Calculate the simple interest on N720,000.00 for 3 years at 5½% per annum. Hence, find the amount at the end of 3 years.

Given: Principal (P) = N720,000, Rate (R) = $5\frac{1}{2}\% = 5.5\%$, Time (T) = 3 years.

The simple interest is N118,800 and the total amount is N838,800.

1. (b) Ratio Problem

Total students = 75. Ratio of boys to girls = 2:3.
Total ratio parts = $2+3 = 5$.
Number of boys = $\frac{2}{5} \times 75 = 30$.
Number of girls = $\frac{3}{5} \times 75 = 45$.

Let $x$ be the number of boys added. The number of girls remains 45.
New number of boys = $30+x$.
The new ratio of boys to girls is 7:9.

Multiply both sides by 45:

5 boys should be added to the class.

2. (a) Solve the Equation

Find a common denominator for the right side, which is $(a+1)(3a-1)$.

Subtract $9a^2$ from both sides:

Subtract $a$ from both sides and add 3 to both sides:

The solution is $\boxed{a = \frac{3}{5}}$.

2. (b) Simplify Expression

Substitute these into the expression:

The simplified expression is $\boxed{-1}$.

3. (a) Shortest Distance from A to C

A man travels 5 km due east from A to B. So, AB = 5 km.
Then he travels 7 km on a bearing of $162^\circ$ from B to C. So, BC = 7 km.
The bearing of $162^\circ$ is measured clockwise from the North line at B.
Since AB is due East, the angle from the North line at B to the East direction is $90^\circ$.
The angle from the East direction (extension of AB) to BC is $162^\circ - 90^\circ = 72^\circ$.
The interior angle $\angle ABC$ in triangle ABC is $180^\circ - 72^\circ = 108^\circ$.

The Cosine Rule states $c^2 = a^2 + b^2 - 2ab \cos C$. Here, $AC^2 = AB^2 + BC^2 - 2(AB)(BC)\cos(\angle ABC)$.

The shortest distance from A to C is $\boxed{10 \text{ km}}$.

3. (b) Bearing of C from A

The Sine Rule states $\frac{a}{\sin A} = \frac{b}{\sin B}$. Here, $\frac{BC}{\sin(\angle BAC)} = \frac{AC}{\sin(\angle ABC)}$.

Since A to B is due East, the bearing of B from A is $090^\circ$.
The angle $\angle BAC$ is measured clockwise from AB.
The bearing of C from A is $90^\circ + \angle BAC$.
Bearing = $90^\circ + 42.9^\circ = 132.9^\circ$.

Bearing $\approx 133^\circ$.
The bearing of C from A is $\boxed{133^\circ}$.

4. (a) Simplify Algebraic Expression

Substitute the factored denominators back into the expression:

The common denominator is $3r(r-4)(r+4)$.

Cancel out the common factor $(r-4)$:

The simplified expression is $\boxed{\frac{4}{3r(r+4)}}$.

4. (b) Evaluate Integral

The evaluated integral is $\boxed{\frac{9}{5}x^5 + 4x^3 + 4x + C}$.