More Difficult Problems Revision Notes for Leaving Cert Mathematics

More Difficult Problems

When dealing with area and volume calculations, some problems require you to think beyond basic formula applications. These more challenging problems often involve flow rates, ratio comparisons, and composite shapes. Understanding these concepts will help you tackle complex real-world applications with confidence.

Flow rate problems

Flow rate refers to the volume of liquid that passes through a pipe or channel per unit of time. This creates an important relationship between the speed of flow and the volume displaced.

Understanding the flow rate concept

When water flows through a cylindrical pipe, the volume of water that passes through in one second equals the volume of a cylinder with the same radius as the pipe and a length equal to the distance the water travels in that time.

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For example, if water flows at 10 cm/sec through a pipe, then in one second, the volume of water equals the volume of a 10 cm length of that pipe. This visualisation helps you understand why we multiply the cross-sectional area by the flow speed.

Key formula for flow problems

The volume per second = $\pi \times r^2 \times \text{(flow rate in cm/sec)}$

Where $r$ is the radius of the pipe's internal diameter.

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Worked Example: Filling a Cylinder

Problem: Water flows through a cylindrical pipe of internal diameter 3 cm at the rate of 14 cm/sec. How long will it take to fill a cylinder of base diameter 14 cm and height 45 cm?

Step 1: Calculate the volume of water flowing per second

Radius of pipe = $3 \div 2 = 1.5$ cm
Volume per second = $\pi \times (1.5)^2 \times 14$
Volume per second = $\pi \times 2.25 \times 14 = \frac{63\pi}{2}$ cm³

Step 2: Calculate the volume of the cylinder to be filled

Radius of cylinder = $14 \div 2 = 7$ cm
Volume = $\pi \times (7)^2 \times 45 = 2205\pi$ cm³

Step 3: Calculate the time needed

Time = Total volume ÷ Volume per second
Time = $2205\pi \div \frac{63\pi}{2} = 2205\pi \times \frac{2}{63\pi} = 70$ seconds

Ratio problems when dimensions are not given

Sometimes you need to compare volumes without knowing the exact measurements. Instead, you're given ratios of corresponding lengths. This technique is particularly useful for comparing similar shapes.

The key principle

When linear dimensions are in a certain ratio, volumes are in the ratio of the cubes of those linear dimensions.

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Critical Rule: If two similar shapes have corresponding lengths in the ratio $a:b$ , then their volumes are in the ratio $a^3:b^3$ .

This is because volume involves three dimensions, so each dimension contributes to the scaling factor.

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Worked Example: Comparing Cylinder and Cone Volumes

Problem: A cylinder and a cone have radii of equal lengths. If the height of the cone is twice the height of the cylinder, find the ratio Volume of cylinder : Volume of cone.

Step 1: Define the variables

Let radius of each figure = $r$
Height of cylinder = $h$
Height of cone = $2h$

Step 2: Calculate both volumes

Volume of cylinder = $\pi r^2 h$
Volume of cone = $\frac{1}{3}\pi r^2 \times (2h) = \frac{2\pi r^2 h}{3}$

Step 3: Form the ratio

Volume of cylinder : Volume of cone = $\pi r^2 h : \frac{2\pi r^2 h}{3}$
Simplify by dividing both sides by $\pi r^2$ : $h : \frac{2h}{3}$
Multiply both sides by 3: $3h : 2h$
Final ratio = 3:2

Complex composite shape problems

More challenging problems often involve composite shapes - combinations of basic 3D shapes like cones sitting on hemispheres, or cylinders with hemispherical ends.

For these problems, you need to develop a systematic approach that breaks down complex shapes into manageable components.

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Key Strategies for Composite Shapes:

Break down complex shapes into familiar components (cylinders, cones, hemispheres, spheres)
Use the same radius for connecting parts (e.g., cone sitting on hemisphere)
Apply conservation principles (e.g., volume is conserved when shapes are melted and recast)
Set up equations when volumes are equal or in given ratios

Remember: For composite shapes, you add volumes of separate components. For hollow sections, you subtract the inner volume from the outer volume.

The key to success with composite shapes is recognising that complex problems become much simpler when you identify each component and calculate systematically.

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Essential Volume Formulas

Make sure you know these formulas by heart:

Cylinder: $V = \pi r^2 h$
Cone: $V = \frac{1}{3}\pi r^2 h$
Sphere: $V = \frac{4}{3}\pi r^3$
Hemisphere: $V = \frac{2}{3}\pi r^3$

Flow Rate Calculations:

Volume per second = $\pi \times r^2 \times \text{(speed in cm/sec)}$
Time to fill = $\frac{\text{Total volume}}{\text{Volume per second}}$
Always check units - convert between cm³ and litres when needed (1 litre = 1000 cm³)

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Key Points to Remember:

Flow rate problems: Volume per second equals the volume of a pipe segment with length equal to the flow speed
Ratio problems: When linear dimensions are in ratio $a:b$ , volumes are in ratio $a^3:b^3$
Composite shapes: Break complex shapes into familiar components and calculate each volume separately
Keep your formulas ready: Memorise the volume formulas for cylinder, cone, sphere, and hemisphere
Unit conversion: Remember that 1 litre = 1000 cm³ for practical problems involving water flow

More Difficult Problems (Leaving Cert Mathematics): Revision Notes