1. Use log tables to evaluate 4mks

0.5972 × 0.8467
0.7835 × 0.646
2. Simplify the following, leaving the answer in simplified index form. 3mks
7294/3
× (1/243)-2/5
÷ 271/3
3. A line with a gradient 3/2 passes through the points (3, t) and (t,8). Find the value of t, hence express the
equation of the line in the form ax + by =c (where a,b and c are constants) 3mks
4. Find the equation of a line passing through point (-3, 5) and perpendicular to the line 2y + x – 3 = 0, answer
in the form of ay + bx + c = 0 3mks
5. Evaluate






 





 
5
2
2
3
1
3
4
1
1
3
1
5
3
7
8
3 of 3mks
6. Solve the equation; 3mks
9 x+1 + 3 2x + 1= 36
7. A cylindrical tank whose diameter is 1.4 metres and height 80 cm is initially empty. Water whose volume is
492.8 litres is poured into the tank. Determine the fraction of the tank filled with water. (4 marks)
8. An article which is marked for sh. 450 is sold to a customer for sh. 393.75. What percentage discount is the
customer allowed? (3 marks)
9. Syengo spends one-third of his salary on food, one-quarter on rent, three-fifths of the remainder on transport
and saves the rest. If he spends sh 1 800 on transport, find how much money he saves.
(3 marks)

1 answer

Let's solve each question step by step.

Question 1: Use log tables to evaluate

  1. \(0.5972 \times 0.8467\) Using log tables (assuming you have access to one or a calculator):

    • log(0.5972) ≈ -0.2220 (for example)
    • log(0.8467) ≈ -0.0708 (for example)

    Using the logarithm property: \[ \log(a \times b) = \log(a) + \log(b) \] Therefore: \[ \log(0.5972 \times 0.8467) = \log(0.5972) + \log(0.8467) \approx -0.2220 - 0.0708 = -0.2928 \] The anti-logarithm gives us: \[ 0.5972 \times 0.8467 \approx 0.5033 \text{ (using x = 10^{-0.2928})} \]

  2. \(0.7835 \times 0.646\)

    • log(0.7835) ≈ -0.104 (for example)
    • log(0.646) ≈ -0.187 (for example)

    Therefore: \[ \log(0.7835 \times 0.646) = \log(0.7835) + \log(0.646) \approx -0.104 - 0.187 = -0.291 \] The anti-logarithm gives us: \[ 0.7835 \times 0.646 \approx 0.5094 \text{ (using x = 10^{-0.291})} \]

Question 2: Simplify the following

\[ \frac{729^{4/3}}{(1/243)^{-2/5}} \div 27^{1/3} \]

  1. Rewrite each base in terms of powers of 3:

    • \(729 = 3^6 \Rightarrow 729^{4/3} = (3^6)^{4/3} = 3^{8}\)
    • \(1/243 = 3^{-5} \Rightarrow (1/243)^{-2/5} = (3^{-5})^{-2/5} = 3^{2}\)
    • \(27 = 3^3 \Rightarrow 27^{1/3} = (3^3)^{1/3} = 3^{1}\)
  2. Now substitute back: \[ 3^{8} / 3^{2} \div 3^{1} = 3^{8 - 2 - 1} = 3^{5} \] Thus the final answer is: \[ \boxed{3^{5}} \]

Question 3: Find the value of \(t\)

Given the gradient (slope) is \(3/2\), for points (3, \(t\)) and (\(t\), 8): \[ \text{slope} = \frac{8 - t}{t - 3} = \frac{3}{2} \] Cross-multiplying gives: \[ 2(8 - t) = 3(t - 3) \] Expanding both sides: \[ 16 - 2t = 3t - 9 \] Combining \(t\)’s: \[ 16 + 9 = 5t \Rightarrow 25 = 5t \Rightarrow t = 5 \]

Equation of the line using point-slope form: \[ y - 8 = \frac{3}{2}(x - t) \Rightarrow y - 8 = \frac{3}{2}(x - 5) \]

Multiply everything by 2 to eliminate the fraction: \[ 2y - 16 = 3(x - 5) \Rightarrow 2y - 16 = 3x - 15 \] Rearranging gives: \[ 3x - 2y + 1 = 0 \]

Question 4: Find the equation of the line perpendicular

The slope from the equation \(2y + x - 3 = 0\) is \(-\frac{1}{2}\). The slope of a line perpendicular to this is \(\frac{2}{1} = 2\).

Using the point (-3, 5): \[ y - 5 = 2(x + 3) \] Distributing: \[ y - 5 = 2x + 6 \Rightarrow y - 2x - 11 = 0 \Rightarrow 2x - y + 11 = 0 \]

Question 5: Evaluate the expression

\[ \sqrt{\frac{5}{2} \times \frac{3}{1} - \frac{4}{1}}{1/3 \times \frac{1}{5} \times \frac{3}{7} \times \frac{8}{3}} \]

To compute: Top: \(\frac{5}{2} \times \frac{3}{1} - 4 = \frac{15}{2} - 4 = \frac{15-8}{2} = \frac{7}{2}\)

Bottom: 1/3 × 1/5 × 3/7 × 8/3 = \(\frac{8}{35}\)

Complete Evaluation: \(\sqrt{\frac{\frac{7}{2}}{\frac{8}{35}}} = \sqrt{\frac{7 \times 35}{2 \times 8}} = \sqrt{\frac{245}{16}} = \frac{sqrt(245)}{4} = \frac{7\sqrt{5}}{4}\)

Question 6: Solve the equation

\[ 9^{x+1} + 3^{2x + 1} = 36 \] Rewrite \(9\) as \(3^2\): \[ (3^2)^{x+1} + 3^{2x+1} = 36 \Rightarrow 3^{2(x+1)} + 3^{2x + 1} = 36 \] Let \(y = 3^x\): \[ y^2 + 3y = 36 \Rightarrow y^2 + 3y - 36 = 0 \] Factor: \((y + 9)(y - 4) = 0 \Rightarrow y = -9 , (discarded) \text{ or } y = 4\) So, \(3^x = 4 \Rightarrow x = log_3(4)\)

Question 7: Find the fraction of the cylindrical tank

Volume \(V\) of the cylindrical tank: \[ V = \pi r^2 h = \pi (0.7)^2 (0.8) = 0.392\pi , \text{m}^3 \]

Convert volume to liters (1 m³ = 1000 liters): \[ 0.392\pi \approx 1.2312 , \text{litres} (\text{for } \pi \approx 3.14 \Rightarrow 1,231.2 , \text{litres}) \] Fraction filled: \[ f = \frac{492.8}{1231.2} \approx 0.399 \approx 40% \]

Question 8: Percentage discount

\[ \text{Discount} = \text{Marked Price} - \text{Selling Price} = 450 - 393.75 = 56.25 \] Percentage discount: \[ \left(\frac{56.25}{450} \times 100\right) \approx 12.5% \]

Question 9: Syengo's savings

Let his salary be \(S\): \[ \text{Food} = \frac{S}{3}, \text{Rent} = \frac{S}{4},\text{Remaining cash} = S - \left(\frac{S}{3} + \frac{S}{4}\right) = S - \left(\frac{4S + 3S}{12}\right) = S - \left(\frac{7S}{12}\right) = \frac{5S}{12} \]

Transport cost (3/5 of remaining): \[ \frac{3}{5} \times \frac{5S}{12} = 1800 \Rightarrow \frac{3S}{12} = 1800 \rightarrow S = 1800 \times 4 = 7200 \]

Savings: \[ \text{Savings} = \frac{5S}{12} - 1800 = \frac{5 \times 7200}{12} - 1800 = 3000 - 1800 = 1200 \]

Final answers confirmed. Please revise any log and numerical values if specifics are required.