Patna, being the educational hub, has a relatively higher performance in Mathematics compared to other districts.

BSEB mandates Mathematics as a core subject till Class 10 for all students under the state syllabus.

This university is named after the mathematician Aryabhata and is located in Patna, focusing on technical and medical education.

Aryabhata, born in Bihar (Kusumapura, now Patna), made pioneering contributions to algebra, arithmetic, and astronomy.

Aryabhata, born in Bihar (Kusumapura, now Patna), made pioneering contributions to algebra, arithmetic, and astronomy.

Urdhva-Tiryagbhyam is a key formula in Vedic Maths, taught in schools across Bihar to enhance arithmetic skills.

Aryabhatiya includes 121 Sanskrit verses covering arithmetic, algebra, plane trigonometry, and astronomy.

Brajendra Nath Prasad, born in Bihar, was a mathematician who made notable contributions to number theory and mathematical research in India.

Aryabhata used zero as a placeholder in the place-value system, laying the foundation for modern mathematics.

Aryabhata was born in Kusumpur (near present-day Patna) and is considered the father of Indian mathematics and astronomy.

The Sulbasutras, part of Vedic literature, were widely used in Magadh (now Bihar) for temple construction and included concepts like the Pythagorean theorem.

Nalanda University was a world-renowned center of learning where subjects like mathematics and astronomy were taught extensively to students from across Asia.

Though Vedic Maths is popular, it was pioneered by Bharathi Krishna Tirtha, who was from Odisha—not Bihar.

Probability of complement = $ 1 - P(A) = 1 - 0.4 = 0.6 $.

Let height = $ h $. Then, $ \tan 30^\circ = \frac{h}{50} $. Since $ \tan 30^\circ = \frac{1}{\sqrt{3}} $, we get $ \frac{h}{50} = \frac{1}{\sqrt{3}} $, so $ h = \frac{50}{\sqrt{3}} = \frac{50\sqrt{3}}{3} \approx 25\sqrt{3} $ m.

$ \cos \theta = \frac{1}{2} $ at $ \theta = 60^\circ $ or $ 300^\circ $. Since $ 0 \leq \theta < 360 $, $ \theta = 60^\circ $.

Perimeter = $ 4 \times \text{side} = 48 $, so side = 12 cm. Area = $ \text{side}^2 = 12^2 = 144 $ cm².

Diagonal = $ \sqrt{\text{length}^2 + \text{width}^2} = \sqrt{12^2 + 5^2} = \sqrt{144 + 25} = \sqrt{169} = 13 $ cm.

Area of an equilateral triangle = $ \frac{\sqrt{3}}{4} \times \text{side}^2 $. For side = 6 cm, area = $ \frac{\sqrt{3}}{4} \times 36 = 9\sqrt{3} $ cm².

Area of sector = $ \frac{\theta}{360} \times \pi r^2 $. Given area = 308 cm² and $ \theta = 45^\circ $, we have $ \frac{45}{360} \times \frac{22}{7} \times r^2 = 308 $. Simplify: $ \frac{1}{8} \times \frac{22}{7} \times r^2 = 308 $. Solving, $ r^2 = 784 $, so $ r = 28 $ cm. An equilateral triangle has sides of length 6 cm. What is its area?

Given $ \frac{1}{x} + \frac{1}{y} = 1 $, we have $ \frac{x + y}{xy} = 1 $. Since $ xy = 12 $, it follows that $ x + y = 12 $. Check with roots of $ t^2 - (x+y)t + xy = 0 $: $ t^2 - 12t + 12 = 0 $. Discriminant = $ 144 - 48 = 96 $, roots are $ t = \frac{12 \pm \sqrt{96}}{2} $. However, directly, $ x + y = 12 $, but testing pairs (e.g., 3, 4) gives $ x + y = 7 $,

Solve $ 2x + 3 = 7x - 2 $. Rearrange: $ 3 + 2 = 7x - 2x $, so $ 5 = 5x $, and $ x = 1 $.

Add the equations: $ (2x + 3y) + (3x - 2y) = 12 + 5 $, giving $ 5x + y = 17 $. Subtract: $ (2x + 3y) - (3x - 2y) = 12 - 5 $, giving $ -x + 5y = 7 $. Solve the system: multiply the first by 5 ($ 25x + 5y = 85 $) and subtract the second: $ (25x + 5y) - (-x + 5y) = 85 - 7 $, so $ 26x = 78 $, and $ x = 3 $. Substitute into $ 5x + y = 17 $: $ 5(3) + y = 17 $, so $ y = 2 $. Thus, $ x + y = 3 + 2 = 4 $.

Let the greater number be $ x $, and the smaller number be $ y $. Given $ x - y = 30 $ and $ y = \frac{x}{2} + 13 $. Substitute $ y $ in the first equation: $ x - \left( \frac{x}{2} + 13 \right) = 30 $. Solving, $ \frac{x}{2} - 13 = 30 $, so $ \frac{x}{2} = 43 $, and $ x = 86 $.

Given $ x + y + z = 0 $, the expression is $ \frac{x^2}{yz} + \frac{y^2}{xz} + \frac{z^2}{xy} $. This simplifies to $ \frac{x^3 + y^3 + z^3}{xyz} $. Since $ x + y + z = 0 $, we use the identity $ x^3 + y^3 + z^3 = 3xyz $. Thus, $ \frac{3xyz}{xyz} = 3 $.

Using the alligation rule: $ (75 - 65) : (65 - 50) = 10 : 15 = 2 : 3 $. Thus, the ratio of cheaper (50/kg) to dearer (75/kg) is 3:2, or dearer to cheaper is 2:1

Let the original price be 100. After a 20% increase, price = $ 100 \times 1.2 = 120 $. After a 20% decrease, price = $ 120 \times 0.8 = 96 $. Net change = $ \frac{96 - 100}{100} \times 100 = -4\% $.

Total distance = 275 + 315 = 590 km. Time for the first part = $ \frac{275}{50} = 5.5 $ hours; second part = $ \frac{315}{70} \approx 4.5 $ hours. Total time = $ 5.5 + 4.5 = 10 $ hours. Average speed = $ \frac{590}{10} = 59 $ km/h (approximately 58.33 km/h for precision).