All practice questions

Question 1Question →

In Khadija Abubakar Jalli's prescribed novel *The Life Changer*, what is the name of the introverted Lafayette resident whose compound is secretly used to detain a kidnapped child, illustrating the narrative function of how innocence can be exploited by criminals?

Show answer & explanation

Answer: Talle; talle; The Silent One; the silent one

Answer

Talle

Talle, popularly known as 'the silent one' in Lafayette, is the minor character whose introverted lifestyle and gullibility make him an unwitting accomplice to Zaki's kidnapping operation.

Step-by-Step Solution

1

Recall the key plot incident in the town of Lafayette from *The Life Changer*.

A young boy is abducted and hidden away in a quiet villager's compound.

This incident introduces the theme of crime creeping into peaceful rural communities.

2

Identify the specific minor character whose personality and home were involved in this event.

Talle, who earned the sobriquet 'the silent one' because of his reticent nature, was duped into harboring the victim by Zaki.

Talle's character functions as a structural catalyst to expose the dangers of gullibility and bad association.

Key Concept

Minor Character Identification and Narrative Function

Question 2Question →

In Khadija Abubakar Jalli's prescribed prose text *The Life Changer*, which minor character serves as the traditional leader and village head of Lafayette?

Show answer & explanation

Answer: Hakimi; The Hakimi; Hakimi the village head

Answer

Hakimi is the minor character who functions as the traditional leader and village head of Lafayette in *The Life Changer*.

Hakimi is the traditional leader and village head of Lafayette in *The Life Changer*, responsible for maintaining order and handling community matters in the village.

Step-by-Step Solution

1

Recall the minor characters from the Lafayette community in *The Life Changer*.

Hakimi is identified as the titleholder governing Lafayette.

When strange occurrences happen in Lafayette (such as Talle's sudden spending and arrest), the community affairs and traditional leadership fall under Hakimi.

Key Concept

Minor Character Identification and Functions

Estimated Time:45s

Question 3Question →

In Khadija Abubakar Jalli's prescribed prose text, The Life Changer, which minor character serves as the personal driver and trusted confidant of Honourable Habib?

Show answer & explanation

Answer: Labaran; Mr. Labaran; Driver Labaran

Answer

Labaran

In Khadija Abubakar Jalli's novella The Life Changer, Labaran is the minor character who functions as Honourable Habib's driver and confidant, assisting him in private matters and executing behind-the-scenes arrangements.

Step-by-Step Solution

1

Identify the character's functional role in relation to Honourable Habib in The Life Changer.

Honourable Habib relies on his personal driver to carry out confidential tasks, private errands, and discreet arrangements.

Understanding minor character identification and narrative functions.

2

Match the role of driver and confidant to the specific character name.

The minor character named Labaran fills this role throughout the narrative.

Direct text recall of character identity.

Key Concept

Minor Character Identification and Functions

Question 4Question →

In Khadija Abubakar Jalli's prescribed prose text, *The Life Changer*, which minor character serves as Tomiwa's hostel roommate hailing from Anambra State, whose background and interactions contribute to the narrative's thematic representation of inter-ethnic unity?

Show answer & explanation

Answer: Nkechi; nkechi

Answer

Nkechi

In *The Life Changer*, Nkechi is the minor character who represents Anambra State (South-Eastern Nigeria) among Tomiwa's room members. Together with Ada from Benue and Clare from Ondo, her inclusion highlights the structural and thematic function of promoting ethnic harmony, accommodation, and peaceful coexistence among tertiary institution students from diverse backgrounds.

Step-by-Step Solution

1

Examine the narrative context surrounding Tomiwa's living arrangements at the university in *The Life Changer*.

Tomiwa shares a hostel room with three other female students from distinct geopolitical regions of Nigeria: Ada, Clare, and Nkechi.

Establishing the composition of Tomiwa's hostel room isolates the candidates for this minor character role.

2

Correlate each roommate with her state of origin and narrative function as described in the text.

Ada hails from Benue State, Clare from Ondo State, and Nkechi represents Anambra State (South-East).

Matching the regional origin specified in the question stem confirms Nkechi as the character who embodies this aspect of national diversity and integration.

Key Concept

Minor Character Identification and Cultural Functions in *The Life Changer*

Question 5Question →

In Khadija Abubakar Jalli's prescribed novel *The Life Changer*, which minor character in Lafayette is commonly referred to as 'the silent one' prior to his involvement in a kidnapping incident?

Show answer & explanation

Answer: Talle; Mallam Talle

Answer

Talle

Talle is explicitly introduced in *The Life Changer* as 'the silent one' in Lafayette because he became introverted and withdrawn after losing his family.

Step-by-Step Solution

1

Identify the character described by the sobriquet 'the silent one' in the Lafayette community context of the novel.

In the text, Talle earns the epithet 'the silent one' due to his reclusive and quiet disposition following the death of his parents.

This specific characterization sets up his vulnerability to being manipulated into harboring a kidnapped child in his house.

Key Concept

Minor Character Identification

Question 6Question →

In Khadija Abubakar Jalli's prescribed novel *The Life Changer*, which minor character is Salma's hostel roommate from Benue State, best known for her generosity in sharing home-cooked food with her roommates?

Show answer & explanation

Answer: Ada; Ada.

Answer

Ada

In *The Life Changer*, Ada is one of Salma's three hostel roommates at the university. Hailing from Benue State, she is characterized by her generosity and habit of bringing food items from home to share with her room members.

Step-by-Step Solution

1

Identify the character based on state of origin and behavioral details in the text.

Ada is identified as the roommate hailing from Benue State who frequently brings food to share in the hostel room.

In *The Life Changer*, the author highlights the harmonious relationship among Salma's diverse roommates, specifically noting Ada's food-sharing habit.

Key Concept

Minor Character Identification and Functions

Question 7Question →

In the prescribed prose text *The Life Changer*, which minor character acts as Salim's close friend and confidant, listening to his narrative regarding an online dating encounter that led to an attempted carjacking?

Show answer & explanation

Answer: Gimba; Gimba.

Answer

Gimba

Gimba is the minor character who acts as Salim's confidant in *The Life Changer*, listening to Salim detail his encounter with Natasha on social media and his narrow escape from carjackers.

Step-by-Step Solution

1

Identify the subplot in *The Life Changer* involving Salim's social media acquaintance and attempted robbery.

Salim recounts his ordeal regarding his meeting with Natasha to his close friend.

Analyzing narrative subplots allows accurate identification of minor characters and their functional roles.

2

Determine the exact identity of the character listening to Salim's experience.

The friend is Gimba.

Gimba serves as a confidant, providing a sounding board that highlights the moral lesson against naive online interactions.

Key Concept

Minor Character Identification and Narrative Function

Question 8Question →

In Khadija Abubakar Jalli's prescribed prose text, *The Life Changer*, what is the name of the exceptionally quiet Lafayette resident who is arrested by the police after unwittingly harboring a kidnapped child in his compound?

Show answer & explanation

Answer: Talle; Tale; Talle (The Silent One); The Silent One

Answer

Talle

In *The Life Changer*, Talle is the minor character known throughout Lafayette as 'The Silent One'. His quiet, secretive lifestyle makes him an easy target for Zaki, who uses Talle's house as a hideout for a kidnapped child, leading to Talle's unexpected arrest.

Step-by-Step Solution

1

Analyze the context of the Lafayette kidnapping subplot in the novel.

The narrative describes a well-known, introverted villager who lost his parents and rarely speaks to community members.

This background establishes the minor character's personality and vulnerability to manipulation.

2

Identify the character manipulated into storing the kidnapped victim.

The character Talle, nicknamed 'The Silent One', was coerced and paid by Zaki to hold the abducted boy in his house.

His naive participation leads directly to the police raid on his home and his subsequent arrest alongside Zaki.

Key Concept

Minor Character Identification and Narrative Functions

Question 9Question →

In *The Life Changer*, what is the name of the university lecturer who makes uncharacteristic advances toward Salma during her registration, only to immediately regret his conduct?

Show answer & explanation

Answer: Dr. Dabo; Dabo; Doctor Dabo; Dr Dabo

Answer

Dr. Dabo

Dr. Dabo is the university lecturer who loses his composure and tries to solicit a personal relationship with Salma during her registration. Overcome with remorse immediately afterward, he asks God for forgiveness and returns to his strict ethical standards.

Step-by-Step Solution

1

Identify the character based on the described narrative event during registration at the university.

The scene involves a traditionally strict lecturer who briefly yields to temptation when meeting Salma.

This event serves as a narrative introduction to Salma's impact on others and tests the lecturer's moral character.

2

Recall the exact name of the lecturer from the prescribed text.

The lecturer's name is Dr. Dabo.

After Salma leaves his office, Dr. Dabo recognizes his lapse in judgment, prays for forgiveness, and vows never to repeat the mistake.

Key Concept

Minor Character Identification and Functions

Question 10Question →

Tayo Enterprises earned a gross profit of $\text{N}150,000$ for the year ended 31st December 2025. The books of the business revealed the following additional items at the end of the accounting period:

- Rent paid: $\text{N}18,000$ (includes $\text{N}3,000$ prepaid for the next accounting period)
- Discount received: $\text{N}2,500$
- Commission received: $\text{N}4,000$ ( $\text{N}1,000$ is accrued and not yet received)
- Salaries paid: $\text{N}35,000$
- Reduction in provision for doubtful debts: $\text{N}1,500$

What is the net profit of Tayo Enterprises for the year?

Show answer & explanation

Answer: N109,000; ��109,000; 109,000; N109000; ₦109000; 109000; N 109,000; ₦ 109,000

Answer

The net profit of Tayo Enterprises for the year is

\text{N}109,000

.

To determine the net profit, add all income earned during the period (Gross Profit

\text{N}150,000

+ Discount received

\text{N}2,500

+ Total Commission earned

\text{N}5,000

+ Decrease in provision for doubtful debts

\text{N}1,500 = \text{N}159,000

) and subtract all operating expenses incurred (Adjusted Rent

\text{N}15,000

+ Salaries

\text{N}35,000 = \text{N}50,000

). This gives

\text{N}159,000 - \text{N}50,000 = \text{N}109,000

.

Step-by-Step Solution

1

Calculate total income to be added to Gross Profit.

Discount received:

\text{N}2,500

; Commission earned:

\text{N}4,000 + \text{N}1,000 = \text{N}5,000

; Reduction in provision for doubtful debts:

\text{N}1,500

. Total additional income =

\text{N}2,500 + \text{N}5,000 + \text{N}1,500 = \text{N}9,000

.

Accrued income is added to income received, and reductions in provision for doubtful debts represent gains credited to the Profit and Loss Account.

2

Calculate total adjusted operating expenses.

Adjusted Rent expense:

\text{N}18,000 - \text{N}3,000 = \text{N}15,000

; Salaries:

\text{N}35,000

. Total expenses =

\text{N}15,000 + \text{N}35,000 = \text{N}50,000

.

Prepaid expenses must be deducted from the total paid to reflect only the expense incurred for the current accounting period.

3

Calculate Net Profit.

\text{Net Profit} = \text{Gross Profit} (\text{N}150,000) + \text{Total Additional Income} (\text{N}9,000) - \text{Total Expenses} (\text{N}50,000) = \text{N}109,000

.

Net profit is computed by adding non-operating/other revenue to gross profit and deducting total operating expenses for the period.

Key Concept

Determination of Net Profit in Profit and Loss Account with year-end adjustments

Estimated Time:2m 0s

Question 11Question →

Consider the unsaturated hydrocarbon 2-methylbut-1-en-3-yne, which has the condensed structural formula $CH_2=C(CH_3)C\equiv CH$ . What is the total number of sigma ( $\sigma$ ) bonds present in one molecule of this compound?

Show answer & explanation

Answer: 10; 10 sigma bonds; 10 bonds

Answer

The total number of sigma (

\sigma

) bonds present in one molecule of 2-methylbut-1-en-3-yne is 10.

In 2-methylbut-1-en-3-yne (

CH_2=C(CH_3)C\equiv CH

), there are 6 carbon-hydrogen single bonds (2 from C1, 3 from the methyl group, and 1 from C4) and 4 carbon-carbon sigma bonds (1 from the C1=C2 double bond, 1 connecting C2 to the methyl carbon, 1 connecting C2 to C3, and 1 from the C3\equiv C4 triple bond). Summing these gives 10 sigma bonds in total.

Step-by-Step Solution

1

Expand the condensed structural formula to identify all individual carbon-hydrogen (C-H) bonds.

The

=CH_2

group contains 2 C-H

\sigma

bonds, the methyl group (

-CH_3

) contains 3 C-H

\sigma

bonds, and the terminal alkynyl group (

\equiv CH

) contains 1 C-H

\sigma

bond, giving a total of 6 C-H

\sigma

bonds.

Every single bond between carbon and hydrogen is a single sigma bond.

2

Identify all carbon-carbon (C-C) sigma bonds in the backbone and side chain.

The

C=C

double bond contributes 1 C-C

\sigma

bond, the single bond to the methyl branch contributes 1 C-C

\sigma

bond, the C2-C3 single bond contributes 1 C-C

\sigma

bond, and the

C\equiv C

triple bond contributes 1 C-C

\sigma

bond, giving a total of 4 C-C

\sigma

bonds.

Multiple bonds (double or triple) contain exactly one sigma bond each, with the remaining bonds being pi (

\pi

) bonds.

3

Sum the total number of C-H and C-C sigma bonds.

6 \text{ (C-H } \sigma\text{ bonds)} + 4 \text{ (C-C } \sigma\text{ bonds)} = 10 \text{ total } \sigma \text{ bonds}

.

Adding all localized

\sigma

bonds yields the total count for the molecule.

Key Concept

Determination of sigma (\sigma) and pi (\pi) bond counts in complex open-chain hydrocarbons

Question 12Question →

Given the universal set $\mathcal{U} = \{x \in \mathbb{Z} : 1 \le x \le 15\}$ , with subsets $P = \{x \in \mathcal{U} : x \text{ is a prime number}\}$ and $Q = \{x \in \mathcal{U} : x \text{ is an odd number}\}$ , what is the number of elements in the set $(P \cup Q)'$ ?

Show answer & explanation

Answer: 6; six; 6 elements

Answer

The number of elements in

(P \cup Q)'

is 6.

The set

P \cup Q

consists of all numbers from 1 to 15 that are either prime or odd:

\{1, 2, 3, 5, 7, 9, 11, 13, 15\}

. The complement

(P \cup Q)'

relative to

\mathcal{U}

contains all elements of

\mathcal{U}

that are neither prime nor odd, which are the even composite numbers:

\{4, 6, 8, 10, 12, 14\}

. Counting these elements gives 6.

Step-by-Step Solution

1

List all elements of the universal set

\mathcal{U}

, subset

P

, and subset

Q

.

\mathcal{U} = \{1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15\}

,

P = \{2, 3, 5, 7, 11, 13\}

, and

Q = \{1, 3, 5, 7, 9, 11, 13, 15\}

.

Listing the explicit elements allows precise execution of set union and complement operations.

2

Find the union

P \cup Q

.

P \cup Q = \{1, 2, 3, 5, 7, 9, 11, 13, 15\}

.

The union combines all unique elements that belong to either set

P

, set

Q

, or both.

3

Determine the complement set

(P \cup Q)'

relative to

\mathcal{U}

and count its elements.

(P \cup Q)' = \{4, 6, 8, 10, 12, 14\}

, which contains 6 elements.

The complement set consists of all elements in

\mathcal{U}

that are not present in

P \cup Q

.

Key Concept

Complement of Set Union

Question 13Question →

Using the first principles of differentiation for the reciprocal function $f(x) = \frac{4}{x}$ , evaluate the limit of the difference quotient as $h \to 0$ : $\lim_{h \to 0} \frac{f(x+h) - f(x)}{h}$ . What is the simplified expression for the derivative $\frac{dy}{dx}$ ?

Show answer & explanation

Answer: -\frac{4}{x^2}; -4/x^2; -4 / x^2; -\frac{4}{x^{2}}

Answer

The derivative

\frac{dy}{dx}

is

-\frac{4}{x^2}

.

Substituting

f(x) = \frac{4}{x}

into the first principles limit formula yields

\lim_{h \to 0} \frac{4x - 4(x+h)}{h \cdot x(x+h)} = \lim_{h \to 0} \frac{-4h}{h \cdot x(x+h)}

. Canceling

h

gives

\lim_{h \to 0} -\frac{4}{x(x+h)} = -\frac{4}{x^2}

.

Step-by-Step Solution

1

Write the first principles formula and substitute

f(x) = \frac{4}{x}

.

\frac{dy}{dx} = \lim_{h \to 0} \frac{\frac{4}{x+h} - \frac{4}{x}}{h}

Apply the definition of differentiation from first principles.

2

Combine the fractions in the numerator using a common denominator

x(x+h)

.

\frac{4}{x+h} - \frac{4}{x} = \frac{4x - 4(x+h)}{x(x+h)} = \frac{4x - 4x - 4h}{x(x+h)} = \frac{-4h}{x(x+h)}

Simplify the numerator into a single fractional expression.

3

Divide the simplified numerator by

h

and cancel the common factor of

h

.

\frac{\frac{-4h}{x(x+h)}}{h} = \frac{-4h}{h \cdot x(x+h)} = -\frac{4}{x(x+h)}

Eliminate the indeterminate factor of

h

from the denominator.

4

Evaluate the limit as

h \to 0

.

\lim_{h \to 0} -\frac{4}{x(x+h)} = -\frac{4}{x(x+0)} = -\frac{4}{x^2}

Substitute

h = 0

into the simplified algebraic expression.

Key Concept

Differentiation from First Principles

Question 14Question →

An experimental effusion cell measures gas diffusion rates through a micro-porous membrane at fixed temperature and pressure. In a calibration run, $120\text{ cm}^3$ of neon gas ( $\text{Ne}$ , atomic mass $= 20\text{ g/mol}$ ) effuses through the membrane in $30\text{ seconds}$ . Calculate the volume (in $\text{cm}^3$ ) of sulfur trioxide gas ( $\text{SO}_3$ , atomic masses: $\text{S} = 32\text{ g/mol}$ , $\text{O} = 16\text{ g/mol}$ ) that will effuse through the exact same membrane in $50\text{ seconds}$ .

Show answer & explanation

Answer: 100; 100 cm3; 100cm3; 100 cm^3; 100 cm³

Answer

100 cm³

The effusion rate of neon is

r_{\text{Ne}} = \frac{120\text{ cm}^3}{30\text{ s}} = 4\text{ cm}^3/\text{s}

. Given

M_{\text{Ne}} = 20\text{ g/mol}

and

M_{\text{SO}_3} = 80\text{ g/mol}

, Graham's law dictates

\frac{r_{\text{Ne}}}{r_{\text{SO}_3}} = \sqrt{\frac{80}{20}} = 2

. Thus,

r_{\text{SO}_3} = \frac{4}{2} = 2\text{ cm}^3/\text{s}

. Over a period of

50\text{ seconds}

, the volume of

\text{SO}_3

effused is

2\text{ cm}^3/\text{s} \times 50\text{ s} = 100\text{ cm}^3

.

Step-by-Step Solution

1

Calculate the rate of effusion of neon gas (

r_{\text{Ne}}

)

r_{\text{Ne}} = \frac{120\text{ cm}^3}{30\text{ s}} = 4\text{ cm}^3/\text{s}

Effusion rate is defined as the volume of gas effusing per unit time.

2

Determine the molar mass of sulfur trioxide (

\text{SO}_3

)

M_{\text{SO}_3} = 32 + 3(16) = 80\text{ g/mol}

The molar mass is calculated from the constituent relative atomic masses of sulfur and oxygen.

3

Apply Graham's Law of Effusion to determine the effusion rate of sulfur trioxide (

r_{\text{SO}_3}

)

\frac{r_{\text{Ne}}}{r_{\text{SO}_3}} = \sqrt{\frac{M_{\text{SO}_3}}{M_{\text{Ne}}}} \implies \frac{4}{r_{\text{SO}_3}} = \sqrt{\frac{80}{20}} = \sqrt{4} = 2 \implies r_{\text{SO}_3} = 2\text{ cm}^3/\text{s}

Graham's law states that the rate of effusion of a gas is inversely proportional to the square root of its molar mass.

4

Compute the total volume of sulfur trioxide effused in

50\text{ seconds}

V_{\text{SO}_3} = r_{\text{SO}_3} \times t = 2\text{ cm}^3/\text{s} \times 50\text{ s} = 100\text{ cm}^3

Multiplying the calculated rate of effusion by the given time yields the total volume.

Key Concept

Graham's Law of Diffusion and Effusion

Question 15Question →

Consider the organic compound 3-methylbut-1-yne, which has the condensed structural formula $HC\equiv C-CH(CH_3)_2$ . How many carbon atoms in a single molecule of this compound are $sp^3$ hybridized?

Show answer & explanation

Answer: 3; three; 3 carbon atoms; 3 carbons

Answer

There are 3

sp^3

hybridized carbon atoms in one molecule of 3-methylbut-1-yne.

In 3-methylbut-1-yne (

HC\equiv C-CH(CH_3)_2

), there are 5 total carbon atoms. The two terminal/alkyne carbons (

C_1

and

C_2

) participate in a triple bond, giving them 2

\sigma

bonds each and an

sp

hybridization state. The central methine carbon (

C_3

) is bonded to four distinct atoms (

C_2

,

H

, and two methyl carbons) via single

\sigma

bonds, making it

sp^3

hybridized. The two methyl group carbons (

C_4

and

C_5

) are each single-bonded to three hydrogen atoms and

C_3

, making them

sp^3

hybridized as well. Thus, exactly 3 carbon atoms are

sp^3

hybridized.

Step-by-Step Solution

1

Expand the condensed structural formula to identify every carbon atom and its bonding environment.

The expanded structure is

H-C_1 \equiv C_2 - C_3H(C_4H_3)(C_5H_3)

, containing a total of 5 carbon atoms.

Expanding the formula clarifies the number of single (

\sigma

) and multiple bonds connected to each carbon atom.

2

Determine the hybridization state of the triply bonded carbon atoms (

C_1

and

C_2

).

C_1

and

C_2

are each involved in one triple bond and one single bond, forming 2

\sigma

bonds and 2

\pi

bonds. Thus, both

C_1

and

C_2

are

sp

hybridized.

A carbon atom with 2 steric domains (linear geometry) uses

sp

hybrid orbitals.

3

Determine the hybridization state of the methine carbon atom (

C_3

) and the two methyl carbon atoms (

C_4

and

C_5

).

C_3

forms four single

\sigma

bonds (one to

C_2

, one to

H

, and two to methyl carbons).

C_4

and

C_5

each form four single

\sigma

bonds (one to

C_3

and three to

H

). Therefore,

C_3

,

C_4

, and

C_5

are all

sp^3

hybridized.

A carbon atom bonded to 4 separate atoms via single

\sigma

bonds has 4 steric domains (tetrahedral geometry) and undergoes

sp^3

hybridization.

4

Count the total number of

sp^3

hybridized carbon atoms.

3 carbon atoms (

C_3

,

C_4

, and

C_5

) are

sp^3

hybridized.

Combining the results from steps 2 and 3 gives 2

sp

carbons and 3

sp^3

carbons.

Key Concept

Identification of carbon hybridization states (

sp^3, sp^2, sp

) in aliphatic molecules

Estimated Time:1m 30s

Question 16Question →

What is the equation of the locus of a point $P(x, y)$ that moves in a plane such that its distance from the origin $(0,0)$ is always 5 units?

Show answer & explanation

Answer: x^2 + y^2 = 25; x^2+y^2=25; x²+y²=25

Answer

x^2 + y^2 = 25

By definition, the locus of a point moving at a fixed distance of 5 units from the origin

(0,0)

is a circle centered at

(0,0)

with radius 5. Substituting into the standard circle equation

x^2 + y^2 = r^2

yields

x^2 + y^2 = 5^2 = 25

.

Step-by-Step Solution

1

Apply the distance formula between a general point

P(x, y)

and the origin

(0,0)

.

d = \sqrt{(x - 0)^2 + (y - 0)^2} = \sqrt{x^2 + y^2}

The distance between two points

(x_1, y_1)

and

(x_2, y_2)

is

\sqrt{(x_2-x_1)^2 + (y_2-y_1)^2}

.

2

Equate the distance formula to the given fixed distance of 5 units and square both sides.

\sqrt{x^2 + y^2} = 5 \implies x^2 + y^2 = 25

Squaring both sides eliminates the square root to give the algebraic equation of the locus.

Key Concept

The locus of points at a constant distance

r

from a fixed point

(h, k)

forms a circle with equation

(x-h)^2 + (y-k)^2 = r^2

.

Estimated Time:45s

Question 17Question →

If set $A = \{a, b, c, d, e\}$ and set $B = \{c, d, e, f, g\}$ , what is the number of elements in the set $(A \setminus B) \cup (B \setminus A)$ ?

Show answer & explanation

Answer: 4; four

Answer

The number of elements in

(A \setminus B) \cup (B \setminus A)

is 4.

The set difference

A \setminus B

consists of elements in

A

that are not in

B

, which gives

\{a, b\}

. Similarly,

B \setminus A

consists of elements in

B

that are not in

A

, giving

\{f, g\}

. The union

(A \setminus B) \cup (B \setminus A)

is

\{a, b, f, g\}

, which has a cardinality of 4.

Step-by-Step Solution

1

Find the relative difference

A \setminus B

A \setminus B = \{a, b\}

Remove elements of

B

present in

A

.

2

Find the relative difference

B \setminus A

B \setminus A = \{f, g\}

Remove elements of

A

present in

B

.

3

Take the union of the two set differences

(A \setminus B) \cup (B \setminus A) = \{a, b, f, g\}

Combine elements from both set differences.

4

Count the number of elements in the resulting set

4 elements

The set

\{a, b, f, g\}

contains 4 distinct elements.

Key Concept

Symmetric Difference of Two Sets

Estimated Time:45s

Question 18Question →

In a group of 100 candidates preparing for an entrance examination, 48 registered for Mathematics, 45 for Physics, and 40 for Chemistry. If 18 candidates registered for both Mathematics and Physics, 15 for both Physics and Chemistry, 20 for both Mathematics and Chemistry, and 8 registered for none of these three subjects, how many candidates registered for all three subjects?

Show answer & explanation

Answer: 12; 12 candidates; 12 students

Answer

12 candidates registered for all three subjects.

By the Principle of Inclusion-Exclusion, the total number of candidates taking at least one subject is

100 - 8 = 92

. Summing the individual totals gives

48 + 45 + 40 = 133

. Subtracting the pairwise intersections gives

133 - (18 + 15 + 20) = 80

. Adding the intersection of all three sets must equal 92, yielding

92 - 80 = 12

.

Step-by-Step Solution

1

Determine the total number of candidates who registered for at least one of the three subjects.

n(M \cup P \cup C) = 100 - 8 = 92

Subtracting the number of candidates who registered for none of the subjects from the universal set gives the cardinality of the union.

2

Apply the Principle of Inclusion-Exclusion for three sets.

n(M \cup P \cup C) = n(M) + n(P) + n(C) - n(M \cap P) - n(P \cap C) - n(M \cap C) + n(M \cap P \cap C)

This formula relates the individual set sizes, pair intersections, and triple intersection to the union.

3

Substitute the known values into the inclusion-exclusion equation.

92 = 48 + 45 + 40 - 18 - 15 - 20 + n(M \cap P \cap C)

Insert the given cardinalities into the formula.

4

Simplify and solve for

n(M \cap P \cap C)

.

92 = 80 + n(M \cap P \cap C) \implies n(M \cap P \cap C) = 92 - 80 = 12

Isolate the unknown triple intersection term.

Key Concept

Principle of Inclusion-Exclusion for Three Sets

Question 19Question →

Find the equation of the locus of a point $P(x, y)$ that moves such that it is equidistant from the fixed points $A(-1, 3)$ and $B(5, -1)$ .

Show answer & explanation

Answer: 3x - 2y - 4 = 0; 3x-2y-4=0; 3x - 2y = 4; 3x-2y=4; 12x - 8y - 16 = 0; y = (3/2)x - 2; y = 1.5x - 2

Answer

The equation of the locus is

3x - 2y - 4 = 0

(or

3x - 2y = 4

).

The locus of a point equidistant from two fixed points

A(-1, 3)

and

B(5, -1)

is the perpendicular bisector of the line segment joining them. Equating the squared distances

(x+1)^2 + (y-3)^2 = (x-5)^2 + (y+1)^2

and simplifying yields the linear equation

3x - 2y - 4 = 0

.

Step-by-Step Solution

1

Set up the distance equality condition using the distance formula.

\\sqrt{(x - (-1))^2 + (y - 3)^2} = \\sqrt{(x - 5)^2 + (y - (-1))^2}

Since point

P(x, y)

is equidistant from

A

and

B

,

PA = PB

.

2

Square both sides to remove the radical signs.

(x + 1)^2 + (y - 3)^2 = (x - 5)^2 + (y + 1)^2

Squaring both sides eliminates square roots and simplifies polynomial expansion.

3

Expand all squared terms on both sides.

x^2 + 2x + 1 + y^2 - 6y + 9 = x^2 - 10x + 25 + y^2 + 2y + 1

Expanding allows gathering like terms.

4

Cancel

x^2

and

y^2

from both sides and collect all terms on one side.

(2x + 10x) + (-6y - 2y) + (10 - 26) = 0 \\Rightarrow 12x - 8y - 16 = 0

Combining like terms simplifies the locus equation into standard linear form.

5

Divide the entire equation by the common factor of 4.

3x - 2y - 4 = 0

Expressing the linear equation in its simplest form gives the perpendicular bisector of line segment

AB

.

Key Concept

Locus equidistant from two fixed points (Perpendicular Bisector)

Estimated Time:2m 0s

Question 20Question →

An acyclic hydrocarbon $X$ with the molecular formula $C_5H_8$ rapidly decolourises bromine water. However, when $X$ is treated with ammoniacal silver nitrate solution, no precipitate is observed. Upon complete catalytic hydrogenation in the presence of a nickel catalyst, $X$ is converted into pentane. What is the IUPAC name of hydrocarbon $X$ ?

Show answer & explanation

Answer: pent-2-yne; 2-pentyne; Pent-2-yne; 2-Pentyne

Answer

Pent-2-yne (or 2-pentyne)

Hydrocarbon

X

has the molecular formula

C_5H_8

, corresponding to two degrees of unsaturation. Complete hydrogenation to pentane confirms an unbranched five-carbon chain. Decolourisation of bromine water verifies unsaturation. Because

X

yields no precipitate with ammoniacal silver nitrate solution, it lacks acidic terminal acetylenic hydrogens (

R-C \equiv C-H

). Therefore, the triple bond must be located internally between carbon-2 and carbon-3, making the compound pent-2-yne.

Step-by-Step Solution

1

Determine the degree of unsaturation and carbon skeleton of hydrocarbon

X

.

Degree of unsaturation is 2, and the carbon skeleton is a straight 5-carbon chain.

The molecular formula

C_5H_8

corresponds to

C_nH_{2n-2}

, indicating two degrees of unsaturation (an alkyne or alkadiene). Complete catalytic hydrogenation yields pentane (

C_5H_{12}

), proving an unbranched five-carbon chain.

2

Analyze the reaction with bromine water.

Hydrocarbon

X

contains carbon-carbon multiple bonds.

Decolourisation of bromine water confirms the presence of unsaturation.

3

Evaluate the test with ammoniacal silver nitrate solution.

Hydrocarbon

X

is an internal (non-terminal) alkyne.

Terminal alkynes possess acidic acetylenic hydrogen atoms (

R-C \equiv C-H

) that react with ammoniacal silver nitrate to form a characteristic white silver acetylide precipitate. The absence of a precipitate rules out pent-1-yne and confirms that the triple bond is located internally at C-2.

4

Deduce the final IUPAC name of hydrocarbon

X

.

pent-2-yne

Combining a straight 5-carbon chain with an internal triple bond between carbon-2 and carbon-3 gives pent-2-yne.

Key Concept

Distinction between terminal and non-terminal alkynes using ammoniacal silver nitrate test and carbon skeleton determination via hydrogenation

Estimated Time:2m 0s

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