Quantitative Aptitude - MCQ Practice Questions

Let numbers be 3k and 5k.

Their HCF is k = 8.

So numbers are 3(8) = 24 and 5(8) = 40.

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

HCF × LCM = Product of numbers. 36 × LCM = 408 × 468. LCM = (408 × 468) ÷ 36 = 190944 ÷ 36 = 5304.

They ring together after LCM(12, 18) minutes. 12 = 2² × 3, 18 = 2 × 3². LCM = 2² × 3² = 36 minutes.

144 = 2⁴ × 3², 180 = 2² × 3² × 5, 192 = 2⁶ × 3.

Common factors: 2² and 3. HCF = 2² × 3 = 12.

$\frac{1}{10}$+$\frac{1}{15}$=$\frac{5}{30}$=$\frac{1}{6}$. Together=6 days.

R=$\frac{100}{8}$=12.5%.

Sum=140. New sum=100. Excluded=40.

CI=10000×1.1²-10000=12100-10000=Rs 2100.

Net=$\frac{1}{6}$-$\frac{1}{12}$=$\frac{1}{12}$. Time=12 hrs.

CP=840×$\frac{100}{120}$=Rs 700.

5!=120.

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.

# 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