Let the numbers be $3x$ and $4x$. LCM = $12x$. Given: $12x = 84 \Rightarrow x = 7$. Numbers: $3 \times 7 = 21$ and $4 \times 7 = 28$. The greater number is 28.

Let HCF = $x$. Then LCM = $90x$. According to the question, $x + 90x = 1456 \Rightarrow 91x = 1456 \Rightarrow x = 16$. Thus, HCF = 16, LCM = $90 \times 16 = 1440$. We know: First number $\times$ Second number = LCM $\times$ HCF $\Rightarrow 160 \times \text{Second number} = 1440 \times 16 \Rightarrow \text{Second number} = \frac{1440 \times 16}{160} = 144$.

Find the HCF of the differences: $277 - 152 = 125$, $427 - 277 = 150$, $427 - 152 = 275$. HCF (125, 150, 275) = 25.

LCM of fractions = $\frac{\text{LCM of numerators}}{\text{HCF of denominators}}$. LCM of numerators (1, 2, 5, 4) = 20. HCF of denominators (3, 9, 6, 27) = 3. Result = $\frac{20}{3}$.

Lengths: 30, 90, 120, 135 cm. HCF (30, 90, 120, 135) = 15.

Factors: $q^2 + 3q = q(q+3)$, $q^2 + 5q + 6 = (q+2)(q+3)$, $q^2 + 4q + 3 = (q+1)(q+3)$. LCM = $q(q+3)(q+2)(q+1)$. (Note: $q$ is missing from options, but (D) is the closest correct answer).

LCM(15, 21, 28) = $3 \times 5 \times 7 \times 4 = 420$. Among the options, 100380 is divisible by 420.

Let the numbers be $7x$ and $11x$. LCM = $77x$. Given: $77x = 154 \Rightarrow x = 2$. First number = $7 \times 2 = 14$.

Formula: $\text{First number} \times \text{Second number} = \text{LCM} \times \text{HCF}$. $25 \times x = 225 \times 5 \Rightarrow x = \frac{225 \times 5}{25} = 45$.

LCM(25, 50, 75) = 150. 43582/150 gives quotient 290, remainder 82. Multiples of 150 are: $150 \times 290 = 43500$ and $150 \times 291 = 43650$. 43582 is closer to 43650.

First, find the HCF of 408 and 1032: $408 = 2^3 \times 3 \times 17$, $1032 = 2^3 \times 3 \times 43$. HCF = $2^3 \times 3 = 24$. Given: $24 = 1032x - 408y \Rightarrow 1 = 43x - 17y$. Equation $x+y=7 \Rightarrow y=7-x$. Substituting the value: $43x - 17(7-x) = 1 \Rightarrow 43x - 119 + 17x = 1 \Rightarrow 60x = 120 \Rightarrow x = 2$. Since $x+y=7$, $2+y=7 \Rightarrow y=5$. Thus, $x-y = 2-5 = -3$.

Let the numbers be $4x, 5x$, and $6x$. LCM = $60x$. According to the question, $60x = 3600$, therefore $x = 60$. The HCF is $x = 60$.

For numbers 15 and 20, the HCF is 5 and the LCM is 60.

LCM of fractions = $\frac{\text{LCM of numerators}}{\text{HCF of denominators}}$. Thus, $\frac{\text{LCM}(1, 2, 1)}{\text{HCF}(4, 3, 2)} = \frac{2}{1} = 2$.

For rooms with equal capacity, we must find the Highest Common Factor (HCF) of 60, 105, and 135. HCF(60, 105, 135) = 15. The total number of rooms required = $\frac{60}{15} + \frac{105}{15} + \frac{135}{15} = 4 + 7 + 9 = 20$.

The time when both devices produce sound is based on the LCM (60, 62). LCM (60, 62) = 1860 seconds. 1860 seconds = $\frac{1860}{60}$ minutes = 31 minutes. If they start at a certain time, they will ring together after 31 minutes.

LCM(6, 8, 10, 12) = 120. Largest 4-digit number = 9999. Dividing 9999 by 120, quotient = 83, remainder = 39. Perfectly divisible number = 9999 - 39 = 9960. Required number = 9960 + 4 = 9964.

Let the numbers be $13a$ and $13b$, where $a$ and $b$ are co-prime. $13a \times 13b = 6760 \Rightarrow a \times b = \frac{6760}{169} = 40$. Pairs of co-prime factors of 40: (1, 40) and (5, 8). Thus, 2 pairs can be formed.