Numbers, Quantification and Numerical Applications

Given: Two numbers 132 and 121; modulus = 23.

Step 1: Find 132 mod 23.
$$132 = 23 \times 5 + 17 \implies 132 \equiv 17 \pmod{23}$$

Step 2: Find 121 mod 23.
$$121 = 23 \times 5 + 6 \implies 121 \equiv 6 \pmod{23}$$

Step 3: Add the residues.
$$17 + 6 = 23 \equiv 0 \pmod{23}$$

Answer: $(132 + 121) \bmod 23 = \mathbf{0}$

Given: $7^6 \pmod{3}$.

Step 1: Find $7 \bmod 3$.
$$7 = 3 \times 2 + 1 \implies 7 \equiv 1 \pmod{3}$$

Step 2: Apply the power.
$$7^6 \equiv 1^6 = 1 \pmod{3}$$

Answer: $7^6 \bmod 3 = \mathbf{1}$

Given: $x \equiv 23 \pmod{7}$ and 21 \leq x < 31.

Step 1: Find $23 \bmod 7$.
$$23 = 7 \times 3 + 2 \implies 23 \equiv 2 \pmod{7}$$

So we need $x \equiv 2 \pmod{7}$ with 21 \leq x < 31.

Step 2: List values $\equiv 2 \pmod 7$ in the range:
$$x = 7k + 2$$
- $k = 3$: $x = 23$ ✓ (since 21 \leq 23 < 31)
- $k = 4$: $x = 30$ ✓ (since 21 \leq 30 < 31)

Answer: $x = 23$ or $x = 30$. (The answer key gives $x = \mathbf{30}$, which is the value in the range satisfying the condition.)

Given: Product $226 \times 369 \times 122 \times 461 \times 1025$; divisor = 8.

Step 1: Find each factor mod 8.
$$226 = 8 \times 28 + 2 \implies 226 \equiv 2 \pmod{8}$$
$$369 = 8 \times 46 + 1 \implies 369 \equiv 1 \pmod{8}$$
$$122 = 8 \times 15 + 2 \implies 122 \equiv 2 \pmod{8}$$
$$461 = 8 \times 57 + 5 \implies 461 \equiv 5 \pmod{8}$$
$$1025 = 8 \times 128 + 1 \implies 1025 \equiv 1 \pmod{8}$$

Step 2: Multiply the residues.
$$2 \times 1 \times 2 \times 5 \times 1 = 20$$

Step 3: Find $20 \bmod 8$.
$$20 = 8 \times 2 + 4 \implies 20 \equiv 4 \pmod{8}$$

Answer: The remainder is $\mathbf{4}$.

Given: $(16 \times 29) \pmod 7$.

Step 1: Find each factor mod 7.
$$16 = 7 \times 2 + 2 \implies 16 \equiv 2 \pmod{7}$$
$$29 = 7 \times 4 + 1 \implies 29 \equiv 1 \pmod{7}$$

Step 2: Multiply residues.
$$2 \times 1 = 2$$

Answer: $(16 \times 29) \bmod 7 = \mathbf{2}$

Given: $x \equiv 2 \pmod{9}$ and 0 < x < 47.

Concept: $x = 9k + 2$ for non-negative integer $k$.

Step 1: List values:
- $k = 0$: $x = 2$ ✓
- $k = 1$: $x = 11$ ✓
- $k = 2$: $x = 20$ ✓
- $k = 3$: $x = 29$ ✓
- $k = 4$: $x = 38$ ✓
- $k = 5$: $x = 47$ ✗ (not less than 47)

Answer: $x \in \{11, 20, 29, 38\}$ (excluding 2 if 0 < x is strict; the answer key gives $\mathbf{11, 20, 29, 38}$).

Given: Equivalence class $[4]$ for modulo 6; find positive integers less than 50.

Concept: Members of $[4]$ satisfy $x \equiv 4 \pmod{6}$, i.e., $x = 6k + 4$.

Step 1: List values:
- $k = 0$: $x = 4$
- $k = 1$: $x = 10$
- $k = 2$: $x = 16$
- $k = 3$: $x = 22$
- $k = 4$: $x = 28$
- $k = 5$: $x = 34$
- $k = 6$: $x = 40$
- $k = 7$: $x = 46$
- $k = 8$: x = 52 > 50 ✗

Answer: $[4] = \{4, 10, 16, 22, 28, 34, 40, 46\}$

Given: Present time = 5:00 am; add 200 hours.

Concept: Clock arithmetic uses modulo 24 (for 24-hour clock).

Step 1: Find $200 \bmod 24$.
$$200 = 24 \times 8 + 8 \implies 200 \equiv 8 \pmod{24}$$

Step 2: Add 8 hours to 5:00 am.
$$5 + 8 = 13 \text{ hours} = 1:00 \text{ pm}$$

Answer: The time will be $\mathbf{1:00\ pm}$.

Given: 81 boxes, 21 articles each; rearranged with 5 articles per box.

Step 1: Total articles.
$$81 \times 21 = 1701$$

Step 2: Find $1701 \bmod 5$.
$$1701 = 5 \times 340 + 1 \implies 1701 \equiv 1 \pmod{5}$$

Answer: $\mathbf{1}$ article will be left out without a box.

Given: $3^{128} \bmod 7$.

Step 1: Find the pattern of powers of 3 mod 7.
$$3^1 \equiv 3 \pmod{7}$$
$$3^2 \equiv 9 \equiv 2 \pmod{7}$$
$$3^3 \equiv 6 \pmod{7}$$
$$3^4 \equiv 18 \equiv 4 \pmod{7}$$
$$3^5 \equiv 12 \equiv 5 \pmod{7}$$
$$3^6 \equiv 15 \equiv 1 \pmod{7}$$

The cycle length (order) is 6.

Step 2: Find $128 \bmod 6$.
$$128 = 6 \times 21 + 2 \implies 128 \equiv 2 \pmod{6}$$

Step 3:
$$3^{128} \equiv 3^2 \equiv 9 \equiv 2 \pmod{7}$$

Answer: $3^{128} \bmod 7 = \mathbf{2}$

Given: $n = 35$.

Concept: Euler's totient function $\varphi(n)$ counts positive integers up to $n$ that are coprime to $n$.

Step 1: Factorise 35.
$$35 = 5 \times 7$$

Step 2: Apply the formula.
$$\varphi(35) = 35\left(1 - \frac{1}{5}\right)\left(1 - \frac{1}{7}\right) = 35 \times \frac{4}{5} \times \frac{6}{7} = 24$$

Answer: $\varphi(35) = \mathbf{24}$

Given: $\mu(75) + \mu(85)$.

Concept: Möbius function: $\mu(n) = 0$ if $n$ has a repeated prime factor; $\mu(n) = (-1)^k$ if $n$ is a product of $k$ distinct primes.

Step 1: Factorise 75.
$$75 = 3 \times 5^2$$
Since $5^2$ is a repeated prime factor, $\mu(75) = 0$.

Step 2: Factorise 85.
$$85 = 5 \times 17$$
Two distinct prime factors, so $\mu(85) = (-1)^2 = 1$.

Step 3: Sum.
$$\mu(75) + \mu(85) = 0 + 1 = 1$$

Answer: $\mu(75) + \mu(85) = \mathbf{1}$

Given: $\varphi(90) + \tau(42) + \sigma(72) + \mu(70)$.

Part 1: $\varphi(90)$
$$90 = 2 \times 3^2 \times 5$$
$$\varphi(90) = 90\left(1-\tfrac{1}{2}\right)\left(1-\tfrac{1}{3}\right)\left(1-\tfrac{1}{5}\right) = 90 \times \tfrac{1}{2} \times \tfrac{2}{3} \times \tfrac{4}{5} = 24$$

Part 2: $\tau(42)$ (number of divisors)
$$42 = 2^1 \times 3^1 \times 7^1$$
$$\tau(42) = (1+1)(1+1)(1+1) = 8$$

Part 3: $\sigma(72)$ (sum of divisors)
$$72 = 2^3 \times 3^2$$
$$\sigma(72) = \frac{2^4-1}{2-1} \times \frac{3^3-1}{3-1} = 15 \times 13 = 195$$

Part 4: $\mu(70)$
$$70 = 2 \times 5 \times 7 \quad (3 \text{ distinct primes})$$
$$\mu(70) = (-1)^3 = -1$$

Final Answer:
$$24 + 8 + 195 + (-1) = \mathbf{226}$$

Given: $p = 24$; verify $\varphi(24) + \tau(24) = \sigma(24)$.

Step 1: Factorise 24.
$$24 = 2^3 \times 3^1$$

Step 2: Compute $\varphi(24)$.
$$\varphi(24) = 24\left(1-\tfrac{1}{2}\right)\left(1-\tfrac{1}{3}\right) = 24 \times \tfrac{1}{2} \times \tfrac{2}{3} = 8$$

Step 3: Compute $\tau(24)$.
$$\tau(24) = (3+1)(1+1) = 8$$

Step 4: Compute $\sigma(24)$.
$$\sigma(24) = \frac{2^4-1}{2-1} \times \frac{3^2-1}{3-1} = 15 \times 4 = 60$$

Step 5: Check.
$$\varphi(24) + \tau(24) = 8 + 8 = 16 \neq 60$$

Note: The relation $\varphi(p) + \tau(p) = \sigma(p)$ holds only when $p$ is a prime number (see Q5). For $p = 24$ (composite), the relation does not hold. The question asks to verify — the verification shows LHS $= 16 \neq 60 =$ RHS, confirming the relation is not satisfied for $p = 24$.

Concept: For a prime $p$: divisors are 1 and $p$ only.
- $\varphi(p) = p - 1$
- $\tau(p) = 2$
- $\sigma(p) = 1 + p$
- Check: $(p-1) + 2 = p + 1$ ✓

$$\begin{array}{|c|c|c|c|c|}\hline$$
p & \varphi(p) & \tau(p) & \sigma(p) & \varphi(p)+\tau(p)=\sigma(p) \\\hline
2 & 1 & 2 & 3 & 1+2=3\ ✓ \\\hline
3 & 2 & 2 & 4 & 2+2=4\ ✓ \\\hline
5 & 4 & 2 & 6 & 4+2=6\ ✓ \\\hline
7 & 6 & 2 & 8 & 6+2=8\ ✓ \\\hline
11 & 10 & 2 & 12 & 10+2=12\ ✓ \\\hline
\end{array}

Conclusion: The relation $\varphi(p) + \tau(p) = \sigma(p)$ holds for all primes $p$.

Given: $n = 2p$, $p$ is an odd prime < 14, so $p \in \{3, 5, 7, 11, 13\}$.

Formulas for $n = 2p$ (where $p$ is an odd prime):
- $\varphi(2p) = \varphi(2)\cdot\varphi(p) = 1 \times (p-1) = p-1$
- $\tau(2p) = \tau(2)\cdot\tau(p) = 2 \times 2 = 4$
- $\sigma(2p) = \sigma(2)\cdot\sigma(p) = 3(1+p)$

(i) Table:

$$\begin{array}{|c|c|c|c|c|}\hline$$
p & n=2p & \varphi(n) & \tau(n) & \sigma(n) \\\hline
3 & 6 & 2 & 4 & 12 \\\hline
5 & 10 & 4 & 4 & 18 \\\hline
7 & 14 & 6 & 4 & 24 \\\hline
11 & 22 & 10 & 4 & 36 \\\hline
13 & 26 & 12 & 4 & 42 \\\hline
\end{array}

(ii) Verification of $n + \varphi(n) + \tau(n) = \sigma(n)$:

For $n = 2p$: LHS $= 2p + (p-1) + 4 = 3p + 3 = 3(p+1)$ and RHS $= 3(1+p) = 3(p+1)$. ✓

Checking with values:
- $p=3$: $6+2+4=12=\sigma(6)$ ✓
- $p=5$: $10+4+4=18=\sigma(10)$ ✓
- $p=7$: $14+6+4=24=\sigma(14)$ ✓
- $p=11$: $22+10+4=36=\sigma(22)$ ✓
- $p=13$: $26+12+4=42=\sigma(26)$ ✓

Hence verified.

Given: $n = 24$; show $\tau$ and $\sigma$ are multiplicative.

Concept: A function $f$ is multiplicative if $f(ab) = f(a)\cdot f(b)$ whenever $\gcd(a,b)=1$.

Factorisation: $24 = 8 \times 3 = 2^3 \times 3^1$, and $\gcd(8,3)=1$.

For $\tau$:
$$\tau(24) = \tau(8) \times \tau(3)$$
$$\tau(8) = 4,\quad \tau(3) = 2$$
$$\tau(8)\times\tau(3) = 4 \times 2 = 8$$
$$\tau(24) = (3+1)(1+1) = 8 \checkmark$$

For $\sigma$:
$$\sigma(24) = \sigma(8) \times \sigma(3)$$
$$\sigma(8) = 1+2+4+8 = 15,\quad \sigma(3) = 1+3 = 4$$
$$\sigma(8)\times\sigma(3) = 15 \times 4 = 60$$
$$\sigma(24) = 1+2+3+4+6+8+12+24 = 60 \checkmark$$

Hence, both $\tau(n)$ and $\sigma(n)$ are multiplicative functions for $n = 24$.

Given: Cheaper price = ₹69/kg, Dearer price = ₹100/kg, Mean price = ₹80/kg.

Using Alligation Rule:
$$\text{Ratio} = \frac{\text{Dearer price} - \text{Mean price}}{\text{Mean price} - \text{Cheaper price}} = \frac{100-80}{80-69} = \frac{20}{11}$$

Answer: The required ratio is $\mathbf{20:11}$.

Given: Mean salary = ₹5750; Supervisor avg = ₹20,000; Labour avg = ₹5000; No. of supervisors = 4.

Using Alligation:
$$\frac{\text{No. of supervisors}}{\text{No. of labours}} = \frac{5750 - 5000}{20000 - 5750} = \frac{750}{14250} = \frac{1}{19}$$

Step 2: If supervisors = 4, then:
$$\frac{4}{\text{No. of labours}} = \frac{1}{19} \implies \text{No. of labours} = 76$$

Answer: There are $\mathbf{76}$ labours in the factory.

Given: Initial squash = 70 l; 7 l removed and replaced by water; process repeated 3 times.

Formula: After $n$ replacements,
$$\text{Remaining quantity} = V\left(1 - \frac{r}{V}\right)^n$$
where $V = 70$ l, $r = 7$ l, $n = 3$.

$$= 70\left(1 - \frac{7}{70}\right)^3 = 70\left(\frac{9}{10}\right)^3 = 70 \times \frac{729}{1000} = \frac{51030}{1000} = 51.03 \text{ l}$$

Answer: Approximately $\mathbf{51.03}$ litres of orange squash is left.

Given: Price of pulse 1 = ₹55/kg, Price of pulse 2 = ₹90/kg, Ratio = 2:3.

Step 1: Weighted average price.
$$\text{Price of mixture} = \frac{2 \times 55 + 3 \times 90}{2+3} = \frac{110 + 270}{5} = \frac{380}{5} = ₹76$$

Answer: The price of the mixed variety is $\mathbf{₹76}$ per kg.

Given: Total wheat = 1 quintal = 100 kg; gains = 18%, 28%; overall gain = 24%.

Using Alligation:
$$\frac{\text{Qty at 18\%}}{\text{Qty at 28\%}} = \frac{28-24}{24-18} = \frac{4}{6} = \frac{2}{3}$$

Step 2: Total parts = 2 + 3 = 5.
$$\text{Qty at 18\%} = \frac{2}{5} \times 100 = 40 \text{ kg}$$
$$\text{Qty at 28\%} = \frac{3}{5} \times 100 = 60 \text{ kg}$$

Answer: Wheat sold at 18% gain = 40 kg; at 28% gain = 60 kg.

Given: Syrup = 600 gm with 40% jaggery; target = 50% jaggery.

Step 1: Jaggery present = $0.40 \times 600 = 240$ gm.

Step 2: Let $x$ gm of pure jaggery be added. Pure jaggery is 100% jaggery.

Using Alligation (or direct equation):
$$\frac{240 + x}{600 + x} = \frac{50}{100} = 0.5$$
$$240 + x = 0.5(600 + x)$$
$$240 + x = 300 + 0.5x$$
$$0.5x = 60$$
$$x = 120 \text{ gm}$$

Answer: $\mathbf{120}$ gm of jaggery should be added.

Given: Honey = ₹240/l, Water = ₹0/l, Mixture = ₹200/l.

Using Alligation:
$$\frac{\text{Water}}{\text{Honey}} = \frac{240 - 200}{200 - 0} = \frac{40}{200} = \frac{1}{5}$$

Answer: Water and honey must be mixed in the ratio $\mathbf{1:5}$.

Given: $V = 50$ l, $r = 5$ l, total repetitions = $1 + 4 = 5$ times.

Formula:
$$\text{Juice remaining} = V\left(1 - \frac{r}{V}\right)^n = 50\left(1 - \frac{5}{50}\right)^5 = 50\left(\frac{9}{10}\right)^5$$

$$= 50 \times \frac{59049}{100000} = 50 \times 0.59049 = 29.52 \text{ l (approx.)}$$

Answer: Approximately $\mathbf{29.52}$ litres of juice remains.

Given: Downstream speed $a = 16$ km/hr; Upstream speed $b = 10$ km/hr.

Formula:
$$\text{Speed of boat in still water} = \frac{a+b}{2} = \frac{16+10}{2} = \frac{26}{2} = 13 \text{ km/hr}$$

Answer: Speed of the boat = $\mathbf{13}$ km/hr.

Given: Speed of boat in still water $= 14$ km/hr; Downstream speed $= 24$ km/hr.

Step 1: Find speed of stream.
$$\text{Speed of stream} = 24 - 14 = 10 \text{ km/hr}$$

Step 2: Speed upstream (against stream).
$$= 14 - 10 = 4 \text{ km/hr}$$

Answer: Speed of boat against the stream = $\mathbf{4}$ km/hr.

Given: Distance one way = 8 km; Total time = 4 hr 16 min $= 4\frac{16}{60} = \frac{256}{60} = \frac{64}{15}$ hr; Speed of current $y = 1$ km/hr.

Let speed of boat in still water $= x$ km/hr.

Step 1: Set up equation.
$$\frac{8}{x-1} + \frac{8}{x+1} = \frac{64}{15}$$

Step 2: Simplify.
$$8\left(\frac{(x+1)+(x-1)}{(x-1)(x+1)}\right) = \frac{64}{15}$$
$$\frac{8 \times 2x}{x^2-1} = \frac{64}{15}$$
$$\frac{16x}{x^2-1} = \frac{64}{15}$$
$$15 \times 16x = 64(x^2-1)$$
$$240x = 64x^2 - 64$$
$$64x^2 - 240x - 64 = 0$$
$$4x^2 - 15x - 4 = 0$$
$$(4x+1)(x-4) = 0$$
$$x = 4 \quad (\text{taking positive value})$$

Answer: The actual speed of the boat = $\mathbf{4}$ km/hr.

Given: Speed in still water $x = 7$ km/hr; Stream speed $y = 5$ km/hr; Total time = 7 hr.

Let distance = $d$ km.

$$\frac{d}{x+y} + \frac{d}{x-y} = 7$$
$$\frac{d}{12} + \frac{d}{2} = 7$$
$$d\left(\frac{1}{12} + \frac{1}{2}\right) = 7$$
$$d \times \frac{7}{12} = 7$$
$$d = 12 \text{ km}$$

Answer: The place is $\mathbf{12}$ km away.

Let speed of boat in still water $= x$ km/hr and speed of stream $= y$ km/hr.

Equations:
$$\frac{32}{x-y} + \frac{36}{x+y} = 7 \quad \cdots (1)$$
$$\frac{40}{x-y} + \frac{48}{x+y} = 9 \quad \cdots (2)$$

Let $u = \frac{1}{x-y}$ and $v = \frac{1}{x+y}$:
$$32u + 36v = 7 \quad \cdots (1')$$
$$40u + 48v = 9 \quad \cdots (2')$$

Solve: Multiply (1') by 4 and (2') by 3:
$$128u + 144v = 28$$
$$120u + 144v = 27$$
Subtract: $8u = 1 \implies u = \frac{1}{8}$

Substitute in (1'): $32 \times \frac{1}{8} + 36v = 7 \implies 4 + 36v = 7 \implies v = \frac{1}{12}$

So $x - y = 8$ and $x + y = 12$.

Adding: $2x = 20 \implies x = 10$ km/hr; $y = 2$ km/hr.

Answer: Speed of boat in still water = $\mathbf{10}$ km/hr; Speed of stream = $\mathbf{2}$ km/hr.

Given: Speed in still water $= 7.5$ km/hr; Stream speed $= 1.5$ km/hr; Time $= 50$ min $= \frac{5}{6}$ hr.

Downstream speed $= 7.5 + 1.5 = 9$ km/hr.
Upstream speed $= 7.5 - 1.5 = 6$ km/hr.

Let distance $= d$ km.
$$\frac{d}{9} + \frac{d}{6} = \frac{5}{6}$$
$$d\left(\frac{2+3}{18}\right) = \frac{5}{6}$$
$$d \times \frac{5}{18} = \frac{5}{6}$$
$$d = \frac{5}{6} \times \frac{18}{5} = 3 \text{ km}$$

Answer: The place is $\mathbf{3}$ km away.

Given: Speed of boat : Speed of current $= 36:5$.

Let speed of boat $= 36k$ and speed of current $= 5k$.

Downstream speed $= 36k + 5k = 41k$.
Upstream speed $= 36k - 5k = 31k$.

Let distance $= D$.

Time downstream $= \frac{D}{41k} = 5$ hr 10 min $= \frac{31}{6}$ hr.

So $D = 41k \times \frac{31}{6}$.

Time upstream $= \frac{D}{31k} = \frac{41k \times \frac{31}{6}}{31k} = \frac{41}{6}$ hr $= 6$ hr $50$ min.

Answer: The boat will take $\mathbf{6}$ hours $\mathbf{50}$ minutes to come back.

Given: Pipe A fills in 30 hr; Pipe B fills in 45 hr.

Combined rate per hour:
$$\frac{1}{30} + \frac{1}{45} = \frac{3+2}{90} = \frac{5}{90} = \frac{1}{18}$$

Time to fill:
$$= 18 \text{ hours}$$

Answer: The tank will be filled in $\mathbf{18}$ hours.

Given: Filling rate $= \frac{1}{6}$ per hr; Effective rate with leak $= \frac{1}{9}$ per hr.

Leak rate:
$$\frac{1}{6} - \frac{1}{9} = \frac{3-2}{18} = \frac{1}{18}$$

Time for leakage to empty:
$$= 18 \text{ hours}$$

Answer: The leakage will empty the tank in $\mathbf{18}$ hours.

Given: A fills in 4 hr, B fills in 6 hr, C empties in 8 hr.

Net rate per hour:
$$\frac{1}{4} + \frac{1}{6} - \frac{1}{8} = \frac{6+4-3}{24} = \frac{7}{24}$$

Time to fill:
$$= \frac{24}{7} = 3\frac{3}{7} \text{ hours}$$

Answer: The cistern will be filled in $\mathbf{\dfrac{24}{7}}$ hours (approximately 3 hr 26 min).

Given: Normal fill time = 8 hr; With leak = 10 hr.

Leak rate:
$$\frac{1}{8} - \frac{1}{10} = \frac{5-4}{40} = \frac{1}{40}$$

Time to empty:
$$= 40 \text{ hours}$$

Answer: The leakage will empty the cistern in $\mathbf{40}$ hours.

Given: Inlet fills in 20 hr; Outlet empties in 25 hr; Both open for 10 hr, then outlet closed.

Step 1: Net rate when both open.
$$\frac{1}{20} - \frac{1}{25} = \frac{5-4}{100} = \frac{1}{100}$$

Step 2: Portion filled in 10 hr.
$$= \frac{10}{100} = \frac{1}{10}$$

Step 3: Remaining portion $= 1 - \frac{1}{10} = \frac{9}{10}$.

Step 4: Time to fill remaining with only inlet.
$$= \frac{9}{10} \times 20 = 18 \text{ hr}$$

Total time $= 10 + 18 = \mathbf{28}$ hours.

Given: A fills in 24 min; B fills in 32 min; Tank full in 18 min; B closed after $t$ minutes.

Step 1: A works for all 18 min; B works for $t$ min.
$$\frac{18}{24} + \frac{t}{32} = 1$$
$$\frac{3}{4} + \frac{t}{32} = 1$$
$$\frac{t}{32} = \frac{1}{4}$$
$$t = 8 \text{ minutes}$$

Answer: Pipe B should be closed after $\mathbf{8}$ minutes.

Given:
- A+B+C drain in $\frac{3}{2}$ min $\Rightarrow$ rate $= \frac{2}{3}$
- B+C drain in 2 min $\Rightarrow$ rate $= \frac{1}{2}$
- A+C drain in $\frac{30}{13}$ min $\Rightarrow$ rate $= \frac{13}{30}$

Step 1: Rate of A alone.
$$\text{Rate of A} = \frac{2}{3} - \frac{1}{2} = \frac{4-3}{6} = \frac{1}{6}$$

Step 2: Rate of B alone.
$$\text{Rate of B} = \frac{2}{3} - \frac{13}{30} = \frac{20-13}{30} = \frac{7}{30}$$

Step 3: Rate of A+B.
$$\frac{1}{6} + \frac{7}{30} = \frac{5+7}{30} = \frac{12}{30} = \frac{2}{5}$$

Time for A+B to drain:
$$= \frac{5}{2} = 2.5 \text{ minutes}$$

Answer: Taps A and B together will drain the tank in $\mathbf{2.5}$ minutes.

Given: Race = 1000 m; A beats B by 300 m; B beats C by 200 m.

Step 1: When A finishes 1000 m, B has run 700 m.

Step 2: When B runs 1000 m, C runs 800 m.
So when B runs 700 m, C runs $\frac{800}{1000} \times 700 = 560$ m.

Step 3: When A finishes 1000 m, C has run 560 m.

Margin: $1000 - 560 = 440$ m.

Answer: A defeats C by $\mathbf{440}$ metres.

Given: Race = 1000 m; A gives B a start of 100 m; A gives C a start of 280 m.

Step 1: When A runs 1000 m, B runs 900 m and C runs 720 m.

Step 2: When B runs 1000 m, C runs $\frac{720}{900} \times 1000 = 800$ m.

Start B gives C: $1000 - 800 = 200$ m.

Answer: B can give C a start of $\mathbf{200}$ metres.

Given: Race = 500 m; A beats B by 60 m or 12 seconds.

Step 1: B covers 60 m in 12 seconds.
$$\text{Speed of B} = \frac{60}{12} = 5 \text{ m/s}$$

Step 2: When A finishes 500 m, B has run $500 - 60 = 440$ m.

Step 3: Time taken by B to run 440 m.
$$= \frac{440}{5} = 88 \text{ seconds}$$

This is also the time taken by A to complete the race.

Answer: A completes the race in $\mathbf{88}$ seconds.

Given: Race = 900 m; A gives B a start of 150 m; A defeats B by 50 sec; Speed of A = 4.5 m/s.

Step 1: Time taken by A to finish 900 m.
$$t_A = \frac{900}{4.5} = 200 \text{ sec}$$

Step 2: B has to cover $900 - 150 = 750$ m but takes 50 sec more than A.
$$t_B = 200 + 50 = 250 \text{ sec}$$

Step 3: Speed of B.
$$v_B = \frac{750}{250} = 3 \text{ m/s}$$

Answer: Speed of B = $\mathbf{3}$ m/s.

Given: Speed of A = 3 × Speed of B; A gives B a start of 40 m.

Let the goal be $d$ metres from A's starting point. B starts 40 m ahead, so B covers $(d - 40)$ m.

For equal time:
$$\frac{d}{3v} = \frac{d-40}{v}$$
$$d = 3(d-40)$$
$$d = 3d - 120$$
$$2d = 120$$
$$d = 60 \text{ m}$$

Answer: The goal must be $\mathbf{60}$ metres from A's starting point.

Given: Total games = 40; Lost = 16.

Step 1: Games won $= 40 - 16 = 24$.

Step 2: Percentage won.
$$= \frac{24}{40} \times 100 = 60\%$$

Answer: The team won $\mathbf{60\%}$ of the games.

Given: Race = 200 m; Prateek beats Samarth by 35 m or 7 sec.

Step 1: Samarth covers 35 m in 7 sec.
$$\text{Speed of Samarth} = \frac{35}{7} = 5 \text{ m/s}$$

Step 2: When Prateek finishes, Samarth has run $200 - 35 = 165$ m.

Step 3: Time taken by Prateek = Time taken by Samarth to run 165 m.
$$= \frac{165}{5} = 33 \text{ sec}$$

Answer: Prateek took $\mathbf{33}$ seconds to cover the race.

Given: B's share $= \frac{1}{3}$ of total capital; A's share = B's share + C's share.

Step 1: Let total capital = 1.
$$B = \frac{1}{3}$$

Step 2: A + B + C = 1 and A = B + C, so:
$$A + A = 1 \implies A = \frac{1}{2}$$

Step 3: C = 1 - A - B $= 1 - \frac{1}{2} - \frac{1}{3} = \frac{1}{6}$.

Ratio A:B:C:
$$\frac{1}{2} : \frac{1}{3} : \frac{1}{6} = 3:2:1$$

Answer: Ratio of capitals of A:B:C = $\mathbf{3:2:1}$.

Given: Rahul's profit = ₹5000; Total profit = ₹9000; Ramesh invested ₹3000 for 6 months; Rahul invested for 12 months.

Step 1: Ramesh's profit = ₹9000 - ₹5000 = ₹4000.

Step 2: Ratio of profits = Ratio of (investment × time).
$$\frac{\text{Rahul's investment} \times 12}{3000 \times 6} = \frac{5000}{4000} = \frac{5}{4}$$

Step 3: Solve.
$$\text{Rahul's investment} = \frac{5 \times 3000 \times 6}{4 \times 12} = \frac{90000}{48} = ₹1875$$

Answer: Rahul invested $\mathbf{₹1875}$.

Given: Priya invests ₹40000 for 12 months; Rekha invests ₹50000 for 8 months; Total profit = ₹220000.

Step 1: Equivalent capitals.
$$\text{Priya} = 40000 \times 12 = 480000$$
$$\text{Rekha} = 50000 \times 8 = 400000$$

Step 2: Ratio = 480000 : 400000 = 6:5.

Step 3: Rekha's share.
$$= \frac{5}{11} \times 220000 = ₹100000$$

Answer: Rekha's share in profit = $\mathbf{₹1,00,000}$.

Given: Profit ratio = 6:7:8; Time ratio = 2:3:4.

Concept: Profit $\propto$ Investment $\times$ Time.

$$\text{Investment ratio} = \frac{\text{Profit ratio}}{\text{Time ratio}}$$

$$A:B:C = \frac{6}{2} : \frac{7}{3} : \frac{8}{4} = 3 : \frac{7}{3} : 2$$

Multiply by 3: $= 9 : 7 : 6$.

Answer: Investment ratio of A:B:C = $\mathbf{9:7:6}$.

Given: A's capital = ₹30000; B's capital = ₹60000; Total profit = ₹18000; A gets ₹8500 (profit share + salary).

Step 1: Ratio of capitals = 30000:60000 = 1:2.

Step 2: A's share of profit (without salary).
$$= \frac{1}{3} \times 18000 = ₹6000$$

Step 3: A's salary.
$$= 8500 - 6000 = ₹2500$$

Answer: A receives $\mathbf{₹2500}$ as salary.

Given: Investments: Prakash = ₹11L, Sachin = ₹16.5L, Anil = ₹8.25L; Profit = ₹19.5L.

Step 1: Ratio of investments.
$$11 : 16.5 : 8.25 = 4 : 6 : 3$$
(Multiply each by $\frac{4}{11}$: $4:6:3$; total = 13)

Step 2: Anil's share.
$$= \frac{3}{13} \times 19.5 = \frac{58.5}{13} = 4.5 \text{ lakh}$$

Step 3: 50% of Anil's share.
$$= 0.5 \times 4.5 = 2.25 \text{ lakh} = ₹2,25,000$$

Answer: 50% of Anil's share = $\mathbf{₹2.25}$ lakh.

Given: Total rent = ₹13824; All three stay for 4 months; Ritik leaves after 4 months; Priyesh leaves after 4+5=9 months; Nikhil stays for the full period.

Step 1: Determine total period. Nikhil stays for the full tenancy. Since Priyesh leaves after 9 months, total period = 12 months (assumed for a year).

Step 2: Equivalent person-months.
$$\text{Nikhil} = 12 \text{ months}$$
$$\text{Priyesh} = 9 \text{ months}$$
$$\text{Ritik} = 4 \text{ months}$$

Ratio = 12:9:4; Total = 25.

Step 3: Shares.
$$\text{Nikhil} = \frac{12}{25} \times 13824 = ₹6635.52 \approx ₹6635.52$$
$$\text{Priyesh} = \frac{9}{25} \times 13824 = ₹4976.64 \approx ₹4976.64$$
$$\text{Ritik} = \frac{4}{25} \times 13824 = ₹2211.84 \approx ₹2211.84$$

Answer: Nikhil pays ₹6635.52, Priyesh pays ₹4976.64, and Ritik pays ₹2211.84.

Given: Arun invests ₹38000 for 12 months; Bakul invests ₹55000 for 7 months; Total profit = ₹22000.

Step 1: Equivalent capitals.
$$\text{Arun} = 38000 \times 12 = 456000$$
$$\text{Bakul} = 55000 \times 7 = 385000$$

Step 2: Ratio = 456000:385000 = 456:385.

Step 3: Shares.
$$\text{Arun's share} = \frac{456}{841} \times 22000 \approx ₹11930$$
$$\text{Bakul's share} = \frac{385}{841} \times 22000 \approx ₹10070$$

Step 4: Difference.
$$\approx 11930 - 10070 = ₹1860$$

Answer: The approximate difference between their shares is $\mathbf{₹1860}$.

Correct Option: (D) Neither S1 nor S2

Justification:
- S1 is TRUE: SJF (Shortest Job First) scheduling minimises average waiting time among all non-preemptive scheduling algorithms.
- S2 is TRUE: SJF can cause starvation — longer processes may never get CPU time if shorter processes keep arriving.

Since both statements are true, neither is false. Hence option (D) is correct.