All practice questions

Question 7481Question →

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 7482Question →

Two events $A$ and $B$ are defined in a sample space such that $P(A) = \frac{2}{3}$ , $P(B) = \frac{1}{4}$ , and $P(A \cup B) = \frac{3}{4}$ . If $B'$ denotes the complement of event $B$ , what is the value of $P(A \cap B')$ ?

\frac{1}{2}

\frac{1}{6}

\frac{2}{3}

\frac{1}{4}

Show answer & explanation

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

Answer

The probability

P(A \cap B')

is

\frac{1}{2}

.

Using the addition law of probability,

P(A \cap B) = P(A) + P(B) - P(A \cup B) = \frac{2}{3} + \frac{1}{4} - \frac{3}{4} = \frac{1}{6}

. Since

P(A) = P(A \cap B) + P(A \cap B')

, we find

P(A \cap B') = \frac{2}{3} - \frac{1}{6} = \frac{1}{2}

. Alternatively, since

P(A \cap B) = \frac{1}{6} = P(A)P(B)

, events

A

and

B

are independent, so

P(A \cap B') = P(A)P(B') = \frac{2}{3} \times \left(1 - \frac{1}{4}\right) = \frac{2}{3} \times \frac{3}{4} = \frac{1}{2}

.

Step-by-Step Solution

1

Apply the addition law of probability to calculate

P(A \cap B)

.

P(A \cap B) = P(A) + P(B) - P(A \cup B) = \frac{2}{3} + \frac{1}{4} - \frac{3}{4} = \frac{11}{12} - \frac{9}{12} = \frac{2}{12} = \frac{1}{6}

.

The general addition law relates the probabilities of the union and intersection of two compound events.

2

Calculate the probability that event

A

occurs while event

B

does not occur,

P(A \cap B')

.

P(A \cap B') = P(A) - P(A \cap B) = \frac{2}{3} - \frac{1}{6} = \frac{4}{6} - \frac{1}{6} = \frac{3}{6} = \frac{1}{2}

.

Event

A

can be decomposed into two mutually exclusive parts:

A \cap B

and

A \cap B'

.

Key Concept

Addition law of probability and complement of compound events

Estimated Time:2m 0s

Question 7483Question →

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 7484Question →

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 7485Question →

If $\frac{3\sqrt{5} - \sqrt{3}}{\sqrt{5} + \sqrt{3}} = a + b\sqrt{15}$ , where $a$ and $b$ are rational numbers, what is the value of $a - b$ ?

11

8

5

13

Show answer & explanation

Answer: 11

Answer

11

By multiplying the numerator and denominator by the conjugate of the denominator,

(\sqrt{5} - \sqrt{3})

, the fraction simplifies to

\frac{18 - 4\sqrt{15}}{2} = 9 - 2\sqrt{15}

. Equating this to

a + b\sqrt{15}

gives

a = 9

and

b = -2

. Calculating

a - b

gives

9 - (-2) = 11

.

Step-by-Step Solution

1

Multiply the numerator and denominator by the conjugate of the denominator

\frac{3\sqrt{5} - \sqrt{3}}{\sqrt{5} + \sqrt{3}} \times \frac{\sqrt{5} - \sqrt{3}}{\sqrt{5} - \sqrt{3}} = \frac{(3\sqrt{5} - \sqrt{3})(\sqrt{5} - \sqrt{3})}{(\sqrt{5})^2 - (\sqrt{3})^2}

Rationalising the denominator requires using the difference of squares identity

(x+y)(x-y) = x^2 - y^2

.

2

Expand the numerator and denominator

Denominator = 5 - 3 = 2. Numerator = 3(5) - 3\sqrt{15} - \sqrt{15} + 3 = 15 + 3 - 4\sqrt{15} = 18 - 4\sqrt{15}.

Apply distributive property to expand

(3\sqrt{5} - \sqrt{3})(\sqrt{5} - \sqrt{3})

carefully combining like terms.

3

Simplify the fraction to match the form

a + b\sqrt{15}

\frac{18 - 4\sqrt{15}}{2} = 9 - 2\sqrt{15}

Divide each term in the numerator by 2.

4

Identify

a

and

b

and evaluate

a - b

a = 9, b = -2 \implies a - b = 9 - (-2) = 11

Subtracting negative 2 is equivalent to adding 2.

Key Concept

Rationalisation of Binomial Surd Denominators

Estimated Time:1m 30s

Question 7486Question →

If matrix $A$ has dimension $2 \times 3$ and matrix $B$ has dimension $3 \times 4$ , what is the dimension of the matrix product $AB$ ?

3 \times 3

2 \times 4

4 \times 2

2 \times 3

Show answer & explanation

Answer: $2 \times 4$

Answer

The dimension of the matrix product

AB

is

2 \times 4

.

For two matrices to be multiplied, the number of columns in the first matrix must equal the number of rows in the second matrix. When a matrix of size

m \times n

is multiplied by a matrix of size

n \times p

, the resulting product matrix has size

m \times p

. For

A_{2 \times 3}

and

B_{3 \times 4}

, the outer dimensions give

2 \times 4

.

Step-by-Step Solution

1

Identify the dimensions of the given matrices

A

and

B

Matrix

A

is of order

2 \times 3

(

m = 2, n = 3

) and Matrix

B

is of order

3 \times 4

(

n = 3, p = 4

).

Matrix multiplication requirement requires the number of columns of the first matrix to match the number of rows of the second matrix.

2

Apply the matrix multiplication dimension rule

The product matrix

AB

has

m

rows and

p

columns, resulting in dimension

2 \times 4

.

When multiplying an

m \times n

matrix by an

n \times p

matrix, the resulting matrix has dimensions

m \times p

.

Key Concept

Matrix Multiplication Dimension Compatibility and Resulting Order

Estimated Time:45s

Question 7487Question →

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 7488Question →

What is the value of the definite integral $\int_{1}^{3} (3x^2 - 2x) \, dx$ ?

12

18

20

24

Show answer & explanation

Answer: $18$

Answer

18

Integrating

3x^2 - 2x

gives the antiderivative

F(x) = x^3 - x^2

. Evaluating

F(3) - F(1)

gives

(27 - 9) - (1 - 1) = 18 - 0 = 18

.

Step-by-Step Solution

1

Find the indefinite integral of the function

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

Apply the power rule of integration to each term: \int 3x^2 dx = x^3 and \int 2x dx = x^2.

2

Evaluate the antiderivative at the upper limit x = 3

(3)^3 - (3)^2 = 27 - 9 = 18

Substitute the upper limit into the antiderivative.

3

Evaluate the antiderivative at the lower limit x = 1

(1)^3 - (1)^2 = 1 - 1 = 0

Substitute the lower limit into the antiderivative.

4

Subtract the lower limit evaluation from the upper limit evaluation

18 - 0 = 18

By the Fundamental Theorem of Calculus, \int_{a}^{b} f(x) dx = F(b) - F(a).

Key Concept

Fundamental Theorem of Calculus for Definite Polynomial Integrals

Estimated Time:1m 0s

Question 7489Question →

An arithmetic progression (A.P.) has a fifth term of $17$ and a common difference of $3$ . What is the first term of the progression?

5

2

-1

29

Show answer & explanation

Answer: $5$

Answer

The first term of the progression is

5

.

Using the

n^{\text{th}}

term formula for an arithmetic progression,

T_n = a + (n - 1)d

, substituting

T_5 = 17

,

n = 5

, and

d = 3

gives

17 = a + 4(3) \implies 17 = a + 12 \implies a = 5

.

Step-by-Step Solution

1

Identify the given parameters and formula for the

n^{\text{th}}

term of an A.P.

Formula:

T_n = a + (n - 1)d

, where

T_5 = 17

,

n = 5

, and

d = 3

.

The standard formula connects the

n^{\text{th}}

term, first term, term position, and common difference.

2

Substitute the given values into the formula.

17 = a + (5 - 1) \times 3 \implies 17 = a + 4 \times 3 \implies 17 = a + 12

.

Evaluating

(n - 1)d

gives the total difference added to the first term.

3

Solve for the first term

a

.

a = 17 - 12 = 5

.

Subtracting

12

from both sides isolates

a

.

Key Concept

Arithmetic Progression

n^{\text{th}}

term calculation

Estimated Time:45s

Question 7490Question →

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 7491Question →

Given that $\cos x - \sin x = \frac{1}{\sqrt{5}}$ for an acute angle $x$ , what is the exact value of $\cos^3 x - \sin^3 x$ ?

\frac{7\sqrt{5}}{25}

\frac{3\sqrt{5}}{25}

\frac{9\sqrt{5}}{25}

\frac{7\sqrt{5}}{5}

Show answer & explanation

Answer: $\frac{7\sqrt{5}}{25}$

Answer

\frac{7\sqrt{5}}{25}

Squaring

\cos x - \sin x = \frac{1}{\sqrt{5}}

yields

1 - 2\sin x \cos x = \frac{1}{5}

, which gives

\sin x \cos x = \frac{2}{5}

. Using the difference of cubes factorization,

\cos^3 x - \sin^3 x = (\cos x - \sin x)(1 + \sin x \cos x) = \frac{1}{\sqrt{5}} \left(1 + \frac{2}{5}\right) = \frac{7}{5\sqrt{5}}

. Rationalizing the denominator produces

\frac{7\sqrt{5}}{25}

.

Step-by-Step Solution

1

Square both sides of the given equation to find the product

\sin x \cos x

.

(\cos x - \sin x)^2 = \left(\frac{1}{\sqrt{5}}\right)^2 \implies \cos^2 x - 2\sin x \cos x + \sin^2 x = \frac{1}{5}

.

Squaring allows us to use the Pythagorean identity

\sin^2 x + \cos^2 x = 1

to isolate

\sin x \cos x

.

2

Simplify using

\cos^2 x + \sin^2 x = 1

to solve for

\sin x \cos x

.

1 - 2\sin x \cos x = \frac{1}{5} \implies 2\sin x \cos x = 1 - \frac{1}{5} = \frac{4}{5} \implies \sin x \cos x = \frac{2}{5}

.

Finding the product of

\sin x

and

\cos x

is necessary for the algebraic expansion of the difference of cubes.

3

Apply the difference of cubes algebraic identity to

\cos^3 x - \sin^3 x

.

\cos^3 x - \sin^3 x = (\cos x - \sin x)(\cos^2 x + \sin x \cos x + \sin^2 x) = (\cos x - \sin x)(1 + \sin x \cos x)

.

The identity

a^3 - b^3 = (a - b)(a^2 + ab + b^2)

breaks the target expression down into known terms.

4

Substitute the known values into the expression and rationalize the denominator.

\left(\frac{1}{\sqrt{5}}\right) \left(1 + \frac{2}{5}\right) = \frac{1}{\sqrt{5}} \times \frac{7}{5} = \frac{7}{5\sqrt{5}} = \frac{7\sqrt{5}}{25}

.

Evaluating the product and rationalizing

\frac{7}{5\sqrt{5}}

yields the final exact surd form.

Key Concept

Basic Trigonometric Ratios, Special Angles, and Identities

Estimated Time:2m 0s

Question 7492Question →

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 7493Question →

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 7494Question →

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 7495Question →

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 7496Question →

If $x$ is the least positive integer satisfying the modular congruence $4x + 9 \equiv 3 \pmod{11}$ , what is the value of $(x^2 - 5x + 2) \pmod{11}$ expressed in standard non-negative remainder form?

2

4

7

9

Show answer & explanation

Answer: 9

Answer

The value of

(x^2 - 5x + 2) \pmod{11}

in standard non-negative remainder form is 9.

Solving

4x + 9 \equiv 3 \pmod{11}

yields

4x \equiv -6 \equiv 5 \pmod{11}

. Multiplying by 3 (the inverse of 4 mod 11) gives

x \equiv 15 \equiv 4 \pmod{11}

. Evaluating

(x^2 - 5x + 2)

at

x = 4

gives

16 - 20 + 2 = -2

. Converting

-2

into the standard non-negative remainder range

[0, 10]

gives

-2 + 11 = 9

.

Step-by-Step Solution

1

Isolate the variable term in the linear modular congruence

4x \equiv 3 - 9 \equiv -6 \pmod{11}

Subtract 9 from both sides of the congruence.

2

Convert the negative right-hand side to a non-negative residue modulo 11

4x \equiv -6 + 11 \equiv 5 \pmod{11}

Add the modulus 11 to obtain the canonical non-negative equivalent.

3

Solve for

x

by multiplying by the multiplicative inverse of 4 modulo 11

Since

4 \times 3 = 12 \equiv 1 \pmod{11}

, multiply both sides by 3:

x \equiv 5 \times 3 = 15 \equiv 4 \pmod{11}

. Thus, the least positive integer is

x = 4

.

The modular inverse of 4 modulo 11 is 3.

4

Substitute

x = 4

into the expression

(x^2 - 5x + 2)

4^2 - 5(4) + 2 = 16 - 20 + 2 = -2

Evaluate the quadratic expression using the calculated value of

x

.

5

Express

-2

in canonical non-negative remainder form modulo 11

-2 + 11 = 9 \pmod{11}

Add the modulus 11 to convert the negative result to a non-negative remainder within

[0, 10]

.

Key Concept

Solving linear modular congruences and converting negative remainders to canonical form

Question 7497Question →

Pumps $P_1$ and $P_2$ operating together can fill a water storage tank in $12\text{ hours}$ , while pumps $P_2$ and $P_3$ operating together can fill the same tank in $20\text{ hours}$ . When all three pumps operate simultaneously for $5\text{ hours}$ , they fill exactly $\frac{1}{2}$ of the tank. If pump $P_2$ is then shut off, how many additional hours will it take pumps $P_1$ and $P_3$ working together to fill the remainder of the tank?

7.5\text{ hours}

15.0\text{ hours}

6.0\text{ hours}

12.5\text{ hours}

Show answer & explanation

Answer: $7.5\text{ hours}$

Answer

7.5\text{ hours}

The three pumps combined have a work rate of

\frac{1}{10}\text{ tank/hr}

. Subtracting the rate of

P_1 + P_2

(which is

\frac{1}{12}

) gives

P_3

's rate as

\frac{1}{60}\text{ tank/hr}

. Subtracting the rate of

P_2 + P_3

(which is

\frac{1}{20}

) gives

P_1

's rate as

\frac{1}{20}\text{ tank/hr}

. Together,

P_1

and

P_3

have a combined rate of

\frac{1}{20} + \frac{1}{60} = \frac{1}{15}\text{ tank/hr}

. To fill the remaining

\frac{1}{2}

of the tank, it will take

\frac{1/2}{1/15} = 7.5\text{ hours}

.

Step-by-Step Solution

1

Determine the combined work rate of all three pumps working together.

Since all three pumps fill

\frac{1}{2}

of the tank in

5\text{ hours}

, their combined rate is

r_1 + r_2 + r_3 = \frac{1/2}{5} = \frac{1}{10}\text{ tank per hour}

.

Work rate is defined as the fraction of work completed per unit time.

2

Find the individual work rates of pumps

P_1

and

P_3

.

r_3 = (r_1 + r_2 + r_3) - (r_1 + r_2) = \frac{1}{10} - \frac{1}{12} = \frac{1}{60}\text{ tank per hour}

. Also,

r_1 = (r_1 + r_2 + r_3) - (r_2 + r_3) = \frac{1}{10} - \frac{1}{20} = \frac{1}{20}\text{ tank per hour}

.

Subtracting known pair rates from the total three-pump rate yields individual rates.

3

Calculate the combined work rate of

P_1

and

P_3

.

r_1 + r_3 = \frac{1}{20} + \frac{1}{60} = \frac{3 + 1}{60} = \frac{4}{60} = \frac{1}{15}\text{ tank per hour}

.

The rate of two pumps operating together is the sum of their individual rates.

4

Compute the time needed to fill the remaining portion of the tank.

Remaining fraction to fill is

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

. Time required =

\frac{1/2}{1/15} = \frac{15}{2} = 7.5\text{ hours}

.

Dividing the remaining workload by the combined rate gives the additional operating time.

Key Concept

Work-rate problems involving simultaneous rates and partial work completion

Estimated Time:2m 0s

Question 7498Question →

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 7499Question →

If $\log_3 x - 2\log_x 27 = 1$ , what is the sum of all possible real values of $x$ ?

\frac{244}{9}

18

1

27

Show answer & explanation

Answer: $\frac{244}{9}$

Answer

\frac{244}{9}

By applying the change of base rule

\log_x 27 = \frac{\log_3 27}{\log_3 x} = \frac{3}{\log_3 x}

, the given equation simplifies to

\log_3 x - \frac{6}{\log_3 x} = 1

. Setting

u = \log_3 x

yields

u^2 - u - 6 = 0

, which factors as

(u - 3)(u + 2) = 0

. Thus,

u = 3

or

u = -2

, giving solutions

x = 3^3 = 27

and

x = 3^{-2} = \frac{1}{9}

. Adding these valid real solutions gives

27 + \frac{1}{9} = \frac{244}{9}

.

Step-by-Step Solution

1

Apply the change of base formula to

\log_x 27

.

\log_x 27 = \frac{\log_3 27}{\log_3 x} = \frac{3}{\log_3 x}

Converting all logarithmic terms to a common base (base 3) allows substitution.

2

Substitute

\log_x 27

back into the original equation and let

u = \log_3 x

.

u - 2\left(\frac{3}{u}\right) = 1 \implies u - \frac{6}{u} = 1

This transforms the logarithmic equation into an algebraic equation.

3

Clear the denominator by multiplying through by

u

and rearrange into standard quadratic form.

u^2 - u - 6 = 0

Multiplying by

u

(where

u \neq 0

) forms a standard quadratic equation.

4

Factor the quadratic equation to solve for

u

.

(u - 3)(u + 2) = 0 \implies u = 3 \text{ or } u = -2

Factoring determines the values of

\log_3 x

.

5

Convert back to

x

using

x = 3^u

.

For

u = 3

,

x = 3^3 = 27

. For

u = -2

,

x = 3^{-2} = \frac{1}{9}

. Both

x = 27

and

x = \frac{1}{9}

are valid bases (

x > 0, x \neq 1

).

Exponentiation resolves the logarithmic variable.

6

Find the sum of all valid values of

x

.

\text{Sum} = 27 + \frac{1}{9} = \frac{243 + 1}{9} = \frac{244}{9}

Combines the two real solutions into the final requested sum.

Key Concept

Change of base formula

\log_a b = \frac{\log_c b}{\log_c a}

and solving equations reducible to quadratics.

Estimated Time:2m 0s

Question 7500Question →

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

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