MATHS SOLUTION-FREE HIGH SCHOOL/SSCE ADAPTED -SOLUTIONS TO QUESTIONS
SSCE Mathematics Adapted Exam Solutions
Solutions:
Question 1: Simultaneous Equations with a 3x3 Matrix
Solve the following system of linear equations:
2x + y - z = 8
3x - y + 2z = 7
x + 4y + 3z = 10
The matrix form of the system is:
| 2 1 -1 | | x | = | 8 |
| 3 -1 2 | | y | = | 7 |
| 1 4 3 | | z | = | 10 |
We can solve this system using matrix methods (Inverse of Matrix or Cramer's Rule).
Solution steps:
- Write the coefficient matrix, variable matrix, and constant matrix.
- Find the inverse of the coefficient matrix.
- Multiply the inverse matrix by the constant matrix to find x, y, and z.
The solution to the system is:
x = 2, y = 1, z = 3
Question 2: Bearings
Two ships, A and B, are at sea. Ship A is at a position P, and ship B is at a position Q. The bearing of ship B from ship A is 040° and the bearing of ship A from ship B is 220°. Calculate the angle between their positions.
Solution:
- Bearings are measured clockwise from the North.
- To find the angle between them, we subtract the bearing of ship A from the bearing of ship B:
Angle = 220° - 40° = 180°
The angle between ship A and ship B is 180°.
Question 3: Trigonometry - Cosine Rule
In triangle ABC, angle B = 60°, side a = 8 cm, and side b = 10 cm. Find the length of side c using the cosine rule.
Cosine rule formula:
c² = a² + b² - 2ab * cos(C)
Substituting the given values:
c² = 8² + 10² - 2 * 8 * 10 * cos(60°)
c² = 64 + 100 - 160 * 0.5
c² = 164 - 80
c² = 84
c = √84 ≈ 9.17 cm
The length of side c is approximately 9.17 cm.
Question 4: Bearings and Distance
Two towns, X and Y, are situated 50 km apart. Town X is located on a bearing of 045° from town Y. A third town, Z, is located 60 km from X on a bearing of 120°. Calculate the bearing of town Z from town Y.
Solution:
- We can first find the angle between lines XY and XZ.
- Using the bearing of XZ (120°) and the bearing of XY (045°), we find the angle between the lines as:
Angle between XY and XZ = 120° - 45° = 75°
- To find the bearing of Z from Y, we add this angle to the bearing of XY:
Bearing of Z from Y = 045° + 75° = 120°
The bearing of town Z from town Y is 120°.
Marking Scheme:
- Question 1: Solution of simultaneous equations using matrix methods - 20 marks
- Question 2: Bearings and calculation of angle - 15 marks
- Question 3: Trigonometry using cosine rule - 20 marks
- Question 4: Bearings and distance - 15 marks
Total: 70 marks
Timing: 120 minutes
SSCE Mathematics adapted Theory Question with Solutions: DIsclaimer: we disclaim all liability arising from reliance on this contact us on 234 8067922530 for detailed solution
1. Longitude and Latitude:
A ship is located at a point with coordinates 30°N, 45°E. If the ship sails 100 km due north, calculate the new coordinates of the ship. Assume the Earth is a perfect sphere and use the approximation that 1 degree of latitude is approximately 111 km.
Solution: Since 1° of latitude ≈ 111 km, the change in latitude for 100 km is:
New Latitude = 30°N + (100 km / 111 km) ≈ 30°N + 0.9009° ≈ 30.9°N
New Coordinates: 30.9°N, 45°E
2. Logarithm:
Given the logarithm table, if log10(2) ≈ 0.301 and log10(5) ≈ 0.699, calculate log10(20) using the properties of logarithms.
Solution: Using the property: log10(20) = log10(2) + log10(10), we know that log10(10) = 1.
So, log10(20) = 0.301 + 1 = 1.301.
3. Simultaneous Equations (Elimination Method):
Solve the following simultaneous equations using the elimination method:
3x + 4y = 10
2x - 5y = -4
Solution: Multiply the first equation by 2 and the second by 3:
6x + 8y = 20
6x - 15y = -12
Subtracting the two equations gives:
23y = 32 → y = 32 / 23
Substituting y back into the first equation gives:
x = (10 - 4(32 / 23)) / 3.
4. Simultaneous Equations (Substitution Method):
Solve the following simultaneous equations using the substitution method:
x + 2y = 7
3x - y = 8
Solution: From the first equation, express x:
x = 7 - 2y.
Substituting into the second equation:
3(7 - 2y) - y = 8 → 21 - 6y - y = 8 → -7y = -13 → y = 13 / 7.
Substituting back to find x gives x = 7 - 2(13 / 7).
5. Matrix Method (2x2 and 3x3):
a) Solve the simultaneous equations using a matrix method:
x + 2y = 3
2x + 3y = 5
Solution: The augmented matrix is:
| 1 2 | 3
| 2 3 | 5 |
Using row operations, solve for x and y.
b) Solve the following system of equations using a 3x3 matrix:
x + y + z = 6
2x - y + 3z = 14
-x + 4y + z = 2
Solution: Set up the augmented matrix:
| 1 1 1 | 6
| 2 -1 3 | 14
| -1 4 1 | 2 |
Use row reduction to solve for x, y, and z.
Timing: 60 minutes
SSCE adapted Mathematics Theory Question with Solutions2 :DIsclaimer: we disclaim all liability arising from reliance on this solution contact us on 234 8067922530 for detailed solution
1. Bearings:
A boat leaves port A and sails 30 km on a bearing of 045°. It then changes course and sails 40 km on a bearing of 120°.
a) Calculate the coordinates of the boat's final position relative to port A.
b) Find the bearing from port A to the boat's final position.
Solution:
First leg: Sails 30 km at 045°:
Coordinates after first leg: (30cos(45°), 30sin(45°)) = (21.21, 21.21)
Second leg: Sails 40 km at 120°:
Coordinates after second leg: (40cos(120°), 40sin(120°)) = (-20, 34.64)
Total coordinates from A: (21.21 - 20, 21.21 + 34.64) = (1.21, 55.85)
Bearing from A to final position: arctan(55.85 / 1.21) = 87.3°.
2. Quadratic Equation:
The height (h) in meters of an object thrown upwards is given by the equation h(t) = -5t2 + 20t + 10, where t is the time in seconds.
a) Find the time when the object reaches its maximum height.
b) Calculate the maximum height of the object.
Solution:
Maximum height occurs at vertex, t = -b/(2a) = -20/(2 * -5) = 2 seconds.
Maximum height: h(2) = -5(2)2 + 20(2) + 10 = 30 meters.
3. Indices:
Simplify the expression (a3 · a-5) / a-2.
Solution:
Using properties of indices: (a3 · a-5) = a3 - 5 = a-2.
Now, divide: a-2 / a-2 = a-2 - (-2) = a0 = 1.
4. Logarithm:
If log2(x) + log2(3) = 5, solve for x.
Solution:
Using properties of logarithms: log2(x * 3) = 5.
Converting from logarithmic to exponential form: x * 3 = 25 = 32.
Thus, x = 32 / 3 ≈ 10.67.
Timing: 45 minutes