The problem consists of five questions: 1. Find the percentage error in measured length, calculated perimeter, and calculated area of a square, given the actual side length and the measured side length.

1. Problem Description

The problem consists of five questions:

1. Find the percentage error in measured length, calculated perimeter, and calculated area of a square, given the actual side length and the measured side length.

2. (a) Determine the initial number of men and women in a factory given that the total number of employees is 80 and doubling the number of men and tripling the number of women yields a total of 190 employees.

(b) Solve for $x$ in the equation $25^{3x+2} = 5^{2x+1}$.

3. (a) Find the base radius of a cone given its volume ($565.7 \, cm^3$) and height ($15 \, cm$). Use $\pi = \frac{22}{7}$.

(b) Find the number of terms in an arithmetic progression, given the first, second, and last terms are $1$, $\frac{1}{4}$, and $3\frac{3}{2}$, respectively.

4. Calculate the mean and variance of a given frequency distribution table.

5. (a) Evaluate the definite integral $\int_1^3 (4x^3 - 3x^2 + 2) \, dx$.

(b) Find the value of $x$ if $\log(5x-6) - \log(2x-3) = 1$.

2. Solution Steps

1. Percentage error in square measurements:

(i) Measured Length:

Actual length = 6.25 cm

Measured length = 6.12 cm

Absolute error = $|6.25 - 6.12| = 0.13 \, cm$

Percentage error = $\frac{0.13}{6.25} \times 100 = 2.08 \%$

(ii) Calculated Perimeter:

Actual perimeter = $4 \times 6.25 = 25 \, cm$

Measured perimeter = $4 \times 6.12 = 24.48 \, cm$

Absolute error = $|25 - 24.48| = 0.52 \, cm$

Percentage error = $\frac{0.52}{25} \times 100 = 2.08 \%$

(iii) Calculated Area:

Actual area = $(6.25)^2 = 39.0625 \, cm^2$

Measured area = $(6.12)^2 = 37.4544 \, cm^2$

Absolute error = $|39.0625 - 37.4544| = 1.6081 \, cm^2$

Percentage error = $\frac{1.6081}{39.0625} \times 100 = 4.116864 \% \approx 4.12 \%$

2. (a) Men and Women in the Factory:

Let $m$ be the initial number of men and $w$ be the initial number of women.

$m + w = 80$

$2m + 3w = 190$

Multiply the first equation by 2: $2m + 2w = 160$

Subtract this from the second equation: $w = 30$

Substitute $w = 30$ into the first equation: $m + 30 = 80$, so $m = 50$

Initial number of men = 50

Initial number of women = 30

(b) Solve for x:

$25^{3x+2} = 5^{2x+1}$

$(5^2)^{3x+2} = 5^{2x+1}$

$5^{6x+4} = 5^{2x+1}$

$6x + 4 = 2x + 1$

$4x = -3$

$x = -\frac{3}{4}$

3. (a) Radius of the Cone:

Volume of a cone, $V = \frac{1}{3} \pi r^2 h$

$565.7 = \frac{1}{3} \times \frac{22}{7} \times r^2 \times 15$

$565.7 = \frac{110}{7} \times r^2$

$r^2 = \frac{565.7 \times 7}{110} = \frac{3959.9}{110} = 35.999 \approx 36$

$r = \sqrt{36} = 6 \, cm$

(b) Number of Terms in Arithmetic Progression:

$a = 1$, $a_2 = \frac{1}{4}$, $l = 3\frac{3}{2} = \frac{9}{2}$

$d = a_2 - a = \frac{1}{4} - 1 = -\frac{3}{4}$

$l = a + (n-1)d$

$\frac{9}{2} = 1 + (n-1)(-\frac{3}{4})$

$\frac{7}{2} = (n-1)(-\frac{3}{4})$

$n-1 = \frac{7}{2} \times (-\frac{4}{3}) = -\frac{14}{3}$

$n = 1 - \frac{14}{3} = \frac{3 - 14}{3} = -\frac{11}{3}$

However, n must be a positive integer, so there must be a mistake in the question. Let's assume last term is 3 1/2 = 7/

2. $\frac{7}{2} = 1+(n-1)(-\frac{3}{4})$

$\frac{5}{2} = (n-1)(-\frac{3}{4})$

$n-1 = \frac{5}{2} * (-\frac{4}{3}) = -\frac{10}{3}$

$n = 1 - \frac{10}{3} = -\frac{7}{3}$

It is still negative and not an integer.

Let us consider the last term as $3\frac{1}{2} = \frac{7}{2}$

$\frac{7}{2} = 1+(n-1)(-\frac{3}{4})$

$\frac{5}{2} = (n-1)(-\frac{3}{4})$

$-\frac{20}{6} = n-1$

$n = -\frac{10}{3} + 1 = -\frac{7}{3}$

4. Mean and Variance of Frequency Distribution:

Midpoints: 5.5, 15.5, 25.5, 35.5, 45.5

Frequencies: 8, 10, 8, 22, 12

Total frequency = 60

Mean = $\frac{(5.5 \times 8) + (15.5 \times 10) + (25.5 \times 8) + (35.5 \times 22) + (45.5 \times 12)}{60} = \frac{44 + 155 + 204 + 781 + 546}{60} = \frac{1730}{60} = 28.8333 \approx 28.83$

Variance = $\frac{\sum f_i (x_i - \bar{x})^2}{\sum f_i}$

Variance = $\frac{8(5.5 - 28.83)^2 + 10(15.5 - 28.83)^2 + 8(25.5 - 28.83)^2 + 22(35.5 - 28.83)^2 + 12(45.5 - 28.83)^2}{60}$

Variance = $\frac{8(-23.33)^2 + 10(-13.33)^2 + 8(-3.33)^2 + 22(6.67)^2 + 12(16.67)^2}{60}$

Variance = $\frac{8(544.29) + 10(177.69) + 8(11.09) + 22(44.49) + 12(277.89)}{60}$

Variance = $\frac{4354.32 + 1776.9 + 88.72 + 978.78 + 3334.68}{60} = \frac{10533.4}{60} = 175.5567 \approx 175.56$

5. (a) Definite Integral:

$\int_1^3 (4x^3 - 3x^2 + 2) \, dx = [x^4 - x^3 + 2x]_1^3$

$= (3^4 - 3^3 + 2(3)) - (1^4 - 1^3 + 2(1)) = (81 - 27 + 6) - (1 - 1 + 2) = 60 - 2 = 58$

(b) Logarithmic Equation:

$\log(5x-6) - \log(2x-3) = 1$

$\log(\frac{5x-6}{2x-3}) = 1$

$\frac{5x-6}{2x-3} = 10$

$5x - 6 = 10(2x - 3)$

$5x - 6 = 20x - 30$

$15x = 24$

$x = \frac{24}{15} = \frac{8}{5} = 1.6$

3. Final Answer

1. (i) 2.08% (ii) 2.08% (iii) 4.12%

2. (a) Men = 50, Women = 30 (b) $x = -\frac{3}{4}$

3. (a) 6 cm (b) Unable to determine

4. Mean = 28.83, Variance = 175.56

5. (a) 58 (b) $x = \frac{8}{5} = 1.6$