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PPSC Lecturer Math Quantitative Math Solved MCQs & Quiz
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PPSC Lecturer Math Quantitative Math Solved MCQs & Quiz
Quantitative Mathematics and General Ability testing form an essential part of competitive examinations administered by the Punjab Public Service Commission (PPSC), Federal Public Service Commission (FPSC), and university entry tests. Mastery over arithmetic, algebraic reasoning, and discrete counting principles allows candidate lecturers to solve high-yield problems rapidly. Advanced preparation at a specialized math learning center emphasizes recognizing fundamental structural formulas, unit conversions, and algebraic shortcuts.
Below is a summary of the fundamental formulas utilized across the exercises in this study guide:
| Domain | Core Concept | Mathematical Formula |
|---|---|---|
| Work & Rate | Combined Work Rate | $\frac{1}{T_{net}} = \frac{1}{T_1} + \frac{1}{T_2} - \frac{1}{T_{leak}}$ |
| Kinematics | Relative Speed (Opposite Directions) | $V_{rel} = v_1 + v_2$; Speed Unit Conversion: $1\text{ km/h} = \frac{5}{18}\text{ m/s}$ |
| Financial Math | Compound Growth / Selling Price | $P_n = P_0\left(1 + \frac{r}{100}\right)^n$; $SP = CP \times \left(1 + \frac{\text{Profit\%}}{100}\right)$ |
| Combinatorics | Permutations with Repetition | $N = \frac{n!}{n_1! n_2! \dots n_k!}$ |
| Number Theory | Fermat's Little Theorem | $a^{p-1} \equiv 1 \pmod p$ where $\gcd(a,p)=1$ and $p$ is prime. |
Question 1 (Part i): Pipe A can fill a tank in $6$ hours, while Pipe B can empty the full tank in $8$ hours. If both pipes are opened simultaneously in an empty tank, calculate the total time required to fill the tank completely.
Solution:
Let the capacity of the tank be $1$ unit (or $W = 1$).
Work rate of Pipe A per hour:
$$R_A = +\frac{1}{6}\text{ tank/hour}$$
Work rate of Pipe B per hour (since it empties, it performs negative work):
$$R_B = -\frac{1}{8}\text{ tank/hour}$$
Net work rate when both pipes operate simultaneously:
$$R_{net} = R_A + R_B = \frac{1}{6} - \frac{1}{8}$$
Find the Least Common Multiple (LCM) of $6$ and $8$, which is $24$:
$$R_{net} = \frac{4 - 3}{24} = \frac{1}{24}\text{ tank/hour}$$
The total time $T$ required to fill the entire tank is given by the inverse of the net rate:
$$T = \frac{1}{R_{net}} = \frac{1}{1/24} = 24\text{ hours}$$
Answer: $$24\text{ hours}$$
Question 1 (Part ii): If $3$ men or $6$ women can complete a piece of work in $16$ days, determine how many days it will take for $12$ men and $8$ women to complete the same work working together.
Solution:
Equate the total work done by men and women:
$$\text{Work} = 3 \text{ Men} \times 16 \text{ days} = 6 \text{ Women} \times 16 \text{ days}$$
$$3 \text{ Men} = 6 \text{ Women} \implies 1 \text{ Man} = 2 \text{ Women}$$
Convert the required combined workforce ($12\text{ men} + 8\text{ women}$) into equivalent women:
$$12 \text{ Men} = 12 \times 2 \text{ Women} = 24 \text{ Women}$$
$$\text{Total equivalent women} = 24 + 8 = 32 \text{ Women}$$
Using the inverse proportion formula $M_1 \times D_1 = M_2 \times D_2$:
$$6 \text{ Women} \times 16 \text{ days} = 32 \text{ Women} \times D_2$$
$$D_2 = \frac{6 \times 16}{32} = \frac{96}{32} = 3\text{ days}$$
Answer: $$3\text{ days}$$
Question 2 (Part i): A train $150\text{ meters}$ long is moving at a uniform speed of $72\text{ km/h}$. Calculate the time in seconds it will take to completely cross a bridge $250\text{ meters}$ long.
Solution:
First, convert the speed from $\text{km/h}$ to $\text{m/s}$:
$$v = 72 \times \frac{5}{18} = 4 \times 5 = 20\text{ m/s}$$
The total distance $D$ covered by the train to clear the bridge completely is the sum of the length of the train and the length of the bridge:
$$D = L_{\text{train}} + L_{\text{bridge}} = 150 + 250 = 400\text{ meters}$$
Using the formula $\text{Time} = \frac{\text{Distance}}{\text{Speed}}$:
$$t = \frac{D}{v} = \frac{400}{20} = 20\text{ seconds}$$
Answer: $$20\text{ seconds}$$
Question 2 (Part ii): Two trains of lengths $180\text{ m}$ and $x\text{ m}$ are moving in opposite directions at speeds of $54\text{ km/h}$ and $36\text{ km/h}$ respectively. If they cross each other completely in $12\text{ seconds}$, find $x$.
Solution:
Convert both speeds to $\text{m/s}$:
$$v_1 = 54 \times \frac{5}{18} = 15\text{ m/s}$$
$$v_2 = 36 \times \frac{5}{18} = 10\text{ m/s}$$
Since the trains are moving in opposite directions, their relative speed $v_{rel}$ is:
$$v_{rel} = v_1 + v_2 = 15 + 10 = 25\text{ m/s}$$
Total distance covered while crossing each other is $D_{total} = L_1 + L_2 = 180 + x$.
Using $D_{total} = v_{rel} \times t$:
$$180 + x = 25 \times 12$$
$$180 + x = 300$$
$$x = 300 - 180 = 120\text{ meters}$$
Answer: $$120\text{ meters}$$
Question 3 (Part i): A shopkeeper purchases an article for $\text{Rs. } 1200$ and sells it at a gain of $15\%$. Find the selling price of the article.
Solution:
Given Cost Price ($CP$) = $1200$, Profit Percentage = $15\%$.
The Profit Amount is calculated as:
$$\text{Profit} = \frac{15}{100} \times 1200 = 15 \times 12 = 180\text{ PKR}$$
Selling Price ($SP$) is given by:
$$SP = CP + \text{Profit} = 1200 + 180 = 1380\text{ PKR}$$
Alternatively:
$$SP = CP \times \left(1 + \frac{15}{100}\right) = 1200 \times 1.15 = 1380\text{ PKR}$$
Answer: $$1380\text{ PKR}$$
Question 3 (Part ii): The population of a city increases at a rate of $5\%$ per annum. If the present population is $80,000$, calculate the population after $2$ years.
Solution:
This follows compound growth:
$$P_n = P_0 \left(1 + \frac{r}{100}\right)^n$$
Substitute $P_0 = 80000$, $r = 5$, and $n = 2$:
$$P_2 = 80000 \times \left(1 + \frac{5}{100}\right)^2$$
$$P_2 = 80000 \times \left(\frac{21}{20}\right)^2$$
$$P_2 = 80000 \times \frac{441}{400}$$
$$P_2 = 200 \times 441 = 88200$$
Answer: $$88,200$$
Question 4 (Part i): Determine the number of distinct arrangements that can be formed using all the letters of the word "LECTURER".
Solution:
Count total letters in "LECTURER": $n = 8$.
Count the frequencies of each repeating letter:
Question 4 (Part ii): A container contains $5$ red balls and $4$ black balls. Two balls are drawn at random simultaneously. What is the probability that both drawn balls are red?
Solution:
Total number of balls = $5 + 4 = 9$.
Total possible outcomes when selecting $2$ balls from $9$:
$$n(S) = \binom{9}{2} = \frac{9 \times 8}{2 \times 1} = 36$$
Favorable outcomes for selecting $2$ red balls from $5$ available red balls:
$$n(E) = \binom{5}{2} = \frac{5 \times 4}{2 \times 1} = 10$$
The required probability $P(E)$ is:
$$P(E) = \frac{n(E)}{n(S)} = \frac{10}{36} = \frac{5}{18}$$
Answer: $$\frac{5}{18}$$
Question 5 (Part i): Two mixtures $A$ and $B$ contain milk and water in ratios $4:3$ and $2:3$ respectively. In what ratio must these two mixtures be combined to produce a new mixture containing milk and water in an equal ratio of $1:1$?
Solution:
Focus on the fraction of milk in each mixture:
Question 5 (Part ii): Divide an amount of $\text{Rs. } 1560$ among $A$, $B$, and $C$ such that $A$ receives $\frac{1}{2}$ of $B$'s share and $B$ receives $\frac{1}{3}$ of $C$'s share. Determine $A$'s exact share.
Solution:
Let $C$'s share be $x$.
Then $B$'s share is $B = \frac{1}{3}x$.
$A$'s share is $A = \frac{1}{2} B = \frac{1}{2}\left(\frac{1}{3}x\right) = \frac{1}{6}x$.
The total sum of shares is $1560$:
$$A + B + C = 1560$$
$$\frac{1}{6}x + \frac{1}{3}x + x = 1560$$
Find a common denominator ($6$):
$$\frac{x + 2x + 6x}{6} = 1560$$
$$\frac{9x}{6} = 1560 \implies \frac{3x}{2} = 1560$$
$$3x = 3120 \implies x = 1040\text{ PKR}$$
Now, compute $A$'s share:
$$A = \frac{1}{6}(1040) = \frac{1040}{6} = 173.33\text{ PKR}$$
Answer: $$173.33\text{ PKR}$$
Question 6 (Part i): Find the remainder when $2^{100}$ is divided by $7$.
Solution:
Using Fermat's Little Theorem, since $7$ is prime and $\gcd(2,7)=1$:
$$2^{7-1} \equiv 1 \pmod 7 \implies 2^6 \equiv 1 \pmod 7$$
Express the exponent $100$ in terms of multiples of $6$:
$$100 = 6 \times 16 + 4$$
Therefore:
$$2^{100} = 2^{6 \times 16 + 4} = (2^6)^{16} \cdot 2^4$$
Apply modular reduction:
$$(2^6)^{16} \cdot 2^4 \equiv (1)^{16} \cdot 16 \pmod 7$$
$$2^{100} \equiv 16 \pmod 7$$
Since $16 = 2 \times 7 + 2$:
$$16 \equiv 2 \pmod 7$$
Hence, the remainder is $2$.
Answer: $$2$$
Question 6 (Part ii): Find the unit digit of the expression $7^{95} - 3^{58}$.
Solution:
Find the unit digit of $7^{95}$ using cyclicity of powers of $7$ (period $4$: $7^1=7, 7^2=9, 7^3=3, 7^4=1$):
$$95 \pmod 4 = 3 \implies \text{Unit digit of } 7^{95} = \text{Unit digit of } 7^3 = 3$$
Find the unit digit of $3^{58}$ using cyclicity of powers of $3$ (period $4$: $3^1=3, 3^2=9, 3^3=7, 3^4=1$):
$$58 \pmod 4 = 2 \implies \text{Unit digit of } 3^{58} = \text{Unit digit of } 3^2 = 9$$
Now calculate the difference of unit digits:
$$\text{Unit digit} = 3 - 9$$
Since $3 < 9$, borrow $10$ from the tens place:
$$\text{Unit digit} = (13 - 9) = 4$$
Answer: $$4$$
Q1: A worker completes a job in $12$ days. With the help of an assistant, the work is finished in $8$ days. How long would the assistant take to finish the work alone?
Q2: If the speed of a boat in still water is $10\text{ km/h}$ and the stream speed is $2\text{ km/h}$, what is the downstream speed?