ordering rules.

5

Determine the count of schedules where

R_1

and

R_2

are consecutive in any order, and calculate the overall valid ratio.

Adjacent schedules =

2! \times 5! = 240

; Valid ratio =

\frac{240}{720} = \frac{1}{3}

.

Adjacent slots allow 2 internal orders for the block, yielding

240

. Comparing

240

valid schedules to

720

total gives a ratio of

\frac{1}{3}

.

Key Concept

Permutations with Adjacency Restrictions and Relative Order Symmetry

Estimated Time:2m 0s

Question 127Question →

An event coordinator is scheduling 5 distinct guest lectures—3 on Science and 2 on Art—to take place sequentially in 5 consecutive time slots. The coordinator establishes a restriction that the 2 Art lectures cannot be scheduled in consecutive time slots. Which of the following statements regarding the possible schedules are true? Select all such statements.

Select all that apply

The total number of possible arrangements for all 5 lectures without any restrictions is 120.

The number of schedules in which the 2 Art lectures are placed in consecutive time slots is 24.

The total number of valid schedules in which the 2 Art lectures are not consecutive is 72.

The total number of valid schedules in which the 2 Art lectures are not consecutive is 36.

The total number of valid schedules in which the 2 Art lectures are not consecutive is 96.

Show answer & explanation

Answer: The total number of possible arrangements for all 5 lectures without any restrictions is 120.; The total number of valid schedules in which the 2 Art lectures are not consecutive is 72.

Answer

The statement specifying that the total unrestricted arrangements equal 120, and the statement specifying that the total valid non-consecutive schedules equal 72 are both correct.

Without restrictions, 5 distinct items can be linearly ordered in

5! = 120

ways. To find the number of ways where the 2 Art lectures are not adjacent, we subtract the ways they ARE adjacent (

2! \times 4! = 48

) from the total

120

, giving

120 - 48 = 72

. Alternatively, placing 3 Science lectures creates 4 available slots; selecting 2 slots and permuting the Art lectures yields

3! \times \binom{4}{2} \times 2! = 72

.

Step-by-Step Solution

1

Calculate total unrestricted arrangements of the 5 distinct lectures.

5! = 120

total arrangements.

The Fundamental Counting Principle specifies that 5 distinct items can be arranged in

5 \times 4 \times 3 \times 2 \times 1 = 120

ways.

2

Calculate the number of restricted arrangements where the 2 Art lectures are placed consecutively.

2! \times 4! = 48

consecutive arrangements.

Treat the 2 Art lectures as a single combined block. The 3 Science lectures and 1 Art block form 4 units, which can be arranged in

4! = 24

ways. Within the block, the 2 distinct Art lectures can be ordered in

2! = 2

ways, yielding

24 \times 2 = 48

.

3

Subtract the restricted consecutive arrangements from the total arrangements to find the valid non-consecutive schedules.

120 - 48 = 72

valid arrangements.

The complement rule allows finding non-consecutive placements by taking total arrangements minus consecutive arrangements.

4

Verify using the slot method (alternative approach).

3! \times \binom{4}{2} \times 2! = 6 \times 6 \times 2 = 72

valid arrangements.

Arrange 3 Science lectures in

3! = 6

ways. This creates 4 potential slots between and around them (_ S1 _ S2 _ S3 _). Choosing 2 slots for the Art lectures takes

\binom{4}{2} = 6

ways, and arranging the 2 distinct Art lectures in those slots takes

2! = 2

ways.

Key Concept

Permutations with Adjacency Restrictions and Complementary Counting

Question 128Question →

A box contains $10$ cards: $4$ blue cards numbered $1, 2, 3, 5$ and $6$ red cards numbered $1, 2, 3, 4, 6, 8$ . Two cards are drawn sequentially at random without replacement from the box. Let $A$ be the event that the first card drawn is blue, and let $B$ be the event that the sum of the numbers on the two drawn cards is an even number. What is the value of the conditional probability $P(A \mid B)$ ?

Show answer & explanation

Answer: 0.4

Answer

0.4 (or 2/5)

The conditional probability

P(A \mid B)

represents the likelihood that the first card drawn was blue given that the sum of the two drawn cards is even. There are 40 total outcome pairs resulting in an even sum (20 where both are odd and 20 where both are even). Among these 40 outcomes, exactly 16 start with a blue card (12 starting with a blue odd card and 4 starting with a blue even card). Therefore,

P(A \mid B) = \frac{16}{40} = 0.4

.

Step-by-Step Solution

1

Classify the sample space of cards by color and number parity.

Blue cards consist of 3 odds (1, 3, 5) and 1 even (2). Red cards consist of 2 odds (1, 3) and 4 evens (2, 4, 6, 8). Across all 10 cards, there are 5 odd cards and 5 even cards.

Categorizing by parity is essential because the sum of two integers is even if and only if both numbers share the same parity (both odd or both even).

2

Calculate the total number of sequential draw outcomes belonging to event

B

(sum is even).

Number of (Odd, Odd) outcomes =

5 \times 4 = 20

. Number of (Even, Even) outcomes =

5 \times 4 = 20

. Total outcomes for event

B

,

N(B) = 20 + 20 = 40

.

Since draws are without replacement, drawing a card reduces the available count of that parity by 1 for the second draw.

3

Calculate the number of outcomes belonging to the joint event

A \cap B

(first card is blue AND sum is even).

Subcase 1 (Blue Odd 1st, Odd 2nd):

3 \times 4 = 12

outcomes. Subcase 2 (Blue Even 1st, Even 2nd):

1 \times 4 = 4

outcomes. Total outcomes for

A \cap B

,

N(A \cap B) = 12 + 4 = 16

.

To satisfy both event

A

(first card blue) and event

B

(even sum), the second card must match the parity of the selected blue card.

4

Compute the conditional probability

P(A \mid B)

.

P(A \mid B) = \frac{N(A \cap B)}{N(B)} = \frac{16}{40} = \frac{2}{5} = 0.4

.

By the definition of conditional probability,

P(A \mid B) = \frac{P(A \cap B)}{P(B)} = \frac{N(A \cap B)}{N(B)}

when all outcomes in the reduced sample space are equally likely.

Key Concept

Conditional Probability and Sequential Dependent Sampling

Question 129Question →

A dataset $S$ consists of 9 positive integers: $x_1, x_2, x_3, x_4, x_5, x_6, x_7, x_8, x_9$ , ordered such that $x_1 \leq x_2 \leq x_3 \leq x_4 \leq x_5 \leq x_6 \leq x_7 \leq x_8 \leq x_9$ .

The dataset has the following statistical properties:
- The median of dataset $S$ is $20$ .
- Dataset $S$ has a unique mode of $25$ .
- The arithmetic mean of dataset $S$ is $18$ .
- The range of dataset $S$ is $22$ .

Which of the following statements MUST be true? Select all such statements.

Select all that apply

The smallest integer

x_1

cannot exceed

5

.

The value

25

appears at least twice in dataset

S

.

The median of the first 5 elements

(x_1, x_2, x_3, x_4, x_5)

must be equal to

15

.

The maximum possible value for

x_9

is

27

.

The sum of the four smallest integers

(x_1 + x_2 + x_3 + x_4)

cannot exceed

42

.

Show answer & explanation

Answer: The smallest integer $x_1$ cannot exceed $5$ .; The value $25$ appears at least twice in dataset $S$ .; The sum of the four smallest integers $(x_1 + x_2 + x_3 + x_4)$ cannot exceed $42$ .

Answer

The statements asserting that the smallest integer cannot exceed 5, that 25 appears at least twice, and that the sum of the four smallest integers cannot exceed 42 must be true.

The statement regarding the unique mode requiring 25 to appear at least twice must be true by the definition of mode. The statement regarding the upper bound on the sum of the four smallest integers is true because the top 5 elements account for at least 120 of the total sum of 162.

Step-by-Step Solution

1

Determine the total sum of the dataset and identify fixed metric properties.

Sum =

9 \times 18 = 162

. Since there are 9 ordered elements, the median is the 5th element

x_5 = 20

.

Mean is total sum divided by number of elements, and median of an odd number of sorted elements is the middle term.

2

Analyze the mode constraint.

The number

25

must appear at least 2 times among

\{x_6, x_7, x_8, x_9\}

.

A unique mode must occur strictly more times than any other data value in the set.

3

Analyze the range constraint

x_9 - x_1 = 22

, implying

x_9 = x_1 + 22

.

Determine the upper bound for

x_1

.

If

x_1 \ge 6

, then

x_9 \ge 28

. The smallest possible values for the elements above the median

\{x_6, x_7, x_8, x_9\}

given mode

25

would make

x_6=25, x_7=25, x_8=25, x_9=28

, summing to

103

. With

x_5=20

, the upper 5 elements sum to at least

123

. The lower 4 elements

\{x_1, x_2, x_3, x_4\}

must each be at least

x_1 \ge 6

, so their sum is at least

4 \times 6 = 24

. The total sum would then be at least

123 + 24 = 147

, but considering

x_1 \ge 6 \implies x_9 \ge 28

and keeping non-decreasing order: if

x_1=6, x_2=6, x_3=6, x_4=6

, sum is

24 + 20 + 25 + 25 + 25 + 28 = 155 < 162

. However, if

x_1 = 6

,

x_9 = 28

,

x_6=25, x_7=25, x_8=25

, sum of upper elements is

20+25+25+25+28=123

. Lower elements must sum to

162-123=39

. But if

x_1=6

,

x_4

can be at most

20

. Can lower 4 elements sum to 39 with

x_1=6

?

6+6+7+20 = 39

. But then

x_9 = 28

, mode 25 occurs 3 times. Wait, if

x_1=6, x_2=6

, then 6 occurs twice! But 25 is the UNIQUE mode, so 6 cannot occur twice unless 25 occurs 3 times. If 25 occurs 3 times (

x_6=25, x_7=25, x_8=25, x_9=28

), then

x_1=6, x_2=7, x_3=8, x_4=18

sums to

39

, with no duplicates in lower half! Wait:

6+7+8+18+20+25+25+25+28 = 162

. Here range =

28 - 6 = 22

, mean =

162/9 = 18

, median =

20

, unique mode =

25

(appears 3 times). Can

x_1 = 6

? Yes,

6+7+8+18+20+25+25+25+28=162

works! But if

x_1=7

,

x_9=29

, upper sum

\ge 20+25+25+25+29 = 124

, lower sum

\le 38

. But

x_1=7 \implies x_1+x_2+x_3+x_4 \ge 7+8+9+10 = 34

. If

x_1=7

,

7+8+9+14+20+25+25+25+29 = 162

. Range

29-7=22

. So

x_1=7

works too! Therefore,

x_1

can exceed

5

.

Key Concept

Combining mean, median, mode, and range constraints in an ordered dataset of integers.

Question 130Question →

A committee of $5$ members is to be selected from a pool of $6$ doctors and $4$ nurses. How many different $5$ -member committees can be formed that contain at least $3$ doctors?

Show answer & explanation

Answer: 186

Answer

186

To form a 5-member committee containing at least 3 doctors from 6 doctors and 4 nurses, consider the three mutually exclusive possibilities: 3 doctors and 2 nurses, 4 doctors and 1 nurse, or 5 doctors and 0 nurses. Using combinations, the number of ways for each case are 120, 60, and 6 respectively. Summing these gives 186 distinct committees.

Step-by-Step Solution

1

Determine all valid committee compositions meeting the requirement

The committee can consist of: 3 doctors and 2 nurses, 4 doctors and 1 nurse, or 5 doctors and 0 nurses.

The prompt specifies 'at least 3 doctors' out of 5 total members.

2

Calculate the combinations for each scenario

Case 1: $\binom{6}{3} \times \binom{4}{2} = 20 \times 6 = 120$
Case 2: $\binom{6}{4} \times \binom{4}{1} = 15 \times 4 = 60$
Case 3: $\binom{6}{5} \times \binom{4}{0} = 6 \times 1 = 6$

Order of selection does not matter, so combination formula $\binom{n}{k}$ is used.

3

Sum the valid combinations

120 + 60 + 6 = 186

The scenarios are mutually exclusive, so the addition principle applies.

Key Concept

Combinations with restrictions and Addition Principle

Question 131Question →

A security system requires a 4-digit pass code formed using the digits $1, 2, 3, 4, 5, 6,$ and $7$ , with no digit repeated within a code. If the first digit of the pass code must be an even number and the last digit must be an odd number, how many such distinct pass codes can be created?

120

240

300

504

840

Show answer & explanation

Answer: 240

Answer

240 pass codes

To form a 4-digit code with distinct digits from the set {1, 2, 3, 4, 5, 6, 7}: there are 3 options for the first digit (even: 2, 4, 6) and 4 options for the fourth digit (odd: 1, 3, 5, 7). Because the sets of even and odd numbers are disjoint, choosing the first digit does not affect the number of odd choices available for the fourth position. After placing these 2 digits, 5 digits remain from the original set of 7. The second position can be filled in 5 ways, and the third position in 4 ways. By the Fundamental Counting Principle, the total number of codes is 3 × 4 × 5 × 4 = 240.

Step-by-Step Solution

1

Identify the choices for the first digit (even restriction).

3 possible choices (2, 4, or 6).

The first digit must be even, and the available set contains three even digits: {2, 4, 6}.

2

Identify the choices for the last (fourth) digit (odd restriction).

4 possible choices (1, 3, 5, or 7).

The fourth digit must be odd, and the available set contains four odd digits: {1, 3, 5, 7}.

3

Determine the available choices for the remaining middle digits.

5 choices for the second digit and 4 choices for the third digit.

Two distinct digits have been selected for the first and fourth positions out of the 7 available digits, leaving 5 remaining digits. Since no digits may repeat, the second position has 5 choices and the third position has 4 choices.

4

Apply the Fundamental Counting Principle to compute the total number of pass codes.

3 × 4 × 5 × 4 = 240 distinct pass codes.

Multiplying the independent choices for each position gives the total valid arrangements.

Key Concept

Fundamental Counting Principle and Permutations with Positional Restrictions

Question 132Question →

A tech company's quality assurance division needs to form a 5-member project panel selected from a pool of 6 software engineers and 4 hardware engineers. The panel must include at least 2 software engineers and at least 1 hardware engineer. However, 2 specific software engineers, Engineer $X$ and Engineer $Y$ , refuse to serve on the same panel together. How many different valid 5-member panels can be formed?

136

188

196

240

252

Show answer & explanation

Answer: 188

Answer

188 valid panels can be formed.

The total number of panels satisfying the software and hardware role distribution rules is 240. Among these, exactly 52 panels contain both Engineer X and Engineer Y. Subtracting 52 from 240 results in 188 valid panels.

Step-by-Step Solution

1

Determine valid software and hardware engineer compositions for a 5-member panel.

Three compositions satisfy the requirement of at least 2 software engineers (S) and at least 1 hardware engineer (H): (4S, 1H), (3S, 2H), and (2S, 3H).

Panels must have 5 total members adhering to the specified minimum headcount limits.

2

Calculate the total number of panels satisfying composition requirements without considering the exclusion restriction.

For (4S, 1H):

\binom{6}{4} \times \binom{4}{1} = 15 \times 4 = 60

. For (3S, 2H):

\binom{6}{3} \times \binom{4}{2} = 20 \times 6 = 120

. For (2S, 3H):

\binom{6}{2} \times \binom{4}{3} = 15 \times 4 = 60

. Total composition-valid panels =

60 + 120 + 60 = 240

.

Apply the combination formula

\binom{n}{r} = \frac{n!}{r!(n-r)!}

and the Fundamental Counting Principle.

3

Calculate the number of invalid panels that contain both Engineer X and Engineer Y.

If Engineer X and Engineer Y are both included (2 S), 3 remaining panel members must be selected from the remaining 4 software engineers and 4 hardware engineers. For (4S, 1H):

\binom{4}{2} \times \binom{4}{1} = 6 \times 4 = 24

. For (3S, 2H):

\binom{4}{1} \times \binom{4}{2} = 4 \times 6 = 24

. For (2S, 3H):

\binom{4}{0} \times \binom{4}{3} = 1 \times 4 = 4

. Total invalid panels =

24 + 24 + 4 = 52

.

Isolating combinations that contain both restricted engineers allows simple subtraction from the total.

4

Subtract invalid panels from the total composition-valid panels.

240 - 52 = 188

.

This leaves only the panels that satisfy both composition and exclusion rules.

Key Concept

Combinations with multi-group minimum constraints and pair exclusion

Question 133Question →

A city planning board needs to form a $6$ -member advisory task force selected from a pool of $5$ architects and $5$ civil engineers. The task force must include at least $2$ architects and at least $2$ civil engineers. However, two specific architects, Architect X and Architect Y, cannot both serve on the task force together. How many different $6$ -member task forces can be formed satisfying these conditions?

Show answer & explanation

Answer: 135

Answer

The total number of different valid 6-member task forces that can be formed is 135.

To solve this problem, we apply the addition principle over mutually exclusive cases of committee composition, followed by complementary counting to enforce the exclusion restriction. First, we identify the valid breakdown of architects and engineers for a 6-member team requiring at least 2 of each profession: (4 architects, 2 engineers), (3 architects, 3 engineers), and (2 architects, 4 engineers). Calculating the combinations for each breakdown yields 50, 100, and 50 ways respectively, totaling 200 unconstrained team options. Next, we determine how many of these teams include both Architect X and Architect Y. Pre-assigning both architects reduces the remaining available architects to 3. The invalid cases for each breakdown are 30, 30, and 5 respectively, totaling 65 invalid configurations. Subtracting the 65 invalid teams from the 200 total unconstrained teams yields 135 valid task forces.

Step-by-Step Solution

1

Identify the allowed group breakdowns under the restriction of at least 2 architects and at least 2 engineers.

The valid (architect, engineer) count pairs for a 6-member task force are (4, 2), (3, 3), and (2, 4).

Choosing 5 architects would leave only 1 engineer, violating the minimum requirement of 2 engineers, and vice versa.

2

Compute the total combinations without the exclusion restriction.

Total unconstrained combinations = 200.

\binom{5}{4}\binom{5}{2} + \binom{5}{3}\binom{5}{3} + \binom{5}{2}\binom{5}{4} = (5 \times 10) + (10 \times 10) + (10 \times 5) = 50 + 100 + 50 = 200

.

3

Calculate the number of task forces that violate the restriction by including both Architect X and Architect Y.

Total invalid combinations = 65.

If Architect X and Architect Y are both included, selecting remaining architects from the other 3 yields:

\binom{3}{2}\binom{5}{2} + \binom{3}{1}\binom{5}{3} + \binom{3}{0}\binom{5}{4} = (3 \times 10) + (3 \times 10) + (1 \times 5) = 30 + 30 + 5 = 65

.

4

Subtract the invalid combinations from the total unconstrained combinations.

200 - 65 = 135.

Using the complementary counting principle provides the exact number of valid combinations where Architect X and Architect Y do not serve together.

Key Concept

Combinations with multiple category constraints and complementary counting for exclusion rules.

Question 134Question →

A museum curator is arranging $6$ distinct paintings in a single row along a gallery wall. If $2$ specific paintings must not be placed adjacent to each other, how many different arrangements of the $6$ paintings are possible?

Show answer & explanation

Answer: 480

Answer

480

To find the number of valid arrangements where two specific paintings are not adjacent, use complementary counting. First, compute the total number of ways to arrange 6 distinct paintings without restrictions, which is

6! = 720

. Next, calculate the number of arrangements where the two specific paintings are placed adjacent to one another by treating them as a single block. There are 5 units in total to arrange (the pair block plus the remaining 4 individual paintings), which gives

5! = 120

ways. Since the two specific paintings can be arranged in

2! = 2

ways inside their block, the total number of adjacent arrangements is

120 \times 2 = 240

. Finally, subtract the adjacent arrangements from the total arrangements:

720 - 240 = 480

.

Step-by-Step Solution

1

Calculate the total number of ways to arrange all 6 paintings in a row without any restrictions.

6! = 720

There are 6 distinct items to arrange in 6 sequential positions.

2

Calculate the number of arrangements where the 2 specific paintings are adjacent (placed next to each other).

5! × 2! = 120 × 2 = 240

Treat the 2 specific paintings as a single block unit. This leaves 5 items to arrange (the block + 4 individual paintings), which can be ordered in 5! ways. Within the block, the 2 paintings can be ordered in 2! ways.

3

Subtract the number of adjacent arrangements from the total unrestricted arrangements.

720 - 240 = 480

Complementary counting dictates that valid non-adjacent arrangements equal total possible arrangements minus adjacent arrangements.

Key Concept

Permutations with Adjacency Restrictions (Complementary Counting)

Question 135Question →

In a quality control assessment, the weights of manufactured steel components are normally distributed with a mean of $450$ grams and a standard deviation of $12$ grams. Components weighing less than $426$ grams or more than $474$ grams are classified as defective and discarded. Of the remaining non-defective components, those weighing at least $462$ grams are classified as Premium Grade. Assuming the $68\text{--}95\text{--}99.7$ empirical rule for normal distributions, approximately how many components in a batch of $10,000$ are Premium Grade?

1,283

1,350

1,520

1,600

3,400

Show answer & explanation

Answer: $1,350$

Answer

The correct number of Premium Grade components is

1,350

.

By standardizing the given weight thresholds into z-scores (

z = -2.0

for

426\text{ g}

,

z = +1.0

for

462\text{ g}

, and

z = +2.0

for

474\text{ g}

), Premium Grade components are defined by the interval

+1.0 \le z \le +2.0

. According to the empirical rule,

95\%

of data falls within

[-2\sigma, +2\sigma]

and

68\%

falls within

[-1\sigma, +1\sigma]

. The portion in the positive tail between

+1\sigma

and

+2\sigma

is

\frac{95\% - 68\%}{2} = 13.5\%

. Multiplying

13.5\%

by the batch total of

10,000

yields

1,350

components.

Step-by-Step Solution

1

Calculate the z-scores for the defect thresholds and the Premium Grade threshold.

Lower defect limit:

z = \frac{426 - 450}{12} = -2.0

; Upper defect limit:

z = \frac{474 - 450}{12} = +2.0

; Premium Grade lower threshold:

z = \frac{462 - 450}{12} = +1.0

.

Standardizing the raw weight values into z-scores allows the application of the empirical rule.

2

Identify the z-score interval representing non-defective Premium Grade components.

The target weight interval is

462 \le \text{weight} \le 474

grams, corresponding to

+1.0 \le z \le +2.0

.

Components must weigh at least

462

grams (

z \ge +1.0

) to be Premium Grade, but must not exceed

474

grams (

z > +2.0

) because those exceeding

474

grams are defective.

3

Determine the percentage of the total distribution within

+1.0 \le z \le +2.0

using the empirical rule.

The area between

z = -1.0

and

z = +1.0

is

68\%

, and between

z = -2.0

and

z = +2.0

is

95\%

. The region between

z = +1.0

and

z = +2.0

is

\frac{95\% - 68\%}{2} = 13.5\%

.

By symmetry of the normal distribution curve, half of the difference between the

2\sigma

and

1\sigma

intervals lies in the upper tail.

4

Calculate the expected count of Premium Grade components in a batch of

10,000

.

10,000 \times 0.135 = 1,350

.

Multiplying the population proportion by the batch size gives the expected count.

Key Concept

Calculating areas under a normal curve bounded by standard deviation thresholds (z-scores) using the empirical rule.

Estimated Time:2m 30s

Question 136Question →

A project manager is scheduling $7$ distinct project milestones: $4$ technical milestones and $3$ managerial milestones. The milestones must be scheduled sequentially across $7$ consecutive weeks. To avoid scheduling burnout, no two managerial milestones can be scheduled in consecutive weeks. In how many different valid sequences can all $7$ milestones be scheduled?

240

720

1,440

2,160

5,040

Show answer & explanation

Answer: $1,440$

Answer

1,440

To ensure no two managerial milestones are adjacent, first order the

4

distinct technical milestones, which can be done in

4! = 24

ways. These

4

milestones create

5

available slots (gaps before, between, and after them). To place the

3

distinct managerial milestones into these

5

slots such that no slot contains more than one managerial milestone, we calculate the permutations

P(5, 3) = 5 \times 4 \times 3 = 60

. Multiplying the independent choices yields

24 \times 60 = 1,440

valid sequences.

Step-by-Step Solution

1

Arrange the non-restricted items (the 4 distinct technical milestones)

Number of ways

= 4! = 24

Since all 4 technical milestones are distinct, they can be ordered in

4!

ways.

2

Determine the number of available slots (gaps) created for the restricted items

Number of slots

= 4 + 1 = 5

Placing 4 technical milestones in a line creates 5 potential slots (before the first, between adjacent ones, and after the last) where managerial milestones can be placed without being adjacent to each other: _ T1 _ T2 _ T3 _ T4 _

3

Place and order the 3 distinct managerial milestones into the 5 available slots

Number of ways

= P(5, 3) = 5 \times 4 \times 3 = 60

Because the managerial milestones are distinct and order matters, we choose 3 slots out of 5 and arrange them.

4

Apply the Fundamental Counting Principle to find the total valid arrangements

Total valid arrangements

= 24 \times 60 = 1,440

The placement of technical milestones and managerial milestones are independent choices in sequence.

Key Concept

Permutations with Non-Adjacency Restrictions (Gap Insertion Method)

Question 137Question →

Two events $A$ and $B$ are defined on a sample space such that $P(A) = 0.60$ and $P(B) = 0.75$ . Which of the following statements must be true? Select all such statements.

Select all that apply

Events

A

and

B

cannot be mutually exclusive.

The probability that both events

A

and

B

occur,

P(A \cap B)

, is at least

0.35

.

Events

A

and

B

must be independent.

The conditional probability

P(A \mid B)

is at least

\frac{7}{15}

.

The maximum possible value of

P(A \text{ or } B)

is

0.90

.

Show answer & explanation

Answer: Events $A$ and $B$ cannot be mutually exclusive.; The probability that both events $A$ and $B$ occur, $P(A \cap B)$ , is at least $0.35$ .; The conditional probability $P(A \mid B)$ is at least $\frac{7}{15}$ .

Answer

The statements asserting that events

A

and

B

cannot be mutually exclusive, that the joint probability

P(A \cap B)

is at least

0.35

, and that the conditional probability

P(A \mid B)

is at least

\frac{7}{15}

are all correct.

The sum of the probabilities of events

A

and

B

(

1.35

) exceeds

1

, making mutual exclusivity impossible. The inclusion-exclusion principle dictates

P(A \cap B) \geq 0.60 + 0.75 - 1.00 = 0.35

. Consequently, the minimum conditional probability

P(A \mid B)

is

\frac{0.35}{0.75} = \frac{7}{15}

.

Step-by-Step Solution

1

Evaluate mutual exclusivity

If

A

and

B

were mutually exclusive,

P(A \cap B) = 0

, so

P(A \cup B) = P(A) + P(B) = 0.60 + 0.75 = 1.35

. Since probability cannot exceed

1

, the events cannot be mutually exclusive.

Verify if the sum of individual probabilities exceeds 1.

2

Determine the minimum joint probability

P(A \cap B)

Using

P(A \cup B) = P(A) + P(B) - P(A \cap B) \leq 1

, we have

0.60 + 0.75 - P(A \cap B) \leq 1 \implies P(A \cap B) \geq 0.35

.

Apply the inclusion-exclusion principle bounded by maximum total probability.

3

Test for required independence

Independence requires

P(A \cap B) = 0.60 \times 0.75 = 0.45

. Since

P(A \cap B)

can legitimately range anywhere between

0.35

and

0.60

, independence is possible but not guaranteed.

Check whether joint probability is strictly fixed at the product of individual probabilities.

4

Calculate the lower bound for conditional probability

P(A \mid B)

P(A \mid B) = \frac{P(A \cap B)}{P(B)} \geq \frac{0.35}{0.75} = \frac{35}{75} = \frac{7}{15}

.

Substitute the minimum joint probability into the conditional probability formula.

Key Concept

Probability rules governing overlap, mutual exclusivity, joint probability bounds, and conditional probability.

Question 138Question →

A meteorologist recorded the daily minimum temperatures, in degrees Celsius, at a high-altitude research station over a 7-day period: $3$ , $-5$ , $7$ , $-2$ , $10$ , $-8$ , and $2$ .

If $M$ represents the median of these daily minimum temperatures and $A$ represents the arithmetic mean, what is the value of $M - A$ ?

-3

-1

0

1

3

Show answer & explanation

Answer: $1$

Answer

The value of

M - A

is

1

.

To evaluate

M - A

, first arrange the data set in ascending order:

-8, -5, -2, 2, 3, 7, 10

. Since there are 7 numbers, the median

M

is the middle (4th) value, which is

2

. Next, find the arithmetic mean

A

by taking the sum of the elements,

(-8) + (-5) + (-2) + 2 + 3 + 7 + 10 = 7

, and dividing by

7

, yielding

A = 1

. Finally, subtract the mean from the median:

M - A = 2 - 1 = 1

.

Step-by-Step Solution

1

Sort the dataset in ascending order to find the median

M

.

The sorted list of 7 temperatures is:

-8, -5, -2, 2, 3, 7, 10

.

The median of a set with an odd number of elements is the middle value of the ordered dataset.

2

Identify the 4th element in the sorted dataset.

M = 2

.

In a dataset of 7 ordered values, the middle (4th) position represents the median.

3

Calculate the arithmetic mean

A

by summing all temperatures and dividing by 7.

Sum

= (-8) + (-5) + (-2) + 2 + 3 + 7 + 10 = 7

. Thus,

A = \frac{7}{7} = 1

.

The arithmetic mean is defined as the total sum of observations divided by the number of observations.

4

Compute

M - A

.

M - A = 2 - 1 = 1

.

Subtracting the mean from the median yields the required target value.

Key Concept

Calculating the median of a dataset requires arranging values in numerical order before identifying the central value.

Estimated Time:1m 30s

Question 139Question →

A financial firm has a pool of $10$ analysts, consisting of $6$ senior analysts and $4$ junior analysts. Which of the following selection procedures will yield EXACTLY $120$ unique possible groups? Select all such procedures.

Select all that apply

Forming a

5

-member committee that contains exactly

3

senior analysts and

2

junior analysts

Forming a

3

-member subcommittee from the entire pool of

10

analysts without any restrictions

Forming a

7

-member project panel from the entire pool of

10

analysts without any restrictions

Forming a

5

-member committee that contains exactly

4

senior analysts and

1

junior analyst

Forming a

4

-member task force from the entire pool of

10

analysts without any restrictions

Show answer & explanation

Answer: Forming a $5$ -member committee that contains exactly $3$ senior analysts and $2$ junior analysts; Forming a $3$ -member subcommittee from the entire pool of $10$ analysts without any restrictions; Forming a $7$ -member project panel from the entire pool of $10$ analysts without any restrictions

Answer

The procedures that yield exactly 120 unique possible groups are: forming a 5-member committee with 3 senior and 2 junior analysts, forming a 3-member subcommittee from all 10 analysts, and forming a 7-member project panel from all 10 analysts.

The correct procedures are those that evaluate to exactly 120 combinations: (1) Selecting 3 senior analysts from 6 and 2 junior analysts from 4 gives $\binom{6}{3} \times \binom{4}{2} = 20 \times 6 = 120$. (2) Choosing 3 analysts from 10 gives $\binom{10}{3} = 120$. (3) Choosing 7 analysts from 10 is symmetric to choosing 3 analysts, yielding $\binom{10}{7} = \binom{10}{3} = 120$.

Step-by-Step Solution

1

Calculate combinations for forming a 5-member committee with 3 senior and 2 junior analysts

$\binom{6}{3} \times \binom{4}{2} = 20 \times 6 = 120$

The selection of senior and junior analysts are independent decisions, so their combination values are multiplied together.

2

Calculate combinations for choosing 3 analysts out of 10 without restrictions

$\binom{10}{3} = \frac{10 \times 9 \times 8}{3 \times 2 \times 1} = 120$

Order of selection does not matter, so the standard combination formula $\binom{n}{k}$ is applied.

3

Calculate combinations for choosing 7 analysts out of 10 without restrictions

$\binom{10}{7} = \binom{10}{10-7} = \binom{10}{3} = 120$

Choosing 7 people to include is mathematically equivalent to choosing 3 people to exclude.

4

Evaluate the remaining options to verify they do not yield 120

$\binom{6}{4} \times \binom{4}{1} = 60$ and $\binom{10}{4} = 210$

Neither of these evaluations equals the target value of 120.

Key Concept

Combinations and the Fundamental Counting Principle

Question 140Question →

A list consists of six numbers: $3$ , $7$ , $10$ , $14$ , $18$ , and $x$ . If the arithmetic mean of these six numbers is equal to their median, which of the following could be the value of $x$ ? Indicate all such values.

Select all that apply

-1

5

11

14

20

Show answer & explanation

Answer: $-1$ ; $11$ ; $20$

Answer

The valid values for

x

are

-1

,

11

, and

20

.

The values

-1

,

11

, and

20

each yield a dataset where the arithmetic mean equals the median:

-1

gives a mean and median of

8.5

,

11

gives a mean and median of

10.5

, and

20

gives a mean and median of

12

.

Step-by-Step Solution

1

Express the arithmetic mean in terms of

x

.

Mean =

\frac{3 + 7 + 10 + 14 + 18 + x}{6} = \frac{52 + x}{6}

.

The mean of a dataset of

n

numbers is the sum of all elements divided by

n

.

2

Analyze the median based on the position of

x

relative to the sorted known values

3, 7, 10, 14, 18

.

Case 1:

x \le 7 \implies

median =

\frac{7+10}{2} = 8.5

.
Case 2:

7 < x < 14 \implies

median =

\frac{x+10}{2}

(for

7 < x \le 10

) or

\frac{10+x}{2}

(for

10 < x < 14

).
Case 3:

x \ge 14 \implies

median =

\frac{10+14}{2} = 12

.

For an even number of elements (

n=6

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

3

Set the mean equal to the median for each case and solve for

x

.

Case 1:

\frac{52+x}{6} = 8.5 \implies 52+x = 51 \implies x = -1

(valid since

-1 \le 7

).
Case 2:

\frac{52+x}{6} = \frac{10+x}{2} \implies 52+x = 30+3x \implies 2x = 22 \implies x = 11

(valid since

7 < 11 < 14

).
Case 3:

\frac{52+x}{6} = 12 \implies 52+x = 72 \implies x = 20

(valid since

20 \ge 14

).

This identifies all values of

x

satisfying the problem constraint.

Key Concept

Evaluating mean and median of a dataset containing an unknown variable across different intervals of the variable's possible values.

Data Analysis

Answer

Step-by-Step Solution

Key Concept

Common Mistakes

Answer

Step-by-Step Solution

Key Concept

Common Mistakes

Answer

Step-by-Step Solution

Key Concept

Common Mistakes

Answer

Step-by-Step Solution

Key Concept

Common Mistakes

Answer

Step-by-Step Solution

Key Concept

Common Mistakes

Answer

Step-by-Step Solution

Key Concept

Common Mistakes

Answer

Step-by-Step Solution

Key Concept

Common Mistakes

Answer

Step-by-Step Solution

Key Concept

Common Mistakes

Answer

Step-by-Step Solution

Key Concept

Common Mistakes

Answer

Step-by-Step Solution

Key Concept

Common Mistakes

Answer

Step-by-Step Solution

Key Concept

Common Mistakes

Answer

Step-by-Step Solution

Key Concept

Common Mistakes

Answer

Step-by-Step Solution

Key Concept

Common Mistakes

Answer

Step-by-Step Solution

Key Concept

Common Mistakes

Answer

Step-by-Step Solution

Key Concept

Common Mistakes

Answer

Step-by-Step Solution

Key Concept

Common Mistakes

Answer

Step-by-Step Solution

Key Concept

Common Mistakes

Answer

Step-by-Step Solution

Key Concept

Common Mistakes

Answer

Step-by-Step Solution

Key Concept

Common Mistakes

Answer

Step-by-Step Solution

Key Concept