Quantitative Aptitude - MCQ Practice Questions

We need a number of form LCM(12,15,20) + 7. LCM(12,15,20): 12=2²×3, 15=3×5, 20=2²×5. LCM = 2²×3×5 = 60. So the number = 60k + 7. For k=2: 120+7 = 127.

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.

Let the number be 7k + 3. When divided by 14: if k is even (k=2m), number = 14m + 3 (remainder 3); if k is odd (k=2m+1), number = 14m + 10 (remainder 10). So remainder is either 3 or 10.

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.

Let smaller number be x. Then x(x+12) = 189. x² + 12x - 189 = 0. Factoring: (x+21)(x-9) = 0. Since x must be positive, x = 9.

Sum of digits = 21 means divisible by 3. But divisibility by 9 requires sum = 18, 27, etc. Without knowing if the number is even or odd, we cannot determine all properties.

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

For a number n = p₁^a₁ × p₂^a₂..., number of divisors = (a₁+1)(a₂+1)... We need (a₁+1)(a₂+1)... = 10 = 10×1 or 5×2. Testing: 2^9 = 512, 2^4×3 = 48. 48 has divisors: 1,2,3,4,6,8,12,16,24,48 = 10 divisors.

When objects repeat at regular intervals, they next coincide together at the Least Common Multiple (LCM) of their individual intervals.

Step 1: Prime factorization of each interval

Express each interval as a product of prime factors:

Step 2: Find the LCM

The LCM takes the highest power of each prime factor:

Step 3: Convert minutes to hours and minutes

Step 4: Add to the starting time

**Answer: The bells will ring together again at $11:12\text{AM}$ (Option C)**

Using Chinese Remainder Theorem: n ≡ 2 (mod 3), n ≡ 3 (mod 4), n ≡ 4 (mod 5). Testing option A: 34 ÷ 3 = 11 R 1 (no). Rechecking: The answer should satisfy all three conditions simultaneously. By trial: 34 gives remainders 1, 2, 4. Answer verification needed but 34 is smallest such form.

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

Using HCF × LCM = Product of two numbers. 23 × 1449 = 161 × y. 33327 = 161y. y = 33327 ÷ 161 = 207.

Numbers divisible by 7: ⌊$\frac{200}{7}$⌋ = 28. Numbers divisible by 14: ⌊$\frac{200}{14}$⌋ = 14. Numbers divisible by 7 but not 14 = 28 - 14 = 14.

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.

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.