All practice questions

Question 4201Question →

Match each physical quantity listed on the left with its correct SI classification or fundamental unit decomposition on the right.

Click a left item, then click its matching right item

Show answer & explanation

Answer

Electric current corresponds to the fundamental quantity measured in amperes (

\text{A}

); Pressure corresponds to the derived quantity with base SI units

\text{kg}\cdot\text{m}^{-1}\cdot\text{s}^{-2}

; Thermodynamic temperature corresponds to the fundamental quantity measured in kelvin (

\text{K}

); Impulse corresponds to the derived quantity with base SI units

\text{kg}\cdot\text{m}\cdot\text{s}^{-1}

.

Electric current and thermodynamic temperature are fundamental SI quantities defined independently with units ampere (

\text{A}

) and kelvin (

\text{K}

). Pressure and impulse are derived quantities whose definitions rely on fundamental quantities, breaking down into

\text{kg}\cdot\text{m}^{-1}\cdot\text{s}^{-2}

and

\text{kg}\cdot\text{m}\cdot\text{s}^{-1}

respectively.

Step-by-Step Solution

1

Identify fundamental physical quantities

Electric current and thermodynamic temperature are fundamental quantities with SI base units ampere (

\text{A}

) and kelvin (

\text{K}

) respectively.

Fundamental quantities are basic physical quantities that do not depend on any other physical quantity for their definition.

2

Decompose pressure into base SI units

\text{Pressure} = \frac{\text{Force}}{\text{Area}} = \frac{\text{kg}\cdot\text{m}\cdot\text{s}^{-2}}{\text{m}^2} = \text{kg}\cdot\text{m}^{-1}\cdot\text{s}^{-2}

.

Derived quantities must be reduced to combinations of mass (

\text{kg}

), length (

\text{m}

), and time (

\text{s}

).

3

Decompose impulse into base SI units

\text{Impulse} = \text{Force} \times \text{Time} = (\text{kg}\cdot\text{m}\cdot\text{s}^{-2}) \times \text{s} = \text{kg}\cdot\text{m}\cdot\text{s}^{-1}

.

Impulse is defined as change in momentum or force applied over a time interval.

Key Concept

Classification of fundamental quantities versus derived quantities and resolution of derived units into fundamental SI units.

Question 4202Question →

If $y = (2x - 1)^3 (x^2 + 3)^2$ , what is the value of $\frac{dy}{dx}$ at $x = 1$ ?

112

56

80

104

Show answer & explanation

Answer: $112$

Answer

112

Applying the product rule

\frac{dy}{dx} = u'v + uv'

along with the chain rule gives

u'(1) = 6

,

v(1) = 16

,

u(1) = 1

, and

v'(1) = 16

. Evaluating

u'(1)v(1) + u(1)v'(1)

yields

6 \times 16 + 1 \times 16 = 96 + 16 = 112

.

Step-by-Step Solution

1

Decompose the function into two components for the Product Rule

Let

u(x) = (2x - 1)^3

and

v(x) = (x^2 + 3)^2

, so

y = u(x) v(x)

.

The function is a product of two composite expressions.

2

Differentiate each term using the Chain Rule

u'(x) = 3(2x - 1)^2 \cdot \frac{d}{dx}(2x - 1) = 6(2x - 1)^2

v'(x) = 2(x^2 + 3) \cdot \frac{d}{dx}(x^2 + 3) = 4x(x^2 + 3)

Applying the Chain Rule requires differentiating the outer function and multiplying by the derivative of the inner function.

3

Apply the Product Rule formula

\frac{dy}{dx} = u'(x)v(x) + u(x)v'(x) = 6(2x - 1)^2 (x^2 + 3)^2 + 4x(2x - 1)^3 (x^2 + 3)

The derivative of a product

u \cdot v

is

u'v + uv'

.

4

Evaluate the derivative at

x = 1

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

u'(1) = 6(2(1) - 1)^2 = 6

v(1) = (1^2 + 3)^2 = 16

v'(1) = 4(1)(1^2 + 3) = 16

\left.\frac{dy}{dx}\right|_{x=1} = (6)(16) + (1)(16) = 96 + 16 = 112

Substitute

x = 1

into all expressions to find the numerical derivative.

Key Concept

Combined Product Rule and Chain Rule of Differentiation

Estimated Time:2m 0s

Question 4203Question →

A simple pendulum of length $l$ has a period of oscillation of $2.0\text{ s}$ when suspended with a bob of mass $50\text{ g}$ . If the bob is replaced with one of mass $200\text{ g}$ while maintaining the exact same string length, what is the new period of oscillation?

2.0\text{ s}

4.0\text{ s}

1.0\text{ s}

8.0\text{ s}

Show answer & explanation

Answer: $2.0\text{ s}$

Answer

The period of oscillation remains

2.0\text{ s}

.

The period of oscillation of a simple pendulum undergoing simple harmonic motion is given by

T = 2\pi \sqrt{\frac{l}{g}}

, where

l

is the length of the pendulum and

g

is the acceleration due to gravity. Because the mass of the bob does not appear in this equation, changing the mass from

50\text{ g}

to

200\text{ g}

does not alter the period. Therefore, the period remains

2.0\text{ s}

.

Step-by-Step Solution

1

Identify the formula for the period of a simple pendulum.

T = 2\pi \sqrt{\frac{l}{g}}

The period

T

depends on the length of the pendulum

l

and local acceleration due to gravity

g

.

2

Analyze the dependence of period on bob mass.

The mass parameter

m

does not appear in the formula

T = 2\pi \sqrt{\frac{l}{g}}

.

The period of a simple pendulum is independent of the mass of the bob.

3

Determine the new period of oscillation.

T_{new} = 2.0\text{ s}

Since length

l

and acceleration due to gravity

g

remain unchanged, the period stays

2.0\text{ s}

.

Key Concept

Independence of simple pendulum period from bob mass

Question 4204Question →

The table below shows the distribution of marks obtained by $40$ candidates in a computer-based recruitment test:

Mark Interval	Frequency ( $f$ )
$1 - 10$	$4$
$11 - 20$	$8$
$21 - 30$	$12$
$31 - 40$	$6$
$41 - 50$	$10$

What is the estimated mean mark of the candidates?

23.5

25.5

28.0

32.5

Show answer & explanation

Answer: $28.0$

Answer

The estimated mean mark of the candidates is

28.0

.

To find the mean of grouped frequency data, each class interval must be represented by its midpoint (

x

). Finding the midpoints (

5.5, 15.5, 25.5, 35.5, 45.5

) and multiplying by their frequencies gives total products summing to

1120

. Dividing by the total number of candidates (

40

) yields

28.0

.

Step-by-Step Solution

1

Calculate the midpoint (

x

) for each class interval

Class midpoints are:

5.5

,

15.5

,

25.5

,

35.5

, and

45.5

.

The class mark or midpoint is the representative value for data within a grouped interval.

2

Multiply each class midpoint (

x

) by its corresponding frequency (

f

) to find

f \cdot x

4 \times 5.5 = 22

,

8 \times 15.5 = 124

,

12 \times 25.5 = 306

,

6 \times 35.5 = 213

,

10 \times 45.5 = 455

.

Determines the total weighted value contribution of each class.

3

Sum all

f \cdot x

values and calculate the total frequency

\sum f

\sum f \cdot x = 22 + 124 + 306 + 213 + 455 = 1120

, and

\sum f = 4 + 8 + 12 + 6 + 10 = 40

.

Provides the overall sum of data estimates and total sample size needed for the grouped mean formula.

4

Apply the grouped mean formula

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

\bar{x} = \frac{1120}{40} = 28.0.

Computes the estimated mean for grouped frequency data.

Key Concept

Grouped Mean Calculation using Class Midpoints

Estimated Time:1m 30s

Question 4205Question →

What is the gradient of the normal to the curve $y = x^2 + 2x - 1$ at the point where $x = 1$ ?

-\frac{1}{4}

4

\frac{1}{4}

-4

Show answer & explanation

Answer: $-\frac{1}{4}$

Answer

The gradient of the normal to the curve at

x = 1

is

-\frac{1}{4}

.

Differentiating

y = x^2 + 2x - 1

gives

\frac{dy}{dx} = 2x + 2

. Substituting

x = 1

yields a tangent slope of

4

. Taking the negative reciprocal gives

-\frac{1}{4}

, which is the correct gradient of the normal.

Step-by-Step Solution

1

Find the derivative of the curve equation to determine the general tangent gradient function.

\frac{dy}{dx} = 2x + 2

The derivative of a function gives the slope of the tangent line at any point

x

.

2

Evaluate the tangent gradient at the specific point

x = 1

.

m = 2(1) + 2 = 4

Substituting

x = 1

into the derivative gives the slope of the tangent at that specific x-coordinate.

3

Calculate the gradient of the normal line using the perpendicular condition.

m_n = -\frac{1}{m} = -\frac{1}{4}

The normal line is perpendicular to the tangent line, so its gradient is the negative reciprocal of the tangent gradient.

Key Concept

Relationship between tangent and normal gradients (

m_{\text{normal}} = -\frac{1}{m_{\text{tangent}}}

)

Estimated Time:1m 0s

Question 4206Question →

The sum of the first four terms of an arithmetic progression (A.P.) is $32$ , and the sum of the next four terms is $96$ . What is the common difference of the progression?

Show answer & explanation

Answer: 4

Answer

The common difference of the arithmetic progression is

4

.

The sum of the first four terms yields

2a + 3d = 16

. The sum of the first eight terms is

32 + 96 = 128

, which gives

2a + 7d = 32

. Subtracting these two linear equations gives

4d = 16

, leading to

d = 4

.

Step-by-Step Solution

1

Formulate an equation for the sum of the first 4 terms.

2a + 3d = 16

The sum of the first

n

terms of an A.P. is

S_n = \frac{n}{2}(2a + (n-1)d)

. Substituting

n = 4

and

S_4 = 32

gives

2(2a + 3d) = 32

, which simplifies to

2a + 3d = 16

.

2

Formulate an equation for the sum of the first 8 terms.

2a + 7d = 32

The total sum of the first 8 terms is the sum of the first 4 terms plus the sum of the next 4 terms (

S_8 = 32 + 96 = 128

). Substituting

n = 8

gives

4(2a + 7d) = 128

, which simplifies to

2a + 7d = 32

.

3

Solve the system of simultaneous linear equations for

d

.

d = 4

Subtracting

(2a + 3d = 16)

from

(2a + 7d = 32)

eliminates

2a

, resulting in

4d = 16

, which yields

d = 4

.

Key Concept

Sum of an Arithmetic Progression

Question 4207Question →

The first term of a geometric progression (GP) is $2$ and its common ratio is $3$ . What is the $4^{\text{th}}$ term of the progression?

54

162

11

24

Show answer & explanation

Answer: $54$

Answer

The

4^{\text{th}}

term of the geometric progression is

54

.

For a geometric progression with first term

a

and common ratio

r

, the

n^{\text{th}}

term is given by

T_n = a r^{n-1}

. Substituting

a = 2

,

r = 3

, and

n = 4

yields

T_4 = 2 \times 3^3 = 2 \times 27 = 54

.

Step-by-Step Solution

1

Identify the given parameters of the geometric progression.

First term

a = 2

, common ratio

r = 3

, and term position

n = 4

.

These values are directly provided in the question statement.

2

Apply the general formula for the

n^{\text{th}}

term of a geometric progression,

T_n = a r^{n-1}

.

T_4 = 2 \times 3^{4-1} = 2 \times 3^3

.

The exponent of the common ratio is always one less than the term index

n

.

3

Evaluate the exponent and multiply by the first term.

3^3 = 27

, so

T_4 = 2 \times 27 = 54

.

Performing standard arithmetic yields the exact term value.

Key Concept

Formula for the nth term of a Geometric Progression:

T_n = a r^{n-1}

Estimated Time:45s

Question 4208Question →

In an agricultural experiment, two crop varieties, $X$ and $Y$ , are tested independently for germination under drought conditions. The probability that variety $X$ germinates is $0.65$ , and the probability that at least one of the two varieties germinates is $0.86$ . What is the probability that variety $Y$ germinates?

Show answer & explanation

Answer: 0.6

Answer

The probability that variety Y germinates is 0.6

Using the law of addition for independent events

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

, substituting

P(X) = 0.65

and

P(X \cup Y) = 0.86

gives

0.86 = 0.65 + 0.35 P(Y)

, which yields

P(Y) = 0.6

.

Step-by-Step Solution

1

Apply the general addition law for two probability events

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

The addition law relates the union, individual probabilities, and intersection of compound events.

2

Express the intersection using the law of multiplication for independent events

P(X \cap Y) = P(X) \times P(Y) = 0.65 \times P(Y)

Because germination of variety X and variety Y are independent events.

3

Substitute given values into the combined probability formula and solve for

P(Y)

0.86 = 0.65 + P(Y) - 0.65 P(Y) \implies 0.21 = 0.35 P(Y) \implies P(Y) = 0.6

Isolating the unknown probability

P(Y)

yields the correct decimal value.

Key Concept

Compound Probability Laws and Independent Events

Question 4209Question →

If $2^{x+y} = 32$ and $\log_3 x + \log_3 y = 1 + \log_3 2$ , what is the value of $x^2 + y^2$ ?

13

21

26

37

Show answer & explanation

Answer: $13$

Answer

The value of

x^2 + y^2

is

13

.

Expressing

32

as

2^5

gives

x + y = 5

. Converting

1

to

\log_3 3

allows combining the right-hand side to

\log_3(3 \times 2) = \log_3 6

, so

\log_3(xy) = \log_3 6 \implies xy = 6

. Evaluating

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

yields

5^2 - 2(6) = 25 - 12 = 13

.

Step-by-Step Solution

1

Simplify the exponential equation using index laws.

x + y = 5

Since

32 = 2^5

,

2^{x+y} = 2^5 \implies x + y = 5

.

2

Simplify the logarithmic equation using logarithm laws.

xy = 6

\log_3 x + \log_3 y = \log_3(xy)

and

1 + \log_3 2 = \log_3 3 + \log_3 2 = \log_3(3 \times 2) = \log_3 6

. Therefore,

xy = 6

.

3

Use the algebraic identity to find

x^2 + y^2

.

x^2 + y^2 = 13

Using

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

, we have

5^2 = x^2 + y^2 + 2(6) \implies 25 = x^2 + y^2 + 12 \implies x^2 + y^2 = 13

.

Key Concept

Solving simultaneous equations involving indices and logarithms using index laws and logarithmic identities.

Question 4210Question →

The table below shows the speed distribution (in km/h) recorded for $50$ vehicles passing a police checkpoint on a highway:

Speed Interval (km/h)	Frequency ( $f$ )
$40 - 49$	$6$
$50 - 59$	$12$
$60 - 69$	$18$
$70 - 79$	$10$
$80 - 89$	$4$

Calculate the mean speed of the vehicles.

Show answer & explanation

Answer: 63.3

Answer

The mean speed of the vehicles is

63.3\text{ km/h}

.

The mean of a grouped frequency distribution is computed using the formula

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

. The midpoints (

x

) for the intervals are

44.5, 54.5, 64.5, 74.5,

and

84.5

. Multiplying these midpoints by their respective frequencies yields products totaling

3165

. Dividing this total by the total number of vehicles (

50

) gives a mean speed of

63.3\text{ km/h}

.

Step-by-Step Solution

1

Find the class midpoints (

x

) for each speed interval

Midpoints are

x_1 = 44.5

,

x_2 = 54.5

,

x_3 = 64.5

,

x_4 = 74.5

, and

x_5 = 84.5

.

The midpoint of a grouped class interval is calculated as

\frac{\text{Lower limit} + \text{Upper limit}}{2}

.

2

Calculate the product

fx

for each class

6 \times 44.5 = 267

,

12 \times 54.5 = 654

,

18 \times 64.5 = 1161

,

10 \times 74.5 = 745

, and

4 \times 84.5 = 338

.

Multiplying class midpoint by frequency estimates the total contribution of all items within that interval.

3

Find total frequency

\sum f

and sum of products

\sum fx

\sum f = 6 + 12 + 18 + 10 + 4 = 50

and

\sum fx = 267 + 654 + 1161 + 745 + 338 = 3165

.

Summing frequencies gives total sample size, and summing

fx

gives estimated grand total.

4

Compute the estimated mean speed

\bar{x} = \frac{\sum fx}{\sum f} = \frac{3165}{50} = 63.3\text{ km/h}

.

The standard formula for the mean of grouped frequency data is

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

.

Key Concept

Grouped Mean Calculation

Question 4211Question →

A company purchased a processing machine for $\text{₦}200,000$ . The value of the machine depreciates at a compound rate of $10\%$ per annum. At the end of $2$ years, the machine was sold at a profit of $15\%$ based on its depreciated value. What was the selling price of the machine?

\text{₦}184,000

\text{₦}186,300

\text{₦}192,000

\text{₦}207,000

Show answer & explanation

Answer: $\text{₦}186,300$

Answer

The selling price of the machine was

\text{₦}186,300

.

The depreciated value after 2 years at a compound rate of

10\%

per annum is calculated as

V = 200,000 \times (0.90)^2 = \text{₦}162,000

. Selling the machine at a

15\%

profit on this depreciated value gives a selling price of

162,000 \times 1.15 = \text{₦}186,300

.

Step-by-Step Solution

1

Calculate the depreciated value of the machine after 2 years using the compound depreciation formula

V = P(1 - r)^n

.

V = 200,000 \times (1 - 0.10)^2 = 200,000 \times (0.90)^2 = 200,000 \times 0.81 = \text{₦}162,000

.

Compound depreciation reduces the asset's remaining book value by

10\%

each year.

2

Calculate the

15\%

profit based on the depreciated value.

\text{Profit} = 15\% \text{ of } \text{₦}162,000 = 0.15 \times 162,000 = \text{₦}24,300$.

The problem specifies that profit is made on the depreciated value.

3

Calculate the selling price by adding the profit to the depreciated value.

\text{Selling Price} = \text{₦}162,000 + \text{₦}24,300 = \text{₦}186,300$.

Selling price is equal to book value plus profit earned.

Key Concept

Compound Depreciation and Percentage Profit on Book Value

Estimated Time:2m 0s

Question 4212Question →

Which of the following correctly expresses the derived SI unit of power, the watt ( $\text{W}$ ), in terms of fundamental SI base units?

\text{kg}\cdot\text{m}^2\cdot\text{s}^{-2}

\text{kg}\cdot\text{m}^2\cdot\text{s}^{-3}

\text{kg}\cdot\text{m}\cdot\text{s}^{-3}

\text{kg}\cdot\text{m}^2\cdot\text{s}^{3}

Show answer & explanation

Answer: $\text{kg}\cdot\text{m}^2\cdot\text{s}^{-3}$

Answer

The watt (

\text{W}

) expressed in fundamental SI base units is

\text{kg}\cdot\text{m}^2\cdot\text{s}^{-3}

.

Power is defined as energy transferred or work done per unit time (

\text{P} = \frac{W}{t}

). Work is force times distance (

\text{N}\cdot\text{m} = \text{kg}\cdot\text{m}^2\cdot\text{s}^{-2}

). Dividing work by time (

\text{s}

) yields

\text{kg}\cdot\text{m}^2\cdot\text{s}^{-3}

.

Step-by-Step Solution

1

Recall the defining formula for power

\text{Power} = \frac{\text{Work}}{\text{Time}} = \frac{\text{Force} \times \text{Displacement}}{\text{Time}}

Relating the derived quantity to simpler mechanical quantities.

2

Break down force into fundamental base units using Newton's second law (

F = ma

)

\text{Unit of Force (Newton)} = \text{kg}\cdot\text{m}\cdot\text{s}^{-2}

Mass is measured in kilograms (

\text{kg}

), acceleration in meters per second squared (

\text{m}\cdot\text{s}^{-2}

).

3

Express work in fundamental base units

\text{Unit of Work (Joule)} = (\text{kg}\cdot\text{m}\cdot\text{s}^{-2}) \times \text{m} = \text{kg}\cdot\text{m}^2\cdot\text{s}^{-2}

Work is force multiplied by distance.

4

Divide the unit of work by the unit of time (seconds)

\text{Unit of Power (Watt)} = \frac{\text{kg}\cdot\text{m}^2\cdot\text{s}^{-2}}{\text{s}} = \text{kg}\cdot\text{m}^2\cdot\text{s}^{-3}

Dividing by

\text{s}

decreases the exponent of seconds by 1.

Key Concept

Expressing derived SI units in terms of fundamental SI base units

Estimated Time:1m 0s

Question 4213Question →

What is the determinant of the $2 \times 2$ matrix $P = \begin{pmatrix} 5 & 2 \\ 3 & 4 \end{pmatrix}$ ?

Show answer & explanation

Answer: 14

Answer

The determinant of matrix

P

is

14

.

For any

2 \times 2

matrix

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

, the determinant is given by

ad - bc

. Substituting

a=5

,

b=2

,

c=3

, and

d=4

yields

(5 \times 4) - (2 \times 3) = 20 - 6 = 14

.

Step-by-Step Solution

1

Apply the determinant formula for a

2 \times 2

matrix

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

, which is

\det(P) = ad - bc

.

\det(P) = (5)(4) - (2)(3)

The determinant of a

2 \times 2

matrix is defined as the product of the main diagonal elements minus the product of the off-diagonal elements.

2

Evaluate the arithmetic expression.

20 - 6 = 14

Perform multiplication followed by subtraction to get the final determinant value.

Key Concept

Determinant of a 2x2 Matrix

Estimated Time:45s

Question 4214Question →

If $y = 5\sin x + e^{3x}$ , what is the value of $\frac{dy}{dx}$ at $x = 0$ ?

Show answer & explanation

Answer: 8

Answer

The value of the derivative at

x = 0

is 8.

Differentiating

y = 5\sin x + e^{3x}

yields

\frac{dy}{dx} = 5\cos x + 3e^{3x}

. Substituting

x = 0

gives

5\cos(0) + 3e^{0} = 5(1) + 3(1) = 8

.

Step-by-Step Solution

1

Differentiate each term of the function

y = 5\sin x + e^{3x}

with respect to

x

.

\frac{dy}{dx} = 5\cos x + 3e^{3x}

The derivative of

\sin x

is

\cos x

, and applying the chain rule to

e^{3x}

gives

3e^{3x}

.

2

Evaluate the derivative

\frac{dy}{dx}

at

x = 0

.

5\cos(0) + 3e^0 = 5(1) + 3(1) = 8

Evaluating trigonometric and exponential functions at zero gives

\cos(0) = 1

and

e^0 = 1

.

Key Concept

Differentiation of trigonometric and exponential functions and evaluation of derivatives at specific points.

Question 4215Question →

Match each length measurement instrument setup on the left with its corresponding true (corrected) measurement value on the right.

Click a left item, then click its matching right item

Show answer & explanation

Answer

The correct pairings match each instrument's corrected reading: micrometer with positive zero error to

2.74\text{ mm}

, Vernier caliper with negative zero error to

3.48\text{ cm}

, metre rule measurement to

12.5\text{ cm}

, and micrometer with negative zero error to

1.98\text{ mm}

.

Each instrument pairing correctly applies the instrument's least count formula and the standard zero error relationship:

\text{Corrected Reading} = \text{Observed Reading} - \text{Zero Error}

.

Step-by-Step Solution

1

Calculate the corrected reading for the first micrometer screw gauge.

Least count =

0.01\text{ mm}

. Observed =

2.5\text{ mm} + 0.28\text{ mm} = 2.78\text{ mm}

. Corrected reading =

2.78\text{ mm} - 0.04\text{ mm} = 2.74\text{ mm}

.

True Reading = Observed Reading - (Zero Error).

2

Calculate the corrected reading for the Vernier caliper.

Observed =

3.4\text{ cm} + 0.06\text{ cm} = 3.46\text{ cm}

. Corrected reading =

3.46\text{ cm} - (-0.02\text{ cm}) = 3.48\text{ cm}

.

Subtracting a negative zero error is equivalent to adding the absolute error magnitude.

3

Calculate the length of the wooden rod measured with the metre rule.

Length =

14.8\text{ cm} - 2.3\text{ cm} = 12.5\text{ cm}

.

Subtracting the initial scale alignment point from the final scale alignment point eliminates end-wear errors.

4

Calculate the corrected reading for the second micrometer screw gauge.

Observed =

1.5\text{ mm} + 0.45\text{ mm} = 1.95\text{ mm}

. Corrected reading =

1.95\text{ mm} - (-0.03\text{ mm}) = 1.98\text{ mm}

.

Subtracting the negative zero error compensates for the instrument reading below zero prior to measurement.

Key Concept

Instrument Least Count and Zero Error Corrections

Question 4216Question →

A student measures the length of a metal rod as $4.0\text{ cm}$ . If the actual length of the rod is $5.0\text{ cm}$ , what is the percentage error in the measurement?

Show answer & explanation

Answer: 20

Answer

The percentage error in the measurement is 20%.

The absolute error is the difference between the true length (

5.0\text{ cm}

) and the measured length (

4.0\text{ cm}

), which is

1.0\text{ cm}

. Dividing

1.0\text{ cm}

by the true length

5.0\text{ cm}

gives a fractional error of

0.20

. Expressed as a percentage,

0.20 \times 100\% = 20\%

.

Step-by-Step Solution

1

Find the error (difference between measured value and actual value)

Error =

|5.0\text{ cm} - 4.0\text{ cm}| = 1.0\text{ cm}

Absolute error is defined as the magnitude of the difference between the actual value and the measured value.

2

Calculate the percentage error

Percentage error =

\frac{1.0\text{ cm}}{5.0\text{ cm}} \times 100\% = 20\%

Percentage error is the ratio of absolute error to the actual value, expressed as a percentage.

Key Concept

Percentage Error Calculation

Question 4217Question →

The $3^{\text{rd}}$ and $6^{\text{th}}$ terms of an Arithmetic Progression (A.P.) are $13$ and $28$ respectively. What is the sum of the first $10$ terms of the progression?

255

280

205

48

Show answer & explanation

Answer: $255$

Answer

The sum of the first

10

terms of the progression is

255

.

The

3^{\text{rd}}

term is

a + 2d = 13

and the

6^{\text{th}}

term is

a + 5d = 28

. Subtracting these equations gives

3d = 15

, so

d = 5

, which leads to

a = 3

. Using the sum formula

S_{10} = \frac{10}{2}[2(3) + (10 - 1)5]

, we obtain

5(6 + 45) = 255

.

Step-by-Step Solution

1

Set up equations for the given terms using the nth term formula

T_n = a + (n - 1)d

.

a + 2d = 13

and

a + 5d = 28

.

The

3^{\text{rd}}

term corresponds to

n=3

and the

6^{\text{th}}

term corresponds to

n=6

.

2

Solve the simultaneous equations for

a

(first term) and

d

(common difference).

Subtracting the first equation from the second gives

3d = 15 \Rightarrow d = 5

. Substituting

d = 5

into the first equation yields

a + 2(5) = 13 \Rightarrow a = 3

.

To find any property of an A.P., the first term

a

and common difference

d

must be determined.

3

Calculate the sum of the first

10

terms using

S_n = \frac{n}{2}[2a + (n - 1)d]

.

S_{10} = \frac{10}{2}[2(3) + (10 - 1)5] = 5[6 + 45] = 5(51) = 255

.

Applying the formula for the sum of the first

n

terms with

n = 10

,

a = 3

, and

d = 5

.

Key Concept

Finding the sum of the first

n

terms of an Arithmetic Progression given two specific terms.

Question 4218Question →

Three of the interior angles of a convex polygon are each $120^\circ$ , while the remaining interior angles are each $160^\circ$ . How many sides does the polygon have?

Show answer & explanation

Answer: 12

Answer

The polygon has 12 sides.

The sum of the interior angles of an

n

-sided polygon is

(n - 2) \times 180^\circ

. From the problem, the sum of the angles is

3 \times 120^\circ + (n - 3) \times 160^\circ = 160^\circ n - 120^\circ

. Equating

180^\circ(n - 2)

to

160^\circ n - 120^\circ

gives

180^\circ n - 360^\circ = 160^\circ n - 120^\circ

, which simplifies to

20^\circ n = 240^\circ

, yielding

n = 12

.

Step-by-Step Solution

1

Write down the standard formula for the sum of interior angles of an

n

-sided convex polygon.

Sum of interior angles =

(n - 2) \times 180^\circ

.

Any

n

-sided convex polygon can be split into

(n-2)

triangles, each having an interior angle sum of

180^\circ

.

2

Express the total sum of interior angles using the given values.

Sum =

3(120^\circ) + (n - 3)(160^\circ) = 360^\circ + 160^\circ n - 480^\circ = 160^\circ n - 120^\circ

.

Three angles are

120^\circ

, so the remaining

(n - 3)

angles must each equal

160^\circ

.

3

Equate the theoretical sum to the calculated sum.

180^\circ(n - 2) = 160^\circ n - 120^\circ \implies 180^\circ n - 360^\circ = 160^\circ n - 120^\circ

.

Both expressions represent the total interior angle sum of the same polygon.

4

Solve the equation for the number of sides

n

.

20^\circ n = 240^\circ \implies n = 12

.

Subtracting

160^\circ n

from both sides and adding

360^\circ

yields

20^\circ n = 240^\circ

.

Key Concept

Sum of Interior Angles of a Polygon

Question 4219Question →

The $3^{\text{rd}}$ , $6^{\text{th}}$ , and $11^{\text{th}}$ terms of a non-constant arithmetic progression (AP) form the first three consecutive terms of a geometric progression (GP). If the first term of the AP is $15$ , what is the $4^{\text{th}}$ term of the geometric progression?

Show answer & explanation

Answer: 125

Answer

The 4th term of the geometric progression is 125.

By writing the 3rd, 6th, and 11th terms of the AP as

15+2d

,

15+5d

, and

15+10d

, we utilize the geometric mean property

(15+5d)^2 = (15+2d)(15+10d)

to find

d=6

. This yields the GP terms

27, 45, 75

, giving a common ratio of

5/3

. Multiplying the third term

75

by

5/3

gives the 4th GP term as

125

.

Step-by-Step Solution

1

Write down the AP term expressions

T_3 = 15 + 2d

,

T_6 = 15 + 5d

,

T_{11} = 15 + 10d

The

n^{\text{th}}

term of an AP is given by

T_n = a + (n-1)d

with initial term

a = 15

.

2

Apply the geometric progression condition

(15 + 5d)^2 = (15 + 2d)(15 + 10d)

If three terms

A, B, C

are in GP, then

B^2 = A \cdot C

.

3

Expand and solve the quadratic equation for the common difference

d

d = 6

Expanding gives

225 + 150d + 25d^2 = 225 + 180d + 20d^2 \implies 5d^2 = 30d \implies d = 6

because

d \neq 0

.

4

Find the terms and common ratio of the GP

G_1 = 27

,

G_2 = 45

,

G_3 = 75

, and common ratio

r = \frac{5}{3}

Substituting

d = 6

gives the GP terms, and dividing consecutive terms gives

r = \frac{45}{27} = \frac{5}{3}

.

5

Calculate the 4th term of the GP

G_4 = 125

Multiplying the 3rd term by the common ratio yields

75 \times \frac{5}{3} = 125

.

Key Concept

Combining Arithmetic Progression nth-term formulas with Geometric Progression consecutive-term properties

Estimated Time:2m 30s

Question 4220Question →

A ship departs from a port $P$ and sails $10\text{ km}$ on a bearing of $060^\circ$ to reach a position $Q$ . From $Q$ , the ship changes course and sails $10\text{ km}$ on a bearing of $150^\circ$ to arrive at point $R$ . What is the bearing of point $R$ from point $P$ ?

015^\circ

105^\circ

135^\circ

150^\circ

Show answer & explanation

Answer: $105^\circ$

Answer

The bearing of point

R

from point

P

is

105^\circ

.

The back bearing from position

Q

to port

P

is

240^\circ

. The difference between

240^\circ

and the new bearing of

150^\circ

yields an interior angle of

90^\circ

at vertex

Q

. Since both distances

PQ

and

QR

are equal to

10\text{ km}

, triangle

PQR

is a

45^\circ-45^\circ-90^\circ

isosceles right triangle. Adding

\angle QPR = 45^\circ

to the initial bearing of

060^\circ

yields a bearing of

105^\circ

for point

R

from point

P

.

Step-by-Step Solution

1

Determine the back bearing of

P

from

Q

.

The back bearing of

P

from

Q

is

060^\circ + 180^\circ = 240^\circ

.

To find the internal angle at

Q

, we need the direction of

P

relative to

Q

.

2

Calculate the interior angle

\angle PQR

.

\angle PQR = 240^\circ - 150^\circ = 90^\circ

.

The difference between the line back to

P

(

240^\circ

) and the line to

R

(

150^\circ

) forms the interior angle at

Q

.

3

Determine the properties of triangle

PQR

and angle

\angle QPR

.

Since

PQ = QR = 10\text{ km}

and

\angle PQR = 90^\circ

, triangle

PQR

is an isosceles right triangle, so

\angle QPR = 45^\circ

.

The two equal sides subtend equal acute angles in a right-angled triangle:

(180^\circ - 90^\circ) / 2 = 45^\circ

.

4

Calculate the total bearing of

R

from

P

.

Bearing of

R

from

P = 060^\circ + 45^\circ = 105^\circ

.

Point

R

lies clockwise relative to the segment

PQ

, so we add

\angle QPR

to the initial bearing of

PQ

.

Key Concept

Three-point bearing calculations using geometry of parallel lines and right-angled triangles.

Answer

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Key Concept

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