Govt Exams
Check 22: 22÷9 = 2 rem 4 ✓, 22÷7 = 3 rem 1 ✗. Check 31: 31÷9 = 3 rem 4 ✓, 31÷7 = 4 rem 3 ✓. Answer is 31 (B, not A). Correction: Option B is correct.
2^a = 32 = 2^5, so a = 5. And 3^b = 81 = 3^4, so b = 4. Therefore a + b = 5 + 4 = 9
Let numbers be x and x+2. Then x(x+2) = 195. x² + 2x - 195 = 0. Using formula: x = 13. Numbers are 13 and 15. Sum = 28
Divisors of 16: 1,2,4,8,16. Product = 1×2×4×8×16 = 1024. Or use formula: if n has d divisors, product = n^(d/2). Here d=5, so product = 16^2.5 = 1024
Let number = x. Then x - x/3 = 30. So 2x/3 = 30, x = 45
Let number be x. x + 1/x = 2.5. Multiply by x: x² - 2.5x + 1 = 0. x = (2.5 ± √(6.25-4))/2 = (2.5 ± 1.5)/2. x = 2 or 0.5.
For divisibility by 11: alternate sum of digits. 12321: (1+3+1) - (2+2) = 5 - 4 = 1. Recheck: (1+3+1) - (2+2) = 5-4=1. Actually 1-2+3-2+1 = 1. Try: 1-2+3-2+1 = 1. Check: 12321/11 = 1120.09... Correct: (2+2) - (1+3+1) = 4-5 = -1, still divisible.
Let numbers be 3k and 5k where HCF(3k, 5k) = k = 4. Numbers are 12 and 20. Sum = 32.
Let numbers be a and b. (a+b)² - 4ab = (a-b)². So (a-b)² = 400 - 384 = 16, thus |a-b| = 4.
Numbers divisible by both 6 and 9 must be divisible by LCM(6,9) = 18. Between 1-100: 18, 36, 54, 72, 90. Count = 5.