Quantitative Aptitude - MCQ Practice Questions

Using dividend = divisor × quotient + remainder: Number = 11 × 9 + 5 = 99 + 5 = 104

51 = 3 × 17 (not prime), 53 is prime (only divisible by 1 and 53), 55 = 5 × 11 (not prime), 57 = 3 × 19 (not prime).

Therefore, 53 is the smallest prime greater than 50.

36 = 2² × 3².

Number of factors = (2+1)(2+1) = 3 × 3 = 9.

The factors are: 1, 2, 3, 4, 6, 9, 12, 18, 36.

12 = 2² × 3, 18 = 2 × 3², 24 = 2³ × 3. LCM = 2³ × 3² = 8 × 9 = 72.

∛512 = ∛(8³) = 8, since 8 × 8 × 8 = 512.

Odd numbers between 10 and 50: 11, 13, 15, ..., 49.

This is an AP with first term 11, last term 49, and common difference 2.

Number of terms = (49-11)/2 + 1 = 19 + 1 = 20.

Let the number be n.

Given: n = 8k + 5 for some integer k.

When divided by 4: n = 8k + 5 = 4(2k) + 4 + 1 = 4(2k + 1) + 1.

Therefore, remainder = 1.

Prime factorization of 360: 360 = 2³ × 3² × 5.

The unique prime factors are 2, 3, and 5.

Product = 2 × 3 × 5 = 30.

Using the formula: HCF × LCM = Product of two numbers.

Therefore, 12 × LCM = 2160, so LCM = 2160 ÷ 12 = 180.

Factors of 18: 1, 2, 3, 6, 9, 18.

Factors of 27: 1, 3, 9, 27.

Common factors: 1, 3, 9.

Highest common factor = 9.

20 = 2² × 5, 30 = 2 × 3 × 5. LCM = 2² × 3 × 5 = 4 × 3 × 5 = 60.

For co-prime numbers, HCF = 1.

Using HCF × LCM = Product: 1 × 221 = Product.

Therefore, product = 221.

12 = 2² × 3, 15 = 3 × 5, 20 = 2² × 5. LCM = 2² × 3 × 5 = 60.

This is the smallest number divisible by all three.

Consecutive even numbers differ by 2.

Examples: HCF(2,4)=2, HCF(6,8)=2, HCF(10,12)=2.

The HCF is always 2.

96 = 2⁵ × 3, 120 = 2³ × 3 × 5.

Common prime factors: 2³ × 3 = 8 × 3 = 24.

Using Euclidean algorithm: 126 = 84 × 1 + 42, then 84 = 42 × 2 + 0.

Therefore, HCF(84, 126) = 42.

Using the formula: HCF × LCM = Product of two numbers. 15 × LCM = 2700, so LCM = 2700 ÷ 15 = 180.

45 = 3² × 5, 60 = 2² × 3 × 5, 75 = 3 × 5².

Common factors: 3 and 5. HCF = 3 × 5 = 15.

20 = 2² × 5, 30 = 2 × 3 × 5, 40 = 2³ × 5. LCM = 2³ × 3 × 5 = 8 × 3 × 5 = 120.

72 = 2³ × 3², 90 = 2 × 3² × 5.

Common factors: 2 and 3². HCF = 2 × 3² = 2 × 9 = 18.