All practice questions

Question 1741Question →

The quadratic equation $(x - a)(x - b) = c$ , where $a$ , $b$ , and $c$ are real constants with $a < b$ and $c > 0$ , has two real roots $x_1$ and $x_2$ such that $x_1 < x_2$ . Which of the following statements must be true? Indicate all such statements.

Select all that apply

x_1 < a

x_2 > b

The distance between the roots,

x_2 - x_1

, is strictly greater than

b - a

x_1 + x_2 > a + b

The product of the roots,

x_1 x_2

, must be negative

Show answer & explanation

Answer: $x_1 < a$ ; $x_2 > b$ ; The distance between the roots, $x_2 - x_1$ , is strictly greater than $b - a$

Answer

The correct statements are that the smaller root is less than

a

(

x_1 < a

), the larger root is greater than

b

(

x_2 > b

), and the distance between the roots is greater than

b - a

(

x_2 - x_1 > b - a

).

Evaluating

f(x) = (x-a)(x-b) - c

at

x = a

and

x = b

yields negative values (

-c

). Because the parabola opens upward, the graph must cross the x-axis to the left of

a

and to the right of

b

. This establishes

x_1 < a

and

x_2 > b

. Combining these inequalities shows that the distance between the roots

x_2 - x_1

must exceed

b - a

.

Step-by-Step Solution

1

Formulate the quadratic function and analyze its values at key points

Let

f(x) = (x - a)(x - b) - c = 0

. Evaluating at

x = a

and

x = b

gives

f(a) = (0)(a-b) - c = -c < 0

and

f(b) = (b-a)(0) - c = -c < 0

.

Evaluating at

a

and

b

reveals the sign of the function inside the interval

[a, b]

.

2

Use the sign of the leading coefficient and intermediate value properties to locate the roots

The coefficient of

x^2

is

1 > 0

, so the parabola opens upward. Since

f(a) < 0

and

f(b) < 0

, and

f(x) \to +\infty

as

x \to \pm\infty

, the two real roots

x_1

and

x_2

must satisfy

x_1 < a < b < x_2

.

A continuous upward-opening parabola must cross the x-axis outside any interval where its values are negative.

3

Analyze the distance between the roots

Subtracting

x_1 < a

from

x_2 > b

gives

x_2 - x_1 > b - a

.

Since

a

and

b

fall strictly between

x_1

and

x_2

, the distance between

x_1

and

x_2

exceeds the distance between

a

and

b

.

4

Check the remaining statements using Vieta's formulas

Expanding

f(x) = x^2 - (a+b)x + (ab - c) = 0

yields sum of roots

x_1 + x_2 = a + b

and product of roots

x_1 x_2 = ab - c

. Thus

x_1 + x_2 > a+b

is false, and

x_1 x_2 < 0

is not necessarily true.

Vieta's formulas give exact values for the sum and product of roots in terms of coefficients.

Key Concept

Geometric interpretation of quadratic functions and root locations relative to evaluated points

Question 1742Question →

Two industrial processing units, Unit X and Unit Y, process liquid solution at constant rates of $40$ liters per hour and $60$ liters per hour, respectively. Unit X begins processing a $1,100$ -liter batch of solution alone. After $5$ hours, Unit Y is activated and joins Unit X, working simultaneously at their respective constant rates until the entire batch is completely processed. What is the total time, in hours, from when Unit X started processing until the entire batch was completely processed?

9

11

14

16

20

Show answer & explanation

Answer: 14

Answer

14 hours

Unit X works alone for 5 hours, processing

40 \times 5 = 200

liters. This leaves

1,100 - 200 = 900

liters of solution remaining. When Unit Y joins, the units process solution at a joint rate of

40 + 60 = 100

liters per hour. The remaining 900 liters require

\frac{900}{100} = 9

hours of combined work. Summing the 5 hours of solo work and 9 hours of combined work gives a total time of 14 hours.

Step-by-Step Solution

1

Calculate the volume processed by Unit X during its solo operation

40 \text{ liters/hour} \times 5 \text{ hours} = 200 \text{ liters}

Unit X operates alone for the first 5 hours at a constant rate of 40 liters per hour.

2

Determine the remaining batch volume to be processed

1,100 \text{ liters} - 200 \text{ liters} = 900 \text{ liters}

Subtract the volume already completed from the total batch size.

3

Calculate the combined processing rate of Unit X and Unit Y

40 \text{ liters/hour} + 60 \text{ liters/hour} = 100 \text{ liters/hour}

Both units work together simultaneously after the first 5 hours.

4

Find the time needed for both units to process the remaining volume

\frac{900 \text{ liters}}{100 \text{ liters/hour}} = 9 \text{ hours}

Divide the remaining volume by the combined rate.

5

Calculate the total elapsed time from the start

5 \text{ hours (solo)} + 9 \text{ hours (combined)} = 14 \text{ hours}

Add the initial solo work time to the combined work time.

Key Concept

Algebraic modeling of combined work rates with staggered start times

Estimated Time:1m 30s

Question 1743Question →

The function $f$ is defined by $f(t) = 2^t$ for all real numbers $t$ , and the binary operation $\odot$ is defined for all non-zero real numbers $m$ and $n$ by $m \odot n = \frac{m}{n} - \frac{n}{m}$ . What is the value of $f(2) \odot f(-2)$ ?

\frac{255}{16}

\frac{15}{16}

\frac{15}{4}

0

-\frac{255}{16}

Show answer & explanation

Answer: $\frac{255}{16}$

Answer

\frac{255}{16}

Evaluating

f(2)

yields

2^2 = 4

, and evaluating

f(-2)

yields

2^{-2} = \frac{1}{4}

. Substituting these values into the binary operation rule

m \odot n = \frac{m}{n} - \frac{n}{m}

gives

\frac{4}{1/4} - \frac{1/4}{4} = 16 - \frac{1}{16} = \frac{255}{16}

.

Step-by-Step Solution

1

Evaluate the function

f(t) = 2^t

at

t = 2

and

t = -2

.

f(2) = 2^2 = 4

and

f(-2) = 2^{-2} = \frac{1}{2^2} = \frac{1}{4}

.

Negative exponents follow the rule

x^{-a} = \frac{1}{x^a}

.

2

Substitute

m = f(2) = 4

and

n = f(-2) = \frac{1}{4}

into the custom operation definition

m \odot n = \frac{m}{n} - \frac{n}{m}

.

4 \odot \frac{1}{4} = \frac{4}{\frac{1}{4}} - \frac{\frac{1}{4}}{4}

.

Apply the defined binary operation rule.

3

Simplify the complex fraction terms.

\frac{4}{\frac{1}{4}} = 4 \times 4 = 16

and

\frac{\frac{1}{4}}{4} = \frac{1}{4 \times 4} = \frac{1}{16}

.

Dividing by a fraction is equivalent to multiplying by its reciprocal.

4

Subtract the two simplified terms.

16 - \frac{1}{16} = \frac{256}{16} - \frac{1}{16} = \frac{255}{16}

.

Find a common denominator to compute the final value.

Key Concept

Function evaluation with exponent rules combined with custom binary symbol operations

Question 1744Question →

In quadrilateral $ABCD$ , diagonal $AC$ divides the figure into two right triangles, $\triangle ABC$ and $\triangle ACD$ . It is given that $\angle ABC = 90^\circ$ , $\angle ACD = 90^\circ$ , $\angle BAC = 30^\circ$ , and $\angle CAD = 45^\circ$ . If $AB = 6\sqrt{3}$ , what is the perimeter of quadrilateral $ABCD$ ?

18 + 6\sqrt{3} + 6\sqrt{2}

18 + 6\sqrt{3} + 12\sqrt{2}

24 + 6\sqrt{3} + 12\sqrt{2}

30 + 6\sqrt{3} + 12\sqrt{2}

18 + 18\sqrt{3} + 12\sqrt{6}

Show answer & explanation

Answer: $18 + 6\sqrt{3} + 12\sqrt{2}$

Answer

18 + 6\sqrt{3} + 12\sqrt{2}

The correct answer is derived by sequentially calculating the side lengths of the two special right triangles that share segment

AC

. In

\triangle ABC

, the given side

AB = 6\sqrt{3}

is adjacent to the

30^\circ

angle, making it the side opposite

60^\circ

. Using the

1 : \sqrt{3} : 2

ratio yields

BC = 6

and hypotenuse

AC = 12

. In

\triangle ACD

, leg

AC = 12

is adjacent to a

45^\circ

angle, making

\triangle ACD

a

45^\circ-45^\circ-90^\circ

triangle with equal leg

CD = 12

and hypotenuse

AD = 12\sqrt{2}

. Summing the outer edges

AB + BC + CD + AD = 6\sqrt{3} + 6 + 12 + 12\sqrt{2} = 18 + 6\sqrt{3} + 12\sqrt{2}

.

Step-by-Step Solution

1

Analyze right triangle

\triangle ABC

using

30^\circ-60^\circ-90^\circ

special right triangle ratios.

BC = 6

and

AC = 12

In a

30^\circ-60^\circ-90^\circ

triangle, the ratio of sides opposite to

30^\circ : 60^\circ : 90^\circ

is

1 : \sqrt{3} : 2

. Since

AB = 6\sqrt{3}

is opposite

60^\circ

, the shorter leg

BC = \frac{6\sqrt{3}}{\sqrt{3}} = 6

. The hypotenuse

AC = 2 \times BC = 12

.

2

Analyze right triangle

\triangle ACD

using

45^\circ-45^\circ-90^\circ

special right triangle ratios.

CD = 12

and

AD = 12\sqrt{2}

In right triangle

\triangle ACD

with

\angle ACD = 90^\circ

and

\angle CAD = 45^\circ

,

\triangle ACD

is an isosceles right triangle with side ratio

1 : 1 : \sqrt{2}

. Since leg

AC = 12

, leg

CD = 12

and hypotenuse

AD = 12\sqrt{2}

.

3

Sum the lengths of the four outer boundary segments to compute the perimeter of quadrilateral

ABCD

.

Perimeter

= 18 + 6\sqrt{3} + 12\sqrt{2}

Perimeter

= AB + BC + CD + AD = 6\sqrt{3} + 6 + 12 + 12\sqrt{2} = 18 + 6\sqrt{3} + 12\sqrt{2}

.

Key Concept

Special Right Triangles (

30^\circ-60^\circ-90^\circ

and

45^\circ-45^\circ-90^\circ

side ratios)

Question 1745Question →

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 1746Question →

For a nonzero real constant $m$ , the quadratic equation $x^2 - \left(m + \frac{6}{m}\right)x + 6 = 0$ has real roots $a$ and $b$ . If $a^2 + b^2 = 25$ , what is the sum of all possible values of $m^2$ ?

13

25

37

49

5

Show answer & explanation

Answer: $25$

Answer

The sum of all possible values of

m^2

is

25

.

Using Vieta's formulas on

x^2 - \left(m + \frac{6}{m}\right)x + 6 = 0

, the sum of the roots is

a + b = m + \frac{6}{m}

and the product is

ab = 6

. Substituting these into the identity

a^2 + b^2 = (a+b)^2 - 2ab = 25

yields

\left(m + \frac{6}{m}\right)^2 - 12 = 25

, which simplifies to

m^2 + \frac{36}{m^2} = 25

. Setting

u = m^2

gives

u^2 - 25u + 36 = 0

. Since this quadratic has two distinct positive real roots for

u

, the sum of all possible values of

m^2

is given by Vieta's formula as

25

.

Step-by-Step Solution

1

Apply Vieta's formulas to relate the roots

a

and

b

to the coefficients of the given quadratic equation.

a + b = m + \frac{6}{m}

and

ab = 6

.

For any quadratic equation

x^2 - Sx + P = 0

, the sum of the roots is

S

and the product of the roots is

P

.

2

Express

a^2 + b^2

using the algebraic identity for the sum of squares.

a^2 + b^2 = (a + b)^2 - 2ab = \left(m + \frac{6}{m}\right)^2 - 2(6) = \left(m + \frac{6}{m}\right)^2 - 12

.

Rewriting the sum of squares allows substitution of the known sum and product of the roots.

3

Set the expression equal to the given value

25

and expand the square.

m^2 + 2(m)\left(\frac{6}{m}\right) + \frac{36}{m^2} - 12 = 25 \implies m^2 + 12 + \frac{36}{m^2} - 12 = 25 \implies m^2 + \frac{36}{m^2} = 25

.

Expanding

\left(m + \frac{6}{m}\right)^2

yields

m^2 + 12 + \frac{36}{m^2}

, which simplifies directly after subtracting

12

.

4

Formulate a quadratic equation in terms of

u = m^2

.

u + \frac{36}{u} = 25 \implies u^2 - 25u + 36 = 0

.

Multiplying through by

u = m^2

transforms the rational equation into a standard polynomial quadratic equation.

5

Find the sum of all valid solutions for

u = m^2

using Vieta's formulas.

The discriminant of

u^2 - 25u + 36 = 0

is

\Delta = (-25)^2 - 4(1)(36) = 625 - 144 = 481 > 0

, and the product of the roots is

36 > 0

, so both roots

u_1, u_2

are positive real numbers. By Vieta's formulas,

u_1 + u_2 = 25

.

Since both roots

u_1

and

u_2

are positive real numbers, they correspond to valid real values of

m^2

, making their sum equal to

25

.

Key Concept

Quadratic Equations, Vieta's Formulas, and Factoring Substitution

Estimated Time:2m 0s

Question 1747Question →

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 1748Question →

In the $xy$ -plane, the graph of a function $g(x)$ is obtained by shifting the graph of $f(x) = x^2 - 6x + 2$ horizontally to the right by $3$ units and then reflecting the resulting graph across the $x$ -axis. What is the value of $g(1)$ ?

-18

-10

0

6

18

Show answer & explanation

Answer: $-18$

Answer

The value of

g(1)

is

-18

.

Shifting the graph of

f(x)

horizontally to the right by

3

units gives the function

h(x) = f(x - 3)

. Reflecting this graph across the

x

-axis negates the entire function output, yielding

g(x) = -h(x) = -f(x - 3)

. To find

g(1)

, evaluate

-f(1 - 3) = -f(-2)

. Substituting

x = -2

into

f(x) = x^2 - 6x + 2

gives

f(-2) = (-2)^2 - 6(-2) + 2 = 4 + 12 + 2 = 18

. Negating this gives

-18

.

Step-by-Step Solution

1

Determine the algebraic expression for the transformed function

g(x)

in terms of

f(x)

.

g(x) = -f(x - 3)

Shifting a graph horizontally to the right by

3

units replaces

x

with

x - 3

, giving

f(x - 3)

. Reflecting across the

x

-axis negates the entire function, resulting in

-f(x - 3)

.

2

Substitute

x = 1

into the expression for

g(x)

.

g(1) = -f(1 - 3) = -f(-2)

Evaluating

g(1)

requires finding the value of

-f(x - 3)

when

x = 1

.

3

Evaluate

f(-2)

using the original definition

f(x) = x^2 - 6x + 2

.

f(-2) = (-2)^2 - 6(-2) + 2 = 4 + 12 + 2 = 18

Substituting

-2

into

f(x)

follows standard algebraic order of operations.

4

Apply the outer negative sign to calculate

g(1)

.

g(1) = -18

Since

g(1) = -f(-2)

and

f(-2) = 18

,

g(1) = -18

.

Key Concept

Graph Transformations of Functions

Question 1749Question →

In the $xy$ -plane, line $k$ is defined by the equation $3x - 4y = 12$ . Line $m$ is parallel to line $k$ , and the perpendicular distance between line $k$ and line $m$ is $5$ units. If the $y$ -intercept of line $m$ is greater than the $y$ -intercept of line $k$ , what is the $y$ -intercept of line $m$ ?

Show answer & explanation

Answer: 3.25

Answer

3.25

Rewriting line

k

as

3x - 4y - 12 = 0

shows its

y

-intercept is

-3

. Line

m

is parallel, so its equation is

3x - 4y + C = 0

. Using the formula for perpendicular distance between parallel lines

d = \frac{|C_1 - C_2|}{\sqrt{A^2 + B^2}}

, we have

\frac{|-12 - C|}{\sqrt{3^2 + (-4)^2}} = 5

. This simplifies to

|-12 - C| = 25

, giving

C = 13

or

C = -37

. Setting

x = 0

for line

m

gives

y = -\frac{C}{-4} = \frac{C}{4}

. For

C = 13

, the

y

-intercept is

\frac{13}{4} = 3.25

. Since

3.25 > -3

, this meets all criteria.

Step-by-Step Solution

1

Find the

y

-intercept of line

k

Line

k

has a

y

-intercept at

(0, -3)

.

Setting

x = 0

in

3x - 4y = 12

gives

-4y = 12 \implies y = -3

.

2

Formulate the general equation for line

m

Line

m

has the equation

3x - 4y + C = 0

.

Parallel lines share the same linear coefficients

A = 3

and

B = -4

.

3

Set up the distance formula between parallel lines

\frac{|-12 - C|}{5} = 5

The distance between

Ax + By + C_1 = 0

and

Ax + By + C_2 = 0

is

d = \frac{|C_1 - C_2|}{\sqrt{A^2 + B^2}}

.

4

Solve for constant

C

C = 13

or

C = -37

|-12 - C| = 25

yields

-12 - C = 25 \implies C = -37

and

-12 - C = -25 \implies C = 13

.

5

Determine the required

y

-intercept

y = 3.25

For

C = 13

, the

y

-intercept is

\frac{13}{4} = 3.25

, which is greater than

-3

.

Key Concept

Perpendicular distance between parallel lines and line intercept calculation

Estimated Time:2m 30s

Question 1750Question →

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 1751Question →

For all non-zero real numbers $x$ and $y$ , the custom binary operation $\star$ is defined by $x \star y = \frac{x}{y} - \frac{y}{x}$ . Which of the following statements must be true for all non-zero real numbers $a$ , $b$ , and $c$ ? Select all that apply.

Select all that apply

a \star b = -(b \star a)

a \star a = 0

(a \star b)^2 = \frac{a^2}{b^2} + \frac{b^2}{a^2} - 2

a \star (b \star c) = (a \star b) \star c

a \star (b + c) = (a \star b) + (a \star c)

Show answer & explanation

Answer: $a \star b = -(b \star a)$ ; $a \star a = 0$ ; $(a \star b)^2 = \frac{a^2}{b^2} + \frac{b^2}{a^2} - 2$

Answer

The statements

a \star b = -(b \star a)

,

a \star a = 0

, and

(a \star b)^2 = \frac{a^2}{b^2} + \frac{b^2}{a^2} - 2

must be true for all non-zero real numbers

a

,

b

, and

c

.

The statements showing anti-commutativity, zero self-operation, and the expanded squared expression are all mathematically valid for all non-zero real numbers upon expanding their definitions using algebra.

Step-by-Step Solution

1

Evaluate

a \star b = -(b \star a)

a \star b = \frac{a}{b} - \frac{b}{a}

and

-(b \star a) = -\left(\frac{b}{a} - \frac{a}{b}\right) = \frac{a}{b} - \frac{b}{a}

.

Verify anti-commutativity property by direct substitution.

2

Evaluate

a \star a = 0

a \star a = \frac{a}{a} - \frac{a}{a} = 1 - 1 = 0

.

Verify identity property for identical inputs.

3

Evaluate

(a \star b)^2

(a \star b)^2 = \left(\frac{a}{b} - \frac{b}{a}\right)^2 = \left(\frac{a}{b}\right)^2 - 2\left(\frac{a}{b}\right)\left(\frac{b}{a}\right) + \left(\frac{b}{a}\right)^2 = \frac{a^2}{b^2} - 2(1) + \frac{b^2}{a^2} = \frac{a^2}{b^2} + \frac{b^2}{a^2} - 2

.

Apply binomial expansion to the squared custom operation.

4

Test associativity statement

a \star (b \star c) = (a \star b) \star c

with a counterexample

Let

a = 4, b = 2, c = 1

. Then

b \star c = \frac{2}{1} - \frac{1}{2} = \frac{3}{2}

.

a \star (b \star c) = 4 \star \frac{3}{2} = \frac{4}{3/2} - \frac{3/2}{4} = \frac{8}{3} - \frac{3}{8} = \frac{55}{24}

. Meanwhile,

a \star b = 4 \star 2 = \frac{4}{2} - \frac{2}{4} = \frac{3}{2}

, and

(a \star b) \star c = \frac{3}{2} \star 1 = \frac{3/2}{1} - \frac{1}{3/2} = \frac{3}{2} - \frac{2}{3} = \frac{5}{6}

. Since

\frac{55}{24} \neq \frac{5}{6}

, associativity fails.

A single counterexample disproves a general identity statement.

5

Test distributivity statement

a \star (b + c) = (a \star b) + (a \star c)

with a counterexample

Let

a = 1, b = 1, c = 1

. Then

a \star (b + c) = 1 \star 2 = \frac{1}{2} - \frac{2}{1} = -\frac{3}{2}

. Meanwhile,

(a \star b) + (a \star c) = (1 \star 1) + (1 \star 1) = 0 + 0 = 0

. Since

-\frac{3}{2} \neq 0

, distributivity fails.

A single counterexample disproves distributivity over addition.

Key Concept

Evaluating algebraic properties and identity statements for custom defined binary operations.

Question 1752Question →

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 1753Question →

In right triangle $ABC$ , the measure of angle $ACB$ is $90^\circ$ . Altitude $CD$ is drawn from vertex $C$ to hypotenuse $AB$ , with point $D$ lying on line segment $AB$ . If $AD = 9$ and $DB = 16$ , what is the perimeter of triangle $ABC$ ?

Show answer & explanation

Answer: 60

Answer

The perimeter of triangle

ABC

is 60.

By the Geometric Mean Theorem for right triangles, the altitude

CD

to hypotenuse

AB

satisfies

CD^2 = AD \cdot DB = 9 \cdot 16 = 144

, giving

CD = 12

. Applying the Pythagorean Theorem to the smaller right triangles

\triangle ADC

and

\triangle BDC

yields

AC = \sqrt{9^2 + 12^2} = 15

and

BC = \sqrt{16^2 + 12^2} = 20

. The hypotenuse

AB = 9 + 16 = 25

. Summing the side lengths gives the perimeter:

15 + 20 + 25 = 60

.

Step-by-Step Solution

1

Calculate the length of altitude

CD

using the Geometric Mean Theorem.

CD = \sqrt{AD \cdot DB} = \sqrt{9 \cdot 16} = \sqrt{144} = 12

In a right triangle, the altitude to the hypotenuse divides the hypotenuse into two segments such that the altitude is the geometric mean of the two segment lengths.

2

Calculate leg

AC

using the Pythagorean Theorem in right triangle

ADC

.

AC = \sqrt{AD^2 + CD^2} = \sqrt{9^2 + 12^2} = \sqrt{81 + 144} = \sqrt{225} = 15

Triangle

ADC

is a right triangle with right angle at

D

(

3-4-5

triple scaled by

3

).

3

Calculate leg

BC

using the Pythagorean Theorem in right triangle

BDC

.

BC = \sqrt{BD^2 + CD^2} = \sqrt{16^2 + 12^2} = \sqrt{256 + 144} = \sqrt{400} = 20

Triangle

BDC

is a right triangle with right angle at

D

(

3-4-5

triple scaled by

4

).

4

Calculate the total perimeter of triangle

ABC

.

Perimeter =

AC + BC + AB = 15 + 20 + (9 + 16) = 15 + 20 + 25 = 60

The perimeter is the sum of the three outer sides of triangle

ABC

.

Key Concept

Right Triangle Altitude Relationships and Pythagorean Triples

Question 1754Question →

A cyclist travels from City $X$ to City $Y$ at a constant speed of $24$ miles per hour. On the return trip from City $Y$ to City $X$ along the exact same route, adverse weather conditions reduce the cyclist's average speed by $25\%$ . If the total time for the entire round trip is $7$ hours, what is the distance, in miles, between City $X$ and City $Y$ ?

73.5

96

72

126

144

Show answer & explanation

Answer: $72$

Answer

The distance between City

X

and City

Y

is

72

miles.

To find the one-way distance

d

, first calculate the return speed:

24 \times 0.75 = 18

mph. Express the total time spent traveling as the sum of the time for each leg:

\frac{d}{24} + \frac{d}{18} = 7

. Combining the fractions over a common denominator of

72

gives

\frac{7d}{72} = 7

, which simplifies to

d = 72

miles.

Step-by-Step Solution

1

Determine the return speed

Return speed =

24 \times (1 - 0.25) = 18

miles per hour.

Adverse weather reduces the outbound speed of

24

mph by

25\%

.

2

Formulate the total time equation in terms of distance

d

\frac{d}{24} + \frac{d}{18} = 7

Time equals distance divided by rate. The sum of the outbound time and return time is

7

hours.

3

Solve the algebraic equation for

d

\frac{3d + 4d}{72} = 7 \implies \frac{7d}{72} = 7 \implies d = 72

Finding a common denominator of

72

allows combining the fractional time expressions.

Key Concept

Distance, Rate, and Time Word Problems

Estimated Time:1m 30s

Question 1755Question →

In the $xy$ -plane, line $L$ passes through the point $(4, 2)$ and has a negative slope $m$ . Line $L$ intersects the positive $x$ -axis at point $P$ and the positive $y$ -axis at point $Q$ . If the area of triangle $OPQ$ , where $O$ is the origin, is equal to $16$ , what is the slope of line $L$ ?

-2

-1

-\frac{1}{2}

-\frac{1}{4}

\frac{1}{2}

Show answer & explanation

Answer: $-\frac{1}{2}$

Answer

The slope of line

L

is

-\frac{1}{2}

.

The line equation in point-slope form is

y - 2 = m(x - 4)

. Setting

y = 0

yields the

x

-intercept

P\left(4 - \frac{2}{m}, 0\right)

, and setting

x = 0

yields the

y

-intercept

Q(0, 2 - 4m)

. The area of right triangle

OPQ

is

\frac{1}{2}\left(4 - \frac{2}{m}\right)(2 - 4m) = 16

. Expanding this equation leads to

4m^2 + 4m + 1 = 0

, which factors as

(2m + 1)^2 = 0

, giving

m = -\frac{1}{2}

.

Step-by-Step Solution

1

Write the point-slope form of line

L

using point

(4, 2)

and slope

m

.

The equation of line

L

is

y - 2 = m(x - 4)

.

Point-slope form allows expressing intercepts directly in terms of the unknown slope

m

.

2

Find the coordinates of the

x

-intercept

P

and

y

-intercept

Q

.

Setting

y = 0

gives

x_P = 4 - \frac{2}{m}

. Setting

x = 0

gives

y_Q = 2 - 4m

.

Intersections with the axes define the base and height of right triangle

OPQ

.

3

Set up the area equation for triangle

OPQ

.

\text{Area} = \frac{1}{2} \cdot x_P \cdot y_Q = \frac{1}{2} \left(4 - \frac{2}{m}\right)(2 - 4m) = 16.

Triangle

OPQ

is a right triangle with base along the

x

-axis and height along the

y

-axis.

4

Solve the algebraic equation for slope

m

.

Expanding gives

\frac{1}{2}\left(8 - 16m - \frac{4}{m} + 8\right) = 16 \implies 16 - 16m - \frac{4}{m} = 32 \implies -16m - 16 - \frac{4}{m} = 0

. Dividing by

-4

and multiplying by

m

yields

4m^2 + 4m + 1 = 0 \implies (2m + 1)^2 = 0 \implies m = -\frac{1}{2}

.

Factoring the perfect square quadratic equation yields the single real value for

m

.

Key Concept

Coordinate Geometry: Intercepts and Area of Triangles

Alternative Method

Use the intercept form of a line

\frac{x}{a} + \frac{y}{b} = 1

. The area condition gives

ab = 32

. Substituting

(4, 2)

gives

\frac{4}{a} + \frac{2}{b} = 1

. Substituting

b = \frac{32}{a}

leads to

\frac{4}{a} + \frac{2a}{32} = 1 \implies \frac{4}{a} + \frac{a}{16} = 1 \implies a^2 - 16a + 64 = 0 \implies (a-8)^2 = 0 \implies a = 8

. Then

b = 4

, and the slope is

m = -\frac{b}{a} = -\frac{4}{8} = -\frac{1}{2}

.

Estimated Time:2m 0s

Question 1756Question →

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 1757Question →

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 1758Question →

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 1759Question →

The quadratic function $f(x) = -2x^2 + kx - 18$ has a maximum value of $14$ , where $k$ is a positive constant. What is the value of $k$ ?

Show answer & explanation

Answer: 16

Answer

The value of

k

is

16

.

The vertex of the parabola

f(x) = -2x^2 + kx - 18

is located at

x = \frac{k}{4}

. Evaluating

f\left(\frac{k}{4}\right)

gives the maximum value

\frac{k^2}{8} - 18

. Setting this expression equal to

14

leads to

\frac{k^2}{8} = 32

, so

k^2 = 256

. Taking the positive root as required by the problem statement yields

k = 16

.

Step-by-Step Solution

1

Find the

x

-coordinate of the vertex of the quadratic function.

For

f(x) = -2x^2 + kx - 18

, we have

a = -2

,

b = k

, and

c = -18

. The vertex occurs at

x = -\frac{b}{2a} = -\frac{k}{2(-2)} = \frac{k}{4}

.

The maximum or minimum of any quadratic function

ax^2 + bx + c

occurs at its vertex, where

x = -\frac{b}{2a}

.

2

Evaluate the function at the vertex to determine the maximum value in terms of

k

.

f\left(\frac{k}{4}\right) = -2\left(\frac{k}{4}\right)^2 + k\left(\frac{k}{4}\right) - 18 = -2\left(\frac{k^2}{16}\right) + \frac{k^2}{4} - 18 = -\frac{k^2}{8} + \frac{k^2}{4} - 18 = \frac{k^2}{8} - 18

.

Substituting the vertex

x

-coordinate into

f(x)

yields the maximum value of the downward-opening parabola.

3

Set the maximum value expression equal to

14

and solve for

k^2

.

\frac{k^2}{8} - 18 = 14 \implies \frac{k^2}{8} = 32 \implies k^2 = 256.

The problem states that the maximum value of

f(x)

is

14

.

4

Solve for the positive constant

k

.

k = \sqrt{256} = 16.

Taking the square root of

256

gives

k = 16

or

k = -16

. Since

k

is given as a positive constant,

k = 16

.

Key Concept

Finding the extreme value of a quadratic function by locating its vertex

Question 1760Question →

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

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