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.

Let the six numbers be \(a_1, a_2, a_3, a_4, a_5, a_6\).

Given that the average of the six numbers is 35, we have:

\[
\frac{a_1 + a_2 + a_3 + a_4 + a_5 + a_6}{6} = 35
\]

Multiplying both sides by 6 gives:

\[
a_1 + a_2 + a_3 + a_4 + a_5 + a_6 = 6 \times 35 = 210
\]

After the operations, the new sum of the numbers will be:

\[
(a_1 + 4) + (a_2 + 4) + (a_3 + 4) + (a_4 – 8) + (a_5 – 8) + (a_6 – 8)
\]

\[
= (a_1 + a_2 + a_3) + 3 \times 4 + (a_4 + a_5 + a_6) – 3 \times 8
\]

\[
= (a_1 + a_2 + a_3) + 12 + (a_4 + a_5 + a_6) – 24
\]

\[
= (a_1 + a_2 + a_3 + a_4 + a_5 + a_6) + 12 – 24
\]

\[
= 210 – 12
\]

\[
= 198
\]

Therefore, the new average is:

\[
\frac{198}{6} = 33
\]

So, the new average is 33.

Let’s break this down step by step:

1. Volume of the Cube : The volume of a cube is given by \( \text{Volume} = a^3 \), where \( a \) is the side length of the cube. Here, \( a = 11 \) cm. So, the volume of the cube is \( 11^3 \) cubic cm.

2. Volume of the Cylinder : The volume of a cylinder is given by \( \text{Volume} = \pi r^2 h \), where \( r \) is the radius and \( h \) is the height. We are given that the height of the cylinder is 7 times the length of the rectangle. The length of the rectangle is not explicitly given, but we can find it using the given information about the width and perimeter of the rectangle.

3. Finding the Length of the Rectangle : The perimeter of a rectangle is given by \( 2(l + w) \), where \( l \) is the length and \( w \) is the width. We are given that the width \( w = 1.5 \) cm and the perimeter is 4 cm. Therefore, \( 2(l + 1.5) = 4 \). Solving for \( l \), we get \( l = \frac{4}{2} – 1.5 = 0.5 \) cm.

4. Height of the Cylinder : Since the height of the cylinder is 7 times the length of the rectangle, the height is \( 7 \times 0.5 = 3.5 \) cm.

5. Volume of the Cylinder (Continued) : Now that we have the height of the cylinder as 3.5 cm, we can find its volume using the formula \( \pi r^2 h \).

6. Equating Volumes : Since the cube is melted to form the cylinder, the volume of the cube should be equal to the volume of the cylinder. Setting these two volumes equal to each other, we can solve for the radius \( r \) of the cylinder.

Let’s calculate the radius \( r \) of the cylinder using the given information.

Given:
Side length of cube, \(a = 11\) cm
Width of rectangle, \(w = 1.5\) cm
Perimeter of rectangle, \(P = 4\) cm

1. Volume of Cube : \(V_{\text{cube}} = a^3 = 11^3\) cm³.

2. Length of Rectangle : Perimeter of rectangle, \(P = 2(l + w)\). We have \(P = 4\) and \(w = 1.5\). Solve for \(l\):

\[4 = 2(l + 1.5) \]
\[2 = l + 1.5 \]
\[l = 0.5\] cm.

3. Height of Cylinder : Height of cylinder, \(h = 7l = 7 \times 0.5\) cm.

4. Volume of Cylinder : Volume of cylinder, \(V_{\text{cylinder}} = \pi r^2 h\).

Since the cube is melted and converted into the cylinder, their volumes are equal:

\[11^3 = \pi r^2 \times 7 \times 0.5\]

\[1331 = 3.5 \pi r^2\]

\[r^2 = \frac{1331}{3.5\pi}\]

\[r = \sqrt{\frac{1331}{3.5\pi}}\]

To find the radius of the cylinder, we first calculate the value inside the square root:

\[ r = \sqrt{\frac{1331}{3.5\pi}} \]

\[ r = \sqrt{\frac{1331}{3.5 \times 3.14159}} \]

\[ r = \sqrt{\frac{1331}{10.99265}} \]

\[ r = \sqrt{121} \]

\[ r = 11 \text{ cm} \]

Therefore, the radius of the cylinder is 11 cm.

Let’s denote the amount of water to be added as \(x\) litres.

Given:
– The cost of 1 litre of milk is Rs. 20.
– The selling price of the mixture is Rs. 20 per litre.
– The desired profit is 25%.

The cost price of 1 litre of milk is Rs. 20, and to achieve a 25% profit, the selling price should be:
\[ \text{Selling price} = \text{Cost price} + 25\% \times \text{Cost price} = 20 + 0.25 \times 20 = Rs. 25 \]

However, the mixture is being sold at Rs. 20 per litre. So, we need to find the volume of the mixture that can be sold for Rs. 25 to achieve the desired profit.

Since the selling price per litre is Rs. 20, to achieve a total selling price of Rs. 25, we need:
\[ \text{Volume of mixture} = \frac{\text{Total selling price}}{\text{Selling price per litre}} = \frac{25}{20} = 1.25 \text{ litres} \]

Therefore, the amount of water that should be added to 1 litre of milk to make the volume of the mixture 1.25 litres is:
\[ \text{Amount of water} = \text{Volume of mixture} – \text{Volume of milk} = 1.25 – 1 = 0.25 \text{ litres} \]

Conclusion

To gain a 25% profit by selling the mixture at Rs. 20 per litre, 0.25 litres of water should be added to 1 litre of milk.

Solution

Given:
\[ \cos^4 A – \sin^4 A = p \]

Step 1: Use the difference of squares formula

\[ \cos^4 A – \sin^4 A = (\cos^2 A + \sin^2 A)(\cos^2 A – \sin^2 A) \]

Step 2: Use the Pythagorean identity

Since \(\sin^2 A + \cos^2 A = 1\), we have:
\[ (\cos^2 A + \sin^2 A)(\cos^2 A – \sin^2 A) = (1)(\cos^2 A – \sin^2 A) \]

Step 3: Use the double angle formula

The double angle formula for cosine is \(\cos 2A = \cos^2 A – \sin^2 A\), so:
\[ \cos^2 A – \sin^2 A = \cos 2A \]

Step 4: Find the value of \(p\)

Therefore, we have:
\[ p = \cos 2A \]

Conclusion

The value of \(p\) is \(\cos 2A\).