All practice questions

Question 7621Question →

A fruit juice mixture is prepared by combining orange extract and water in the ratio $3 : 5$ . If a bottle contains $40\text{ litres}$ of this mixture, how many litres of water does it contain?

Show answer & explanation

Answer: 25

Answer

The volume of water in the mixture is 25 litres.

The mixture ratio of orange extract to water is

3 : 5

, giving a total of

3 + 5 = 8

equal parts. Dividing the total volume of

40\text{ litres}

by

8

parts gives

5\text{ litres}

per part. Since water makes up

5

parts of the ratio, its volume is

5 \times 5 = 25\text{ litres}

.

Step-by-Step Solution

1

Determine the total number of parts in the ratio.

3 + 5 = 8 parts

The full volume of the mixture is divided into equal portions represented by the sum of the ratio parts.

2

Calculate the volume per ratio part.

40 / 8 = 5 litres per part

Dividing the total volume by the total number of parts gives the size of a single part.

3

Find the quantity of water.

5 * 5 = 25 litres

Water accounts for 5 parts of the ratio, so multiply 5 parts by 5 litres per part.

Key Concept

Ratio Sharing and Proportional Division

Question 7622Question →

Given the permutation equation ${^{n}P_4} = 42 \cdot {^{n}P_2}$ , what is the positive integer value of $n$ ?

Show answer & explanation

Answer: 9

Answer

The positive integer value of

n

is

9

.

Expanding

{^{n}P_4}

as

n(n-1)(n-2)(n-3)

and

{^{n}P_2}

as

n(n-1)

transforms the equation to

n(n-1)(n-2)(n-3) = 42n(n-1)

. Dividing both sides by

n(n-1)

gives

(n-2)(n-3) = 42

, which simplifies to

n^2 - 5n - 36 = 0

. Factoring this quadratic gives

(n-9)(n+4) = 0

. Since

n

must be a positive integer greater than or equal to

4

, the correct answer is

9

.

Step-by-Step Solution

1

Write out the expanded expressions for

{^{n}P_4}

and

{^{n}P_2}

.

{^{n}P_4} = n(n-1)(n-2)(n-3)

and

{^{n}P_2} = n(n-1)

.

By definition,

{^{n}P_r} = \frac{n!}{(n-r)!} = n(n-1)\dots(n-r+1)

.

2

Set up the algebraic equation based on the given problem statement.

n(n-1)(n-2)(n-3) = 42n(n-1)

.

Substitute the expanded permutation formulas into the given identity.

3

Simplify the equation by dividing both sides by the non-zero common product

n(n-1)

.

(n-2)(n-3) = 42

.

Since

n \ge 4

,

n(n-1) \neq 0

, so we can cancel these terms from both sides.

4

Expand and rearrange the resulting expression into a standard quadratic equation.

n^2 - 5n - 36 = 0

.

Expanding

(n-2)(n-3)

gives

n^2 - 5n + 6

; subtracting

42

yields

n^2 - 5n - 36 = 0

.

5

Solve the quadratic equation for

n

and discard non-physical roots.

n = 9

.

Factoring yields

(n-9)(n+4) = 0

, so

n = 9

or

n = -4

. A permutation requires

n

to be a positive integer

\ge 4

, so

n = 9

.

Key Concept

Solving algebraic equations involving permutations

{^{n}P_r} = \frac{n!}{(n-r)!}

Question 7623Question →

Two independent events $A$ and $B$ in a sample space satisfy $P(A') = \frac{2}{5}$ and $P(A \cup B) = \frac{7}{10}$ , where $A'$ denotes the complement of event $A$ . What is the probability that exactly one of the two events occurs?

\frac{11}{20}

\frac{7}{10}

\frac{17}{20}

\frac{3}{20}

Show answer & explanation

Answer: $\frac{11}{20}$

Answer

\frac{11}{20}

First, find

P(A) = 1 - \frac{2}{5} = \frac{3}{5}

. Since

A

and

B

are independent,

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

. Using the addition law

P(A \cup B) = P(A) + P(B) - P(A)P(B)

, we substitute the values to get

\frac{7}{10} = \frac{3}{5} + P(B)\left(1 - \frac{3}{5}\right)

, which gives

P(B) = \frac{1}{4}

. The probability of exactly one event occurring is

P(A)P(B') + P(A')P(B) = \frac{3}{5} \times \frac{3}{4} + \frac{2}{5} \times \frac{1}{4} = \frac{11}{20}

.

Step-by-Step Solution

1

Calculate the probability of event A

P(A) = 1 - P(A') = 1 - \frac{2}{5} = \frac{3}{5}

The sum of the probabilities of an event and its complement equals 1.

2

Apply the addition law for independent events to find P(B)

P(A \cup B) = P(A) + P(B) - P(A)P(B) \implies \frac{7}{10} = \frac{3}{5} + P(B)\left(1 - \frac{3}{5}\right) \implies P(B) = \frac{1}{4}

For independent events, the intersection probability is

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

.

3

Calculate the complement of event B

P(B') = 1 - P(B) = 1 - \frac{1}{4} = \frac{3}{4}

The complement of event B represents the event that B does not occur.

4

Compute the probability that exactly one event occurs

P(\text{exactly one}) = P(A \cap B') + P(A' \cap B) = P(A)P(B') + P(A')P(B) = \left(\frac{3}{5} \times \frac{3}{4}\right) + \left(\frac{2}{5} \times \frac{1}{4}\right) = \frac{9}{20} + \frac{2}{20} = \frac{11}{20}

The occurrence of exactly one event means either A occurs and B does not, or A does not occur and B occurs.

Key Concept

Compound Probability and Probability Laws for Independent Events

Estimated Time:2m 0s

Question 7624Question →

A trader bought a quantity of palm oil. He sold $\frac{1}{3}$ of the total quantity at a profit of $20\%$ , $\frac{1}{2}$ of the remaining palm oil at a loss of $10\%$ , and the rest of the palm oil at a profit of $x\%$ . If his overall profit on the entire transaction was $15\%$ , what is the value of $x$ ?

Show answer & explanation

Answer: 35

Answer

The value of

x

is

35

.

The palm oil is divided into three equal portions of

\frac{1}{3}

each. The first portion contributes

\frac{20}{3}\%

profit, the second portion contributes a loss of

-\frac{10}{3}\%

, and the final portion contributes

\frac{x}{3}\%

profit. Summing these contributions yields

\frac{10 + x}{3}\%

. Equating this to the target overall profit of

15\%

gives

10 + x = 45

, which solves to

x = 35

.

Step-by-Step Solution

1

Determine the fractional portion of palm oil sold at each stage.

First portion =

\frac{1}{3}

. Remaining after first sale =

1 - \frac{1}{3} = \frac{2}{3}

. Second portion =

\frac{1}{2} \times \frac{2}{3} = \frac{1}{3}

. Final portion =

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

.

Finding the exact fraction of the total quantity sold at each step is necessary to weight their respective profit/loss rates correctly.

2

Formulate the net percentage profit equation.

\text{Net Profit } \% = \left(\frac{1}{3} \times 20\%\right) + \left(\frac{1}{3} \times (-10\%)\right) + \left(\frac{1}{3} \times x\%\right) = \frac{20 - 10 + x}{3}\% = \frac{10 + x}{3}\%

The overall profit percentage is the sum of individual fractional contributions, treating loss as a negative percentage.

3

Set the net profit equal to

15\%

and solve for

x

.

\frac{10 + x}{3} = 15 \implies 10 + x = 45 \implies x = 35

Solving this linear equation gives the required profit percentage for the final portion.

Key Concept

Weighted Average Profit and Loss across fractional components of an asset

Estimated Time:2m 0s

Question 7625Question →

The physical quantity work is defined as the product of force and displacement. What are the dimensional exponents $a$ , $b$ , and $c$ for mass, length, and time respectively in the dimensional formula for work, $[M^a L^b T^c]$ ?

1, 2, -2

1, 1, -2

1, 2, -1

1, -3, 0

Show answer & explanation

Answer: $1, 2, -2$

Answer

The dimensional exponents for mass, length, and time in the expression for work are

a = 1

,

b = 2

, and

c = -2

.

Work is calculated as force multiplied by displacement. Since force has dimensions

[M L T^{-2}]

and displacement has dimension

[L]

, the resulting dimensional formula for work is

[M^1 L^2 T^{-2}]

. The exponents of

M

,

L

, and

T

are therefore

1

,

2

, and

-2

respectively.

Step-by-Step Solution

1

Express the dimensions of force using base quantities

\text{Force} = \text{mass} \times \text{acceleration} = [M] \times [L T^{-2}] = [M L T^{-2}]

Acceleration has the dimensions of length divided by time squared.

2

Multiply force dimensions by displacement dimension to find the dimensions of work

\text{Work} = \text{Force} \times \text{Displacement} = [M L T^{-2}] \times [L] = [M^1 L^2 T^{-2}]

Displacement is a measure of length, contributing an additional factor of

[L]

.

3

Extract the exponents

a

,

b

, and

c

a = 1, b = 2, c = -2

Comparing

[M^a L^b T^c]

to

[M^1 L^2 T^{-2}]

gives the values of

a

,

b

, and

c

.

Key Concept

Dimensional analysis of work

Question 7626Question →

Two point charges produce mutually perpendicular electric fields at a point $P$ in a vacuum. If the magnitudes of the electric field intensities at $P$ due to the charges individually are $30\text{ N C}^{-1}$ and $40\text{ N C}^{-1}$ , what is the magnitude of the net electric field intensity at point $P$ ?

50\text{ N C}^{-1}

70\text{ N C}^{-1}

10\text{ N C}^{-1}

35\text{ N C}^{-1}

Show answer & explanation

Answer: $50\text{ N C}^{-1}$

Answer

The magnitude of the net electric field intensity at point

P

is

50\text{ N C}^{-1}

.

Because electric field intensity is a vector quantity, two mutually perpendicular electric fields

E_1 = 30\text{ N C}^{-1}

and

E_2 = 40\text{ N C}^{-1}

combine vectorially according to

E_{\text{net}} = \sqrt{E_1^2 + E_2^2} = \sqrt{30^2 + 40^2} = 50\text{ N C}^{-1}

.

Step-by-Step Solution

1

Identify the vector nature and orientation of the given electric fields.

The two component fields

E_1 = 30\text{ N C}^{-1}

and

E_2 = 40\text{ N C}^{-1}

are perpendicular to each other (

\theta = 90^\circ

).

Electric field intensity is a vector quantity, so perpendicular vectors must be combined using vector addition.

2

Apply the Pythagorean theorem to calculate the resultant vector magnitude.

E_{\text{net}} = \sqrt{E_1^2 + E_2^2} = \sqrt{30^2 + 40^2} = \sqrt{900 + 1600} = \sqrt{2500} = 50\text{ N C}^{-1}

.

When two vectors meet at a right angle, the magnitude of their resultant is the hypotenuse of the right-angled triangle formed by the vector components.

Key Concept

Vector Addition of Electric Fields (Superposition Principle)

Estimated Time:1m 0s

Question 7627Question →

What is the canonical non-negative remainder when $3^{45} - 11$ is divided by $7$ ?

Show answer & explanation

Answer: 2

Answer

The correct answer is 2.

Using modular exponentiation,

3^3 = 27 \equiv -1 \pmod{7}

, so

3^{45} = (3^3)^{15} \equiv (-1)^{15} = -1 \pmod{7}

. Subtracting 11 yields

-1 - 11 = -12 \pmod{7}

. Adding 14 (a multiple of 7) gives

-12 + 14 = 2

, which is the canonical non-negative remainder.

Step-by-Step Solution

1

Evaluate

3^{45} \pmod{7}

using power rules

3^{45} \equiv -1 \pmod{7}

Since

3^3 = 27 \equiv -1 \pmod{7}

, raising both sides to the power of 15 gives

(3^3)^{15} \equiv (-1)^{15} = -1 \pmod{7}

.

2

Subtract 11 from the modular result

-12 \pmod{7}

Replacing

3^{45}

with

-1

gives

-1 - 11 = -12

.

3

Reduce -12 to its canonical non-negative remainder in

[0, 6]

2

Adding the smallest multiple of 7 that makes the result non-negative gives

-12 + 14 = 2

.

Key Concept

Modular Exponentiation and Negative Remainder Reduction

Question 7628Question →

In a survey of $120$ tech entrepreneurs at an innovation hub, $65$ secured Angel Investment ( $A$ ), $55$ received Venture Capital ( $V$ ), and $50$ obtained Government Grants ( $G$ ). Furthermore, $25$ received both $A$ and $V$ , $20$ received both $V$ and $G$ , $22$ received both $A$ and $G$ , while $12$ received no funding from any of these three sources. How many entrepreneurs secured funding from exactly two of these sources?

Show answer & explanation

Answer: 52

Answer

The number of entrepreneurs who secured funding from exactly two sources is 52.

Subtracting the

12

unfunded entrepreneurs from the total

120

yields

108

funded entrepreneurs. Applying inclusion-exclusion gives

108 = 65 + 55 + 50 - (25 + 20 + 22) + n(A \cap V \cap G)

, which gives

n(A \cap V \cap G) = 5

. Subtracting

5

from each pairwise intersection yields

20

,

15

, and

17

for entrepreneurs receiving funding from exactly two sources. Summing these gives

52

.

Step-by-Step Solution

1

Determine the cardinality of the union of all three funding sets

n(A \cup V \cup G) = 108

Subtracting the 12 unfunded entrepreneurs from the universal set of 120 gives the total number of entrepreneurs who received at least one form of funding.

2

Solve for the number of entrepreneurs who received funding from all three sources using inclusion-exclusion

n(A \cap V \cap G) = 5

Using

n(A \cup V \cup G) = n(A) + n(V) + n(G) - n(A \cap V) - n(V \cap G) - n(A \cap G) + n(A \cap V \cap G)

, we get

108 = 65 + 55 + 50 - 25 - 20 - 22 + x

, which simplifies to

108 = 103 + x

, giving

x = 5

.

3

Calculate the count for each region representing exactly two funding sources

Only

A \cap V = 20

, Only

V \cap G = 15

, Only

A \cap G = 17

Subtracting the 3-set intersection (

x = 5

) from each pairwise intersection isolates the elements belonging to strictly two sets.

4

Sum the three strictly two-set regions

20 + 15 + 17 = 52

Adding the individual counts for the three disjoint regions gives the total number of entrepreneurs who received funding from exactly two sources.

Key Concept

Principle of Inclusion-Exclusion for three sets and cardinal partitioning of Venn diagrams

Question 7629Question →

The frequency table below shows the fuel efficiency, measured in $\text{km/L}$ , for a fleet of $50$ delivery vans operated by a logistics company:

Fuel Efficiency (km/L)	Frequency ( $f$ )
10 – 14	8
15 – 19	12
20 – 24	20
25 – 29	10

What is the mean fuel efficiency of the fleet, in $\text{km/L}$ ?

Show answer & explanation

Answer: 20.2

Answer

The mean fuel efficiency of the fleet is 20.2 km/L.

The estimated mean is found by calculating the midpoint of each class interval, multiplying each midpoint by its class frequency, summing these products to get 1010, and dividing by the total frequency of 50, resulting in 20.2 km/L.

Step-by-Step Solution

1

Determine the midpoint (x) for each class interval.

Midpoints are 12 for 10–14, 17 for 15–19, 22 for 20–24, and 27 for 25��29.

For grouped frequency data, each class interval is represented by its midpoint x = (lower limit + upper limit) / 2.

2

Compute the product of frequency and midpoint (f * x) for each interval.

Products: (8 * 12) = 96, (12 * 17) = 204, (20 * 22) = 440, (10 * 27) = 270.

To calculate the total contribution of values from each class interval.

3

Sum all products (sum of f * x) and divide by the total number of delivery vans (sum of f).

sum of f * x = 96 + 204 + 440 + 270 = 1010. sum of f = 50. Mean = 1010 / 50 = 20.2.

The formula for the estimated mean of grouped data is mean = (sum of f * x) / (sum of f).

Key Concept

Measures of Central Tendency for Grouped Data (Grouped Mean)

Question 7630Question →

Using differentiation from first principles, which expression represents the derivative $\frac{\mathrm{d}y}{\mathrm{d}x}$ of the function $f(x) = \frac{2}{x}$ , where $x \neq 0$ ?

-\frac{2}{x^2}

\frac{2}{x^2}

-\frac{2}{x}

-\frac{1}{x^2}

Show answer & explanation

Answer: $-\frac{2}{x^2}$

Answer

-\frac{2}{x^2}

The option stating

-\frac{2}{x^2}

is correct because applying the first principles limit formula

\lim_{h \to 0} \frac{\frac{2}{x+h} - \frac{2}{x}}{h}

gives

\lim_{h \to 0} \frac{-2h}{h x(x+h)} = \lim_{h \to 0} \frac{-2}{x(x+h)} = -\frac{2}{x^2}

.

Step-by-Step Solution

1

Set up the difference quotient definition of derivative from first principles.

\frac{\mathrm{d}y}{\mathrm{d}x} = \lim_{h \to 0} \frac{f(x+h) - f(x)}{h}

First principles definition states that the derivative is the limit of the rate of change as

h

approaches zero.

2

Substitute

f(x) = \frac{2}{x}

and

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

into the formula.

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

Replacing the function expressions yields the explicit algebraic fraction.

3

Combine the fractions in the numerator over a common denominator.

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

Simplifying the numerator yields a factor of

h

that can be cancelled with the denominator.

4

Cancel the common factor

h

and evaluate the limit as

h \to 0

.

\lim_{h \to 0} \frac{-2}{x(x+h)} = \frac{-2}{x(x+0)} = -\frac{2}{x^2}

Dividing out

h

allows direct substitution of

h = 0

without producing an indeterminate form.

Key Concept

Differentiation from First Principles for Reciprocal Functions

Estimated Time:1m 30s

Question 7631Question →

A car moves $80\text{ km}$ due East along a straight highway and then turns to travel $60\text{ km}$ due North. What is the magnitude of the displacement of the car from its starting position?

100\text{ km}

140\text{ km}

20\text{ km}

70\text{ km}

Show answer & explanation

Answer: $100\text{ km}$

Answer

The magnitude of the displacement is

100\text{ km}

.

Because displacement is a vector quantity, perpendicular components must be combined vectorially using the Pythagorean theorem rather than scalar addition. The calculation

\sqrt{80^2 + 60^2} = 100\text{ km}

correctly yields the magnitude of the net displacement vector.

Step-by-Step Solution

1

Identify the vector components and their directions.

Eastward displacement

x = 80\text{ km}

, Northward displacement

y = 60\text{ km}

. The directions are mutually perpendicular (

90^\circ

to each other).

Displacement is a vector quantity, so direction must be accounted for when combining components.

2

Apply the Pythagorean theorem to calculate the resultant vector magnitude.

R = \sqrt{x^2 + y^2} = \sqrt{80^2 + 60^2} = \sqrt{6400 + 3600} = \sqrt{10000} = 100\text{ km}

.

When two vector components act at right angles (

90^\circ

), the magnitude of their resultant is given by the hypotenuse of the right-angled triangle formed by the vectors.

Key Concept

Vector Addition of Perpendicular Components

Question 7632Question →

What are the coordinates of the point on the curve $y = 2x^2 - 5x + 1$ where the tangent line is perpendicular to the line $x + 3y - 4 = 0$ ?

(2, -1)

(2, 1)

(1, -2)

(-2, 19)

Show answer & explanation

Answer: $(2, -1)$

Answer

The point on the curve is

(2, -1)

.

Rearranging the line equation

x + 3y - 4 = 0

gives a gradient of

-\frac{1}{3}

. The tangent line is perpendicular, so its gradient must be

3

. Differentiating

y = 2x^2 - 5x + 1

gives

\frac{dy}{dx} = 4x - 5

. Setting

4x - 5 = 3

gives

x = 2

. Substituting

x = 2

into the curve equation yields

y = 2(2)^2 - 5(2) + 1 = -1

. Thus, the point is

(2, -1)

.

Step-by-Step Solution

1

Determine the gradient of the given line.

Rearranging

x + 3y - 4 = 0

into slope-intercept form gives

y = -\frac{1}{3}x + \frac{4}{3}

, so the gradient is

m_1 = -\frac{1}{3}

.

The slope of a linear equation

Ax + By + C = 0

is

-\frac{A}{B}

.

2

Calculate the gradient of the tangent line.

Since the tangent line is perpendicular to the given line, its gradient is

m_T = -\frac{1}{m_1} = 3

.

Perpendicular lines have gradients whose product is

-1

(

m_1 \cdot m_2 = -1

).

3

Find the derivative of the curve and set it equal to the tangent gradient.

\frac{dy}{dx} = \frac{d}{dx}(2x^2 - 5x + 1) = 4x - 5

. Setting

4x - 5 = 3

yields

4x = 8 \implies x = 2

.

The derivative

\frac{dy}{dx}

gives the gradient of the tangent to the curve at any point

x

.

4

Substitute the

x

-coordinate into the original curve equation to find

y

.

y = 2(2)^2 - 5(2) + 1 = 8 - 10 + 1 = -1

.

The point lies on the curve, so its coordinates must satisfy the curve's equation.

Key Concept

Tangents and Normals to Curves

Question 7633Question →

Two identical point charges, $q_1 = +5.0 \times 10^{-6}\text{ C}$ and $q_2 = +5.0 \times 10^{-6}\text{ C}$ , are fixed in a vacuum at Cartesian coordinates $(0\text{ m}, 3.0\text{ m})$ and $(0\text{ m}, -3.0\text{ m})$ , respectively. A third point charge $q_3 = +2.0 \times 10^{-6}\text{ C}$ is placed on the x-axis at $(4.0\text{ m}, 0\text{ m})$ . Taking Coulomb's constant $k = 9.0 \times 10^9\text{ N m}^2\text{ C}^{-2}$ , what is the magnitude of the net electrostatic force exerted on $q_3$ ?

5.76 \times 10^{-3}\text{ N}

7.20 \times 10^{-3}\text{ N}

4.32 \times 10^{-3}\text{ N}

3.60 \times 10^{-3}\text{ N}

Show answer & explanation

Answer: $5.76 \times 10^{-3}\text{ N}$

Answer

The magnitude of the net electrostatic force exerted on

q_3

is

5.76 \times 10^{-3}\text{ N}

.

Each charge exerts a repulsive force of magnitude

3.60 \times 10^{-3}\text{ N}

along the line connecting it to the test charge at

(4.0\text{ m}, 0\text{ m})

. Because of symmetry, the y-components of the two forces cancel out completely while their x-components add constructively. Multiplying the individual force magnitude by the cosine of the angle with the x-axis (

\cos\theta = 0.8

) and doubling for both charges gives a net force of

5.76 \times 10^{-3}\text{ N}

.

Step-by-Step Solution

1

Calculate the straight-line distance

r

from

q_1

(or

q_2

) to

q_3

.

r = \sqrt{(4.0 - 0)^2 + (0 - 3.0)^2} = \sqrt{16 + 9} = 5.0\text{ m}

.

Coulomb's Law requires the straight-line separation distance between interacting point charges.

2

Calculate the magnitude of the electrostatic force

F_1

exerted on

q_3

by

q_1

.

F_1 = \frac{k \cdot q_1 \cdot q_3}{r^2} = \frac{(9.0 \times 10^9) \cdot (5.0 \times 10^{-6}) \cdot (2.0 \times 10^{-6})}{5.0^2} = \frac{9.0 \times 10^{-2}}{25} = 3.60 \times 10^{-3}\text{ N}

.

By symmetry, the force magnitude

F_2

from

q_2

on

q_3

is also equal to

3.60 \times 10^{-3}\text{ N}

.

3

Determine the vector components of the forces along the axes.

The cosine of the angle

\theta

with the positive x-axis is

\cos\theta = \frac{4.0}{5.0} = 0.8

. The sine is

\sin\theta = \frac{3.0}{5.0} = 0.6

. The vertical y-components are equal in magnitude and opposite in direction (

F_{1y} = -F_{2y}

), canceling to zero.

Forces are vector quantities; symmetrically placed identical charges produce opposing vertical components and reinforcing horizontal components.

4

Sum the horizontal x-components to obtain the net force.

F_{\text{net}} = F_{1x} + F_{2x} = 2 \cdot F_1 \cos\theta = 2 \cdot (3.60 \times 10^{-3}\text{ N}) \cdot 0.8 = 5.76 \times 10^{-3}\text{ N}

.

Both x-components point along the positive x-axis, so their magnitudes add directly.

Key Concept

Vector superposition of electric forces

Question 7634Question →

What is the determinant of the matrix $A = \begin{pmatrix} 6 & -3 \\ 2 & 1 \end{pmatrix}$ ?

12

0

-12

-36

Show answer & explanation

Answer: 12

Answer

12

For any

2 \times 2

matrix

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

, the determinant is given by

ad - bc

. Substituting

a = 6

,

b = -3

,

c = 2

, and

d = 1

yields

(6)(1) - (-3)(2) = 6 - (-6) = 12

.

Step-by-Step Solution

1

Identify the elements of the 2x2 matrix

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

a = 6

,

b = -3

,

c = 2

,

d = 1

Standard entries for determinant formula

\det(A) = ad - bc

.

2

Compute the product of the main diagonal elements (

ad

) and off-diagonal elements (

bc

)

ad = (6)(1) = 6

and

bc = (-3)(2) = -6

These form the components of the 2x2 determinant formula.

3

Subtract

bc

from

ad

\det(A) = 6 - (-6) = 6 + 6 = 12

Subtracting a negative quantity is equivalent to addition.

Key Concept

Determinant of a 2x2 Matrix

Estimated Time:45s

Question 7635Question →

A laboratory technician recorded the temperature changes (in °C) of a chemical reaction across five trials as $3$ , $6$ , $7$ , $9$ , and $15$ . What is the variance of this set of data?

Show answer & explanation

Answer: 16

Answer

The variance of the temperature changes is 16.

The mean of the given numbers is

\bar{x} = \frac{3 + 6 + 7 + 9 + 15}{5} = 8

. The squared deviations from the mean are

(3-8)^2 = 25

,

(6-8)^2 = 4

,

(7-8)^2 = 1

,

(9-8)^2 = 1

, and

(15-8)^2 = 49

. Summing these squared deviations gives

25 + 4 + 1 + 1 + 49 = 80

. Dividing by the number of observations

N = 5

yields the variance:

\frac{80}{5} = 16

.

Step-by-Step Solution

1

Calculate the arithmetic mean (\bar{x}) of the dataset.

\bar{x} = \frac{3 + 6 + 7 + 9 + 15}{5} = 8

The mean is required to determine the deviation of each individual value from the central value.

2

Compute the square of the deviation for each data point from the mean.

(3-8)^2 = 25, (6-8)^2 = 4, (7-8)^2 = 1, (9-8)^2 = 1, (15-8)^2 = 49

Squaring deviations ensures all values are positive and emphasizes larger departures from the mean.

3

Sum the squared deviations and divide by the total number of observations (N = 5).

\text{Variance} = \frac{25 + 4 + 1 + 1 + 49}{5} = \frac{80}{5} = 16

Variance measures the average of the squared deviations from the mean.

Key Concept

Population Variance for Ungrouped Data

Question 7636Question →

In a survey of $100$ book club members regarding their reading preferences among Science Fiction ( $S$ ), Mystery ( $M$ ), and Historical Fiction ( $H$ ), it was found that $48$ read Science Fiction, $42$ read Mystery, and $38$ read Historical Fiction. Furthermore, $18$ read both Science Fiction and Mystery, $15$ read both Science Fiction and Historical Fiction, $14$ read both Mystery and Historical Fiction, and $8$ read all three genres. How many of the members read exactly one of these three genres?

Show answer & explanation

Answer: 58

Answer

The number of members who read exactly one of the three genres is

58

.

By isolating the disjoint regions of the 3-set Venn diagram, the number of members reading only Science Fiction is

23

, only Mystery is

18

, and only Historical Fiction is

17

. Adding these disjoint sets yields

58

.

Step-by-Step Solution

1

Determine exclusive two-set intersection regions

n(S \cap M \text{ only}) = 10

,

n(S \cap H \text{ only}) = 7

,

n(M \cap H \text{ only}) = 6

The given pairwise totals include members who read all three genres, so subtracting

n(S \cap M \cap H) = 8

isolates those in exactly two sets.

2

Determine exclusive single-set regions

n(S \text{ only}) = 23

,

n(M \text{ only}) = 18

,

n(H \text{ only}) = 17

Subtracting all overlapping regions within each set's boundary gives the number of members reading only that specific genre.

3

Sum the single-set regions

23 + 18 + 17 = 58

The set of members reading exactly one genre is the disjoint union of the three exclusive single-set regions.

Key Concept

3-Set Venn Diagram Cardinality and Disjoint Region Analysis

Question 7637Question →

A fair six-sided die is rolled $150$ times in a probability experiment. If the number $4$ lands face up $35$ times, what is the positive difference between the experimental probability and the theoretical probability of rolling a $4$ ?

\frac{1}{15}

\frac{7}{30}

\frac{1}{6}

\frac{2}{5}

Show answer & explanation

Answer: $\frac{1}{15}$

Answer

The positive difference between the experimental probability and the theoretical probability is

\frac{1}{15}

.

The theoretical probability of getting a 4 on a fair die is

\frac{1}{6}

, which simplifies to

\frac{5}{30}

. The experimental probability from

35

successes out of

150

trials is

\frac{35}{150} = \frac{7}{30}

. Taking the positive difference yields

\frac{7}{30} - \frac{5}{30} = \frac{2}{30} = \frac{1}{15}

.

Step-by-Step Solution

1

Calculate the theoretical probability of rolling a 4 on a fair six-sided die

P(\text{theoretical}) = \frac{1}{6}

A fair six-sided die has 6 equally likely outcomes, and exactly one outcome is 4.

2

Calculate the experimental probability based on the experiment results

P(\text{experimental}) = \frac{35}{150} = \frac{7}{30}

Experimental probability is the ratio of successful trials (35) to the total number of trials (150).

3

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

\text{Difference} = \frac{7}{30} - \frac{1}{6} = \frac{7}{30} - \frac{5}{30} = \frac{2}{30} = \frac{1}{15}

Converting to a common denominator of 30 allows straightforward subtraction.

Key Concept

Comparing experimental probability (relative frequency from trials) to theoretical probability (expected frequency based on equally likely outcomes).

Question 7638Question →

Find the smallest positive value of $\theta$ , in degrees, that satisfies the trigonometric equation $2\sin(3\theta - 30^\circ) = \sqrt{3}$ .

Show answer & explanation

Answer: 30

Answer

The smallest positive angle

\theta

is

30^\circ

.

To find the smallest positive value of

\theta

, first isolate the sine function by dividing both sides by 2 to obtain

\sin(3\theta - 30^\circ) = \frac{\sqrt{3}}{2}

. The smallest positive angle with a sine of

\frac{\sqrt{3}}{2}

is

60^\circ

. Setting

3\theta - 30^\circ = 60^\circ

yields

3\theta = 90^\circ

, which gives

\theta = 30^\circ

.

Step-by-Step Solution

1

Isolate the trigonometric ratio

\sin(3\theta - 30^\circ) = \frac{\sqrt{3}}{2}

Dividing both sides of

2\sin(3\theta - 30^\circ) = \sqrt{3}

by 2 simplifies the equation into standard form.

2

Determine the primary angle solution

3\theta - 30^\circ = 60^\circ

The smallest positive angle whose sine equals

\frac{\sqrt{3}}{2}

is

60^\circ

.

3

Solve the linear equation for

\theta

\theta = 30^\circ

Adding

30^\circ

to both sides gives

3\theta = 90^\circ

, and dividing by 3 yields

\theta = 30^\circ

.

Key Concept

Solving Trigonometric Equations with Linear Argument Transformations

Question 7639Question →

Three point charges $q_1 = +2.0 \times 10^{-6}\text{ C}$ , $q_2 = +2.0 \times 10^{-6}\text{ C}$ , and $q_3 = -4.0 \times 10^{-6}\text{ C}$ are placed along a straight line at positions $x = 0\text{ m}$ , $x = 0.30\text{ m}$ , and $x = 0.60\text{ m}$ , respectively. Taking Coulomb's constant $k = 9.0 \times 10^9\text{ N m}^2\text{ C}^{-2}$ , calculate the magnitude of the net electrostatic force acting on charge $q_2$ in newtons.

Show answer & explanation

Answer: 1.2

Answer

The magnitude of the net electrostatic force acting on charge

q_2

is

1.20\text{ N}

.

The force exerted on

q_2

by

q_1

is repulsive (

0.40\text{ N}

directed to the right) because both charges are positive. The force exerted on

q_2

by

q_3

is attractive (

0.80\text{ N}

directed to the right) because

q_2

is positive and

q_3

is negative. Since both component forces act in the same direction, the total net force magnitude is

0.40\text{ N} + 0.80\text{ N} = 1.20\text{ N}

.

Step-by-Step Solution

1

Calculate the repulsive force exerted by

q_1

on

q_2

F_{12} = 0.40\text{ N}

pointing to the right

Like charges repel each other, so

q_1

pushes

q_2

away along the

+x

-axis.

2

Calculate the attractive force exerted by

q_3

on

q_2

F_{32} = 0.80\text{ N}

pointing to the right

Unlike charges attract each other, so

q_3

pulls

q_2

towards itself along the

+x

-axis.

3

Sum the component electrostatic forces acting on

q_2

F_{\text{net}} = 0.40\text{ N} + 0.80\text{ N} = 1.20\text{ N}

Because both forces act in the exact same direction along the line, their magnitudes add directly.

Key Concept

Coulomb's Law and Principle of Superposition for Electrostatic Forces

Question 7640Question →

A binary operation $\Delta$ is defined on the set of real numbers $\mathbb{R}$ by $a \Delta b = a + 3b - 2ab$ . If $(3 \Delta x) \Delta 1 = 7$ , what is the value of $x$ ?

\frac{7}{3}

\frac{3}{5}

-\frac{7}{9}

-\frac{4}{3}

Show answer & explanation

Answer: $\frac{7}{3}$

Answer

\frac{7}{3}

Evaluating

3 \Delta x

gives

3 - 3x

. Then operating

(3 - 3x) \Delta 1

yields

(3 - 3x) + 3 - 2(3 - 3x) = 3x

. Equating

3x = 7

gives

x = \frac{7}{3}

.

Step-by-Step Solution

1

Evaluate the inner expression

3 \Delta x

using the definition

a \Delta b = a + 3b - 2ab

.

3 \Delta x = 3 + 3x - 2(3)(x) = 3 + 3x - 6x = 3 - 3x

.

Substitute

a = 3

and

b = x

into the operation rule.

2

Substitute the result

(3 - 3x)

as the first operand in the outer expression

(3 - 3x) \Delta 1

.

(3 - 3x) \Delta 1 = (3 - 3x) + 3(1) - 2(3 - 3x)(1)

.

Apply the binary operation definition with

a = 3 - 3x

and

b = 1

.

3

Expand and simplify the algebraic expression.

(3 - 3x) \Delta 1 = 3 - 3x + 3 - 6 + 6x = 3x

.

Distribute

-2

across

(3 - 3x)

to get

-6 + 6x

, then collect like terms.

4

Set the simplified expression equal to

7

and solve for

x

.

3x = 7 \implies x = \frac{7}{3}

.

Divide both sides by

3

to isolate

x

.

Key Concept

Non-commutative binary operation composition and algebraic equation solving

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