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.

To find the minimum value of \(16\tan^2\theta + 25\cot^2\theta\), we can use the formula for the minimum value of the sum of two positive numbers \(a\) and \(b\), which is \(2\sqrt{ab}\) when \(a\) and \(b\) are positive. In this case, \(a = 16\) and \(b = 25\). So, the minimum value is \(2\sqrt{16 \times 25} = 2 \times 4 \times 5 = 40\).

Therefore, the minimum value of \(16\tan^2\theta + 25\cot^2\theta\) is \(40\).

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 \(\cos 2A\).

Solution

Given:
– A can complete the work in 15 days.
– B can complete the work in 10 days.
– C can complete the work in 12 days.
– The total payment for the work is Rs. 9000.

Step 1: Calculate the total work done by A, B, and C

Let the total work be \(W\) units.
– A’s work rate: \(\frac{W}{15}\) units/day
– B’s work rate: \(\frac{W}{10}\) units/day
– C’s work rate: \(\frac{W}{12}\) units/day

Step 2: Calculate the total work done by A, B, and C together in one day

\[ \text{Total work rate} = \frac{W}{15} + \frac{W}{10} + \frac{W}{12} = \frac{4W}{60} + \frac{6W}{60} + \frac{5W}{60} = \frac{15W}{60} = \frac{W}{4} \text{ units/day} \]

Step 3: Calculate the share of B

Since the total payment is for 1 day of work, B’s share is proportional to his work rate:
\[ \text{B’s share} = \frac{\text{B’s work rate}}{\text{Total work rate}} \times \text{Total payment} = \frac{\frac{W}{10}}{\frac{W}{4}} \times 9000 = \frac{4}{10} \times 9000 = Rs. 3600 \]

Step 4: Calculate F’s share

B’s share is divided among D, E, and F in the ratio of 1:5:3. The total parts in the ratio are \(1 + 5 + 3 = 9\).
– F’s share = \(\frac{3}{9} \times \text{B’s share} = \frac{3}{9} \times 3600 = Rs. 1200\)

Conclusion

The share of F is Rs. 1200.

Solution

Given:
– The correct ratio for dividing the gold coins among Abhay, Vishal, and Kishore is 2:3:4.
– The mistaken ratio used for division is 1/2:1/3:1/4.
– There are a total of 351 gold coins.

Step 1: Calculate the number of coins each person should have received

Total parts in the correct ratio = 2 + 3 + 4 = 9 parts

– Abhay’s share in the correct ratio: \(\frac{2}{9} \times 351 = 78\) coins
– Vishal’s share in the correct ratio: \(\frac{3}{9} \times 351 = 117\) coins
– Kishore’s share in the correct ratio: \(\frac{4}{9} \times 351 = 156\) coins

Step 2: Calculate the number of coins each person received due to the mistake

Total parts in the mistaken ratio = 1/2 + 1/3 + 1/4 = 6/12 + 4/12 + 3/12 = 13/12 parts

– Abhay’s share in the mistaken ratio: \(\frac{1/2}{13/12} \times 351 = \frac{6}{13} \times 351 = 162\) coins
– Vishal’s share in the mistaken ratio: \(\frac{1/3}{13/12} \times 351 = \frac{4}{13} \times 351 = 108\) coins
– Kishore’s share in the mistaken ratio: \(\frac{1/4}{13/12} \times 351 = \frac{3}{13} \times 351 = 81\) coins

Step 3: Calculate the number of extra/deficit gold coins incurred to Abhay due to the mistake

Extra/deficit coins for Abhay = Abhay’s share in the mistaken ratio – Abhay’s share in the correct ratio
\[ = 162 – 78 = 84 \text{ coins} \]

Conclusion

The number of extra gold coins incurred to Abhay due to the mistake is 84 coins.