Let’s denote the number of buffaloes as \(b\) and the number of ducks as \(d\).

Since each buffalo has 4 legs and each duck has 2 legs, the total number of legs in the group is \(4b + 2d\). The total number of heads, which is also the total number of animals, is \(b + d\).

According to the problem, the number of legs is 24 more than twice the number of heads. Therefore, we can write the equation:

\[4b + 2d = 2(b + d) + 24\]

Simplifying this equation:

\[4b + 2d = 2b + 2d + 24\]
\[2b = 24\]
\[b = 12\]

So, there are 12 buffaloes in the group.

The equation \(|x – 2| = 5\) can be solved by considering the two possible cases for the absolute value:

1. When \(x – 2 \geq 0\):
\[x – 2 = 5\]
\[x = 7\]

2. When \(x – 2 < 0\):
\[-(x – 2) = 5\]
\[x – 2 = -5\]
\[x = -3\]

Therefore, the solutions of the equation \(|x – 2| = 5\) are \(x = 7\) and \(x = -3\), which corresponds to option (b) \(-3, 7\).

To find the value of the Roman numeral MCDLXIV, we can break it down into its components:

– M = 1000
– CD = 400 (500 – 100)
– LX = 60 (50 + 10)
– IV = 4 (5 – 1)

Adding these values together, we get:

1000 + 400 + 60 + 4 = 1464

Therefore, the value of the numeral MCDLXIV is 1464, which corresponds to option (c) 1464.

Given the equation:
\[7 \sin^2 \theta + 3 \cos^2 \theta = 4\]

Since \(\cos^2 \theta = 1 – \sin^2 \theta\), we can substitute this into the equation:
\[7 \sin^2 \theta + 3 (1 – \sin^2 \theta) = 4\]

Expanding and simplifying:
\[7 \sin^2 \theta + 3 – 3 \sin^2 \theta = 4\]
\[4 \sin^2 \theta = 1\]
\[\sin^2 \theta = \frac{1}{4}\]
\[\sin \theta = \pm \frac{1}{2}\]

Since \(0^\circ < \theta < 90^\circ\), we know that \(\sin \theta\) is positive in this interval. Therefore, we can disregard the negative solution, leaving us with:
\[\sin \theta = \frac{1}{2}\]

The value of \(\theta\) that satisfies this equation in the given interval is:
\[\theta = 30^\circ \text{ or } \frac{\pi}{6} \text{ radians}\]

So, the value of \(\theta\) is \(30^\circ\) or \(\frac{\pi}{6}\) radians.

Let the efficiency of pipe \(B\) be \(x\). Then the efficiency of pipe \(A\) is \(5x\) since it is 4 times faster than pipe \(B\).

Since pipe \(B\) takes 30 minutes to fill the tank, its filling rate is \(\frac{1}{30}\) of the tank per minute.

Therefore, we can say:
\[x = \frac{1}{30}\]

Now, let’s find the combined filling rate of both pipes when they work together. The combined rate is the sum of the individual rates of pipes \(A\) and \(B\):

\[\text{Combined rate} = \text{Rate of A} + \text{Rate of B} = 5x + x = 6x\]

Substituting the value of \(x\):
\[6x = 6 \times \frac{1}{30} = \frac{1}{5}\]

This means that when both pipes \(A\) and \(B\) work together, they fill \(\frac{1}{5}\) of the tank per minute.

To find the time taken to fill the tank together, we can take the reciprocal of the combined rate:
\[\text{Time taken} = \frac{1}{\text{Combined rate}} = \frac{1}{\frac{1}{5}} = 5 \text{ minutes}\]

So, it takes 5 minutes for both pipes \(A\) and \(B\) to fill the tank together.