All practice questions

Question 7341Question →

The ratio of the measure of each interior angle to each exterior angle of a regular convex polygon is $7 : 2$ . What is the total number of diagonals of this polygon?

27

9

36

54

Show answer & explanation

Answer: 27

Answer

The total number of diagonals of the polygon is 27.

Since interior and exterior angles are supplementary, their ratio

7 : 2

divides

180^\circ

into 9 equal parts of

20^\circ

. The exterior angle is

2 \times 20^\circ = 40^\circ

. The number of sides

n

is

360^\circ / 40^\circ = 9

. Using the formula for diagonals

D = \frac{n(n-3)}{2}

, we obtain

D = \frac{9 \times 6}{2} = 27

.

Step-by-Step Solution

1

Calculate the measure of the exterior angle

Exterior angle

= \frac{2}{7 + 2} \times 180^\circ = 40^\circ

Interior and exterior angles at a vertex are supplementary, summing to

180^\circ

.

2

Determine the number of sides

n

n = \frac{360^\circ}{40^\circ} = 9

The sum of exterior angles of any convex polygon is

360^\circ

, so

n = \frac{360^\circ}{\text{exterior angle}}

.

3

Calculate the number of diagonals using

D = \frac{n(n - 3)}{2}

D = \frac{9(9 - 3)}{2} = \frac{9 \times 6}{2} = 27

Each vertex connects to

n - 3

non-adjacent vertices, and dividing by 2 avoids double-counting.

Key Concept

Interior and exterior angles of regular polygons and the polygon diagonal count formula.

Estimated Time:1m 30s

Question 7342Question →

The set of four numbers $\{x, 6, 8, 10\}$ has a variance of $5$ . Given that $x < 6$ , find the value of $x$ .

Show answer & explanation

Answer: 4

Answer

The value of

x

is 4.

By expressing the mean as

\bar{x} = \frac{x+24}{4}

and setting the sum of squared deviations divided by 4 equal to 5, we arrive at the quadratic equation

x^2 - 16x + 48 = 0

. Solving this equation gives two possible values,

x = 4

and

x = 12

. Applying the restriction

x < 6

uniquely identifies

x = 4

as the correct solution.

Step-by-Step Solution

1

Calculate the mean of the dataset in terms of

x

\bar{x} = \frac{x + 24}{4}

The mean of a dataset is the sum of all values divided by the total number of items.

2

Write the variance equation using the formula

\sigma^2 = \frac{\sum (x_i - \bar{x})^2}{N}

\frac{(x - \bar{x})^2 + (6 - \bar{x})^2 + (8 - \bar{x})^2 + (10 - \bar{x})^2}{4} = 5

The given variance is 5 and the number of observations

N = 4

.

3

Substitute

\bar{x} = \frac{x + 24}{4}

and expand the squared deviations

\frac{3}{16}x^2 - 3x + 14 = 5 \implies x^2 - 16x + 48 = 0

Expanding the squared terms and multiplying through by

\frac{16}{3}

yields a standard quadratic equation.

4

Solve the quadratic equation

x^2 - 16x + 48 = 0

for

x

(x - 4)(x - 12) = 0 \implies x = 4 \text{ or } x = 12

Factoring the quadratic equation gives two possible roots.

5

Apply the given condition

x < 6

x = 4

Since

x < 6

, we reject

x = 12

and select

x = 4

.

Key Concept

Variance of Ungrouped Data and Algebraic Problem Solving

Question 7343Question →

The frequency distribution table below shows the mass, in grams, of $50$ industrial steel bearings measured during a precision manufacturing audit:

Mass (g)	Frequency ( $f$ )
$10 - 19$	$5$
$20 - 29$	$12$
$30 - 39$	$x$
$40 - 49$	$y$
$50 - 59$	$8$

If the mean mass of the bearings is $34.7\text{ g}$ , what is the value of the missing frequency $x$ ?

Show answer & explanation

Answer: 18

Answer

18

The value of

x

is 18 because setting up the total frequency sum gives

x + y = 25

, and using class midpoints to compute the mean yields

1915 - 10x = 1735

, which solves to

x = 18

.

Step-by-Step Solution

1

Express the relationship between the missing frequencies using total frequency.

x + y = 25

or

y = 25 - x

The total number of industrial steel bearings is 50, so

5 + 12 + x + y + 8 = 50

.

2

Determine the midpoint (

m

) of each class interval.

Midpoints are 14.5, 24.5, 34.5, 44.5, and 54.5 respectively.

The class midpoint is calculated as

\frac{\text{lower limit} + \text{upper limit}}{2}

.

3

Formulate the equation for the sum of products of frequencies and midpoints.

\sum fm = 802.5 + 34.5x + 44.5y

Multiply each class midpoint by its corresponding frequency and sum the results.

4

Substitute

y = 25 - x

and solve for

x

using the mean formula.

x = 18

Setting

\frac{1915 - 10x}{50} = 34.7

yields

1915 - 10x = 1735

, which gives

10x = 180

and thus

x = 18

.

Key Concept

Measures of Central Tendency for Grouped Data - Mean with Unknown Frequencies

Estimated Time:3m 0s

Question 7344Question →

A composite plane figure is formed from a rectangle $ABCD$ measuring $20\text{ cm}$ by $14\text{ cm}$ . A semicircle with diameter $BC = 14\text{ cm}$ is attached externally along side $BC$ . At the opposite end, an isosceles triangle $AED$ with base $AD = 14\text{ cm}$ and perpendicular height $6\text{ cm}$ is cut out from the interior of the rectangle. Taking $\pi = \frac{22}{7}$ , what is the total area of the figure in $\text{cm}^2$ ?

Show answer & explanation

Answer: 315

Answer

The total area of the composite figure is 315 cm².

The area of the complete shape is found by adding the area of the external semicircle (

77\text{ cm}^2

) to the area of the rectangle (

280\text{ cm}^2

) and subtracting the area of the removed triangle (

42\text{ cm}^2

), yielding a net area of

315\text{ cm}^2

.

Step-by-Step Solution

1

Calculate the area of the rectangular portion

ABCD

Area of rectangle

= 20\text{ cm} \times 14\text{ cm} = 280\text{ cm}^2

The main section of the figure before modifications is a rectangle with length

20\text{ cm}

and width

14\text{ cm}

.

2

Calculate the area of the externally attached semicircle

Area of semicircle

= \frac{1}{2} \times \frac{22}{7} \times 7^2 = 77\text{ cm}^2

The diameter of the semicircle is

BC = 14\text{ cm}

, giving a radius

r = 7\text{ cm}

. Area is given by

\frac{1}{2}\pi r^2

.

3

Calculate the area of the removed internal triangle

AED

Area of triangle

= \frac{1}{2} \times 14\text{ cm} \times 6\text{ cm} = 42\text{ cm}^2

The triangle has base

AD = 14\text{ cm}

and height

6\text{ cm}

. Area is calculated as

\frac{1}{2} \times \text{base} \times \text{height}

.

4

Combine the areas to find the total remaining area of the figure

Total Area

= 280\text{ cm}^2 + 77\text{ cm}^2 - 42\text{ cm}^2 = 315\text{ cm}^2

The external semicircle increases the total area while the internal cut-out triangle reduces it.

Key Concept

Area of Composite Plane Figures

Estimated Time:2m 0s

Question 7345Question →

Two parallel lines, $L_1$ and $L_2$ , are situated in a plane with $L_1$ above $L_2$ . A regular polygon of $n$ sides has one of its sides, $AB$ , lying entirely on line $L_1$ . An adjacent side, $BC$ , extends downwards into the region between $L_1$ and $L_2$ . A line segment $BD$ is drawn from vertex $B$ perpendicular to line $L_2$ , meeting $L_2$ at point $D$ . The segment $BD$ lies inside the interior angle $\angle ABC$ of the polygon and divides $\angle ABC$ into two angles, $\angle ABD$ and $\angle DBC$ , such that $\angle ABD : \angle DBC = 3 : 2$ . What is the total number of diagonals of this regular polygon?

Show answer & explanation

Answer: 54

Answer

54

Because line L_1 is parallel to line L_2 and BD is perpendicular to L_2, BD is also perpendicular to L_1, making angle ABD equal to 90°. Based on the 3:2 ratio, angle DBC equals 60°, giving a total interior angle of 150°. Solving (n - 2) × 180° / n = 150° yields n = 12 sides. Substituting n = 12 into the diagonal formula n(n - 3) / 2 gives 54 diagonals.

Step-by-Step Solution

1

Determine the measure of angle ABD using parallel line properties

\angle ABD = 90^\circ

Because line L_1 is parallel to line L_2 and line segment BD is perpendicular to L_2, BD is also perpendicular to L_1.

2

Calculate the measure of angle DBC from the given angle ratio

\angle DBC = 60^\circ

With \angle ABD : \angle DBC = 3 : 2 and \angle ABD = 90^\circ, setting 3 ratio units equal to 90^\circ gives 1 unit = 30^\circ, so 2 units = 60^\circ.

3

Find the total interior angle measure of the regular polygon

\angle ABC = 150^\circ

The interior angle \angle ABC is the sum of \angle ABD and \angle DBC, which gives 90^\circ + 60^\circ = 150^\circ.

4

Determine the number of sides n of the regular polygon

n = 12

Using the regular polygon interior angle formula \frac{(n - 2) \times 180^\circ}{n} = 150^\circ, solving for n yields 180n - 360 = 150n \implies 30n = 360 \implies n = 12.

5

Calculate the total number of diagonals

54

Applying the polygon diagonal formula D = \frac{n(n - 3)}{2} with n = 12 gives D = \frac{12 \times (12 - 3)}{2} = \frac{12 \times 9}{2} = 54.

Key Concept

Combining parallel line perpendicularity properties, angle partitioning, regular polygon interior angle measure, and diagonal count calculation

Question 7346Question →

The total cost $C$ of printing a school magazine is partly constant and partly varies directly as the number of copies $n$ printed. If it costs $\text{N}70,000$ to print $500$ copies and $\text{N}140,000$ to print $1,200$ copies, what is the cost of printing $2,000$ copies?

\text{N}200,000

\text{N}220,000

\text{N}240,000

\text{N}280,000

Show answer & explanation

Answer: $\text{N}220,000$

Answer

\text{N}220,000

In partial variation, the total cost

C

is represented by

C = k_1 + k_2 n

. Subtracting the two linear equations

70,000 = k_1 + 500 k_2

and

140,000 = k_1 + 1,200 k_2

gives

700 k_2 = 70,000

, so

k_2 = 100

. Substituting

k_2 = 100

yields the fixed constant

k_1 = 20,000

. Substituting

n = 2,000

into

C = 20,000 + 100n

yields

20,000 + 200,000 = \text{N}220,000

.

Step-by-Step Solution

1

Set up the partial variation equation

C = k_1 + k_2 n

, where

k_1

is the fixed cost and

k_2

is the rate per copy

Partial variation consists of a constant term and a term that varies directly with the independent variable.

2

Substitute given values to form simultaneous equations

Equation 1:

70,000 = k_1 + 500 k_2

Equation 2:

140,000 = k_1 + 1,200 k_2

Using the two given data points

(n=500, C=70,000)

and

(n=1,200, C=140,000)

creates a system of linear equations.

3

Solve for the variation constants

k_1

and

k_2

Subtracting Equation 1 from Equation 2 gives

70,000 = 700 k_2 \implies k_2 = 100

.
Substituting

k_2 = 100

into Equation 1 gives

70,000 = k_1 + 500(100) \implies k_1 = 20,000

.

Determines the specific values for the fixed overhead and rate per copy.

4

Calculate the total cost for

2,000

copies

C = 20,000 + 100(2,000) = 20,000 + 200,000 = \text{N}220,000

Evaluates the completed formula

C = 20,000 + 100n

at

n = 2,000

.

Key Concept

Partial Variation and Simultaneous Equations

Estimated Time:1m 30s

Question 7347Question →

A set of $7$ numbers: $8, 12, 14, 15, 17, 18, 21$ has a mean of $15$ . When two additional numbers, $x$ and $y$ (where $x < y$ ), are included in the dataset, the mean of all $9$ numbers becomes $17$ . Given that the mode of the $9$ numbers is $18$ , what is the median of the combined set of $9$ numbers?

15

17

18

24

Show answer & explanation

Answer: $17$

Answer

The median of the combined set of

9

numbers is

17

.

The sum of the original 7 numbers is 105. For 9 numbers with a mean of 17, the sum is 153, meaning the two new numbers add up to 48. Since 18 is the mode, it must appear more than once, so one of the new numbers is 18 and the other is 30. Arranging the 9 numbers in order yields 8, 12, 14, 15, 17, 18, 18, 21, 30. The middle (5th) term is 17.

Step-by-Step Solution

1

Calculate the sum of the original 7 numbers and the total sum required for 9 numbers

Sum of 7 numbers =

8 + 12 + 14 + 15 + 17 + 18 + 21 = 105

. Sum of 9 numbers =

9 \times 17 = 153

.

The mean formula

\text{Mean} = \frac{\sum x}{n}

gives total sum =

\text{Mean} \times n

.

2

Determine the values of the two added numbers

x

and

y

x + y = 153 - 105 = 48

. Since

18

is the mode,

x = 18

and

y = 30

.

The original set has all distinct numbers. For

18

to be the mode,

18

must repeat, so one of the added values must be

18

.

3

Order all 9 numbers and find the median

Ordered set:

8, 12, 14, 15, 17, 18, 18, 21, 30

. Median (

5^{\text{th}}

term) =

17

.

For an odd number of items

n = 9

, the median is the

\frac{n+1}{2} = 5^{\text{th}}

term in ascending order.

Key Concept

Combining mean, mode, and median properties for ungrouped data

Estimated Time:1m 30s

Question 7348Question →

A student calculated the value of $\frac{0.048}{0.006}$ as $8.5$ . Calculate the percentage error in the student's calculation.

Show answer & explanation

Answer: 6.25

Answer

The percentage error is

6.25\%

.

Evaluating

0.048 \div 0.006

yields a true value of

8

. The error in estimation is

|8.5 - 8| = 0.5

. Dividing this error by the true value of

8

and multiplying by

100\%

gives a percentage error of

6.25\%

.

Step-by-Step Solution

1

Calculate the true value of the expression

True value =

\frac{0.048}{0.006} = 8

The exact result is required to establish the baseline for percentage error calculation.

2

Determine the magnitude of the error

Error =

|8.5 - 8| = 0.5

Error is defined as the absolute difference between the estimated value and the true value.

3

Compute the percentage error

Percentage error =

\frac{0.5}{8} \times 100\% = 6.25\%

Percentage error expresses the error as a percentage of the true value.

Key Concept

Percentage Error

Estimated Time:45s

Question 7349Question →

What is the numerical value of the limit:

\lim_{x \to 0} \frac{1 - \cos(4x)}{x \sin(2x)}

?

4

2

8

0

Show answer & explanation

Answer: 4

Answer

The numerical value of the limit is 4.

Applying the double-angle trigonometric identity

1 - \cos(4x) = 2\sin^2(2x)

reduces the expression to

\frac{2\sin(2x)}{x}

. Rewriting this as

4 \cdot \frac{\sin(2x)}{2x}

and taking the limit as

x \to 0

using the standard limit

\lim_{\theta \to 0} \frac{\sin\theta}{\theta} = 1

yields 4.

Step-by-Step Solution

1

Identify the form of the limit

Direct substitution of

x = 0

gives

\frac{1 - \cos(0)}{0 \cdot \sin(0)} = \frac{0}{0}

, which is an indeterminate form.

Indeterminate forms require algebraic simplification or trigonometric identities before evaluating the limit.

2

Apply trigonometric identity

Use

1 - \cos(4x) = 2\sin^2(2x)

to rewrite the numerator.

This transforms the numerator into a form containing sine terms matching the denominator.

3

Simplify the algebraic expression

\lim_{x \to 0} \frac{2\sin^2(2x)}{x\sin(2x)} = \lim_{x \to 0} \frac{2\sin(2x)}{x}

Cancel out the common

\sin(2x)

factor from numerator and denominator for

x \neq 0

.

4

Evaluate using the standard trigonometric limit

\lim_{x \to 0} 4 \cdot \frac{\sin(2x)}{2x} = 4 \cdot 1 = 4

Since

\lim_{\theta \to 0} \frac{\sin\theta}{\theta} = 1

, setting

\theta = 2x

gives

\lim_{x \to 0} \frac{\sin(2x)}{2x} = 1

.

Key Concept

Limits of Trigonometric Functions and Indeterminate Forms

Question 7350Question →

If $\log_2 x - 3\log_x 16 = 1$ , what is the product of all real values of $x$ that satisfy the equation?

2

16

-12

8

Show answer & explanation

Answer: $2$

Answer

The product of all real values of

x

satisfying the equation is

2

.

Using the change of base formula,

\log_x 16 = \frac{4}{\log_2 x}

. Substituting

u = \log_2 x

yields

u^2 - u - 12 = 0

, giving

u = 4

or

u = -3

. Converting back to

x

gives

x = 16

or

x = \frac{1}{8}

. Their product is

16 \times \frac{1}{8} = 2

.

Step-by-Step Solution

1

Apply the change of base formula to express

\log_x 16

in base 2.

\log_x 16 = \frac{\log_2 16}{\log_2 x} = \frac{4}{\log_2 x}

Standard logarithmic base conversion allows all terms to share base 2.

2

Substitute the change of base result into the original equation.

\log_2 x - 3 \left(\frac{4}{\log_2 x}\right) = 1 \implies \log_2 x - \frac{12}{\log_2 x} = 1

Simplifies the algebraic expression into a single logarithmic variable.

3

Substitute

u = \log_2 x

and transform into a quadratic equation.

u - \frac{12}{u} = 1 \implies u^2 - u - 12 = 0

Clears the fraction by multiplying through by

u

(where

u \neq 0

).

4

Factor the quadratic equation to find values of

u

.

(u - 4)(u + 3) = 0 \implies u = 4 \text{ or } u = -3

Determines the logarithmic roots.

5

Solve for

x

using the exponential form

x = 2^u

.

x_1 = 2^4 = 16

and

x_2 = 2^{-3} = \frac{1}{8}

Converts the logarithmic solution back to the original variable

x

.

6

Calculate the product of the roots

x_1 \cdot x_2

.

16 \times \frac{1}{8} = 2

Fulfills the specific evaluation requirement asked in the stem.

Key Concept

Logarithmic Change of Base and Quadratic Reducible Logarithmic Equations

Estimated Time:2m 0s

Question 7351Question →

In a survey of 120 final year secondary school students in Lagos, 47 offer Physics ( $P$ ), 57 offer Chemistry ( $C$ ), and 48 offer Biology ( $B$ ). Records show that 17 students offer both Physics and Chemistry, 20 offer both Chemistry and Biology, and 15 offer both Physics and Biology. If the number of students who offer none of these three subjects is three times the number of students who offer all three subjects, find the total number of students who offer exactly one of the three subjects.

Show answer & explanation

Answer: 63

Answer

63 students offer exactly one of the three subjects.

By applying the principle of inclusion-exclusion for three sets, the total union is

|P \cup C \cup B| = 47 + 57 + 48 - (17 + 20 + 15) + x = 100 + x

, where

x

is the number of students taking all three subjects. Setting the universal set total to 120 gives

(100 + x) + 3x = 120

, solving to

x = 5

. Subtracting the relevant intersections yields 20 students for Physics only, 25 for Chemistry only, and 18 for Biology only. Summing these gives 63.

Step-by-Step Solution

1

Define unknown variables and set up the inclusion-exclusion formula for the union of three sets.

|P \cup C \cup B| = 100 + x

, where

x = |P \cap C \cap B|

.

The Principle of Inclusion-Exclusion states that

|P \cup C \cup B| = |P| + |C| + |B| - (|P \cap C| + |C \cap B| + |P \cap B|) + |P \cap C \cap B|

.

2

Formulate and solve an equation for the total number of students in the universal set.

120 = (100 + x) + 3x \implies 4x = 20 \implies x = 5

.

The total number of students equals those taking at least one subject plus those taking none (

3x

).

3

Determine the number of students taking exactly two subjects.

Physics and Chemistry only =

17 - 5 = 12

; Chemistry and Biology only =

20 - 5 = 15

; Physics and Biology only =

15 - 5 = 10

.

Subtract the triple intersection count (

x = 5

) from each pairwise intersection count.

4

Calculate the number of students offering only Physics, only Chemistry, and only Biology.

Physics only =

47 - (12 + 10 + 5) = 20

; Chemistry only =

57 - (12 + 15 + 5) = 25

; Biology only =

48 - (10 + 15 + 5) = 18

.

Subtract the sum of the two-subject-only regions and the three-subject region from each set's total cardinality.

5

Sum the counts of students offering exactly one subject.

20 + 25 + 18 = 63

.

The total offering exactly one subject is the sum of the three disjoint single-subject regions.

Key Concept

3-Set Principle of Inclusion-Exclusion and Venn Diagram Region Decomposition

Question 7352Question →

A point $P(x, y)$ moves in the Cartesian plane such that its distance from the origin $O(0, 0)$ is always half of its distance from the fixed point $Q(6, 0)$ . Which of the following equations represents the locus of $P$ ?

x^2 + y^2 + 4x - 12 = 0

x^2 + y^2 - 4x - 12 = 0

x^2 + y^2 + 12x - 36 = 0

x^2 + y^2 - 16x + 48 = 0

Show answer & explanation

Answer: $x^2 + y^2 + 4x - 12 = 0$

Answer

The equation of the locus of

P

is

x^2 + y^2 + 4x - 12 = 0

.

The correct equation

x^2 + y^2 + 4x - 12 = 0

is derived by expressing the condition

PO = \frac{1}{2} PQ

as

2 \cdot PO = PQ

, squaring both sides to get

4(x^2 + y^2) = (x - 6)^2 + y^2

, and simplifying to standard circle form.

Step-by-Step Solution

1

Express the distance condition algebraically.

The distance from

P(x,y)

to

O(0,0)

is

PO = \sqrt{x^2 + y^2}

, and the distance from

P(x,y)

to

Q(6,0)

is

PQ = \sqrt{(x-6)^2 + y^2}

. Given

PO = \frac{1}{2} PQ

, we have

2 \cdot PO = PQ

.

Translate the geometric distance description into algebraic expressions.

2

Square both sides of the equation to eliminate square roots.

4(x^2 + y^2) = (x - 6)^2 + y^2

.

Squaring removes the radicals; note that

(2 \cdot PO)^2 = 4 \cdot PO^2

.

3

Expand and group like terms.

4x^2 + 4y^2 = x^2 - 12x + 36 + y^2 \implies 3x^2 + 12x + 3y^2 - 36 = 0

.

Expand

(x - 6)^2 = x^2 - 12x + 36

and collect terms on one side.

4

Divide the entire equation by 3 to standard form.

x^2 + y^2 + 4x - 12 = 0

.

Simplifying by the common factor of 3 gives the equation of a circle representing the Circle of Apollonius.

Key Concept

Locus of a point with a constant ratio of distances from two fixed points (Circle of Apollonius)

Estimated Time:2m 0s

Question 7353Question →

The electrical resistance $R$ of a wire varies directly as its length $L$ and inversely as the square of its diameter $d$ . A wire of length $50\text{ m}$ and diameter $2\text{ mm}$ has a resistance of $5\text{ ohms}$ . Calculate the resistance (in ohms) of a wire made of the same material with a length of $80\text{ m}$ and a diameter of $4\text{ mm}$ .

Show answer & explanation

Answer: 2

Answer

The resistance of the wire is

2\text{ ohms}

.

The variation model is

R = \frac{k L}{d^2}

. Substituting

R = 5\text{ ohms}

,

L = 50\text{ m}

, and

d = 2\text{ mm}

gives

5 = \frac{50k}{4}

, so

k = 0.4

. Substituting

k = 0.4

,

L = 80\text{ m}

, and

d = 4\text{ mm}

yields

R = \frac{0.4 \times 80}{4^2} = \frac{32}{16} = 2\text{ ohms}

.

Step-by-Step Solution

1

Set up the variation formula

R = \frac{k L}{d^2}

Resistance varies directly as length

L

and inversely as the square of diameter

d

.

2

Calculate the constant of variation

k

k = 0.4

Substitute the initial values

R = 5

,

L = 50

, and

d = 2

into the variation equation.

3

Compute the new resistance

R

R = 2\text{ ohms}

Substitute

k = 0.4

,

L = 80

, and

d = 4

into the formula.

Key Concept

Direct and Inverse Joint Variation

Estimated Time:1m 30s

Question 7354Question →

Five packages delivered by a courier service have masses of $5\text{ kg}$ , $8\text{ kg}$ , $11\text{ kg}$ , $12\text{ kg}$ , and $14\text{ kg}$ . What is the mean deviation of the masses of these packages?

0.0\text{ kg}

2.8\text{ kg}

10.0\text{ kg}

14.0\text{ kg}

Show answer & explanation

Answer: $2.8\text{ kg}$

Answer

2.8\text{ kg}

The mean of the data set is

10\text{ kg}

. The distances of each data value from the mean are

5

,

2

,

1

,

2

, and

4

. The sum of these distances is

14

, and dividing by

5

yields a mean deviation of

2.8\text{ kg}

.

Step-by-Step Solution

1

Calculate the arithmetic mean

(\bar{x})

of the dataset

\bar{x} = \frac{5 + 8 + 11 + 12 + 14}{5} = \frac{50}{5} = 10\text{ kg}

Mean deviation requires the central mean value as a reference point for all deviations.

2

Find the absolute deviation

|x - \bar{x}|

for each value in the dataset

|5 - 10| = 5, |8 - 10| = 2, |11 - 10| = 1, |12 - 10| = 2, |14 - 10| = 4

Mean deviation measures the average distance of values from the mean regardless of sign.

3

Sum the absolute deviations and divide by the sample size

(n = 5)

\text{Mean Deviation} = \frac{5 + 2 + 1 + 2 + 4}{5} = \frac{14}{5} = 2.8\text{ kg}

Dividing total absolute deviation by the total count yields the mean deviation.

Key Concept

Mean Deviation

Question 7355Question →

The cumulative frequency distribution table below shows the completion times (in minutes) for $120$ software engineers during a technical skill test:

Time Interval (min)	Frequency ( $f$ )	Cumulative Frequency ( $cf$ )
$10 - 19$	$12$	$12$
$20 - 29$	$28$	$40$
$30 - 39$	$40$	$80$
$40 - 49$	$24$	$104$
$50 - 59$	$16$	$120$

Using linear interpolation from the cumulative frequency data, what is the $65^{\text{th}}$ percentile completion time?

39.0\text{ minutes}

29.0\text{ minutes}

39.5\text{ minutes}

34.5\text{ minutes}

Show answer & explanation

Answer: $39.0\text{ minutes}$

Answer

39.0\text{ minutes}

To find the

65^{\text{th}}

percentile (

P_{65}

), we calculate

\frac{65}{100} \times 120 = 78

. The cumulative frequency table shows that rank

78

falls within the class interval

30 - 39

. The true lower boundary for this class is

L = 29.5

, the cumulative frequency of the preceding class is

F = 40

, the frequency of the percentile class is

f = 40

, and the class width is

c = 10

. Substituting into

P_{65} = L + \left(\frac{78 - F}{f}\right)c

yields

29.5 + \left(\frac{38}{40}\right) \times 10 = 39.0\text{ minutes}

.

Step-by-Step Solution

1

Determine the rank of the

65^{\text{th}}

percentile (

P_{65}

).

Rank

= \frac{65}{100} \times 120 = 78

.

The percentile rank identifies the position of the data point within the total frequency

N = 120

.

2

Identify the percentile class and its boundaries.

Percentile class is

30 - 39

, with lower boundary

L = 29.5

, upper boundary

= 39.5

, and class width

c = 10

.

The cumulative frequency increases from

40

to

80

across this interval, containing rank

78

.

3

Apply the linear interpolation formula for percentiles on grouped data.

P_{65} = L + \left(\frac{\frac{65N}{100} - F}{f}\right) \times c = 29.5 + \left(\frac{78 - 40}{40}\right) \times 10 = 29.5 + 9.5 = 39.0\text{ minutes}

.

Where

F = 40

is the cumulative frequency prior to the class and

f = 40

is the frequency of the class.

Key Concept

Linear Interpolation of Percentiles from Cumulative Frequency Distributions

Question 7356Question →

What is the derivative of the function $f(x) = 3x^2 + 5x$ with respect to $x$ , obtained using differentiation from first principles?

6x + 5

6x - 5

3x + 5

6x + 5 + 3h

Show answer & explanation

Answer: $6x + 5$

Answer

The derivative of the function is

6x + 5

.

Using the first-principles formula

f'(x) = \lim_{h \to 0} \frac{f(x+h) - f(x)}{h}

, we expand

f(x+h) = 3(x+h)^2 + 5(x+h) = 3x^2 + 6xh + 3h^2 + 5x + 5h

. Subtracting

f(x) = 3x^2 + 5x

yields

6xh + 3h^2 + 5h

. Dividing by

h

produces

6x + 3h + 5

. Taking the limit as

h \to 0

gives

6x + 5

.

Step-by-Step Solution

1

Evaluate

f(x+h)

for

f(x) = 3x^2 + 5x

f(x+h) = 3(x+h)^2 + 5(x+h) = 3(x^2 + 2xh + h^2) + 5x + 5h = 3x^2 + 6xh + 3h^2 + 5x + 5h

Substitute

(x+h)

into the original function expression and expand algebraically.

2

Find the difference

f(x+h) - f(x)

f(x+h) - f(x) = (3x^2 + 6xh + 3h^2 + 5x + 5h) - (3x^2 + 5x) = 6xh + 3h^2 + 5h

Subtract the original function

f(x)

to find the net change in output.

3

Divide the difference by

h

to set up the difference quotient

\frac{f(x+h) - f(x)}{h} = \frac{6xh + 3h^2 + 5h}{h} = 6x + 3h + 5

Cancel out

h

from each term in the numerator.

4

Take the limit as

h \to 0

f'(x) = \lim_{h \to 0} (6x + 3h + 5) = 6x + 5

Evaluate the expression as

h

approaches 0 to determine the instantaneous rate of change.

Key Concept

Differentiation from First Principles

Estimated Time:1m 0s

Question 7357Question →

Given that $\tan \theta + \cot \theta = 4$ for an acute angle $\theta$ , what is the exact decimal value of $\sin^4 \theta + \cos^4 \theta$ ?

Show answer & explanation

Answer: 0.875

Answer

The exact decimal value of

\sin^4 \theta + \cos^4 \theta

is 0.875.

By writing

\tan \theta + \cot \theta

as

\frac{\sin \theta}{\cos \theta} + \frac{\cos \theta}{\sin \theta} = \frac{\sin^2 \theta + \cos^2 \theta}{\sin \theta \cos \theta} = 4

, we find that

\sin \theta \cos \theta = \frac{1}{4}

. Squaring the fundamental identity

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

gives

\sin^4 \theta + 2\sin^2 \theta \cos^2 \theta + \cos^4 \theta = 1

. Isolating

\sin^4 \theta + \cos^4 \theta

yields

1 - 2(\sin \theta \cos \theta)^2 = 1 - 2\left(\frac{1}{16}\right) = 1 - \frac{1}{8} = \frac{7}{8} = 0.875

.

Step-by-Step Solution

1

Rewrite the expression

\tan \theta + \cot \theta = 4

using sine and cosine ratios

\frac{\sin \theta}{\cos \theta} + \frac{\cos \theta}{\sin \theta} = 4

Applies the fundamental quotient identities for tangent and cotangent.

2

Combine fractions over a common denominator and apply the Pythagorean identity

\frac{\sin^2 \theta + \cos^2 \theta}{\sin \theta \cos \theta} = \frac{1}{\sin \theta \cos \theta} = 4 \implies \sin \theta \cos \theta = \frac{1}{4}

Utilizes the identity

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

.

3

Relate

\sin^4 \theta + \cos^4 \theta

to

(\sin^2 \theta + \cos^2 \theta)^2

\sin^4 \theta + \cos^4 \theta = (\sin^2 \theta + \cos^2 \theta)^2 - 2\sin^2 \theta \cos^2 \theta = 1 - 2(\sin \theta \cos \theta)^2

Uses the algebraic identity

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

where

a = \sin^2 \theta

and

b = \cos^2 \theta

.

4

Substitute

\sin \theta \cos \theta = \frac{1}{4}

into the algebraic relation

1 - 2\left(\frac{1}{4}\right)^2 = 1 - 2\left(\frac{1}{16}\right) = 1 - \frac{1}{8} = \frac{7}{8} = 0.875

Evaluates the expression to obtain the final decimal result.

Key Concept

Trigonometric Identities and Algebraic Polynomial Expansion

Estimated Time:2m 0s

Question 7358Question →

Find the real value of $x$ that satisfies the equation $\log_2 (2^{2x} + 12) - x = 3 - \log_2 (2^x - 1)$ .

1

2

3

0

Show answer & explanation

Answer: 1

Answer

1

By bringing all terms involving

x

together, the equation becomes

\log_2(2^{2x}+12) + \log_2(2^x-1) - \log_2(2^x) = 3

. Combining these logarithmic terms yields

\log_2\left(\frac{(2^{2x}+12)(2^x-1)}{2^x}\right) = 3

. Converting to exponential form gives

\frac{(2^{2x}+12)(2^x-1)}{2^x} = 8

. Setting

u = 2^x

converts this into the cubic equation

u^3 - u^2 + 4u - 12 = 0

. Factoring out

(u - 2)

leaves

(u - 2)(u^2 + u + 6) = 0

. Since

u^2 + u + 6 = 0

has no real solutions,

u = 2

is the only real root. Hence

2^x = 2

, which gives

x = 1

.

Step-by-Step Solution

1

Rearrange the equation to group all logarithmic terms on one side.

\log_2 (2^{2x} + 12) + \log_2 (2^x - 1) - x = 3

Grouping logarithmic terms allows the application of logarithm combination laws.

2

Express

x

as

\log_2 (2^x)

and combine the logarithms using log laws.

\log_2 \left( \frac{(2^{2x} + 12)(2^x - 1)}{2^x} \right) = 3

Product and quotient laws of logarithms allow single logarithmic expression formation:

\log_2 A + \log_2 B - \log_2 C = \log_2\left(\frac{AB}{C}\right)

.

3

Convert from logarithmic to exponential form.

\frac{(2^{2x} + 12)(2^x - 1)}{2^x} = 2^3 = 8

Definition of logarithm:

\log_b(Y) = Z \implies Y = b^Z

.

4

Substitute

u = 2^x

(where

u > 1

) to form an algebraic equation.

\frac{(u^2 + 12)(u - 1)}{u} = 8 \implies (u^2 + 12)(u - 1) = 8u

Since

2^{2x} = (2^x)^2 = u^2

, substitution simplifies the exponential expression into a polynomial.

5

Expand and simplify into a standard cubic equation.

u^3 - u^2 + 12u - 12 = 8u \implies u^3 - u^2 + 4u - 12 = 0

Standard polynomial form enables root searching via polynomial division and factoring.

6

Factor the cubic equation.

(u - 2)(u^2 + u + 6) = 0

Testing

u = 2

gives

2^3 - 2^2 + 4(2) - 12 = 8 - 4 + 8 - 12 = 0

, confirming

(u - 2)

is a factor.

7

Solve for real values of

u

and

x

.

u = 2 \implies 2^x = 2 \implies x = 1

The quadratic factor

u^2 + u + 6 = 0

has discriminant

\Delta = 1 - 24 = -23 < 0

(no real roots). Thus

u = 2

is the unique solution.

Key Concept

Logarithmic equations requiring exponential substitution and polynomial factorization

Estimated Time:2m 30s

Question 7359Question →

What is the $y$ -intercept of the tangent line to the curve $y = x^3 - 2x + 4$ at the point where $x = 1$ ?

Show answer & explanation

Answer: 2

Answer

The y-intercept of the tangent line is 2.

The curve evaluated at

x=1

gives point

(1,3)

. The derivative

y'=3x^2-2

gives slope

m=1

at

x=1

. The tangent line equation is

y-3=1(x-1)

, which simplifies to

y=x+2

. The

y

-intercept occurs at

x=0

, giving

y=2

.

Step-by-Step Solution

1

Find the y-coordinate of the point on the curve at

x = 1

.

At

x = 1

,

y = (1)^3 - 2(1) + 4 = 3

. The point of tangency is

(1, 3)

.

The point of tangency lies on the curve.

2

Find the gradient function of the curve by differentiation.

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

.

The first derivative represents the slope of the tangent line.

3

Calculate the slope of the tangent line at

x = 1

.

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

.

Substituting

x = 1

into the derivative yields the slope at that specific point.

4

Formulate the equation of the tangent line.

y - 3 = 1(x - 1) \implies y = x + 2

.

Use point-slope form

y - y_1 = m(x - x_1)

with point

(1,3)

and slope

m=1

.

5

Find the

y

-intercept of the tangent line.

Setting

x = 0

gives

y = 2

.

The

y

-intercept is the value of

y

where the line crosses the vertical axis.

Key Concept

Tangents and Normals to Curves

Question 7360Question →

The table below presents the distribution of weekly expenditure (in thousands of Naira) of $20$ small-scale farmers in a rural community:

Weekly Expenditure (\text{N}'000)	Frequency ( $f$ )
10 – 19	3
20 – 29	7
30 – 39	6
40 – 49	4

What is the mean weekly expenditure of the farmers?

\text{N}25.5\text{ thousand}

\text{N}29.5\text{ thousand}

\text{N}30.0\text{ thousand}

\text{N}34.5\text{ thousand}

Show answer & explanation

Answer: $\text{N}30.0\text{ thousand}$

Answer

The mean weekly expenditure is

\text{N}30.0\text{ thousand}

(or

\text{N}30,000

).

The mean of grouped data is computed using the formula

\bar{x} = \frac{\sum fx}{\sum f}

. Finding the midpoints (

14.5, 24.5, 34.5, 44.5

), multiplying by their respective frequencies (

3, 7, 6, 4

), and dividing the sum (

600

) by the total frequency (

20

) yields

\text{N}30.0\text{ thousand}

.

Step-by-Step Solution

1

Calculate the midpoint (

x

) for each class interval.

Midpoints are:

14.5

for 10–19;

24.5

for 20–29;

34.5

for 30–39; and

44.5

for 40–49.

Grouped data mean calculations require representing each interval by its class mark (midpoint).

2

Multiply each class midpoint (

x

) by its corresponding frequency (

f

) to get

f \cdot x

, and sum these values.

\sum (f \cdot x) = (3 \times 14.5) + (7 \times 24.5) + (6 \times 34.5) + (4 \times 44.5) = 43.5 + 171.5 + 207.0 + 178.0 = 600

.

This yields the total estimated sum of all expenditures across all observations.

3

Divide the total sum

\sum (f \cdot x)

by the total frequency

\sum f

.

Mean

\bar{x} = \frac{600}{20} = 30.0

.

The mean formula for grouped data is

\bar{x} = \frac{\sum fx}{\sum f}

.

Key Concept

Calculation of Mean for Grouped Frequency Data

Estimated Time:1m 0s

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