Quantitative Aptitude - MCQ Practice Questions

6³ = 216. Thus 216 is a perfect cube.

91 = 7 × 13, so it's composite. 97, 89, and 83 are all prime numbers.

24 = 2³ × 3, 36 = 2² × 3². LCM = 2³ × 3² = 8 × 9 = 72.

If number is x, then (x × $\frac{3}{4}$) ÷ ($\frac{2}{3}$) = (3x/4) × ($\frac{3}{2}$) = 9x/8. The factor is $\frac{9}{8}$.

# Sum of Prime Numbers Between 10 and 25

To find the sum of prime numbers, we must first identify all prime numbers in the given range, then add them together.

Step 1: Identify All Prime Numbers Between 10 and 25

A prime number is a natural number greater than 1 that has no positive divisors other than 1 and itself. We check each number in the range for primality.

(Note: 12, 14, 15, 16, 18, 20, 21, 22, 24 are composite; 10 and 25 are composite)

Step 2: Calculate the Sum

Adding all identified prime numbers together:

The sum of all prime numbers between 10 and 25 is 83.

Answer: (C) 83

48 = 2⁴ × 3¹. Number of factors = (4+1)(1+1) = 5 × 2 = 10. The factors are: 1, 2, 3, 4, 6, 8, 12, 16, 24, 48.

√500 ≈ 22.36. The next integer is 23. 23² = 529. This is the smallest perfect square greater than 500.

Let the numbers be 4x and 5x. Sum = 4x + 5x = 9x = 180. So x = 20. The larger number = 5x = 5(20) = 100.

If a number is divisible by 9, then: (A) It must be divisible by 3 (since 9 = 3²). (B) Sum of its digits must be divisible by 9 (divisibility rule). Both A and B are true.

Even numbers from 1 to 100: 2, 4, 6, ..., 100. This is an AP with first term 2, last term 100, common difference 2. Number of terms = (100-2)/2 + 1 = 50.

Let the integers be x and x+1. Sum = x + (x+1) = 51. 2x + 1 = 51. 2x = 50. x = 25. The larger integer = 26.

Using formula for sum of first n natural numbers: S = n(n+1)/2 where n=99. S = 99(100)/2 = $\frac{9900}{2}$ = 4950.

For divisibility by both 6 and 8, the number must be divisible by LCM(6,8) = 24. Check: 48 ÷ 24 = 2 ✓. Option A is correct.

Using the formula: HCF × LCM = Product of two numbers. Therefore, 13 × LCM = 2028. LCM = 2028 ÷ 13 = 156.

Let the number be x. According to problem: 3x - 4 = 17. Therefore, 3x = 21, x = 7.

Let the three consecutive odd numbers be (x-2), x, and (x+2). Their sum: (x-2) + x + (x+2) = 51, so 3x = 51, x = 17.

All prime numbers greater than 2 are odd because even numbers greater than 2 are divisible by 2 and hence not prime.

2^5 = 2×2×2×2×2 = 32. Therefore, x = 5.

Prime numbers less than 20: 2, 3, 5, 7, 11, 13, 17, 19. Sum = 2+3+5+7+11+13+17+19 = 77.

Since 9 and 16 are coprime (GCD = 1), LCM(9,16) = 9×16 = 144. Any number divisible by both must be divisible by 144.