JSS 3 Mathematics Scheme of Work for Second Term

Mathematics at the Junior Secondary School (JSS) level builds a foundation that supports not just academic learning, but real-life problem-solving skills. In the second term, students will focus on key mathematical concepts ranging from solving simultaneous linear equations to exploring the relationship between volume and capacity. This article provides a comprehensive breakdown of the second-term scheme of work for JSS 3, explaining each topic in detail with examples for better understanding.

Scheme of Work for JSS 3 Second Term

1. Revision of Work Done in the First Term

Before diving into new topics, it’s crucial for students to revisit concepts covered in the first term. This provides a foundation for understanding more advanced topics. Students should review the basics of algebra, number systems, geometry, and other areas to refresh their memory.

2. Simultaneous Linear Equations: Graphical Solution

Concept Explanation: Simultaneous linear equations are two or more linear equations that share common variables. The solution to these equations is the point where the lines intersect when plotted on a graph.

Key Learning Points:

• Plotting the graphs of linear equations.
• Compilation of tables of values for linear equations.
• Graphical solution of simultaneous linear equations in two variables.

Examples:

1. Solve the system of equations:
   $2x+3y=6$
   $x-y=1$
2. Graph the equations and find the point of intersection.
3. Determine the coordinates where the lines meet.
4. Use a table of values to plot points for the equation $y=2x+1$.
5. Solve the system using graphical methods for equations $3x+2y=8$ and $x-y=2$.
6. Use graphing to find the solution to $4x-y=3$ and $x+y=7$.

3. Solving Simultaneous Linear Equations by Substitution and Elimination

Concept Explanation: Simultaneous equations can be solved using two primary methods: substitution and elimination. The substitution method involves solving one equation for one variable and substituting it into the other equation. The elimination method involves adding or subtracting the equations to eliminate one variable.

Key Learning Points:

• Substitution method for solving simultaneous equations.
• Elimination method for solving simultaneous equations.
• Application of word problems.

Examples:

1. Solve the system by substitution:
   $x+y=5$
   $x-y=1$
2. Solve the system by elimination:
   $2x+3y=10$
   $4x-2y=6$
3. Solve for $x$ and $y$ in the equations $3x+4y=12$ and $x-2y=2$.
4. Word problem: Two trains leave at the same time from different stations. If the distance between them is described by two linear equations, find their point of meeting.
5. Word problem: A school sells tickets for a concert. If the total revenue from ticket sales is represented by simultaneous equations, find the number of tickets sold.
6. Solve the system of equations $x+2y=9$ and $2x-y=4$ using both methods.

4. Variations

Concept Explanation: Variation refers to how one quantity changes in relation to another. There are different types of variations:

• Direct Variation: $y\propto x$, meaning $y=kx$.
• Inverse Variation: y∝1xy \propto \frac{1}{x}y∝x1​, meaning y=kxy = \frac{k}{x}y=xk​.
• Joint Variation: $y\propto xz$, meaning $y=kxz$.
• Partial Variation: Combination of direct and inverse variations.

Key Learning Points:

• Understanding direct, inverse, joint, and partial variation.
• Formulas and real-life examples of variations.

Examples:

1. Direct variation: If $y=3x$, find $y$ when $x=5$.
2. Inverse variation: If y=12xy = \frac{12}{x}y=x12​, find $y$ when $x=4$.
3. Joint variation: If $y=4xz$, find $y$ when $x=2$ and $z=3$.
4. Partial variation: If y=2x+5xy = 2x + \frac{5}{x}y=2x+x5​, find $y$ when $x=3$.
5. Solve for $k$ in the equation $y=kx$, given that $y=8$ when $x=2$.
6. Solve for $k$ in the inverse variation y=kxy = \frac{k}{x}y=xk​, given that $y=5$ when $x=2$.

5. Construction

Concept Explanation: Construction involves using geometrical tools like compasses and rulers to draw shapes accurately. This section focuses on constructing line segments, angles, and geometric shapes.

Key Learning Points:

• Bisection of line segments and angles.
• Construction of triangles and quadrilaterals.

Examples:

1. Bisect a line segment of 6 cm using a compass and ruler.
2. Construct a right-angled triangle with sides 3 cm, 4 cm, and 5 cm.
3. Bisect an angle of 60°.
4. Construct a quadrilateral given the lengths of the sides.
5. Construct a regular hexagon using a compass.
6. Bisect the angle between two intersecting lines.

6. Similar Shapes and Enlargement

Concept Explanation: Similar shapes have the same shape but differ in size. The enlargement and scale factor refer to resizing shapes while maintaining their proportionality.

Key Learning Points:

• Identifying similar shapes.
• Understanding enlargement and scale factors.

Examples:

1. Determine if two triangles are similar.
2. Enlarge a square with a scale factor of 2.
3. Find the scale factor of an enlargement of a rectangle.
4. Determine the missing side length of a similar triangle.
5. Solve for the new area of a shape after enlargement.
6. Find the volume of an enlarged shape using the scale factor.

7. Similar Shapes: Scale Factor in Calculating Lengths, Areas, and Volumes

Concept Explanation: The scale factor applies not just to lengths, but also to areas and volumes. When shapes are enlarged, the lengths, areas, and volumes increase proportionally.

Key Learning Points:

• Scale factor applied to lengths, areas, and volumes.
• Calculating the new area and volume after enlargement.

Examples:

1. Calculate the area of a square after enlargement with a scale factor of 3.
2. Find the volume of a cone after enlargement with a scale factor of 2.
3. Determine the new length of a side of a cube after a scale factor of 4.
4. Find the new area of a rectangle after enlargement.
5. Solve for the area of a triangle after applying a scale factor of 2.
6. Calculate the volume of a cylinder after enlargement with a scale factor of 2.

8. Measurement of Solids

Concept Explanation: This topic deals with calculating the area and volume of 3D shapes like cubes, cuboids, cylinders, and cones. It also covers the relationship between volume and capacity.

Key Learning Points:

• Surface areas and volumes of cubes, cuboids, cylinders, and cones.
• Understanding the relationship between volume and capacity.

Examples:

1. Calculate the surface area of a cube with side length 4 cm.
2. Find the volume of a cuboid with dimensions 3 cm, 4 cm, and 5 cm.
3. Calculate the curved surface area of a cone with radius 3 cm and height 4 cm.
4. Determine the volume of a cylinder with radius 2 cm and height 6 cm.
5. Convert the volume of a solid from cubic centimeters to liters.
6. Solve for the total surface area of a sphere with radius 5 cm.

Conclusion

The JSS 3 second-term mathematics curriculum provides a rich and engaging study of fundamental mathematical concepts that will help students understand and apply these ideas in real-world scenarios. By following this scheme of work, students will develop critical thinking, problem-solving, and reasoning skills.