Quantitative Aptitude - MCQ Practice Questions

# 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.

# 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

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

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.

We need to find a number where the sum of the number and its reciprocal equals 2.5.

Step 1: Set up the equation

Let the number be $x$. The reciprocal is $\frac{1}{x}$.

Step 2: Clear the fraction

Multiply both sides by $x$:

Rearrange to standard quadratic form:

Step 3: Apply the quadratic formula

Using $x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}$ where $a=1$, $b=-2.5$, $c=1$:

Step 4: Find both solutions

Verification:

**Answer: The number is $2$ or $\frac{1}{2}$ (Option B)**

Let number = x. Then x - x/3 = 30. So 2x/3 = 30, x = 45

Divisors of 16: 1,2,4,8,16. Product = 1×2×4×8×16 = 1024. Or use formula: if n has d divisors, product = n^(d/2). Here d=5, so product = 16^2.5 = 1024

Let numbers be x and x+2. Then x(x+2) = 195. x² + 2x - 195 = 0. Using formula: x = 13. Numbers are 13 and 15. Sum = 28

2^a = 32 = 2^5, so a = 5. And 3^b = 81 = 3^4, so b = 4. Therefore a + b = 5 + 4 = 9

Check 22: 22÷9 = 2 rem 4 ✓, 22÷7 = 3 rem 1 ✗. Check 31: 31÷9 = 3 rem 4 ✓, 31÷7 = 4 rem 3 ✓. Answer is 31 (B, not A). Correction: Option B is correct.

144 = 2⁴ × 3². Odd divisors come only from 3² = (2+1) = 3 odd divisors: 1, 3, 9.

We know that GCD × LCM = Product of the two numbers. So 4 × LCM = 120. Therefore, LCM = $\frac{120}{4}$ = 30.

Number of divisors = (3+1)(2+1)(1+1) = 4 × 3 × 2 = 24.

20 = 2² × 5. Divisors are: 1, 2, 4, 5, 10, 20. Sum = 1 + 2 + 4 + 5 + 10 + 20 = 42.

For n ≡ 3 (mod 7): possible numbers are 3, 10, 17, 24, 31, 38, 45, 52, 59... For n ≡ 5 (mod 11): possible numbers are 5, 16, 27, 38, 49, 60... Common number is 58. Check: 58 = 7(8) + 2... Let me recheck: $\frac{58}{7}$ = 8 rem 2, not 3. Try 38: $\frac{38}{7}$ = 5 rem 3 ✓, $\frac{38}{11}$ = 3 rem 5 ✓. Answer is A=38.

To find the remainder when $2^{100}$ is divided by 7, we use Fermat's Little Theorem, which states that if $p$ is prime and $\gcd (a,p)=1$, then $a^{p-1}\equiv 1(\mathrm{mod}p)$.

Step 1: Apply Fermat's Little Theorem

Since 7 is prime and $\gcd (2,7)=1$:

Step 2: Express the exponent in terms of 6

Divide 100 by 6:

Therefore:

Step 3: Simplify using the congruence

**Step 4: Calculate $2^4\mathrm{mod}7$

Therefore:

**Answer: The remainder is $2$ (Option B)**

# Unit Digit of 7^2019

To find the unit digit of any power, we identify the cyclical pattern of unit digits for that base number.

Step 1: Find the Cyclical Pattern of Unit Digits for Powers of 7

The unit digit of powers of 7 repeats in a cycle. Let's calculate the first few powers:

The pattern repeats: 7, 9, 3, 1, 7, 9, 3, 1, ...

The cycle length is 4.

Step 2: Find the Position of 7^2019 in the Cycle

Divide the exponent 2019 by the cycle length 4 to find which position in the pattern it corresponds to:

Since the remainder is 3, we need the 3rd unit digit in our cycle pattern (7, 9, 3, 1).

The 3rd position corresponds to unit digit 3.

The unit digit of 7^2019 is 3.

Answer: (A) 3

Using the property: HCF × LCM = Product of two numbers. 6 × 60 = 12 × x. 360 = 12x. x = 30.

To find $3^5\mathrm{mod}11$, we calculate successive powers of 3 and reduce modulo 11 at each step to keep numbers manageable.

**Step 1: Calculate $3^2\mathrm{mod}11$

Since $9<11$, we have $3^2\equiv 9(\mathrm{mod}11)$

**Step 2: Calculate $3^4\mathrm{mod}11$

Dividing: $81=7\times 11+4$, so $3^4\equiv 4(\mathrm{mod}11)$

**Step 3: Calculate $3^5\mathrm{mod}11$

Step 4: Final reduction

**Answer: $3^5\equiv 1(\mathrm{mod}11)$ (Option A)**

HCF takes minimum power of each prime: HCF = 2^min(4,3) × 3^min(3,2) × 5^min(1,2) = 2^3 × 3^2 × 5.