Given the equation C = 2πr + b, make b the subject. A. b = C – 2πr, B. b = C + 2πr, C. b = 2πr (C – r) D. b = C/2πr

here are the solutions to the mathematics questions:

SECTION A: OBJECTIVE TEST QUESTIONS

1. The number 709,045 in words is "Seven hundred nine thousand, forty five".

• A) Seven hundred nine thousand, forty five

2. In the number 28.356, the digit 5 is in the hundredths place.

• C) Hundredths

3. To express 0.00205 in standard form, move the decimal point 3 places to the right. This gives $2.05 \times 10^{-3}$.

• D) $2.05 \times 10^{-3}$

4. To convert $101101_2$ to base ten:

$$1 \times 2^5 + 0 \times 2^4 + 1 \times 2^3 + 1 \times 2^2 + 0 \times 2^1 + 1 \times 2^0$$ $$= 1 \times 32 + 0 \times 16 + 1 \times 8 + 1 \times 4 + 0 \times 2 + 1 \times 1$$ $$= 32 + 0 + 8 + 4 + 0 + 1 = 45$$

• A) 45

5. A prime number is a natural number greater than 1 that has no positive divisors other than 1 and itself.

• 2 is prime.
• 17 is prime.
• 39 is not prime because $39 = 3 \times 13$.
• 1 is not considered a prime number.
• In this context, 39 is the composite number, and thus not prime.
• C) 39

6. To approximate 14.897 to 2 decimal places, look at the third decimal place (7). Since it is 5 or greater, round up the second decimal place. The 9 in the hundredths place rounds up to 10, so the 8 in the tenths place becomes 9, and the hundredths place becomes 0.

$$14.897 \approx 14.90$$

• B) 14.90

7. To find the L.C.M of 9, 15, and 20:

• Prime factorization:
  • $9 = 3^2$
  • $15 = 3 \times 5$
  • $20 = 2^2 \times 5$
• L.C.M = $2^2 \times 3^2 \times 5 = 4 \times 9 \times 5 = 180$
• D) 180

8. Simplify $(\frac{1}{2} + \frac{1}{3}) \div (\frac{1}{2} - \frac{1}{3})$:

• First, calculate the sum: $\frac{1}{2} + \frac{1}{3} = \frac{3}{6} + \frac{2}{6} = \frac{5}{6}$
• Next, calculate the difference: $\frac{1}{2} - \frac{1}{3} = \frac{3}{6} - \frac{2}{6} = \frac{1}{6}$
• Now, divide: $\frac{5/6}{1/6} = \frac{5}{6} \times \frac{6}{1} = 5$
• A) 5

9. To express 15 m as a percentage of 1.20 km:

• Convert 1.20 km to meters: $1.20 \text{ km} = 1.20 \times 1000 \text{ m} = 1200 \text{ m}$
• Percentage = $\frac{15 \text{ m}}{1200 \text{ m}} \times 100\%$
• Percentage = $\frac{15}{1200} \times 100 = \frac{15}{12} = \frac{5}{4} = 1.25\%$
• A) 1.25%

10. To arrange $\frac{1}{3}, \frac{2}{5}, \frac{3}{4}, \frac{1}{2}$ in ascending order, convert them to decimals:

• $\frac{1}{3} \approx 0.333$
• $\frac{2}{5} = 0.4$
• $\frac{3}{4} = 0.75$
• $\frac{1}{2} = 0.5$
• In ascending order: $0.333, 0.4, 0.5, 0.75$, which corresponds to $\frac{1}{3}, \frac{2}{5}, \frac{1}{2}, \frac{3}{4}$.
• A) $\frac{1}{3}, \frac{2}{5}, \frac{1}{2}, \frac{3}{4}$

11. The perimeter of a rectangle with length 15 cm and width 8 cm is:

• Perimeter = $2 \times (\text{length} + \text{width})$
• Perimeter = $2 \times (15 \text{ cm} + 8 \text{ cm}) = 2 \times 23 \text{ cm} = 46 \text{ cm}$
• C) 46 cm

12. The area of a triangle is given by $\frac{1}{2} \times \text{base} \times \text{height}$.

• Given Area = 60 cm$^2$ and height = 12 cm.
• $60 = \frac{1}{2} \times \text{base} \times 12$
• $60 = 6 \times \text{base}$
• $\text{base} = \frac{60}{6} = 10 \text{ cm}$
• B) 10 cm

13. The total surface area of a cube with side length 3 cm is:

• TSA = $6 \times (\text{side})^2$
• TSA = $6 \times (3 \text{ cm})^2 = 6 \times 9 \text{ cm}^2 = 54 \text{ cm}^2$
• A) 54 cm$^2$

14. A triangular prism has 2 triangular faces, 3 rectangular faces, 6 vertices, and 9 edges.

• C) 9

15. A parallelogram, in general, has no lines of symmetry.

• A) 0

16. Simplify $\frac{a^2b^3 - ab^2}{ab(b-1)}$

• Factor out $ab^2$ from the numerator: $ab^2(ab-1)$
• The expression becomes $\frac{ab^2(ab-1)}{ab(b-1)} = \frac{b(ab-1)}{b-1}$.
• Note: Assuming there might be a typo in the question and the denominator was intended to be $ab(ab-1)$ or $a^2b^2 - ab$, the expression would simplify to $b$. Given the options, this is the most likely intended answer if a typo exists.
• If the denominator was $a^2b^2 - ab$, then $\frac{ab^2(ab-1)}{ab(ab-1)} = b$.
• B) $b$

17. To find the coefficient of x in the expansion of $(2x-3)(x-5)$:

• Expand the expression: $$(2x-3)(x-5) = 2x(x) + 2x(-5) - 3(x) - 3(-5)$$ $$= 2x^2 - 10x - 3x + 15$$ $$= 2x^2 - 13x + 15$$
• The coefficient of x is -13.
• C) -13

SECTION B: THEORY

1. (a) (i) Simplify $2\frac{1}{2} + 1\frac{1}{4} - \frac{1}{10}$