The dimensions of a square is given as 6.25 cm. A student measured one side of the square as 6.12 cm to calculate the perimeter and the area of the square. Find the percentage error in the:
SOLUTION:
i. Measured length = 6.12 cm
Finding percentage error in measured length: Formula: Percentage error = |Actual value - Measured value|/Actual value × 100%
Actual length = 6.25 cm Measured length = 6.12 cm
Error = |6.25 - 6.12|/6.25 × 100% Error = 0.13/6.25 × 100% Error = 0.0208 × 100% Percentage error in measured length = 2.08%
ii. Calculated perimeter: Perimeter of square = 4 × side length = 4 × 6.12 = 24.48 cm
Finding percentage error in perimeter: Actual perimeter = 4 × 6.25 = 25.0 cm Error = |25.0 - 24.48|/25.0 × 100% Error = 0.52/25.0 × 100% Error = 0.0208 × 100% Percentage error in perimeter = 2.08%
iii. Calculated area: Area of square = side² = 6.12² = 6.12 × 6.12 = 37.4544 cm²
Finding percentage error in area: Actual area = 6.25² = 6.25 × 6.25 = 39.0625 cm² Error = |39.0625 - 37.4544|/39.0625 × 100% Error = 1.6081/39.0625 × 100% Error = 0.0412 × 100% Percentage error in area = 4.12%
(a) A factory has 80 employees. If the number of men is doubled and the women tripled, the staff strength would be 190. Find the initial number of men and women working in the factory.
(b) Solve for x if 25³ˣ⁺² = 5²ˣ⁺¹
SOLUTION:
(a) Factory problem: Let initial number of men = x Let initial number of women = y
Setting up equations: Total employees = 80 x + y = 80 ... (1)
After changes:
2x + 3y = 190 ... (2)
Solving: From equation (1): y = 80 - x Substitute into equation (2): 2x + 3(80 - x) = 190 2x + 240 - 3x = 190 -x = 190 - 240 -x = -50 x = 50
From equation (1): y = 80 - 50 = 30
Answer: Initial men = 50, Initial women = 30
(b) Solve 25³ˣ⁺² = 5²ˣ⁺¹
Step 1: Express both sides with same base 25 = 5² So: 25³ˣ⁺² = (5²)³ˣ⁺²
Step 2: Simplify left side (5²)³ˣ⁺² = 5²⁽³ˣ⁺²⁾ = 5⁶ˣ⁺⁴
Step 3: Equation becomes 5⁶ˣ⁺⁴ = 5²ˣ⁺¹
Step 4: Since bases are equal, equate exponents 6x + 4 = 2x + 1 6x - 2x = 1 - 4 4x = -3 x = -3/4
(a) The volume of a cone is 565.7 cm³. If the cone is 15 cm high, find its base radius correct to the nearest whole number. (Take π = 22/7)
(b) The 1st, 2nd and last terms of an Arithmetic Progression (A.P.) are 1, 1⅓ and 3⅓ respectively. Find the number of terms in the A.P.
SOLUTION:
(a) Volume of cone = 565.7 cm³, height = 15 cm
Formula: V = 1/3 πr²h
Substituting values: 565.7 = 1/3 × 22/7 × r² × 15 565.7 = 22 × r² × 15/(3 × 7) 565.7 = 330r²/21 565.7 = 110r²/7
Solving for r²: r² = 565.7 × 7/110 r² = 3959.9/110 r² = 35.999 ≈ 36
Finding r: r = √36 = 6 cm
(b) A.P. with a = 1, second term = 1⅓, last term = 3⅓
Step 1: Find common difference Second term = a + d = 1⅓ 1 + d = 1⅓ = 4/3 d = 4/3 - 1 = 4/3 - 3/3 = 1/3
Step 2: Convert last term 3⅓ = 3 + 1/3 = 9/3 + 1/3 = 10/3
Step 3: Use formula aₙ = a + (n-1)d 10/3 = 1 + (n-1) × 1/3 10/3 = 1 + (n-1)/3 10/3 - 1 = (n-1)/3 10/3 - 3/3 = (n-1)/3 7/3 = (n-1)/3
Step 4: Solve 7 = n - 1 n = 8 terms
Table 2.1 shows the distribution of marks scored by 60 students in an examination.
Table 2.1: Frequency distribution
| Mark | 1-10 | 11-20 | 21-30 | 31-40 | 41-50 |
|---|---|---|---|---|---|
| Frequency | 8 | 10 | 8 | 22 | 12 |
Calculate and correct to 2 decimal places, the:
SOLUTION:
i. Mean:
Step 1: Find midpoints
Step 2: Calculate Σfx
| x | f | fx |
|---|---|---|
| 5.5 | 8 | 44 |
| 15.5 | 10 | 155 |
| 25.5 | 8 | 204 |
| 35.5 | 22 | 781 |
| 45.5 | 12 | 546 |
| Total | 60 | 1730 |
Step 3: Calculate mean Mean = Σfx/Σf = 1730/60 = 28.83
ii. Variance:
Step 1: Calculate Σfx²
| x | f | fx² |
|---|---|---|
| 5.5 | 8 | 8(30.25) = 242 |
| 15.5 | 10 | 10(240.25) = 2402.5 |
| 25.5 | 8 | 8(650.25) = 5202 |
| 35.5 | 22 | 22(1260.25) = 27725.5 |
| 45.5 | 12 | 12(2070.25) = 24843 |
| Total | 60 | 60415 |
Step 2: Calculate variance Variance = Σfx²/Σf - (mean)² = 60415/60 - (28.83)² = 1006.92 - 831.37 Variance = 175.55
(a) Evaluate ∫₂⁵(4x³ - 3x² + 2)dx
(b) Find the value of x if log(5x - 6) - log(2x - 3) = 1
SOLUTION:
(a) Evaluate ∫₂⁵(4x³ - 3x² + 2)dx
Step 1: Find antiderivative ∫(4x³ - 3x² + 2)dx = 4x⁴/4 - 3x³/3 + 2x + C = x⁴ - x³ + 2x + C
Step 2: Evaluate definite integral [x⁴ - x³ + 2x]₂⁵
Step 3: Substitute upper limit (x = 5) 5⁴ - 5³ + 2(5) = 625 - 125 + 10 = 510
Step 4: Substitute lower limit (x = 2) 2⁴ - 2³ + 2(2) = 16 - 8 + 4 = 12
Step 5: Calculate final answer 510 - 12 = 498
(b) Solve log(5x - 6) - log(2x - 3) = 1
Step 1: Use log properties log(5x - 6) - log(2x - 3) = log[(5x - 6)/(2x - 3)]
Step 2: Equation becomes log[(5x - 6)/(2x - 3)] = 1
Step 3: Convert to exponential form (5x - 6)/(2x - 3) = 10¹ = 10
Step 4: Cross multiply 5x - 6 = 10(2x - 3) 5x - 6 = 20x - 30
Step 5: Solve for x 5x - 20x = -30 + 6 -15x = -24 x = 24/15 = 8/5 = 1.6
Given the matrices A = [3 2 4; 3 1 3; 1 4 1] and B = [4 2 3; 1 5 2; 2 1 4]
(a) Find the determinant of:
(b) Find the inverse of A
SOLUTION:
(a) Find determinants:
i. det A: A = [3 2 4; 3 1 3; 1 4 1]
det A = 3|1 3; 4 1| - 2|3 3; 1 1| + 4|3 1; 1 4|
Calculating 2×2 determinants: |1 3; 4 1| = 1(1) - 3(4) = 1 - 12 = -11 |3 3; 1 1| = 3(1) - 3(1) = 3 - 3 = 0 |3 1; 1 4| = 3(4) - 1(1) = 12 - 1 = 11
Final calculation: det A = 3(-11) - 2(0) + 4(11) = -33 + 0 + 44 det A = 11
ii. det B: B = [4 2 3; 1 5 2; 2 1 4]
det B = 4|5 2; 1 4| - 2|1 2; 2 4| + 3|1 5; 2 1|
Calculating 2×2 determinants: |5 2; 1 4| = 5(4) - 2(1) = 20 - 2 = 18 |1 2; 2 4| = 1(4) - 2(2) = 4 - 4 = 0 |1 5; 2 1| = 1(1) - 5(2) = 1 - 10 = -9
Final calculation: det B = 4(18) - 2(0) + 3(-9) = 72 - 0 - 27 det B = 45
(b) Find inverse of A:
Step 1: Find cofactor matrix C₁₁ = +|1 3; 4 1| = 1 - 12 = -11 C₁₂ = -|3 3; 1 1| = -(3 - 3) = 0 C₁₃ = +|3 1; 1 4| = 12 - 1 = 11 C₂₁ = -|2 4; 4 1| = -(2 - 16) = 14 C₂₂ = +|3 4; 1 1| = 3 - 4 = -1 C₂₃ = -|3 2; 1 4| = -(12 - 2) = -10 C₃₁ = +|2 4; 1 3| = 6 - 4 = 2 C₃₂ = -|3 4; 3 3| = -(9 - 12) = 3 C₃₃ = +|3 2; 3 1| = 3 - 6 = -3
Step 2: Form cofactor matrix Cof A = [-11 0 11; 14 -1 -10; 2 3 -3]
Step 3: Adjugate matrix (transpose of cofactor) adj A = [-11 14 2; 0 -1 3; 11 -10 -3]
Step 4: Calculate inverse A⁻¹ = (1/det A) × adj A A⁻¹ = (1/11) × [-11 14 2; 0 -1 3; 11 -10 -3]
(a)i Use completing the square method to solve the equation ax² + bx + c = 0
ii Hence, use the result in 7.(a)i above to find the roots of 4x² + 7x - 2 = 0
(b) The mean of 6 numbers is 40. Three of the numbers sum up to be 90. What is the mean of the remaining 3 numbers?
SOLUTION:
(a)i. Completing the square method:
For general equation ax² + bx + c = 0:
Step 1: Divide by a x² + (b/a)x + c/a = 0
Step 2: Move constant to right x² + (b/a)x = -c/a
Step 3: Complete the square Add (b/2a)² to both sides: x² + (b/a)x + (b/2a)² = -c/a + (b/2a)²
Step 4: Factor left side (x + b/2a)² = -c/a + b²/4a²
Step 5: Combine right side (x + b/2a)² = (-4ac + b²)/4a² (x + b/2a)² = (b² - 4ac)/4a²
Step 6: Take square root x + b/2a = ±√(b² - 4ac)/2a
Step 7: Solve for x x = -b/2a ± √(b² - 4ac)/2a x = [-b ± √(b² - 4ac)]/2a
(a)ii. Solve 4x² + 7x - 2 = 0:
Using quadratic formula: a = 4, b = 7, c = -2
x = [-7 ± √(7² - 4(4)(-2))]/2(4) x = [-7 ± √(49 + 32)]/8 x = [-7 ± √81]/8 x = [-7 ± 9]/8
Two solutions: x₁ = (-7 + 9)/8 = 2/8 = 1/4 x₂ = (-7 - 9)/8 = -16/8 = -2
(b) Mean problem:
Given:
Step 1: Find sum of all 6 numbers Sum = Mean × Number of values Sum = 40 × 6 = 240
Step 2: Find sum of remaining 3 numbers Sum of remaining 3 = 240 - 90 = 150
Step 3: Find mean of remaining 3 Mean = 150/3 = 50
(a) Calculate the area of the shaded portion in Fig. 2.1 [Figure shows a sector with radius 8 cm and angle 54°]
(b) A tin of milk has a radius 6.5 cm and height 13 cm. Find, correct to 2 decimal places, the:
SOLUTION:
(a) Area of shaded sector:
Given: r = 8 cm, θ = 54°
Formula: Area of sector = (θ/360°) × πr²
Substituting: Area = (54°/360°) × π × 8² Area = (54/360) × π × 64 Area = (3/20) × 64π Area = 192π/20 Area = 9.6π
Using π = 22/7: Area = 9.6 × 22/7 = 211.2/7 = 30.17 cm²
(b) Cylindrical tin:
Given: r = 6.5 cm, h = 13 cm
i. Total surface area: Formula: TSA = 2πr² + 2πrh
Calculating each part:
Total: TSA = 169π + 84.5π = 253.5π
Using π = 22/7: TSA = 253.5 × 22/7 = 5577/7 = 796.71 cm²
ii. Volume in litres: Formula: V = πr²h
Calculating: V = π(6.5)²(13) V = π(42.25)(13) V = 549.25π
Using π = 22/7: V = 549.25 × 22/7 = 12083.5/7 = 1726.21 cm³
Converting to litres: 1 litre = 1000 cm³ V = 1726.21/1000 = 1.73 litres
(a) An aircraft took off from an airstrip with an average speed of 35 km/h on a bearing of 015° for 2 hours. It then changed course and flew on a bearing of 100° with an average speed of 22 km/h for another 2½ hours. Find its:
(b) Differentiate (1+x)(1+2x²)/x with respect to x
SOLUTION:
(a) Aircraft navigation:
Given:
i. Distance from starting point:
Step 1: Calculate distances
Step 2: Find angle between paths Angle = 100° - 15° = 85°
Step 3: Use law of cosines d² = d₁² + d₂² - 2d₁d₂cos(85°) d² = 70² + 55² - 2(70)(55)cos(85°) d² = 4900 + 3025 - 7700cos(85°) d² = 7925 - 7700(0.0872) d² = 7925 - 671.44 d² = 7253.56 d = 85.17 km
ii. Bearing from airstrip:
Using law of sines: sin A/55 = sin 85°/85.17 sin A = 55 × sin 85°/85.17 sin A = 55 × 0.9962/85.17 sin A = 0.6430 A = 39.94° ≈ 40°
Final bearing = 15° + 40° = 55°
(b) Differentiate (1+x)(1+2x²)/x:
Step 1: Expand numerator (1+x)(1+2x²) = 1 + 2x² + x + 2x³ = 1 + x + 2x² + 2x³
Step 2: Rewrite function y = (1 + x + 2x² + 2x³)/x y = 1/x + x/x + 2x²/x + 2x³/x y = x⁻¹ + 1 + 2x + 2x²
Step 3: Differentiate dy/dx = -1·x⁻² + 0 + 2·1 + 2·2x dy/dx = -x⁻² + 2 + 4x dy/dx = -1/x² + 2 + 4x
(a) Using a ruler and a pair of compasses only, construct:
(b) Label the point where L₁ and L₂ intersect as X and measure |QX|
SOLUTION:
(a) Construction steps:
i. Construct triangle PQR:
ii. Construct locus L₁ (perpendicular bisector of PQ):
iii. Construct locus L₂ (angle bisector of ∠QPR):
(b) Measure |QX|: Mark intersection point of L₁ and L₂ as X Using ruler, measure distance from Q to X Answer: |QX| = [measurement value from actual construction]
(a) Copy and complete Table 2.2 for y = 2x² - 4x - 3
Table 2.2: Table of values
| x | -2 | -1 | 0 | 1 | 2 | 3 | 4 | 5 | 6 |
|---|---|---|---|---|---|---|---|---|---|
| y | 3 | -3 | 45 |
(b) Using a scale of 2 cm to 1 unit on x-axis and 2 cm to 5 units on y-axis, draw the graph of y = 2x² - 4x - 3 for -2 ≤ x ≤ 6
(c) Use your graph to find the:
SOLUTION:
(a) Complete table for y = 2x² - 4x - 3:
For each x value, substitute into equation:
x = -2: y = 2(-2)² - 4(-2) - 3 = 2(4) + 8 - 3 = 8 + 8 - 3 = 13
x = -1: y = 2(-1)² - 4(-1) - 3 = 2(1) + 4 - 3 = 2 + 4 - 3 = 3 ✓
x = 0: y = 2(0)² - 4(0) - 3 = 0 - 0 - 3 = -3
x = 1: y = 2(1)² - 4(1) - 3 = 2 - 4 - 3 = -5
x = 2: y = 2(2)² - 4(2) - 3 = 8 - 8 - 3 = -3 ✓
x = 3: y = 2(3)² - 4(3) - 3 = 18 - 12 - 3 = 3
x = 4: y = 2(4)² - 4(4) - 3 = 32 - 16 - 3 = 13
x = 5: y = 2(5)² - 4(5) - 3 = 50 - 20 - 3 = 27
x = 6: y = 2(6)² - 4(6) - 3 = 72 - 24 - 3 = 45 ✓
Completed table:
| x | -2 | -1 | 0 | 1 | 2 | 3 | 4 | 5 | 6 |
|---|---|---|---|---|---|---|---|---|---|
| y | 13 | 3 | -3 | -5 | -3 | 3 | 13 | 27 | 45 |
(b) Graph plotting: Using scale: 2cm = 1 unit on x-axis, 2cm = 5 units on y-axis Plot all coordinate points and draw smooth parabolic curve
(c) From graph:
i. Roots of 2x² - 4x - 3 = 4: This means solving 2x² - 4x - 3 = 4 Or 2x² - 4x - 7 = 0 Draw horizontal line y = 4 and find where it intersects the curve Reading from graph: x = -1.0, x = 4.0
ii. Gradient at x = 4: Method 1: Draw tangent line at point (4, 13) and calculate slope Method 2: Use calculus: dy/dx = d/dx(2x² - 4x - 3) = 4x - 4 At x = 4: dy/dx = 4(4) - 4 = 16 - 4 = 12
Table 2.3 shows the frequency distribution of marks obtained by 50 students in an examination.
Table 2.3: Frequency distribution
| Mark | 40-44 | 45-49 | 50-54 | 55-59 | 60-64 | 65-69 | 70-74 | 75-79 |
|---|---|---|---|---|---|---|---|---|
| Frequency | 3 | 5 | 8 | 11 | 9 | 7 | 5 | 2 |
(a) Using Table 2.3, construct the cumulative frequency table for the distribution and draw the cumulative frequency curve
(b) Use the cumulative frequency curve in 12.(a) to find the:
(c) Determine the mode of the distribution
SOLUTION:
(a) Cumulative frequency table:
Step 1: Calculate cumulative frequencies
Cumulative frequency table: | Mark | 40-44 | 45-49 | 50-54 | 55-59 | 60-64