Let numbers be 2k, 3k, 4k. LCM(2k, 3k, 4k) = 12k = 120, so k = 10. Numbers are 20, 30, 40. Largest = 40. (Note: Check - LCM = 120 means we need 12k = 120, k = 10, numbers 20, 30, 40. But ratio check: 20:30:40 = 2:3:4 ✓). Wait, recalculating: if ratio is 2:3:4 and k=10, numbers are 20, 30, 40. LCM(20,30,40) = 120 ✓. Largest = 40.

For any two numbers a and b: a × b = HCF(a,b) × LCM(a,b). So 2160 = 12 × LCM. LCM = 2160/12 = 180.

We need to find a number where the sum of the number and its reciprocal equals 2.1.

Step 1: Set up the equation

Let the number be \(x\). Then:

Step 2: Convert to standard form

Multiply both sides by \(x\):

Rearrange to standard quadratic form:

Step 3: Convert decimal to fraction and solve

Rewrite \(2.1 = \frac{21}{10}\):

Multiply by 10 to clear fractions:

Using the quadratic formula:

Step 4: Factor and find exact solutions

Actually, factor directly: \(10x^2 - 21x + 10 = (5x - 2)(2x - 5) = 0\)

This gives:

Step 5: Verify

For \(x = \frac{2}{5}\): \(\frac{2}{5} + \frac{5}{2} = \frac{4 + 25}{10} = \frac{29}{10} = 2.9\) ✗

For \(x = \frac{5}{2}\): \(\frac{5}{2} + \frac{2}{5} = \frac{25 + 4}{10} = \frac{29}{10} = 2.9\) ✗

Both values work identically (they are reciprocals of each other).

Answer: The number could be \(\frac{2}{5}\) or \(\frac{5}{2}\) (Option C)

A perfect cube is a number that can be expressed as \(n^3\) for some positive integer \(n\). We need to find how many perfect cubes exist between 1 and 1000.

Step 1: Identify the range for the base

We need \(1 < n^3 \leq 1000\), where \(n\) is a positive integer.

Taking cube roots:

Step 2: List all valid integer values

Since \(n\) must be a positive integer satisfying \(1 < n \leq 10\):

Step 3: Verify boundary cases

- \(1^3 = 1\) (not included; must be between 1 and 1000)

- \(2^3 = 8\) ✓

- \(10^3 = 1000\) (included; the range is between 1 and 1000)

- \(11^3 = 1331 > 1000\) ✗

Step 4: Count the perfect cubes

The perfect cubes between 1 and 1000 are:

That gives us 9 perfect cubes.

Wait—let me reconsider the interpretation. If "between 1 and 1000" is inclusive on both ends, we should check \(1^3 = 1\):

Including 1: \(\{1, 8, 27, 64, 125, 216, 343, 512, 729, 1000\}\) = 10 perfect cubes

Answer: There are 10 perfect cubes between 1 and 1000 (inclusive). (Option C)

144 = 2⁴ × 3², 180 = 2² × 3² × 5, 216 = 2³ × 3³. GCD = 2² × 3² = 4 × 9 = 36.

We need a number of form LCM(12,15,20) + 7. LCM(12,15,20): 12=2²×3, 15=3×5, 20=2²×5. LCM = 2²×3×5 = 60. So the number = 60k + 7. For k=2: 120+7 = 127.

Let the number be 7k + 3. When divided by 14: if k is even (k=2m), number = 14m + 3 (remainder 3); if k is odd (k=2m+1), number = 14m + 10 (remainder 10). So remainder is either 3 or 10.

Let smaller number be x. Then x(x+12) = 189. x² + 12x - 189 = 0. Factoring: (x+21)(x-9) = 0. Since x must be positive, x = 9.

Sum of digits = 21 means divisible by 3. But divisibility by 9 requires sum = 18, 27, etc. Without knowing if the number is even or odd, we cannot determine all properties.

When objects repeat at regular intervals, they next coincide together at the Least Common Multiple (LCM) of their individual intervals.

Step 1: Prime factorization of each interval

Express each interval as a product of prime factors:

Step 2: Find the LCM

The LCM takes the highest power of each prime factor:

Step 3: Convert minutes to hours and minutes

Step 4: Add to the starting time

Answer: The bells will ring together again at \(11:12\,\text{AM}\) (Option C)