Paper II, Part I. Answer all questions in this part. 1. A rectangular garden is 18.6 m long and 12.4 m. A student measured the length as 18.2 m and the width as 12.7 m. Find the percentage error in the: i. Measured length ii. Calculated perimeter iii. Calculated area

$$
\text{Error in length} = |L_M - L_T| = |18.2 - 18.6| = |-0.4| = 0.4 \text{ m}
$$

$$
\text{Percentage Error (Length)} = \frac{0.4}{18.6} \times 100\% \approx 2.1505\%
$$
Rounded to two decimal places:
$$
\boxed{\text{2.15%}}
$$

$$
P_T = 2(L_T + W_T) = 2(18.6 + 12.4) = 2(31.0) = 62.0 \text{ m}
$$
$$
P_M = 2(L_M + W_M) = 2(18.2 + 12.7) = 2(30.9) = 61.8 \text{ m}
$$

$$
\text{Error in perimeter} = |P_M - P_T| = |61.8 - 62.0| = |-0.2| = 0.2 \text{ m}
$$

$$
\text{Percentage Error (Perimeter)} = \frac{0.2}{62.0} \times 100\% \approx 0.3226\%
$$
Rounded to two decimal places:
$$
\boxed{\text{0.32%}}
$$

$$
A_T = L_T \times W_T = 18.6 \times 12.4 = 230.64 \text{ m}^2
$$
$$
A_M = L_M \times W_M = 18.2 \times 12.7 = 231.14 \text{ m}^2
$$

$$
\text{Error in area} = |A_M - A_T| = |231.14 - 230.64| = 0.50 \text{ m}^2
$$

$$
\text{Percentage Error (Area)} = \frac{0.50}{230.64} \times 100\% \approx 0.2168\%
$$
Rounded to two decimal places:
$$
\boxed{\text{0.22%}}
$$

From the first sentence, "In a class of 45 students, the number of boys is three times the number of girls."
$$
B + G = 45 \quad (1)
$$
$$
B = 3G \quad (2)
$$

$$
3G + G = 45
$$
$$
4G = 45
$$
$$
G = \frac{45}{4} = 11.25
$$

$$
B = 3 \times 11.25 = 33.75
$$
Note: The number of students must be whole numbers. This problem as stated leads to non-integer values, indicating a potential flaw in the question's numerical values. However, based on the direct interpretation of the initial conditions, these are the calculated values.  
Initial number of boys: $\boxed{\text{33.75}}$  
Initial number of girls: $\boxed{\text{11.25}}$

$$
7^{2x-1} = (7^2)^{x+2}
$$
$$
7^{2x-1} = 7^{2(x+2)}
$$

$$
2x-1 = 2(x+2)
$$
$$
2x-1 = 2x+4
$$

$$
2x - 2x = 4 + 1
$$
$$
0 = 5
$$
This is a contradiction, which means there is no value of $x$ that satisfies the equation.  
The equation has $\boxed{\text{no solution}}$.

$$
V = \pi r^2 h
$$

$$
1386 = \frac{22}{7} \times r^2 \times 14
$$

$$
1386 = 22 \times r^2 \times 2
$$
$$
1386 = 44 r^2
$$
$$
r^2 = \frac{1386}{44}
$$
$$
r^2 = 31.5
$$

$$
r = \sqrt{31.5}
$$
$$
r \approx 5.61248... \text{ cm}
$$
Rounded to two decimal places:
$$
r \approx \boxed{\text{5.61 cm}}
$$

$$
a_1 = 5
$$
$$
a_2 = 11
$$
$$
a_n = 59
$$

$$
d = a_2 - a_1 = 11 - 5 = 6
$$

$$
59 = 5 + (n-1)6
$$

$$
59 - 5 = (n-1)6
$$
$$
54 = 6(n-1)
$$
$$
\frac{54}{6} = n-1
$$
$$
9 = n-1
$$
$$
n = 9 + 1
$$
$$
n = \boxed{\text{10}}
$$

$$
\bar{x} = \frac{\sum f_i x_i}{\sum f_i}
$$

$$
\bar{x} = \frac{2350}{60} = \frac{235}{6}
$$
$$
\bar{x} \approx 39.1666... \text{ years}
$$
Rounded to two decimal places:
$$
\bar{x} = \boxed{\text{39.17 years}}
$$

$$
\sigma = \sqrt{\frac{\sum f_i x_i^2}{\sum f_i} - \left(\frac{\sum f_i x_i}{\sum f_i}\right)^2}
$$

$$
\sigma = \sqrt{\frac{103700}{60} - \left(\frac{2350}{60}\right)^2}
$$

$$
\sigma = \sqrt{1728.3333... - (39.1666...)^2}
$$
$$
\sigma = \sqrt{1728.3333... - 1533.9166...}
$$
$$
\sigma = \sqrt{194.4166...}
$$

$$
\sigma \approx 13.9433... \text{ years}
$$
Rounded to two decimal places:
$$
\sigma = \boxed{\text{13.94 years}}
$$

$$
\int (3x^2 - 2x + 5) dx = \frac{3x^{2+1}}{2+1} - \frac{2x^{1+1}}{1+1} + 5x + C
$$
$$
= \frac{3x^3}{3} - \frac{2x^2}{2} + 5x + C
$$
$$
= x^3 - x^2 + 5x + C
$$

$$
[x^3 - x^2 + 5x]_{0}^{2} = (2^3 - 2^2 + 5(2)) - (0^3 - 0^2 + 5(0))
$$
$$
= (8 - 4 + 10) - (0)
$$
$$
= 14 - 0
$$
$$
= \boxed{\text{14}}
$$

$$
\log((x+2)(2x-1)) = \log 15
$$

$$
(x+2)(2x-1) = 15
$$

$$
2x^2 - x + 4x - 2 = 15
$$
$$
2x^2 + 3x - 2 = 15
$$
$$
2x^2 + 3x - 17 = 0
$$

$$
x = \frac{-3 \pm \sqrt{3^2 - 4(2)(-17)}}{2(2)}
$$
$$
x = \frac{-3 \pm \sqrt{9 + 136}}{4}
$$
$$
x = \frac{-3 \pm \sqrt{145}}{4}
$$

$$
(\sqrt{2x+6})^2 = 10^2
$$
$$
2x+6 = 100
$$

$$
2x = 100 - 6
$$
$$
2x = 94
$$
$$
x = \frac{94}{2}
$$
$$
x = \boxed{\text{47}}
$$

$$
(\sqrt{3x+1})^2 = 4^2
$$
$$
3x+1 = 16
$$

$$
3x = 16 - 1
$$
$$
3x = 15
$$
$$
x = \frac{15}{3}
$$
$$
x = \boxed{\text{5}}
$$