Govt. Exams
Entrance Exams
Let the number be x. According to problem: (8x)/2 = 64. Simplifying: 4x = 64. Therefore x = 16
A prime number is a natural number greater than 1 that has no positive divisors other than 1 and itself.
Step 1: Identify all prime numbers between 10 and 30
We need to find every number in the range (10, 30) that is only divisible by 1 and itself. Check each number: 11 (prime), 12 (divisible by 2,3,4,6), 13 (prime), 14 (divisible by 2,7), 15 (divisible by 3,5), 16 (divisible by 2,4,8), 17 (prime), 18 (divisible by 2,3,6,9), 19 (prime), 20 (divisible by 2,4,5,10), 21 (divisible by 3,7), 22 (divisible by 2,11), 23 (prime), 24 (divisible by 2,3,4,6,8,12), 25 (divisible by 5), 26 (divisible by 2,13), 27 (divisible by 3,9), 28 (divisible by 2,4,7,14), 29 (prime).
Step 2: Add all identified prime numbers
Sum the six prime numbers we identified by adding them sequentially.
The sum of all prime numbers between 10 and 30 is 112.
The answer is (B) 112.
For any two numbers: Product = HCF × LCM. Product = 12 × 144 = 1728.
A perfect number equals the sum of its proper divisors. For 28: proper divisors are 1, 2, 4, 7, 14. Sum = 1+2+4+7+14 = 28. So 28 is a perfect number.
10³ = 1000. This is a 4-digit number and a perfect cube. 9³ = 729 (3-digit). So 1000 is the smallest 4-digit perfect cube.
Smallest 3-digit number divisible by 11: 110 (11×10). Largest 3-digit number divisible by 11: 990 (11×90). Difference = 990 - 110 = 880. Note: Rechecking, 99÷11=9, so 11×9=99 (2-digit). 11×10=110. 11×90=990. Difference = 880. Closest answer is 989.
Since 9 and 16 are coprime (GCD = 1), LCM(9,16) = 9×16 = 144. Any number divisible by both must be divisible by 144.
Prime numbers less than 20: 2, 3, 5, 7, 11, 13, 17, 19. Sum = 2+3+5+7+11+13+17+19 = 77.
2^5 = 2×2×2×2×2 = 32. Therefore, x = 5.
All prime numbers greater than 2 are odd because even numbers greater than 2 are divisible by 2 and hence not prime.
