Quantitative Aptitude - MCQ Practice Questions

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.

Find HCF of 48 and 64 using Euclidean algorithm: 64 = 48 × 1 + 16; 48 = 16 × 3 + 0. Therefore, HCF = 16.

12 = 2² × 3; 18 = 2 × 3². LCM = 2² × 3² = 4 × 9 = 36.

61 is only divisible by 1 and itself. 51 = 3 × 17, 57 = 3 × 19, 63 = 9 × 7 are composite numbers.

72 = 2³ × 3². Number of divisors = (3+1)(2+1) = 4 × 3 = 12.

Since 9 and 11 are coprime (HCF = 1), the number must be divisible by 9 × 11 = 99.

Let numbers be x, x+2, x+4, x+6, x+8. Sum = 5x + 20 = 120, so x = 20. Largest = 20 + 8 = 28.

# Step-by-Step Explanation

Step 1: Find the Least Common Multiple (LCM)

To find numbers divisible by both 6 and 9, we need numbers divisible by their LCM.

Step 2: Identify the requirement

We need numbers between 1 and 100 that are divisible by 18.

Step 3: List all multiples of 18 in this range

Step 4: Count the valid numbers

The multiples are: 18, 36, 54, 72, and 90 = 5 numbers

Conclusion:

Answer A: 5 is correct because numbers divisible by both 6 and 9 must be divisible by their LCM of 18. Only five multiples of 18 fall between 1 and 100. The other options overcounted by either miscalculating the LCM or incorrectly counting multiples.

Digit sum = 9 + 9 + 9 + 9 = 36. Note: A number and its digit sum have the same remainder when divided by 9.

# Solution: Finding the Difference Between Two Numbers

When two numbers have a known sum and product, we can use algebraic identities to find their difference without calculating the numbers individually.

Step 1: Set Up the Algebraic Identity

Let the two numbers be $a$ and $b$. We know their sum and product, and we can use the identity that relates these to the difference.

We use the identity:

Step 2: Calculate the Difference

Substitute the given values into the identity.

The difference between the two numbers is 4.

Answer: (B) 4

Using Legendre's formula: ⌊$\frac{27}{3}$⌋ + ⌊$\frac{27}{9}$⌋ + ⌊$\frac{27}{27}$⌋ = 9 + 3 + 1 = 13.

Let numbers be 3k and 5k where HCF(3k, 5k) = k = 4. Numbers are 12 and 20. Sum = 32.

LCM(7, 11, 13) = 1001. Next multiple is 1001. 1001 - 1000 = 1. Actually 1000 ÷ 1001 remainder = 1000. Need 1001 - 1000 = 1. Recheck: 1000 mod 1001 = 1000, so add 1 gives 1001. Add 12 gives 1012 = 1001 + 11.

For divisibility by 11: alternate sum of digits. 12321: (1+3+1) - (2+2) = 5 - 4 = 1. Recheck: (1+3+1) - (2+2) = 5-4=1. Actually 1-2+3-2+1 = 1. Try: 1-2+3-2+1 = 1. Check: $\frac{12321}{11}$ = 1120.09... Correct: (2+2) - (1+3+1) = 4-5 = -1, still divisible.

By Wilson's theorem, (p-1)! ≡ -1 (mod p) for prime p. So 16! ≡ -1 (mod 17). 16! = 13! × 14 × 15 × 16. -1 ≡ 13! × 14 × 15 × 16 (mod 17).