All practice questions

Question 701Question →

A contractor estimates that $20$ workers, working $8\text{ hours}$ per day, can complete a road construction project in $30\text{ days}$ . All workers work at the same constant rate. After working for $10\text{ days}$ , $4$ workers leave the site. To make up for the loss, the daily working duration for each remaining worker is increased to $10\text{ hours}$ per day. However, due to fatigue, the work efficiency of each remaining worker drops by $20\%$ . How many additional days will be required to complete the remaining work?

Show answer & explanation

Answer: 25

Answer

It will take 25 additional days to complete the remaining work.

The total project requires

4,800

worker-hours (

20 \times 8 \times 30

). In the first

10

days,

1,600

worker-hours are completed, leaving

3,200

worker-hours. After

4

workers leave,

16

workers remain. Working

10

hours a day at

80\%

efficiency, the

16

workers produce

16 \times 10 \times 0.8 = 128

effective worker-hours each day. Dividing the remaining

3,200

worker-hours by

128

gives exactly

25

additional days.

Step-by-Step Solution

1

Calculate total work units required for the entire project

Total work =

4800\text{ worker-hours}

Work rate is proportional to workers multiplied by total hours worked.

2

Calculate completed work and remaining work

Work done =

1600\text{ worker-hours}

, Remaining work =

3200\text{ worker-hours}

Subtracting completed work from total work gives the remaining required effort.

3

Determine the effective rate per day after conditions change

Effective daily output =

128\text{ worker-hours/day}

16 remaining workers working 10 hours daily at 80% efficiency yield

16 \times 10 \times 0.8 = 128

worker-hours per day.

4

Divide remaining work by the new effective daily rate

Number of additional days =

25\text{ days}

Dividing

3200\text{ worker-hours}

by

128\text{ worker-hours/day}

gives

25\text{ days}

.

Key Concept

Compound Proportion and Work Efficiency

Question 702Question →

Given the matrices $A = \begin{pmatrix} 2 & 1 \\ 4 & 3 \end{pmatrix}$ and $B = \begin{pmatrix} 1 & -1 \\ 2 & 5 \end{pmatrix}$ , find the determinant of the matrix $C = 2A - B$ .

Show answer & explanation

Answer: -15

Answer

The determinant of the matrix

C = 2A - B

is

-15

.

Scalar multiplication gives

2A = \begin{pmatrix} 4 & 2 \\ 8 & 6 \end{pmatrix}

. Subtracting

B

yields

C = \begin{pmatrix} 3 & 3 \\ 6 & 1 \end{pmatrix}

. Evaluating the determinant gives

\det(C) = (3)(1) - (3)(6) = 3 - 18 = -15

.

Step-by-Step Solution

1

Multiply matrix

A

by scalar

2

2A = \begin{pmatrix} 4 & 2 \\ 8 & 6 \end{pmatrix}

Scalar multiplication requires multiplying each entry of matrix

A

by

2

.

2

Subtract matrix

B

from matrix

2A

entry-wise to find matrix

C

C = \begin{pmatrix} 4 - 1 & 2 - (-1) \\ 8 - 2 & 6 - 5 \end{pmatrix} = \begin{pmatrix} 3 & 3 \\ 6 & 1 \end{pmatrix}

Subtract corresponding entries of matrix

B

from

2A

.

3

Calculate the determinant of matrix

C

\det(C) = (3)(1) - (3)(6) = 3 - 18 = -15

The determinant of a

2 \times 2

matrix

\begin{pmatrix} a & b \\ c & d \end{pmatrix}

is

ad - bc

.

Key Concept

Matrix Operations and Determinants

Question 703Question →

In a survey of 160 music enthusiasts, 80 listen to Afrobeat, 70 listen to Highlife, and 65 listen to Reggae. It was found that 30 listen to both Afrobeat and Highlife, 25 listen to both Highlife and Reggae, and 28 listen to both Afrobeat and Reggae. If 15 enthusiasts listen to none of these three genres, how many enthusiasts listen to exactly two of these genres?

Show answer & explanation

Answer: 44

Answer

The number of enthusiasts who listen to exactly two of the three genres is 44.

The correct answer is 44. By subtracting the 15 non-listeners from 160, the union of the three sets contains 145 members. Solving the 3-set inclusion-exclusion equation yields 13 enthusiasts who listen to all three genres. Subtracting 13 from each given pairwise intersection gives 17, 12, and 15 for the regions representing listeners of exactly two genres. Summing these three counts yields 44.

Step-by-Step Solution

1

Calculate the cardinality of the union of all three sets.

n(A \cup H \cup R) = 160 - 15 = 145

Subtracting the number of enthusiasts listening to none of the genres from the universal set.

2

Apply the Principle of Inclusion-Exclusion for 3 sets to find the triple intersection.

n(A \cap H \cap R) = 13

Substituting known cardinalities into

n(A \cup H \cup R) = n(A)+n(H)+n(R) - [n(A \cap H)+n(H \cap R)+n(A \cap R)] + n(A \cap H \cap R)

.

3

Calculate the count of enthusiasts in each 'exactly two genres' region.

Afrobeat and Highlife only = 17; Highlife and Reggae only = 12; Afrobeat and Reggae only = 15.

Subtracting the triple intersection count (

13

) from each pairwise intersection.

4

Sum the counts of the three distinct two-genre regions.

17 + 12 + 15 = 44

Combining all mutually exclusive regions representing enthusiasts of exactly two genres.

Key Concept

Principle of Inclusion-Exclusion for three sets and Venn diagram region partitioning.

Question 704Question →

A machine produces electronic components. Based on long-term specifications, the theoretical probability of producing a defective component is $0.04$ . In a quality control inspection, a random sample of $400$ components produced by the machine is tested, and $22$ are found to be defective. What is the absolute difference between the experimental probability and the theoretical probability of selecting a defective component from this sample?

Show answer & explanation

Answer: 0.015

Answer

The absolute difference between the experimental probability and theoretical probability is 0.015.

The experimental probability is calculated from the trial outcomes as 22 divided by 400, which equals 0.055. Taking the absolute difference between this value and the theoretical probability of 0.04 gives |0.055 - 0.04| = 0.015.

Step-by-Step Solution

1

Determine the experimental probability from the sample results

P(Experimental) = \frac{22}{400} = 0.055

Experimental probability is calculated as the ratio of observed successful outcomes (defective items found) to total trials (sample size).

2

Identify the given theoretical probability

P(Theoretical) = 0.04

Theoretical probability represents the expected likelihood under model specifications prior to sampling.

3

Subtract the theoretical probability from the experimental probability to find the absolute difference

|0.055 - 0.04| = 0.015

The absolute difference measures how much the observed empirical frequency deviates from the predicted model probability.

Key Concept

Experimental vs Theoretical Probability

Question 705Question →

A sports delegation of $6$ athletes is to be selected from a pool of $7$ sprinters and $5$ distance runners. If the delegation must contain at least $4$ sprinters, in how many different ways can the delegation be formed?

Show answer & explanation

Answer: 462

Answer

The total number of different ways to form the delegation is 462.

To select at least 4 sprinters in a delegation of 6 athletes, we must evaluate three mutually exclusive cases: selecting 4 sprinters and 2 distance runners (

{}^7\text{C}_4 \times {}^5\text{C}_2 = 350

), selecting 5 sprinters and 1 distance runner (

{}^7\text{C}_5 \times {}^5\text{C}_1 = 105

), and selecting 6 sprinters and 0 distance runners (

{}^7\text{C}_6 \times {}^5\text{C}_0 = 7

). Adding these yields

350 + 105 + 7 = 462

total ways.

Step-by-Step Solution

1

Determine all valid combinations of sprinters and distance runners satisfying the condition of having at least 4 sprinters in a group of 6.

Three valid cases: (4 sprinters, 2 distance runners), (5 sprinters, 1 distance runner), and (6 sprinters, 0 distance runners).

The delegation requires 6 members and at least 4 sprinters.

2

Calculate the combinations for each case using

{}^n\text{C}_r = \frac{n!}{r!(n-r)!}

.

Case 1:

{}^7\text{C}_4 \times {}^5\text{C}_2 = 350

; Case 2:

{}^7\text{C}_5 \times {}^5\text{C}_1 = 105

; Case 3:

{}^7\text{C}_6 \times {}^5\text{C}_0 = 7

.

Apply the product rule of counting for selecting sprinters and distance runners independently within each case.

3

Sum the results of the mutually exclusive cases.

350 + 105 + 7 = 462.

Apply the addition principle of counting for mutually exclusive scenarios.

Key Concept

Combinations with restrictions and the addition principle of counting

Estimated Time:2m 0s

Question 706Question →

A normal line is drawn to the curve $y = \frac{x^2 + 3}{x - 1}$ at the point where $x = 2$ . Calculate the $x$ -intercept of this normal line.

Show answer & explanation

Answer: -19

Answer

The x-intercept of the normal line is -19.

Substituting

x = 2

into the curve function gives

y = 7

. Differentiating via the quotient rule yields

\frac{dy}{dx} = \frac{x^2 - 2x - 3}{(x - 1)^2}

, which equals

-3

at

x = 2

. The perpendicular slope of the normal line is

\frac{1}{3}

. Constructing the line equation through

(2, 7)

gives

x - 3y + 19 = 0

. Setting

y = 0

produces

x = -19

.

Step-by-Step Solution

1

Find the point of contact by substituting

x = 2

into the curve equation

y = \frac{2^2 + 3}{2 - 1} = \frac{7}{1} = 7

, so the point is

(2, 7)

The line is drawn at

x = 2

, so we need the full coordinate pair

(x_1, y_1)

2

Differentiate the curve

y = \frac{x^2 + 3}{x - 1}

using the quotient rule

\frac{dy}{dx} = \frac{(x - 1)(2x) - (x^2 + 3)(1)}{(x - 1)^2} = \frac{x^2 - 2x - 3}{(x - 1)^2}

The derivative gives the gradient function of the curve

3

Evaluate the tangent slope

m_t

at

x = 2

m_t = \frac{2^2 - 2(2) - 3}{(2 - 1)^2} = \frac{-3}{1} = -3

Evaluating the derivative yields the gradient of the tangent at the point

4

Determine the slope of the normal line

m_n

m_n = -\frac{1}{m_t} = -\frac{1}{-3} = \frac{1}{3}

Normal lines are perpendicular to tangent lines, so

m_n \cdot m_t = -1

5

Formulate the equation of the normal line passing through

(2, 7)

y - 7 = \frac{1}{3}(x - 2) \implies 3y - 21 = x - 2 \implies x - 3y + 19 = 0

Use the point-slope form

y - y_1 = m(x - x_1)

6

Find the

x

-intercept by setting

y = 0

x - 3(0) + 19 = 0 \implies x = -19

The

x

-intercept occurs where the line crosses the

x

-axis (

y = 0

)

Key Concept

Finding the equation and x-intercept of a normal line to a curve

Estimated Time:2m 0s

Question 707Question →

A continuous grouped frequency distribution consists of four class intervals: $10 - 14$ , $15 - 24$ , $25 - 29$ , and $30 - 44$ . The total frequency of the distribution is $160$ , and the frequency of the class interval $25 - 29$ is $22$ .

In a histogram representing this data, the height of the rectangle for the interval $15 - 24$ corresponds to a frequency density of $6$ . In a pie chart representing the same distribution, the sector angle for the class interval $30 - 44$ is $108^\circ$ .

What is the frequency density of the class interval $10 - 14$ ?

Show answer & explanation

Answer: 6

Answer

The frequency density of the class interval

10 - 14

is

6

.

By converting the pie chart sector angle of

108^\circ

into a frequency of

48

out of

160

, and using the frequency density of

6

with class width

10

to find a frequency of

60

for

15 - 24

, the remaining frequency for

10 - 14

is

30

. Dividing this by the true class width of

5

(from boundaries

9.5

to

14.5

) gives a frequency density of

6

.

Step-by-Step Solution

1

Calculate the frequency of the class interval

30 - 44

from the pie chart sector angle.

Frequency

f_{30-44} = \frac{108^\circ}{360^\circ} \times 160 = 48

.

The sector angle in a pie chart is directly proportional to the frequency relative to the

360^\circ

total.

2

Determine the class width and frequency of the class interval

15 - 24

.

Class boundaries are

14.5

and

24.5

, so width

w = 10

. Frequency

f_{15-24} = 6 \times 10 = 60

.

Frequency density is defined as frequency divided by class width, so frequency equals frequency density multiplied by class width.

3

Determine the frequency of the class interval

10 - 14

.

Frequency

f_{10-14} = 160 - (60 + 22 + 48) = 30

.

The sum of all class frequencies must equal the total frequency of

160

.

4

Calculate the class width and frequency density of

10 - 14

.

Class width

w_{10-14} = 14.5 - 9.5 = 5

. Frequency density

= \frac{30}{5} = 6

.

Dividing the frequency of the class (

30

) by its exact class boundary width (

5

) yields the frequency density.

Key Concept

Integration of Frequency Density and Pie Chart Sector Angles

Question 708Question →

On an international flight carrying $150$ passengers, each passenger was offered three meal options: Chicken ( $C$ ), Fish ( $F$ ), and Vegetarian ( $V$ ). A survey of their choices showed that $75$ passengers chose Chicken, $60$ chose Fish, and $50$ chose Vegetarian. Additionally, $15$ passengers chose both Chicken and Fish, $12$ chose both Fish and Vegetarian, $18$ chose both Chicken and Vegetarian, while $8$ passengers chose all three meals. How many passengers chose none of the three meal options?

Show answer & explanation

Answer: 2

Answer

2 passengers chose none of the meal options.

Using the inclusion-exclusion principle for three overlapping sets, the total number of passengers taking at least one meal is calculated as

|C \cup F \cup V| = 75 + 60 + 50 - (15 + 12 + 18) + 8 = 148

. Subtracting this value from the total count of

150

passengers yields

150 - 148 = 2

passengers who selected none of the meal options.

Step-by-Step Solution

1

Identify the cardinalities of the individual sets, pairwise intersections, triple intersection, and the universal set.

N(U) = 150

,

|C| = 75

,

|F| = 60

,

|V| = 50

,

|C \cap F| = 15

,

|F \cap V| = 12

,

|C \cap V| = 18

, and

|C \cap F \cap V| = 8

.

Organizing the given information allows for direct application of set cardinality formulas.

2

Calculate the total number of passengers who selected at least one meal using the Principle of Inclusion-Exclusion for three sets.

|C \cup F \cup V| = 75 + 60 + 50 - (15 + 12 + 18) + 8 = 185 - 45 + 8 = 148

.

Adding individual set totals overcounts elements in pairwise intersections, and subtracting pairwise intersections subtracts the triple intersection one too many times, so it must be added back.

3

Find the number of passengers who selected none of the meal choices by taking the complement of the union with respect to the universal set.

N(\text{None}) = N(U) - |C \cup F \cup V| = 150 - 148 = 2

.

Passengers choosing none of the meal options correspond to the region outside all three sets within the universal set.

Key Concept

Principle of Inclusion-Exclusion for Three Sets and Complement of Union

Estimated Time:1m 30s

Question 709Question →

If

x > 0

satisfies the exponential equation

3^{x+1} + 3^{1-x} = 10

find the value of

8^x + 4^{x-1}

.

Show answer & explanation

Answer: 9

Answer

The value of

8^x + 4^{x-1}

is 9.

Applying index laws to

3^{x+1} + 3^{1-x} = 10

yields

3(3^x) + \frac{3}{3^x} = 10

. Substituting

u = 3^x

gives

3u^2 - 10u + 3 = 0

, which factors to

(3u - 1)(u - 3) = 0

, yielding

u = 3

or

u = \frac{1}{3}

. Thus

x = 1

or

x = -1

. Given

x > 0

,

x = 1

. Substituting

x = 1

into

8^x + 4^{x-1}

gives

8^1 + 4^0 = 8 + 1 = 9

.

Step-by-Step Solution

1

Apply the product and negative power laws of indices to separate the terms in the given equation.

3^{x+1} = 3^1 \cdot 3^x = 3(3^x)

and

3^{1-x} = 3^1 \cdot 3^{-x} = \frac{3}{3^x}

, making the equation

3(3^x) + \frac{3}{3^x} = 10

.

According to the laws of indices,

a^{m+n} = a^m \cdot a^n

and

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

.

2

Substitute

u = 3^x

into the equation and clear the fraction to form a standard quadratic equation.

3u + \frac{3}{u} = 10 \implies 3u^2 - 10u + 3 = 0

.

Multiplying through by

u

eliminates the fraction and forms a quadratic in terms of

u

.

3

Factor the quadratic equation

3u^2 - 10u + 3 = 0

to find the values of

u

.

(3u - 1)(u - 3) = 0 \implies u = 3

or

u = \frac{1}{3}

.

Factoring by splitting the middle term gives the linear factors.

4

Equate

3^x

to the values of

u

and apply the constraint

x > 0

.

3^x = 3^1 \implies x = 1

and

3^x = 3^{-1} \implies x = -1

. Selecting the positive root gives

x = 1

.

Equating exponents with matching base 3 gives the solutions for

x

.

5

Substitute

x = 1

into the target expression

8^x + 4^{x-1}

and simplify.

8^1 + 4^{1-1} = 8 + 4^0 = 8 + 1 = 9

.

By the zero index law, any non-zero base raised to the power 0 equals 1 (

a^0 = 1

).

Key Concept

Solving quadratic-form exponential equations using index laws and applying the zero index rule.

Question 710Question →

Given that $\int_{1}^{a} (3x^2 - 2x) \, dx = 48$ , where $a > 1$ is a constant, find the value of $a$ .

Show answer & explanation

Answer: 4

Answer

The value of the constant upper limit is 4.

Integrating

3x^2 - 2x

yields

x^3 - x^2

. Applying limits from

1

to

a

gives

(a^3 - a^2) - (1 - 1) = a^3 - a^2

. Setting

a^3 - a^2 = 48

, solving for

a > 1

gives

a = 4

because

4^3 - 4^2 = 64 - 16 = 48

.

Step-by-Step Solution

1

Integrate the polynomial function with respect to

x

\int (3x^2 - 2x) \, dx = x^3 - x^2 + C

Using the power rule of integration

\int x^n \, dx = \frac{x^{n+1}}{n+1}

for each term.

2

Evaluate the antiderivative between the upper limit

a

and lower limit

1

[x^3 - x^2]_1^a = (a^3 - a^2) - (1^3 - 1^2) = a^3 - a^2

By the Fundamental Theorem of Calculus,

\int_{b}^{c} f(x)dx = F(c) - F(b)

.

3

Equate the expression to the given total integral value and solve for

a

a^3 - a^2 = 48 \implies a = 4

Substituting

a=4

yields

4^3 - 4^2 = 64 - 16 = 48

, which satisfies the equation.

Key Concept

Definite Integrals with Unknown Limits

Question 711Question →

If the determinant of the matrix $M = \begin{pmatrix} x & 3 \\ 2 & 5 \end{pmatrix}$ is $14$ , find the value of $x$ .

Show answer & explanation

Answer: 4

Answer

The value of

x

is

4

.

For matrix

M = \begin{pmatrix} x & 3 \\ 2 & 5 \end{pmatrix}

, the determinant is calculated as

(x)(5) - (3)(2) = 5x - 6

. Equating this to

14

gives

5x - 6 = 14

, which simplifies to

5x = 20

, yielding

x = 4

.

Step-by-Step Solution

1

Apply the

2 \times 2

determinant formula

\det = ad - bc

\det(M) = (x \times 5) - (3 \times 2) = 5x - 6

The determinant of a

2 \times 2

matrix is the product of the main diagonal minus the product of the anti-diagonal.

2

Set the determinant equal to the given value

14

5x - 6 = 14

The problem states that the determinant is equal to 14.

3

Solve the linear equation for

x

5x = 20 \implies x = 4

Adding 6 to both sides gives

5x = 20

, and dividing by 5 yields

x = 4

.

Key Concept

Determinant of a 2x2 Matrix

Question 712Question →

The $3^{\text{rd}}$ term of a geometric progression (G.P.) is $18$ and its common ratio is $3$ . What is the first term of the progression?

Show answer & explanation

Answer: 2

Answer

The first term of the geometric progression is 2.

In a geometric progression, the

n^{\text{th}}

term is given by

T_n = a r^{n-1}

. For the

3^{\text{rd}}

term (

n = 3

) with common ratio

r = 3

and term value

18

, the equation is

18 = a \cdot 3^{2} = 9a

. Dividing by

9

yields the first term

a = 2

.

Step-by-Step Solution

1

Identify the formula for the

n^{\text{th}}

term of a geometric progression.

T_n = a r^{n-1}

This formula connects the

n^{\text{th}}

term

T_n

to the first term

a

, common ratio

r

, and term index

n

.

2

Substitute

T_3 = 18

,

r = 3

, and

n = 3

into the formula.

18 = a \cdot 3^{3-1} \implies 18 = 9a

Evaluating

3^{3-1} = 3^2 = 9

simplifies the equation.

3

Solve for the first term

a

.

a = \frac{18}{9} = 2

Dividing both sides of the equation by 9 isolates the first term.

Key Concept

Geometric Progression nth term calculation

Question 713Question →

Given the matrices $A = \begin{pmatrix} 2 & -1 \\ 3 & 4 \end{pmatrix}$ and $B = \begin{pmatrix} 1 & x \\ -2 & 5 \end{pmatrix}$ , if the determinant of the product matrix $AB$ is equal to $121$ , find the value of $x$ .

Show answer & explanation

Answer: 3

Answer

The value of

x

is

3

.

By applying the property

\det(AB) = \det(A) \cdot \det(B)

, we compute

\det(A) = (2)(4) - (-1)(3) = 11

and

\det(B) = (1)(5) - (x)(-2) = 5 + 2x

. Substituting into the equation gives

11(5 + 2x) = 121

, which yields

5 + 2x = 11

and leads to

x = 3

.

Step-by-Step Solution

1

Find the determinant of matrix

A

\det(A) = (2)(4) - (-1)(3) = 11

The determinant of a

2 \times 2

matrix

\begin{pmatrix} a & b \\ c & d \end{pmatrix}

is calculated as

ad - bc

.

2

Express the determinant of matrix

B

in terms of

x

\det(B) = (1)(5) - (x)(-2) = 5 + 2x

Apply the

2 \times 2

determinant formula to matrix

B

.

3

Apply the determinant product rule

\det(AB) = \det(A) \cdot \det(B) = 11(5 + 2x) = 121

For any square matrices

A

and

B

,

\det(AB) = \det(A) \det(B)

.

4

Solve the resulting linear equation for

x

x = 3

Dividing

121

by

11

gives

5 + 2x = 11

, which simplifies to

2x = 6

and

x = 3

.

Key Concept

Determinant of a Matrix Product

Estimated Time:1m 30s

Question 714Question →

A container contains $6$ red counters, $4$ blue counters, and $3$ green counters. In how many different ways can a selection of $5$ counters be made if the selection must contain at least $2$ red counters, at least $1$ blue counter, and at most $1$ green counter?

Show answer & explanation

Answer: 750

Answer

The total number of different ways to make the selection is 750.

Breaking the problem into disjoint cases based on the number of green counters (0 or 1) and applying the combination formula for red, blue, and green counters in each valid configuration yields

240 + 510 = 750

total ways.

Step-by-Step Solution

1

Determine the allowable counts for Green (

G

), Red (

R

), and Blue (

B

) counters

Green counters can be 0 or 1. If

G=0

,

R+B=5

with

R \geq 2, B \geq 1

. If

G=1

,

R+B=4

with

R \geq 2, B \geq 1

.

The constraints state

G \leq 1

,

R \geq 2

, and

B \geq 1

for a total of 5 counters.

2

Calculate combinations for Case 1 (

G = 0

)

Ways for

(2R, 3B, 0G) = 60

;

(3R, 2B, 0G) = 120

;

(4R, 1B, 0G) = 60

. Total for Case 1 = 240.

Apply the combinations formula

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

to each sub-case.

3

Calculate combinations for Case 2 (

G = 1

)

Ways for

(2R, 2B, 1G) = 270

;

(3R, 1B, 1G) = 240

. Total for Case 2 = 510.

Apply combinations to choose 1 Green counter along with the valid Red and Blue combinations.

4

Sum all mutually exclusive cases

Total selection ways =

240 + 510 = 750

.

According to the addition principle of counting, the totals of mutually exclusive cases are added together.

Key Concept

Combinations with Multiple Conditional Constraints

Question 715Question →

Given that $\theta$ is an acute angle such that $\tan \theta = \frac{4}{3}$ , calculate the numerical value of the expression $\frac{3\sin \theta + 2\cos \theta}{3\sin \theta - \cos \theta}$ .

Show answer & explanation

Answer: 2

Answer

The exact numerical value of the given expression is 2.

Dividing both the numerator and denominator of

\frac{3\sin \theta + 2\cos \theta}{3\sin \theta - \cos \theta}

by

\cos \theta

gives

\frac{3\tan \theta + 2}{3\tan \theta - 1}

. Substituting

\tan \theta = \frac{4}{3}

yields

\frac{3(4/3) + 2}{3(4/3) - 1} = \frac{4 + 2}{4 - 1} = \frac{6}{3} = 2

.

Step-by-Step Solution

1

Express sine and cosine terms in terms of tangent or find individual ratio values

Divide every term in the numerator and denominator by

\cos \theta

to obtain

\frac{3\tan \theta + 2}{3\tan \theta - 1}

. Alternatively, using a right triangle with opposite side = 4 and adjacent side = 3 gives hypotenuse = 5, so

\sin \theta = \frac{4}{5}

and

\cos \theta = \frac{3}{5}

.

Converting to

\tan \theta

simplifies the calculation directly without evaluating square roots or hypotenuse.

2

Substitute the value of

\tan \theta = \frac{4}{3}

into the expression

Numerator:

3\left(\frac{4}{3}\right) + 2 = 4 + 2 = 6

. Denominator:

3\left(\frac{4}{3}\right) - 1 = 4 - 1 = 3

.

Simplifies numerical fractions in both parts of the fraction.

3

Divide numerator by denominator

\frac{6}{3} = 2.

Yields the final integer solution.

Key Concept

Basic Trigonometric Ratios and Quotients

Question 716Question →

The cumulative frequency distribution of the operational lifespans (in hours) for a batch of $100$ precision LED modules tested in a laboratory is summarized below:

Lifespan Interval (hours)	Class Boundaries	Cumulative Frequency
$100 - 119$	$99.5 - 119.5$	$10$
$120 - 139$	$119.5 - 139.5$	$25$
$140 - 159$	$139.5 - 159.5$	$60$
$160 - 179$	$159.5 - 179.5$	$85$
$180 - 199$	$179.5 - 199.5$	$100$

Using linear interpolation for cumulative frequency distributions, calculate the 75th percentile ( $P_{75}$ ) of the lifespan of these modules in hours.

Show answer & explanation

Answer: 171.5

Answer

The 75th percentile of operational lifespan is 171.5 hours.

The 75th percentile rank position is

0.75 \times 100 = 75

. The percentile falls within the class boundary

159.5 - 179.5

. Substituting lower class boundary

L = 159.5

, preceding cumulative frequency

c.f. = 60

, class frequency

f = 25

, and class width

c = 20

into

P_{75} = L + \left(\frac{75 - c.f.}{f}\right) \times c

gives

159.5 + \left(\frac{15}{25}\right) \times 20 = 171.5

hours.

Step-by-Step Solution

1

Determine the rank position of the 75th percentile.

Rank position = 75th value out of 100.

The 75th percentile corresponds to 75% of the total frequency N = 100.

2

Identify the percentile class interval and extract relevant parameters.

Class interval is 159.5 - 179.5, with L = 159.5, c.f. = 60, f = 25, and c = 20.

The cumulative frequency increases from 60 to 85 across the boundary 159.5 to 179.5, which contains the 75th value.

3

Compute the percentile value using ogive linear interpolation.

P_75 = 159.5 + [(75 - 60) / 25] * 20 = 171.5 hours.

Applying the cumulative frequency interpolation formula yields the exact value.

Key Concept

Calculating Percentiles from Cumulative Frequency / Ogives

Question 717Question →

By evaluating the limit of the difference quotient $\lim_{h \to 0} \frac{f(x+h) - f(x)}{h}$ , determine the value of the derivative of the function $f(x) = 2x^2 - 4x + 5$ at the point where $x = 3$ .

Show answer & explanation

Answer: 8

Answer

The value of the derivative of

f(x) = 2x^2 - 4x + 5

at

x = 3

is 8.

Applying first principles,

f(x+h) - f(x) = 4xh + 2h^2 - 4h

. Dividing by

h

gives

4x + 2h - 4

. Taking the limit as

h \to 0

produces

f'(x) = 4x - 4

. Substituting

x = 3

yields

4(3) - 4 = 8

.

Step-by-Step Solution

1

Substitute

(x+h)

into the function

f(x) = 2x^2 - 4x + 5

and expand

f(x+h) = 2(x^2 + 2xh + h^2) - 4x - 4h + 5 = 2x^2 + 4xh + 2h^2 - 4x - 4h + 5

Apply algebraic expansion to determine the value of the function at

x+h

.

2

Form the difference

f(x+h) - f(x)

f(x+h) - f(x) = 4xh + 2h^2 - 4h

Subtract the original function terms to leave only terms containing

h

.

3

Divide the difference by

h

\frac{f(x+h) - f(x)}{h} = 4x + 2h - 4

Simplify the difference quotient prior to taking the limit.

4

Compute the limit as

h \to 0

f'(x) = 4x - 4

Taking the limit yields the general derivative function

f'(x)

.

5

Evaluate

f'(x)

at

x = 3

f'(3) = 4(3) - 4 = 8

Substitute

x = 3

to find the numerical rate of change at the given point.

Key Concept

Differentiation from First Principles

Question 718Question →

In a survey of $200$ agricultural exporters regarding three major commodities—Cocoa ( $C$ ), Palm Oil ( $P$ ), and Rubber ( $R$ )—it was found that $110$ export Cocoa, $90$ export Palm Oil, and $75$ export Rubber. Exactly $20$ exporters export none of the three commodities, and $45$ export Cocoa only. Furthermore, the number of exporters who export Cocoa and Palm Oil only is twice the number of exporters who export all three commodities; the number who export Palm Oil and Rubber only is equal to the number who export all three; and the number who export Cocoa and Rubber only is $5$ more than the number who export all three. Find the total number of exporters who export at least two of the three commodities.

Show answer & explanation

Answer: 80

Answer

The total number of exporters who export at least two of the three commodities is 80.

The correct answer of 80 is obtained by solving for the number of exporters trading in all three commodities (

x = 15

) using the Cocoa set equation

45 + 2x + (x + 5) + x = 110

, and then evaluating the region sum for at least two commodities (

2x + x + (x + 5) + x = 5x + 5 = 80

).

Step-by-Step Solution

1

Assign a variable to the triple intersection

Let

x = n(C \cap P \cap R)

be the number of exporters of all three commodities.

The intersection of all three sets serves as the common parameter for all double-intersection regions.

2

Write algebraic expressions for the three pairwise-only intersections

n(C \cap P \text{ only}) = 2x

,

n(P \cap R \text{ only}) = x

, and

n(C \cap R \text{ only}) = x + 5

.

These expressions are derived directly from the relationships given in the problem statement.

3

Formulate and solve an equation using the set of Cocoa exporters

45 + 2x + (x + 5) + x = 110 \implies 50 + 4x = 110 \implies x = 15

.

The set of Cocoa exporters consists of four mutually exclusive regions whose cardinalities sum to 110.

4

Sum the regions corresponding to 'at least two commodities'

(2x) + (x) + (x + 5) + x = 5x + 5 = 5(15) + 5 = 80

.

'At least two' encompasses everyone who exports exactly two commodities plus those who export all three.

Key Concept

Three-set principle of inclusion-exclusion and cardinal region decomposition

Question 719Question →

The operating cost $C$ (in Naira per hour) of a speed boat is partly constant and partly varies directly as the square of its speed $v$ (in km/h). At a speed of $20\text{ km/h}$ , the operating cost is $₦18,000$ per hour, and at a speed of $30\text{ km/h}$ , the operating cost is $₦33,000$ per hour. What is the operating cost per hour (in Naira) when the boat travels at a speed of $40\text{ km/h}$ ?

Show answer & explanation

Answer: 54000

Answer

The operating cost per hour when the speed boat travels at

40\text{ km/h}

is

54,000

Naira.

By writing the partial variation relationship as

C = k_1 + k_2 v^2

and substituting the given pairs

(20, 18000)

and

(30, 33000)

, we set up simultaneous equations

k_1 + 400k_2 = 18000

and

k_1 + 900k_2 = 33000

. Solving these yields

k_2 = 30

and

k_1 = 6000

. Evaluating the model at

v = 40\text{ km/h}

gives

C = 6000 + 30(40^2) = 54,000

Naira per hour.

Step-by-Step Solution

1

Set up the general formula for partial variation

C = k_1 + k_2 v^2

The total cost consists of a fixed constant component

k_1

and a variable component

k_2 v^2

that varies directly with speed squared.

2

Form simultaneous linear equations from the given conditions

18,000 = k_1 + 400 k_2

and

33,000 = k_1 + 900 k_2

Substituting

v = 20, C = 18,000

and

v = 30, C = 33,000

creates a solvable system of equations in

k_1

and

k_2

.

3

Solve for the variation constants

k_1

and

k_2

k_2 = 30

and

k_1 = 6,000

Subtracting the first equation from the second eliminates

k_1

, giving

500 k_2 = 15,000 \implies k_2 = 30

. Substituting

k_2 = 30

into the first equation yields

k_1 = 6,000

.

4

Calculate the operating cost at

v = 40\text{ km/h}

C = 54,000

Evaluating

C = 6,000 + 30(40)^2 = 6,000 + 30(1,600) = 6,000 + 48,000 = 54,000

.

Key Concept

Partial Variation with Simultaneous Equations

Question 720Question →

What is the determinant of the matrix $P = \begin{pmatrix} 5 & 3 \\ 2 & 7 \end{pmatrix}$ ?

Show answer & explanation

Answer: 29

Answer

29

For a

2 \times 2

matrix

\begin{pmatrix} a & b \\ c & d \end{pmatrix}

, the determinant is given by

ad - bc

. Substituting the values

a = 5

,

b = 3

,

c = 2

, and

d = 7

yields

(5 \times 7) - (3 \times 2) = 35 - 6 = 29

.

Step-by-Step Solution

1

Identify the values of

a, b, c,

and

d

from the given

2 \times 2

matrix

P = \begin{pmatrix} 5 & 3 \\ 2 & 7 \end{pmatrix}

.

a = 5, b = 3, c = 2, d = 7

To set up the values for the

2 \times 2

determinant formula.

2

Multiply the elements of the main diagonal and the secondary diagonal.

Main diagonal product:

5 \times 7 = 35

; Secondary diagonal product:

3 \times 2 = 6

.

The determinant of a

2 \times 2

matrix is the difference between the main diagonal product and the secondary diagonal product.

3

Subtract the product of the secondary diagonal from the product of the main diagonal.

Determinant =

35 - 6 = 29

.

Completes the formula

\det(P) = ad - bc

.

Key Concept

Determinant of a 2x2 Matrix

Answer

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