Let’s denote the original expenditure of the mess as \(E\) rupees per day and the original average expenditure per person as \(A\) rupees per person per day.

According to the given information:

– Original number of personnel = 45
– New number of personnel = 45 + 9 = 54
– Increase in expenses of mess = Rs. 54 per day
– Decrease in average expenditure per person = Rs. 1 per person per day

Using the definition of average expenditure per person, we can write the following equations for the original and new scenarios:

1. Original scenario:

\[ A = \frac{E}{45} \]

2. New scenario:

\[ A – 1 = \frac{E + 54}{54} \]

From equation (1), we can express \(E\) in terms of \(A\):

\[ E = 45A \]

Substituting \(E\) in equation (2):

\[ A – 1 = \frac{45A + 54}{54} \]

Multiplying through by 54:

\[ 54A – 54 = 45A + 54 \]

Rearranging:

\[ 9A = 108 \]

\[ A = 12 \]

Now, using the value of \(A\) to find \(E\):

\[ E = 45A = 45 \times 12 = 540 \]

Therefore, the original expenditure of the mess is Rs. 540 per day.

To find the length of the ladder, we can use the Pythagorean theorem, as the ladder, the wall, and the ground form a right-angled triangle.

Let’s denote:
– The length of the ladder as \(L\).
– The distance of the foot of the ladder from the wall as 2.5 m.
– The height of the window above the ground as 6 m.

According to the Pythagorean theorem:

\[
L^2 = 6^2 + 2.5^2
\]

\[
L^2 = 36 + 6.25
\]

\[
L^2 = 42.25
\]

\[
L = \sqrt{42.25}
\]

\[
L = 6.5 \text{ m}
\]

Therefore, the length of the ladder is 6.5 meters.

Let’s denote the principal amount as \(P\), the rate of compound interest as \(r\), and the number of years as \(n\).

According to the given information, the amount becomes fivefold in 7 years. Therefore, we can write the compound interest formula as follows:

\[ A = P(1 + r)^n \]

For the amount to be fivefold:

\[ 5P = P(1 + r)^7 \]

Simplifying:

\[ (1 + r)^7 = 5 \]

Now, we need to find in how many years the amount will be 25 times itself. We can set up a similar equation:

\[ 25P = P(1 + r)^n \]

\[ (1 + r)^n = 25 \]

We know that \((1 + r)^7 = 5\), so we can express 25 in terms of 5:

\[ (1 + r)^n = 5^2 \]

\[ (1 + r)^n = ((1 + r)^7)^2 \]

\[ (1 + r)^n = (1 + r)^{14} \]

Thus, we can see that \(n = 14\).

Therefore, it will take 14 years for the sum of money to become 25 times itself at the same rate of compound interest.

Let’s denote the side length of the square as \(s\) cm.

The area of the square is given as 578 sq cm, so we can write:

\[ \text{Area} = s^2 = 578 \text{ sq cm} \]

To find the side length \(s\), we take the square root of the area:

\[ s = \sqrt{578} \text{ cm} \]

The diagonal of a square is given by \(d = s\sqrt{2}\), so the length of the diagonal is:

\[ d = \sqrt{578} \times \sqrt{2} = \sqrt{1156} \text{ cm} = 34 \text{ cm} \]

Therefore, the measure of the diagonal of the square is 34 cm.

– Distance covered by both Amit and Suresh is 30 km.
– Amit takes 2 hours more than Suresh to cover this distance.
– If Amit doubles his speed, he takes 1 hour less than Suresh.

Let’s denote:
– Amit’s original speed as \(s\) km/h.
– Suresh’s time to cover 30 km as \(t\) hours.

From the first condition, Amit’s time to cover 30 km is \(t + 2\) hours, so we can write Amit’s speed as:

\[ s = \frac{30}{t + 2} \]

From the second condition, if Amit doubles his speed, his time to cover 30 km becomes \(t – 1\) hours. So, we can write:

\[ 2s = \frac{30}{t – 1} \]

Substituting the expression for \(s\) from the first equation into the second equation:

\[ 2 \times \frac{30}{t + 2} = \frac{30}{t – 1} \]

Cross-multiplying:

\[ 60(t – 1) = 30(t + 2) \]

Expanding:

\[ 60t – 60 = 30t + 60 \]

Rearranging:

\[ 30t = 120 \]

\[ t = 4 \text{ hours} \]

Now, we can find Amit’s original speed:

\[ s = \frac{30}{t + 2} = \frac{30}{4 + 2} = \frac{30}{6} = 5 \text{ km/h} \]

So, Amit’s original speed is 5 km/h.