All practice questions

Question 7501Question →

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 7502Question →

An arithmetic progression has a first term of $5$ and a common difference of $4$ . What is the $12^{\text{th}}$ term of this progression?

49

53

57

45

Show answer & explanation

Answer: $49$

Answer

49

The

n^{\text{th}}

term of an arithmetic progression is determined using the formula

T_n = a + (n - 1)d

. Substituting

a = 5

,

d = 4

, and

n = 12

yields

T_{12} = 5 + 11 \times 4 = 49

, making

49

the correct value.

Step-by-Step Solution

1

Identify the given values from the problem statement

First term

a = 5

, common difference

d = 4

, and position

n = 12

These are the standard variables required for calculating terms in an arithmetic progression.

2

Apply the

n^{\text{th}}

term formula for an arithmetic progression

T_{12} = 5 + (12 - 1) \times 4

The standard formula for the

n^{\text{th}}

term of an AP is

T_n = a + (n - 1)d

.

3

Simplify the expression to find the final value

T_{12} = 5 + 11 \times 4 = 5 + 44 = 49

Perform multiplication before addition according to standard order of operations.

Key Concept

Arithmetic Progression

n^{\text{th}}

Term Formula

Estimated Time:45s

Question 7503Question →

If $\left( \frac{16}{81} \right)^{x-1} \times \left( \frac{27}{8} \right)^{x+2} = \frac{9}{4}$ , find the value of $x$ .

12

8

-4

10

Show answer & explanation

Answer: $8$

Answer

The value of

x

is

8

.

By writing

\frac{16}{81}

as

\left(\frac{2}{3}\right)^4

,

\frac{27}{8}

as

\left(\frac{2}{3}\right)^{-3}

, and

\frac{9}{4}

as

\left(\frac{2}{3}\right)^{-2}

, the equation simplifies via exponent addition to

\left(\frac{2}{3}\right)^{4x - 4 - 3x - 6} = \left(\frac{2}{3}\right)^{-2}

. Equating indices gives

x - 10 = -2

, which yields the correct solution

x = 8

.

Step-by-Step Solution

1

Express all fractional terms with a common base of

\frac{2}{3}

\frac{16}{81} = \left(\frac{2}{3}\right)^4

,

\frac{27}{8} = \left(\frac{3}{2}\right)^3 = \left(\frac{2}{3}\right)^{-3}

, and

\frac{9}{4} = \left(\frac{3}{2}\right)^2 = \left(\frac{2}{3}\right)^{-2}

Converting all terms to a single common base allows exponents to be combined using the laws of indices.

2

Substitute the common base expressions back into the original equation

\left( \left(\frac{2}{3}\right)^4 \right)^{x-1} \times \left( \left(\frac{2}{3}\right)^{-3} \right)^{x+2} = \left(\frac{2}{3}\right)^{-2}

Applying the power of a power law

(a^m)^n = a^{mn}

to simplify each term.

3

Apply the power law and multiplication law of indices

\left(\frac{2}{3}\right)^{4(x-1)} \times \left(\frac{2}{3}\right)^{-3(x+2)} = \left(\frac{2}{3}\right)^{-2} \implies \left(\frac{2}{3}\right)^{4(x-1) - 3(x+2)} = \left(\frac{2}{3}\right)^{-2}

When multiplying exponential terms with identical bases, add their exponents:

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

.

4

Equate the exponents and solve for

x

4(x-1) - 3(x+2) = -2 \implies 4x - 4 - 3x - 6 = -2 \implies x - 10 = -2 \implies x = 8

Since the bases on both sides are equal and non-zero, their indices must be equal.

Key Concept

Laws of Indices: Base Conversion and Exponential Equations

Estimated Time:2m 0s

Question 7504Question →

Let the universal set be $\mathcal{U} = \{x \in \mathbb{Z} : 1 \le x \le 25\}$ . If $A = \{x \in \mathcal{U} : x \text{ is a multiple of } 3\}$ and $B = \{x \in \mathcal{U} : x \text{ is a perfect square}\}$ , what is the cardinality of $(A \cup B)'$ ?

13

12

14

17

Show answer & explanation

Answer: 13

Answer

13

The total number of elements in the universal set is 25. Set A has 8 elements (multiples of 3) and set B has 5 elements (perfect squares). The element 9 belongs to both sets. Therefore, the union of A and B contains 8 + 5 - 1 = 12 elements. Subtracting this from the universal set size gives 25 - 12 = 13 elements in the complement.

Step-by-Step Solution

1

Determine the elements of the universal set and its cardinality.

\mathcal{U} = \{1, 2, 3, \dots, 25\}

, so

n(\mathcal{U}) = 25$.

The universal set contains all integers from 1 to 25 inclusive.

2

List the elements of set A and set B.

A = \{3, 6, 9, 12, 15, 18, 21, 24\}

(so

n(A) = 8

) and

B = \{1, 4, 9, 16, 25\}

(so

n(B) = 5

).

Set A contains multiples of 3 within the domain, and set B contains perfect squares within the domain.

3

Find the intersection

A \cap B

and calculate the cardinality of the union

A \cup B

.

A \cap B = \{9\}

, so

n(A \cap B) = 1

. Thus,

n(A \cup B) = n(A) + n(B) - n(A \cap B) = 8 + 5 - 1 = 12

.

The principle of inclusion-exclusion avoids double-counting the common element 9.

4

Calculate the cardinality of the complement

(A \cup B)'

.

n((A \cup B)') = n(\mathcal{U}) - n(A \cup B) = 25 - 12 = 13

.

The complement consists of all elements in the universal set that do not belong to the union of A and B.

Key Concept

Complement of Set Union and Inclusion-Exclusion Principle

Question 7505Question →

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 7506Question →

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 7507Question →

Given the matrices $A = \begin{pmatrix} 2 & x \\ -1 & 3 \end{pmatrix}$ and $B = \begin{pmatrix} 1 & 0 \\ 4 & 2 \end{pmatrix}$ , if $\det(A^{-1} B^T) = \frac{1}{5}$ , what is the value of $x$ ?

4

-4

14

-1

Show answer & explanation

Answer: 4

Answer

The value of

x

is 4.

Using the properties

\det(B^T) = \det(B) = 2

and

\det(A^{-1}) = \frac{1}{\det(A)} = \frac{1}{6+x}

, the product rule gives

\det(A^{-1}B^T) = \frac{2}{6+x}

. Equating this to

\frac{1}{5}

gives

6+x = 10

, so

x = 4

.

Step-by-Step Solution

1

Calculate the determinants of matrices

A

and

B

.

\det(A) = (2)(3) - (-1)(x) = 6 + x

and

\det(B) = (1)(2) - (0)(4) = 2$.

The determinant of a 2x2 matrix is computed using

ad - bc

.

2

Apply determinant rules for inverse and transpose matrices.

\det(B^T) = \det(B) = 2

and

\det(A^{-1}) = \frac{1}{\det(A)} = \frac{1}{6+x}$.

The determinant of a transpose equals the original determinant, and the determinant of an inverse is its reciprocal.

3

Express

\det(A^{-1} B^T)

using the product rule of determinants.

\det(A^{-1} B^T) = \det(A^{-1}) \cdot \det(B^T) = \frac{1}{6+x} \cdot 2 = \frac{2}{6+x}$.

For square matrices,

\det(MN) = \det(M)\det(N)

.

4

Equate the expression to

\frac{1}{5}

and solve for

x

.

\frac{2}{6+x} = \frac{1}{5} \implies 10 = 6 + x \implies x = 4$.

Cross-multiplying and simplifying gives

x = 4

.

Key Concept

Properties of Determinants (Product, Inverse, and Transpose)

Question 7508Question →

A box contains $6$ red balls and $4$ blue balls. A ball is drawn at random, its color is recorded, and it is then returned to the box. A second ball is subsequently drawn at random. What is the probability of selecting a red ball on the first draw and a blue ball on the second draw?

\frac{6}{25}

\frac{4}{15}

\frac{1}{10}

1

Show answer & explanation

Answer: $\frac{6}{25}$

Answer

\frac{6}{25}

Because the first ball is replaced after being recorded, the two draws are independent compound events. The probability of drawing a red ball first is

\frac{6}{10} = \frac{3}{5}

, and the probability of drawing a blue ball second is

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

. Multiplying these probabilities yields

P(\text{Red and Blue}) = \frac{3}{5} \times \frac{2}{5} = \frac{6}{25}

.

Step-by-Step Solution

1

Calculate the probability of drawing a red ball on the first draw

P(\text{Red}) = \frac{6}{6 + 4} = \frac{6}{10} = \frac{3}{5}

There are

6

red balls out of a total of

10

balls.

2

Determine independence and calculate the probability of drawing a blue ball on the second draw

P(\text{Blue}) = \frac{4}{10} = \frac{2}{5}

Because the first ball is replaced before the second draw, the total number of balls and their composition remain unchanged.

3

Apply the multiplication law for independent compound events

P(\text{Red and then Blue}) = P(\text{Red}) \times P(\text{Blue}) = \frac{3}{5} \times \frac{2}{5} = \frac{6}{25}

For independent events

A

and

B

,

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

.

Key Concept

Multiplication Law of Probability for Independent Events

Estimated Time:45s

Question 7509Question →

Given the universal set $U = \{1, 2, 3, 4, 5, 6, 7, 8, 9, 10\}$ and a subset $P = \{x \in U : x \text{ is a prime number}\}$ , which of the following sets represents the complement of $P$ , denoted as $P'$ ?

\{1, 4, 6, 8, 9, 10\}

\{4, 6, 8, 9, 10\}

\{1, 3, 5, 7, 9\}

\{2, 4, 6, 8, 10\}

Show answer & explanation

Answer: $\{1, 4, 6, 8, 9, 10\}$

Answer

The set

\{1, 4, 6, 8, 9, 10\}

The prime numbers in

U

are

2, 3, 5,

and

7

, making

P = \{2, 3, 5, 7\}

. The complement

P'

consists of all elements in

U

that do not belong to

P

, which yields

\{1, 4, 6, 8, 9, 10\}

.

Step-by-Step Solution

1

Identify the elements of the universal set

U

U = \{1, 2, 3, 4, 5, 6, 7, 8, 9, 10\}

The universal set defines the boundary of all possible elements under consideration.

2

Identify the elements belonging to set

P

(prime numbers in

U

)

P = \{2, 3, 5, 7\}

Prime numbers have exactly two distinct positive divisors: 1 and themselves. Note that 1 is not a prime number.

3

Compute the complement

P' = U \setminus P

P' = \{1, 4, 6, 8, 9, 10\}

The complement of set

P

consists of all elements present in

U

that are not present in

P

.

Key Concept

Set Complement and Universal Set Boundaries

Estimated Time:45s

Question 7510Question →

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 7511Question →

If $y = \ln(\cos(3x))$ , what is $\frac{dy}{dx}$ ?

-3\tan(3x)

3\tan(3x)

-\tan(3x)

-3\cot(3x)

Show answer & explanation

Answer: $-3\tan(3x)$

Answer

-3\tan(3x)

Differentiating

y = \ln(\cos(3x))

requires using the chain rule twice: first for

\ln(u)

, giving

\frac{1}{\cos(3x)}

, and then for

\cos(3x)

, giving

-3\sin(3x)

. Multiplying these results yields

\frac{-3\sin(3x)}{\cos(3x)}

, which simplifies to

-3\tan(3x)

.

Step-by-Step Solution

1

Apply the chain rule formula for differentiating logarithmic functions

\frac{d}{dx}[\ln(u)] = \frac{1}{u}\frac{du}{dx}

\frac{dy}{dx} = \frac{1}{\cos(3x)} \cdot \frac{d}{dx}(\cos(3x))

The outer function is

\ln(u)

where

u = \cos(3x)

.

2

Differentiate the inner trigonometric function

\cos(3x)

\frac{d}{dx}(\cos(3x)) = -3\sin(3x)

By the chain rule,

\frac{d}{dx}[\cos(kx)] = -k\sin(kx)

.

3

Multiply the terms and simplify using trigonometric identities

\frac{dy}{dx} = \frac{-3\sin(3x)}{\cos(3x)} = -3\tan(3x)

Using the identity

\frac{\sin(\theta)}{\cos(\theta)} = \tan(\theta)

, the derivative simplifies to

-3\tan(3x)

.

Key Concept

Differentiation of Logarithmic and Trigonometric Functions via Chain Rule

Estimated Time:1m 30s

Question 7512Question →

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 7513Question →

Two independent security systems, $X$ and $Y$ , operate in a facility. The probability that system $Y$ detects an intrusion is $0.20$ greater than the probability that system $X$ detects it. If the probability that at least one of the systems detects an intrusion is $0.92$ , what is the probability that exactly one of the systems detects the intrusion?

0.44

0.48

0.52

0.32

Show answer & explanation

Answer: $0.44$

Answer

The probability that exactly one of the systems detects the intrusion is

0.44

.

The correct answer is

0.44

. Setting

P(X) = p

and

P(Y) = p + 0.20

, we use the independence rule

P(X \cap Y) = P(X)P(Y)

in the addition law

P(X \cup Y) = P(X) + P(Y) - P(X)P(Y) = 0.92

. Solving

p^2 - 1.80p + 0.72 = 0

gives

P(X) = 0.60

and

P(Y) = 0.80

. The probability of both detecting the intrusion is

0.60 \times 0.80 = 0.48

. Subtracting the probability of both from the probability of at least one (

0.92 - 0.48

) gives

0.44

for exactly one system detecting the intrusion.

Step-by-Step Solution

1

Define variables for the individual probabilities

Let

P(X) = p

. Then

P(Y) = p + 0.20

.

System

Y

's probability is given as

0.20

greater than System

X

's probability.

2

Apply the addition law for independent events

P(X \cup Y) = P(X) + P(Y) - P(X \cap Y) = p + (p + 0.20) - p(p + 0.20) = 0.92

Since

X

and

Y

are independent,

P(X \cap Y) = P(X)P(Y)

.

3

Solve the quadratic equation for

p

2p + 0.20 - p^2 - 0.20p = 0.92 \implies -p^2 + 1.80p - 0.72 = 0 \implies p^2 - 1.80p + 0.72 = 0

. Factoring yields

(p - 0.60)(p - 1.20) = 0

. Since

p \le 1

,

p = 0.60

.

Probability values cannot exceed

1

, so

p = 0.60

is the valid root.

4

Calculate individual probabilities and the probability of both occurring

P(X) = 0.60

,

P(Y) = 0.80

, and

P(X \cap Y) = 0.60 \times 0.80 = 0.48

.

These are needed to evaluate the compound probability of exactly one event occurring.

5

Calculate the probability that exactly one system detects the intrusion

P(\text{exactly one}) = P(X \cup Y) - P(X \cap Y) = 0.92 - 0.48 = 0.44

.

The probability of exactly one event occurring is the probability of at least one minus the probability of both.

Key Concept

Probability laws for independent and compound events

Question 7514Question →

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 7515Question →

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

Question 7516Question →

When the polynomial $P(x) = 3x^4 - 2x^3 + ax^2 + bx - 12$ is divided by $(x^2 - 4)$ , the remainder is $5x - 4$ . What is the value of $a + b$ ?

Show answer & explanation

Answer: 3

Answer

The value of

a + b

is

3

.

By using the Remainder Theorem for the quadratic divisor

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

, evaluating

P(2) = 6

gives

2a + b = -7

, and evaluating

P(-2) = -14

gives

2a - b = -33

. Solving these linear equations simultaneously yields

a = -10

and

b = 13

, which sums to

a + b = 3

.

Step-by-Step Solution

1

Set up the polynomial division relation.

P(x) = (x - 2)(x + 2)Q(x) + (5x - 4)

By the Remainder Theorem and Division Algorithm, dividing by

(x^2 - 4)

yields a remainder of

R(x) = 5x - 4

.

2

Find the values of

P(2)

and

P(-2)

from the remainder.

P(2) = 6

and

P(-2) = -14

Substituting the roots of the divisor

x = 2

and

x = -2

eliminates the quotient term

(x^2 - 4)Q(x)

.

3

Substitute

x = 2

into the polynomial

P(x)

and set equal to

6

.

2a + b = -7

3(16) - 2(8) + 4a + 2b - 12 = 20 + 4a + 2b = 6 \implies 4a + 2b = -14

.

4

Substitute

x = -2

into the polynomial

P(x)

and set equal to

-14

.

2a - b = -33

3(16) - 2(-8) + 4a - 2b - 12 = 52 + 4a - 2b = -14 \implies 4a - 2b = -66

.

5

Solve the system of equations for

a

and

b

.

a = -10

,

b = 13

, and

a + b = 3

Adding the two linear equations gives

4a = -40 \implies a = -10

. Substituting

a = -10

into

2a + b = -7

gives

b = 13

.

Key Concept

Polynomial Remainder Theorem for Non-Linear Divisors

Estimated Time:2m 30s

Question 7517Question →

An arithmetic progression (A.P.) has a first term of $7$ and a common difference of $5$ . What is the $12^{\text{th}}$ term of the progression?

Show answer & explanation

Answer: 62

Answer

The

12^{\text{th}}

term of the progression is

62

.

Using the formula for the

n^{\text{th}}

term of an arithmetic progression,

T_n = a + (n - 1)d

, with

a = 7

,

d = 5

, and

n = 12

, we calculate

T_{12} = 7 + (12 - 1) \times 5 = 7 + 55 = 62

.

Step-by-Step Solution

1

Identify known parameters from the question

a = 7

,

d = 5

,

n = 12

These parameters are given in the problem statement.

2

Use the general formula for the

n^{\text{th}}

term of an arithmetic progression

T_n = a + (n - 1)d

This formula connects the first term, common difference, and term number to the value of the term.

3

Substitute values and evaluate

T_{12} = 7 + (12 - 1) \times 5 = 7 + 55 = 62

Performing basic arithmetic gives the final result.

Key Concept

Finding the

n^{\text{th}}

term of an Arithmetic Progression

Estimated Time:45s

Question 7518Question →

Which of the following types of waves requires a physical material medium for its propagation?

Sound waves

Radio waves

Infrared rays

Gamma rays

Show answer & explanation

Answer: Sound waves

Answer

Sound waves are mechanical waves and require a material medium to propagate.

Sound waves are mechanical longitudinal waves. They propagate through compressions and rarefactions of particles in a medium, meaning they cannot travel through a vacuum.

Step-by-Step Solution

1

Classify waves based on their propagation requirements.

Waves are classified broadly into mechanical waves (which require a medium) and electromagnetic waves (which do not require a medium).

Understanding the fundamental distinction between mechanical and electromagnetic propagation.

2

Identify the nature of each given wave option.

Radio waves, infrared rays, and gamma rays are electromagnetic waves. Sound waves are mechanical waves.

Mechanical waves depend on the elastic interactions of physical particles to transmit energy.

Key Concept

Classification of waves into mechanical and electromagnetic based on medium requirement

Estimated Time:45s

Question 7519Question →

An isosceles triangle has a perimeter of $36\text{ cm}$ and a base of length $16\text{ cm}$ . Calculate the area of the triangle in $\text{cm}^2$ .

Show answer & explanation

Answer: 48

Answer

The area of the isosceles triangle is

48\text{ cm}^2

.

The two equal sides of the isosceles triangle measure

\frac{36 - 16}{2} = 10\text{ cm}

each. An altitude dropped perpendicularly to the base bisects the

16\text{ cm}

base into two

8\text{ cm}

segments. By the Pythagorean theorem, the perpendicular height is

h = \sqrt{10^2 - 8^2} = 6\text{ cm}

. Therefore, the area is

\frac{1}{2} \times 16 \times 6 = 48\text{ cm}^2

.

Step-by-Step Solution

1

Determine the length of the two equal sides

Each equal side is

10\text{ cm}

Subtract the base length from the total perimeter (

36 - 16 = 20\text{ cm}

) and divide by

2

.

2

Calculate the perpendicular height (altitude) to the base

Height

h = 6\text{ cm}

The perpendicular altitude bisects the base into two

8\text{ cm}

segments, creating right-angled triangles with hypotenuse

10\text{ cm}

. Use Pythagoras' theorem:

h = \sqrt{10^2 - 8^2} = 6\text{ cm}

.

3

Calculate the area of the triangle

Area =

48\text{ cm}^2

Multiply half the base by the perpendicular height:

\frac{1}{2} \times 16 \times 6 = 48\text{ cm}^2

.

Key Concept

Perimeter and Area of Isosceles Triangles using Pythagorean Theorem

Estimated Time:1m 30s

Question 7520Question →

A binary operation $\star$ defined on the set of real numbers $\mathbb{R}$ is given by $a \star b = 2a - 3b + ab$ . If $(2 \star x) \star 3 = 16$ , what is the value of $x$ ?

\frac{11}{4}

-1

-\frac{1}{5}

-\frac{3}{5}

Show answer & explanation

Answer: $-1$

Answer

-1

Applying the binary operation rule

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

step-by-step gives

2 \star x = 4 - x

. Operating on this with

3

yields

(4 - x) \star 3 = 2(4 - x) - 3(3) + 3(4 - x) = 11 - 5x

. Setting

11 - 5x = 16

gives

x = -1

.

Step-by-Step Solution

1

Evaluate the inner expression

2 \star x

using the given definition

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

.

2 \star x = 2(2) - 3(x) + (2)(x) = 4 - 3x + 2x = 4 - x

The expression inside the parentheses must be simplified first.

2

Substitute

4 - x

into the outer operation

(4 - x) \star 3

.

(4 - x) \star 3 = 2(4 - x) - 3(3) + (4 - x)(3) = 8 - 2x - 9 + 12 - 3x = 11 - 5x

Apply the operation rule with first element

a = 4 - x

and second element

b = 3

.

3

Equate the resulting expression to 16 and solve for

x

.

11 - 5x = 16 \implies -5x = 5 \implies x = -1

Solve the linear equation to determine the value of

x

.

Key Concept

Non-commutative nested binary operation evaluation

Estimated Time:2m 0s

Answer

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