The initial concentration of the strong acid \(HCl\) is \([H^+]_{initial} = 1.0 \times 10^{-4} M\).


When diluted 10 times, the final concentration \([H^+]_{final}\) is:

\([H^+]_{final} = \frac{10^{-4}}{10} = 1.0 \times 10^{-5} M\).


The \(pH\) is calculated as \(pH = -\log[H^+]\).

\(pH = -\log(10^{-5}) = 5\).


The \(pH\) of the final solution is 5, which falls between 4 and 5.
Quick Tip: Dilution reduces the concentration proportionally. A 10-fold dilution increases the \(pH\) by \(\log(10) = 1\) unit for strong acids and bases, provided the resulting concentration is significantly greater than \(10^{-7} M\).

The initial concentration of \(OH^-\) is \([OH^-]_{initial} = 10^{-6} M\).


After 100 times dilution, the \([OH^-]\) from the base is:

\([OH^-]_{base} = \frac{10^{-6}}{100} = 10^{-8} M\).


Since this concentration is comparable to \([OH^-]\) from water auto-ionization (\(10^{-7} M\)), the total \([OH^-]\) must be considered.


Total \([OH^-]\) will be slightly greater than \(10^{-7} M\).


If \([OH^-] = 10^{-7} M\), \(pOH = 7\) and \(pH = 7\).


Since the total \([OH^-]\) is greater than \(10^{-7} M\) (due to the added base), the \(pOH\) will be slightly less than 7.


Therefore, the \(pH = 14 - pOH\) will be slightly greater than 7.


The \(pH\) of the diluted solution is between 7 and 8.
Quick Tip: When the acid or base concentration falls below \(10^{-6} M\), the self-ionization of water must be accounted for. The final \(pH\) of any aqueous solution at \(25^{\circ}C\) cannot exceed the range \([0, 14]\) and cannot be far from 7 for ultra-dilute solutions.

We use the definition \(pH = -\log[H^+]\).


Given \([H^+] = 3.2 \times 10^{-3} M\).

\(pH = -\log(3.2 \times 10^{-3})\).

\(pH = -(\log(3.2) + \log(10^{-3}))\).

\(pH = -(\log(3.2) - 3) = 3 - \log(3.2)\).


We use the approximation \(\log(3.2) \approx 0.5\).

\(pH \approx 3 - 0.5 = 2.5\).


The exact value is \(3 - 0.505 = 2.495\), which is closest to 2.5.
Quick Tip: To estimate \(pH\) without a calculator for \([H^+] = C \times 10^{-N}\), remember \(pH = N - \log(C)\). Since \(\log(C)\) is positive, the \(pH\) must be slightly less than \(N\).

At \(25^{\circ}C\), the ion product of water \(K_w\) is \(1.0 \times 10^{-14}\).


In pure water, the concentrations of \(H^+\) and \(OH^-\) are equal.

\([H^+] = \sqrt{K_w} = \sqrt{1.0 \times 10^{-14}}\).

\([H^+] = 1.0 \times 10^{-7} M\).


The \(pH\) is calculated as:
\(pH = -\log(10^{-7}) = 7\).


Pure water is neutral, and at \(25^{\circ}C\), its \(pH\) is 7.
Quick Tip: Neutrality means \([H^+] = [OH^-]\). This condition occurs when \(pH = pOH = 7\) at \(25^{\circ}C\). Remember that \(pH\) is temperature-dependent because \(K_w\) is temperature-dependent.

Acidity is inversely related to the \(pH\) value.


A lower \(pH\) corresponds to a higher concentration of hydrogen ions \([H^+]\) and thus greater acidity.


The options are compared: \(4 < 6 < 7 < 9\).


The smallest \(pH\) value listed is \(pH=4\).


Therefore, the solution with \(pH=4\) is the most acidic.
Quick Tip: Solutions with \(pH < 7\) are acidic, \(pH > 7\) are basic, and \(pH = 7\) is neutral. Acidity increases exponentially as \(pH\) approaches 0.

"Abundant" means existing or available in large quantities; ample.


We look for the word with the same meaning (synonym).


(A) Scarce, (C) Rare, and (D) Limited all imply a small or restricted quantity (antonyms).


(B) Plentiful means existing in great abundance.


Plentiful is the correct synonym.
Quick Tip: Focus on the core definition of the word. "Abundant" implies high quantity; "Plentiful" confirms this meaning.

"Expand" means to increase in size, extent, or volume.


We look for the word with the opposite meaning (antonym).


(A) Increase, (B) Enlarge, and (C) Spread are synonyms for "Expand".


(D) Contract means to decrease in size or draw together.


Contract is the correct antonym.
Quick Tip: In selecting antonyms, first identify the action/state implied by the word (e.g., expansion/growth) and then seek the direct opposite (e.g., contraction/reduction).

The word "Accommodation" is often misspelled regarding the number of 'c's and 'm's.


The correct spelling is A-C-C-O-M-M-O-D-A-T-I-O-N, containing double C and double M.


Option (C) matches this spelling.
Quick Tip: Memorize the tricky parts of commonly misspelled words. For "accommodation," remember the structure \(ACC\)...\(OMM\)...

The phrase "good at" is the standard idiomatic expression used to indicate proficiency or skill in a particular subject or activity.


The correct grammatical usage is: \(Subject + is/are + good/bad + at + skill/subject\).


Therefore, the correct preposition is "at".
Quick Tip: Phrasal verbs and idiomatic expressions often dictate preposition usage. Skill or competence always takes the preposition 'at' (e.g., adept at, talented at, poor at).

The active sentence is: "The teacher (Subject) explained (Verb: Simple Past) the lesson (Object)."


To convert to passive voice, the object becomes the subject: "The lesson".


The verb tense (Simple Past) changes to the passive structure: was/were + past participle.

"The lesson was explained".


The original subject becomes the agent: "by the teacher".


The correct passive sentence is: "The lesson was explained by the teacher."
Quick Tip: Always match the tense of the auxiliary verb ('to be') in the passive voice to the tense of the main verb in the active voice. Since 'explained' is Simple Past, we use 'was explained'.

The formula relating average, sum, and count is: \(Sum = Average \times Count\).


Given: \(Average = 18\).

Given: \(Count = 5\).


Total Sum \(= 18 \times 5\).


Total Sum \(= 90\).
Quick Tip: Multiplication is the fastest route to finding the total sum when the average and count are known. \(18 \times 5 = (20-2) \times 5 = 100 - 10 = 90\).

Speed is calculated by dividing distance by time.
\(Speed = \frac{Distance}{Time}\).


Given: \(Distance = 120 km\).

Given: \(Time = 2 hours\).

\(Speed = \frac{120 km}{2 hours}\).

\(Speed = 60 km/h\).
Quick Tip: Ensure units are correct. If the distance unit is kilometer (\(km\)) and time unit is hour (\(h\)), the resulting speed is in \(km/h\).

Cost Price (\(CP\)) \(= 500\).

Profit percentage (\(P%\)) \(= 10%\).


The selling price (\(SP\)) is the \(CP\) plus the profit amount.


Profit amount \(= 10% of 500 = 0.10 \times 500 = 50\).

\(SP = CP + Profit\).

\(SP = 500 + 50\).

\(SP = 550\).
Quick Tip: A profit of \(10%\) means the selling price is \(110%\) of the cost price. \(SP = 500 \times 1.10 = 550\).

We use the Simple Interest (\(SI\)) formula: \(SI = \frac{P \times R \times T}{100}\).


Given: Principal (\(P\)) \(= 2000\).

Rate (\(R\)) \(= 5\).

Time (\(T\)) \(= 2\).

\(SI = \frac{2000 \times 5 \times 2}{100}\).

\(SI = \frac{20000}{100}\).

\(SI = 200\).
Quick Tip: Simple interest calculations are straightforward multiplication. \(5%\) per year for 2 years means a total interest rate of \(10%\). \(10%\) of \(2000\) is \(200\).

First, we find the cost of a single pen (unit cost).


Cost of 1 pen \(= \frac{Total Cost}{Quantity} = \frac{45}{3}\).


Cost of 1 pen \(= 15\).


Now, calculate the cost of 10 pens.


Cost of 10 pens \(= 10 \times 15\).


Cost of 10 pens \(= 150\).
Quick Tip: Proportionality problems require establishing the unit rate first. \(3:45\) implies a ratio of \(1:15\). Therefore, \(10\) pens correspond to \(10 \times 15 = 150\).