Quantitative Aptitude - MCQ Practice Questions

We need LCM(12, 15, 20). 12 = 2² × 3, 15 = 3 × 5, 20 = 2² × 5. LCM = 2² × 3 × 5 = 4 × 3 × 5 = 60.

To count how many times digit 7 appears in numbers from 1 to 100, we organize by position: units place and tens place.

Step 1: Count 7 in the units place

The digit 7 appears in the units place in: 7, 17, 27, 37, 47, 57, 67, 77, 87, 97

Step 2: Count 7 in the tens place

The digit 7 appears in the tens place in: 70, 71, 72, 73, 74, 75, 76, 77, 78, 79

Step 3: Account for overlaps

The number 77 contains the digit 7 twice (once in units place, once in tens place). We've counted it correctly in both steps above—no double-counting error since we're counting occurrences of the digit, not distinct numbers.

Step 4: Total count

Answer: The digit 7 appears 20 times in numbers from 1 to 100. (Option D)

Last digits of powers of 3: 3¹=3, 3²=9, 3³=27(7), 3⁴=81(1), 3⁵=243(3)...

Pattern: 3,9,7,1 repeats every 4. 2023 = 4×505 + 3, so 3^2023 has same last digit as 3³, which is 7.

First 20 multiples of 3: 3, 6, 9, ..., 60.

This is AP with a=3, d=3, n=20, l=60.

Sum = n(a+l)/2 = 20(3+60)/2 = 20×$\frac{63}{2}$ = 10×63 = 630.

100 = 2² × 5².

Sum of all divisors = (1+2+4)(1+5+25) = 7 × 31 = 217.

Sum excluding 100 = 217 - 100 = 117.

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.

Using HCF × LCM = Product of two numbers: 15 × 180 = 45 × other number.

Therefore, other number = 2700 ÷ 45 = 60.

144 = 2⁴ × 3². 196 = 2² × 7².

Common prime factors: 2².

Therefore, HCF = 4.

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.

Let numbers be 4k and 5k. HCF = k = 6.

Therefore, numbers are 24 and 30.

The larger number is 30.

56 = 2³ × 7, 84 = 2² × 3 × 7, 140 = 2² × 5 × 7.

Common prime factors: 2 × 7 = 14.

16 = 2⁴, 24 = 2³ × 3, 36 = 2² × 3². LCM = 2⁴ × 3² = 16 × 9 = 144.

HCF × LCM = Product of numbers. 18 × 324 = 36 × other.

Therefore, other = 5832 ÷ 36 = 162.

Consecutive even numbers differ by 2.

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

The HCF is always 2.

110 = 2 × 5 × 11, 165 = 3 × 5 × 11, 220 = 2² × 5 × 11.

Common factors: 5 × 11 = 55.