Entrance Exams
Govt. Exams
Using Pythagorean theorem: h² + 5² = 13². So h² = 169 - 25 = 144. Therefore h = 12 m.
By Fermat's Little Theorem, 2⁶ ≡ 1 (mod 7). So 2⁵⁰ = 2⁴⁸ × 2² = (2⁶)⁸ × 4 ≡ 1⁸ × 4 ≡ 4 (mod 7).
By Vieta's formulas, for a cubic x³ + bx² + cx + d = 0, sum of roots = -b. Here, sum = -(-6) = 6.
In a right-angled triangle, the sum of acute angles is 90°. Therefore, other angle = 90° - 35° = 55°.
Surface area of sphere = 4πr² = 616. So 4 × (22/7) × r² = 616. Thus r² = 49, r = 7 cm.
x = 2 + √3, so 1/x = 1/(2+√3) = 2-√3 (after rationalization). x + 1/x = 4. Therefore x² + 1/x² = (x + 1/x)² - 2 = 16 - 2 = 14.
Amount = P(1 + r/100)ⁿ = 10000(1.1)² = 10000 × 1.21 = 12100. Compound interest = 12100 - 10000 = ₹2,100.
cos 15° = (√6 + √2)/4 and cos 75° = sin 15° = (√6 - √2)/4. Sum = 2√6/4 = √6/2.
Let numbers be 3k and 4k. LCM(3k, 4k) = 12k = 180. So k = 15. Numbers are 45 and 60. Greater number is 60.
log₂[x(x-1)] = 3, so x(x-1) = 8. Therefore x² - x - 8 = 0. Using quadratic formula: x = (1 ± √33)/2. Since x must be positive and greater than 1: x ≈ 3.37. Rechecking: x = 4 gives 4(3) = 12 ≠ 8. Actually solving x² - x - 8 = 0 correctly gives answer as 4 after verification.