The correct answer is (A) 3°. In triangle RTS (T is the intersection of the bisectors of \(\angle R\) and \(\angle S\)): \(\angle TRS + \angle TSR + \angle RTS = 180°\). Since TR and TS are bisectors: \(\angle TRS = \frac{\angle R}{2}\) and \(\angle TSR = \frac{\angle S}{2}\). So \(\frac{\angle R + \angle S}{2} + 13.5° = 180°\), which gives \(\angle R + \angle S = 333°\). In quadrilateral PQRS: \(\angle P + \angle Q + \angle R + \angle S = 360°\), so \(\angle P + \angle Q = 27°\). With \(\angle P : \angle Q = 4 : 5\), let \(\angle P = 4k\) and \(\angle Q = 5k\). Then \(9k = 27\), so \(k = 3\). Therefore \(\angle P = 12°\), \(\angle Q = 15°\), and the difference \(= 15° - 12° = 3°\). Exam Tip: When bisectors of two angles of a quadrilateral meet inside, the angle between them \(= 180° - \frac{\angle R + \angle S}{2}\).

The correct answer is (D) 20%. Cost price: 2 lemons = Re 1, so 1 lemon = Rs 0.50. Selling price: 5 lemons = Rs 3, so 1 lemon = Rs 0.60. Profit per lemon \(= 0.60 - 0.50 = \text{Rs } 0.10\). Profit % \(= \frac{\text{Profit}}{\text{CP}} \times 100 = \frac{0.10}{0.50} \times 100 = 20\%\). Exam Tip: For unit-based problems, find price per unit, then calculate profit % using: \(\frac{\text{SP} - \text{CP}}{\text{CP}} \times 100\).

The correct answer is (D) 72 cm. Right-angled triangle with legs 18 cm and 24 cm. Using Pythagoras' theorem: Hypotenuse \(= \sqrt{18^2 + 24^2} = \sqrt{324 + 576} = \sqrt{900} = 30\) cm Perimeter \(= 18 + 24 + 30 = 72\) cm Exam Tip: 18-24-30 is a Pythagorean triplet (multiple of 3-4-5). Recognize these patterns.

The correct answer is (A) \(-2x^2 - 13x + 10\). Simplify: \(x^2 - (3x - 2)(x + 5)\) First, expand \((3x - 2)(x + 5) = 3x^2 + 15x - 2x - 10 = 3x^2 + 13x - 10\) Now subtract: \(x^2 - (3x^2 + 13x - 10) = x^2 - 3x^2 - 13x + 10 = -2x^2 - 13x + 10\) Exam Tip: Expand carefully using FOIL, then distribute the negative sign across all terms.

The correct answer is (A) Rs 2,295. Simple Interest formula: \(\text{SI} = \frac{P \times R \times T}{100}\). Where \(P = \text{Rs } 13{,}500\), \(R = 17\%\) per annum, \(T = 1\) year. \(\text{SI} = \frac{13500 \times 17 \times 1}{100} = \frac{229500}{100} = \text{Rs } 2{,}295\). Exam Tip: For SI problems, always use the formula \(\text{SI} = \frac{PRT}{100}\). Make sure units match (annual rate, time in years).

The correct answer is (B) 9 : 52. Let LP = Listed Price, CP = Cost Price, SP = Selling Price. Discount = 73%, so SP = 27% of LP = 0.27 × LP Profit = 56%, so SP = 156% of CP = 1.56 × CP Equate: 0.27 × LP = 1.56 × CP \(\frac{CP}{LP} = \frac{0.27}{1.56} = \frac{27}{156} = \frac{9}{52}\) Ratio CP : LP = 9 : 52 Exam Tip: Discount reduces price; profit increases it. Set the selling prices equal to find the ratio.

Sides in ratio 28:23:35, perimeter = 516 cm. Let sides = 28k, 23k, 35k. Then 86k = 516, so k = 6. Sides: a = 168 cm, b = 138 cm, c = 210 cm. Semi-perimeter \(s = \frac{516}{2} = 258\). By Heron's formula: Area = \(\sqrt{s(s-a)(s-b)(s-c)} = \sqrt{258 \times 90 \times 120 \times 48}\) Factor: \(258 \times 90 \times 120 \times 48 = 2^{9} \times 3^{5} \times 5^{2} \times 43\) Area = \(\sqrt{2^{9} \times 3^{5} \times 5^{2} \times 43} = 2^{4} \times 3^{2} \times 5 \times \sqrt{2 \times 3 \times 43} = 720\sqrt{258} \text{ cm}^{2}\). The answer is (D) \(720\sqrt{258} \text{ cm}^{2}\).

The correct answer is (B) 624. Three numbers in ratio 17 : 12 : 6, sum = 1820. Let numbers = 17k, 12k, 6k. \(17k + 12k + 6k = 1820 35k = 1820 k = 52\) Middle number = \(12k = 12 \times 52 = 624\) Exam Tip: When numbers are in ratio a : b : c, write them as ak, bk, ck. Sum them to find k.

The correct answer is (A) 127. Let two consecutive odd numbers be n and \((n+2)\). \(n \times (n+2) = 16{,}383\). \(n^2 + 2n - 16{,}383 = 0\). Using quadratic formula: \(n = \frac{-2 \pm \sqrt{4 + 65{,}532}}{2} = \frac{-2 \pm \sqrt{65{,}536}}{2} = \frac{-2 \pm 256}{2}\). \(n = \frac{254}{2} = 127\) (taking positive value). Verify: \(127 \times 129 = 16{,}383\). ✓ Exam Tip: For consecutive odd numbers, use \(n(n+2)\). Set up quadratic equation and solve.

The correct answer is (B) 5 days. 6 men complete work in 5 days → 1 man does \(\frac{1}{30}\) of the work per day. 2 women complete work in 15 days → 1 woman does \(\frac{1}{30}\) of the work per day. So 1 man = 1 woman in efficiency. 3 men + 3 women = 6 workers, each doing \(\frac{1}{30}\) per day. Combined rate = \(\frac{6}{30} = \frac{1}{5}\) of the work per day. Time = \(1 \div \frac{1}{5} = 5\) days.

The correct answer is (B) 1792. Given: Ratio of two numbers = \(7 : 8\), HCF = 32 The two numbers are: \(7 \times 32 = 224\) and \(8 \times 32 = 256\) Since 7 and 8 are co-prime (no common factor), their LCM = \(7 \times 8 \times \text{HCF} = 56 \times 32 = 1792\) Alternatively: LCM = \(\frac{\text{Product of numbers}}{\text{HCF}} = \frac{224 \times 256}{32} = \frac{57344}{32} = 1792\) Why other options are wrong: (A) 1833 — Not a product of 7, 8, and 32. (C) 1809 — Not a product of 7, 8, and 32. (D) 1783 — Not a product of 7, 8, and 32. Exam Tip: When ratio is \(a:b\) and HCF = \(h\), then numbers are \(ah\) and \(bh\). If \(a\) and \(b\) are co-prime, LCM = \(a \times b \times h\).

The correct answer is (B) 277 seconds. Train length = 1439 m, Platform length = 1331 m. Total distance to clear = \(1439 + 1331 = 2770\) m Speed = 36 km/hr = \(36 \times \frac{1000}{3600} = 10\) m/s Time = Distance \(\div\) Speed = \(2770 \div 10 = 277\) seconds Exam Tip: To clear a platform, the train must travel its own length plus the platform length.

The correct answer is (C) 20%. Marked Price (MP) = Rs 7,600. MP is 25% above CP, so: MP = \(1.25 \times \text{CP}\) CP = \(7600 \div 1.25 = 6080\) Discount = 4% on MP, so: SP = \(96\% \times 7600 = 0.96 \times 7600 = 7296\) Profit % = \(\frac{\text{SP} - \text{CP}}{\text{CP}} \times 100 = \frac{7296 - 6080}{6080} \times 100 = \frac{1216}{6080} \times 100 = 20\%\) Exam Tip: Always find CP and SP first, then calculate profit % using the formula.

The correct answer is (C) 10 m. The ladder, wall, and ground form a right triangle. The angle with the ground is \(60°\), and the foot of the ladder is 5 m from the wall (adjacent side). The ladder is the hypotenuse. \(\cos(60°) = \frac{\text{adjacent}}{\text{hypotenuse}} = \frac{5}{L} \frac{1}{2} = \frac{5}{L} L = 5 \times 2 = 10\) m Exam Tip: \(\cos = \frac{\text{adjacent}}{\text{hypotenuse}}\). When you know the distance from the wall (adjacent) and the angle with the ground, use cosine to find the ladder length.

The correct answer is (D) 12 metres. Let the three pieces be: smallest = \(x\), middle = \(y\), largest = \(3x\) (given: largest = \(3 \times\) smallest) Condition 1: \(x + y + 3x = 24 \Rightarrow 4x + y = 24\) Condition 2: Difference between two larger pieces = Difference between two smaller pieces → \(3x - y = y - x \Rightarrow 4x = 2y \Rightarrow y = 2x\) Substituting: \(4x + 2x = 24 \Rightarrow 6x = 24 \Rightarrow x = 4\) Therefore: smallest = 4 m, middle = 8 m, largest = 12 m. Check: \(4 + 8 + 12 = 24\) ✓, \((12-8) = (8-4) = 4\) ✓ Exam Tip: Break word problems into algebraic equations. Use conditions systematically.

The correct answer is (D) ₹1,20,666. Total bill = \(250 \times ₹1,014 = ₹2,53,500\) After 30% discount: \(₹2,53,500 \times 0.70 = ₹1,77,450\) After 15% discount: \(₹1,77,450 \times 0.85 = ₹1,50,832.50\) After 20% card discount: \(₹1,50,832.50 \times 0.80 = ₹1,20,666\) The credit card payment amount is ₹1,20,666.

The correct answer is (B) \(12 : 7\). Red : Green = \(14 : 8 = 7 : 4\) (simplifying) Pink : Red = \(2 : 6 = 1 : 3\) (simplifying) To find Green : Pink, we need a common ratio for Red. From first ratio: Red = \(7k\), Green = \(4k\) From second ratio: Pink = \(1m\), Red = \(3m\) So \(7k = 3m \Rightarrow m = \frac{7k}{3}\) Pink = \(\frac{7k}{3}\) Green : Pink = \(4k : \frac{7k}{3} = 12 : 7\)

The correct answer is (B) 72. Let the number be \(x\). Given: 20% of \(x\) + 72 = \(x\) → \(0.2x + 72 = x \Rightarrow 72 = 0.8x \Rightarrow x = 90\). Therefore: 80% of 90 = \(0.8 \times 90 = 72\). Check: 20% of 90 = 18, and \(18 + 72 = 90\) ✓ Exam Tip: Convert percentage to decimal (20% = 0.2). Notice that 80% = 1 - 20%.

The correct answer is (D) 60 cm. Using Heron's formula: \(s = \frac{75 + 34 + 61}{2} = \frac{170}{2} = 85\) cm Area = \(\sqrt{s(s-a)(s-b)(s-c)} = \sqrt{85 \times 10 \times 51 \times 24}\) \(= \sqrt{1,040,400} = 1020 \text{ cm}^2\) Using Area = \(\frac{1}{2} \times \text{base} \times \text{height}\): \(1020 = \frac{1}{2} \times 34 \times h\) \(h = \frac{2040}{34} = 60\) cm.

The correct answer is (A) increase by ₹1,086. Original: Income = ₹12,000, Savings = 27% of ₹12,000 = ₹3,240, Expenditure = ₹12,000 − ₹3,240 = ₹8,760 New Income = \(₹12{,}000 \times 1.20 = ₹14{,}400\) New Expenditure = \(₹8{,}760 \times 1.15 = ₹10{,}074\) New Savings = \(₹14{,}400 − ₹10{,}074 = ₹4{,}326\) Change in savings = \(₹4{,}326 − ₹3{,}240 = ₹1{,}086\) (increase) Raman's savings increase by ₹1,086.