Question One

a) Evaluate:
[tex]\[ 2 \frac{1}{2} + 3 \frac{1}{8} \div 1 \frac{1}{4} \][/tex]
(2 marks)

b) A sum of money is divided into four parts in the ratio 1:2:3:4. If the largest part is K10,000, find the value of the other parts. (2 marks)

c) Solve for [tex]\( x \)[/tex]:
[tex]\[ \log_5(x^2 - 3x + 2) = 2 + \log_5(x - 1) \][/tex]
(6 marks)

d) The progression is given as follows:
[tex]\[ 800, 1600, 3200, \ldots \][/tex]
Calculate the sum of the first five terms of the progression. (4 marks)



Answer :

Certainly! Let's go through each part of the question step-by-step.

### Part (a)
Evaluate: [tex]\(2 \frac{1}{2} + 3 \frac{1}{8} \div 1 \frac{1}{4}\)[/tex]

First, we need to convert the mixed numbers to improper fractions:
- [tex]\(2 \frac{1}{2} = 2 + \frac{1}{2} = \frac{4}{2} + \frac{1}{2} = \frac{5}{2}\)[/tex]
- [tex]\(3 \frac{1}{8} = 3 + \frac{1}{8} = \frac{24}{8} + \frac{1}{8} = \frac{25}{8}\)[/tex]
- [tex]\(1 \frac{1}{4} = 1 + \frac{1}{4} = \frac{4}{4} + \frac{1}{4} = \frac{5}{4}\)[/tex]

We evaluate the division first: [tex]\(\frac{25}{8} \div \frac{5}{4}\)[/tex]. When we divide by a fraction, we multiply by its reciprocal:
[tex]\[ \frac{25}{8} \div \frac{5}{4} = \frac{25}{8} \times \frac{4}{5} = \frac{25 \times 4}{8 \times 5} = \frac{100}{40} = \frac{5}{2} \][/tex]

Now, add this result to [tex]\(\frac{5}{2}\)[/tex]:
[tex]\[ \frac{5}{2} + \frac{5}{2} = \frac{5+5}{2} = \frac{10}{2} = 5.0 \][/tex]

So, the result for part (a) is [tex]\(5.0\)[/tex].

### Part (b)
The sum of money is divided into four parts in the ratio [tex]\(1:2:3:4\)[/tex]. If the largest part is 10,000, find the value of the other parts.

Let the parts be [tex]\( x, 2x, 3x, 4x \)[/tex]. According to the problem, the largest part [tex]\(4x = 10,000\)[/tex]:
[tex]\[ 4x = 10,000 \implies x = \frac{10,000}{4} = 2,500 \][/tex]

Therefore, the parts are:
[tex]\[ \begin{align*} \text{Part 1} &= x = 2,500 \\ \text{Part 2} &= 2x = 2(2,500) = 5,000 \\ \text{Part 3} &= 3x = 3(2,500) = 7,500 \\ \text{Part 4} &= 4x = 10,000 \end{align*} \][/tex]

So, the values of the parts are 2,500, 5,000, 7,500, and 10,000.

### Part (c)
Solve the equation: [tex]\(\log_5(x^2 - 3x + 2) = 2 + \log_5(x-1)\)[/tex].

To solve this, we use logarithmic properties:
[tex]\[ \log_5(x^2 - 3x + 2) = 2 + \log_5(x-1) \][/tex]
Rewriting [tex]\(2\)[/tex] in terms of the log base 5:
[tex]\[ 2 = \log_5(25) \][/tex]
Thus, the equation becomes:
[tex]\[ \log_5(x^2 - 3x + 2) = \log_5(25) + \log_5(x-1) \][/tex]
Using the property [tex]\(\log_b(a) + \log_b(c) = \log_b(ac)\)[/tex]:
[tex]\[ \log_5(x^2 - 3x + 2) = \log_5[25(x-1)] \][/tex]
Since the bases and the logarithms are identical, we can equate the arguments:
[tex]\[ x^2 - 3x + 2 = 25(x - 1) \][/tex]
Simplify the equation:
[tex]\[ x^2 - 3x + 2 = 25x - 25 \implies x^2 - 3x + 2 - 25x + 25 = 0 \implies x^2 - 28x + 27 = 0 \][/tex]

Solve the quadratic equation using the quadratic formula [tex]\(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\)[/tex]:
[tex]\[ x = \frac{28 \pm \sqrt{(-28)^2 - 4 \cdot 1 \cdot 27}}{2 \cdot 1} = \frac{28 \pm \sqrt{784 - 108}}{2} = \frac{28 \pm \sqrt{676}}{2} = \frac{28 \pm 26}{2} \][/tex]

So, the solutions are:
[tex]\[ x = \frac{28 + 26}{2} = 27, \quad x = \frac{28 - 26}{2} = 1 \][/tex]

However, [tex]\( x = 1 \)[/tex] does not satisfy the logarithmic constraints (since [tex]\(\log_5(0)\)[/tex] is undefined). Therefore, the solution is:
[tex]\[ x = 27 \][/tex]

### Part (d)
Calculate the sum of the first five terms of the progression:
[tex]\[ 800, 1600, 3200, \ldots \][/tex]

This is a geometric progression where the first term [tex]\(a = 800\)[/tex] and the common ratio [tex]\(r = 2\)[/tex]. The sum of the first [tex]\(n\)[/tex] terms of a geometric progression is given by:
[tex]\[ S_n = a \frac{1 - r^n}{1 - r} \][/tex]

Given [tex]\(n = 5\)[/tex]:
[tex]\[ S_5 = 800 \frac{1 - 2^5}{1 - 2} = 800 \frac{1 - 32}{-1} = 800 \frac{-31}{-1} = 800 \times 31 = 24,800 \][/tex]

Therefore, the sum of the first five terms is [tex]\(24,800\)[/tex].

In conclusion, the answers to the question parts are as follows:
[tex]\[ \begin{align*} (a) & \quad 5.0 \\ (b) & \quad 2,500, 5,000, 7,500, 10,000 \\ (c) & \quad x = 27 \\ (d) & \quad 24,800 \end{align*} \][/tex]