Maths642Updated 20 Oct 202620 pages

National 5 Applications of Maths 2023 Paper 2 Marked Example

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You've got an Applications of Mathematics exam paper here that covers loads of real-world maths problems. These questions show how you'll actually use maths in everyday situations like gardening, home improvements, and travel - making it way more relevant than you might think!
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Exam Paper Information

This is your N5 Applications of Mathematics Paper 2 from 2023. You've got 1 hour and 40 minutes to tackle all the questions, and you can definitely use a calculator - result!

The paper's worth 55 marks total, so that's roughly a mark per minute. Make sure you show all your working clearly because that's where you'll pick up marks even if your final answer isn't perfect.

Top tip: Always include units in your answers when they're needed. It's an easy way to lose marks if you forget them, and the questions will usually make it obvious what units you should use.

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Essential Formulae You Need

Here's your formula toolkit - these are the mathematical weapons you'll need for the exam! You don't need to memorise these as they're given to you, but knowing when to use each one is crucial.

The circle formulas (circumference and area) pop up loads in real-world problems. Pythagoras' theorem is your best mate for any right-angled triangle work, especially in construction and design questions.

Volume formulas for cylinders, cones, and spheres are dead useful for practical problems involving containers, building materials, or any 3D shapes. The standard deviation formula looks scary but it's just measuring how spread out your data is.

Remember: The gradient formula is simply rise over run - vertical distance divided by horizontal distance. Dead simple when you think about it as climbing a hill!

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Percentage Decrease Problems

This lake problem shows you how to handle compound percentage changes over multiple years. When something decreases by 2.8%, you're left with 97.2% of the original amount 100100% - 2.8% = 97.2%.

The key trick is converting percentages to decimals: 97.2% becomes 0.972. Then you multiply by this decimal three times because it's happening over three years.

So you get: 14,730,000 × 0.972 × 0.972 × 0.972 = 13,500,000 litres (to 3 significant figures). Notice how you round at the very end, not during your calculations.

Pro tip: For compound changes, you can use the power function on your calculator: 14,730,000 × (0.972)³ is much quicker!

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Complex Perimeter Calculations

This window problem combines rectangles and quarter circles - exactly the sort of mixed shape you'll see in real construction work. You need to find the total length of edging needed.

Start by identifying what makes up the perimeter: the straight edges of rectangles plus the curved edges of quarter circles. For the rectangles, you only count the outer edges that actually form the perimeter.

For the quarter circles, use the circumference formula C = πd, then divide by 4 since you only want a quarter. Add everything together: rectangle perimeters + quarter circle arcs = total edging length.

The working shown gives 7,399.11mm, which makes sense when you think about the size of a large window.

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Pythagoras in Garden Design

This fencing problem is Pythagoras in action - perfect for working out diagonal distances when you're planning layouts. You need to find the lengths AB and BC to calculate fencing costs.

For triangle with sides 5m and 7m: a² + b² = c², so 5² + 7² = 25 + 49 = 74, giving you c = √74 = 8.60m. That's your distance from A to B.

For the second triangle, you use the fact that one side is 21m total, then work backwards with Pythagoras to find the other distances. Total fencing needed = 27.75m, which needs 14 rolls at £22 each = £308.

Smart move: Always round UP when buying materials - you can't buy part of a roll!

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Complex Area Calculations

This mirror problem tests your ability to handle composite shapes - a square with four semicircles attached. You calculate each part separately, then add them together.

Square area = 1.2 × 1.2 = 1.44 m². For the semicircles, be careful with the radius - if diameter is 0.7m, then radius is 0.35m. Each semicircle area = (π × 0.35²) ÷ 2 = 0.1924 m².

Since you've got four identical semicircles, multiply by 4: 0.1924 × 4 = 0.7696 m². Total mirror area = 1.44 + 0.7696 = 2.21 m².

The key here is reading the question carefully - diameter vs radius trips up loads of students!

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Multi-Step Costing Problems

This mirror costing question tests your ability to extract information from tables and calculate total costs. You need to identify each component cost for a 3m² mirror.

From the table: 4mm silver glass = £38 per m², so 3 × £38 = £114. Anti-glare coating = £16 per m², so 3 × £16 = £48. Standard fixings = £32 (fixed cost), foil backing = £20 (fixed cost).

Total cost = £114 + £48 + £32 + £20 = £214. Notice how some costs are per square metre (glass, coating) while others are fixed per mirror (fixings, backing).

Watch out: Always check whether costs are per unit area or per item - it's easy to mix these up under exam pressure!

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Statistics: Mean and Standard Deviation

These chlorine level calculations show you practical statistics in action. The mean is straightforward: add all values 0.8+1.9+1.1+2.6+3.1+2.4+2.1=140.8 + 1.9 + 1.1 + 2.6 + 3.1 + 2.4 + 2.1 = 14 and divide by 7 = 2.

Standard deviation measures how spread out the data is. You need Σx = 14, Σx² = 32, and n = 7. Using the formula: s = √(32−142/7)/(7−1)(32 - 14²/7)/(7-1) = √4/64/6 = 0.816.

The key is being systematic with your calculations - make a table with x and x² columns to avoid mistakes. Standard deviation tells you about consistency - lower values mean more predictable chlorine levels.

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Comparing Data and Time Calculations

For data comparison, you need to compare both averages and spread. Colin's mean (2.2) is higher than Stuart's (2.0), so "Colin's hot tub has higher average chlorine levels." Colin's standard deviation (1.4) is higher than Stuart's (0.816), so "Colin's levels are more variable."

The time calculation uses inverse proportion: if 4 workers take 12 hours, that's 48 worker-hours total. With 5 workers: 48 ÷ 5 = 9.6 hours = 9 hours 36 minutes.

Starting at 08:00, add 9 hours 36 minutes work time plus 30 minutes lunch break = 18:06 (6:06 PM). Always include breaks in your time calculations!

Remember: In inverse proportion, more workers = less time, but the total work (worker-hours) stays constant.

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Currency Conversion Problems

This multi-step currency problem tests your ability to track money through different conversions. Start with £640 → Polish zlotys: £640 × 4.94 = 3,161.60 zlotys.

After spending 340 zlotys per day for 4 days (340 × 4 = 1,360 zlotys), Lorna has 3,161.60 - 1,360 = 1,801.60 zlotys remaining.

Convert back to pounds first: 1,801.60 ÷ 4.94 = £364.70, then convert to euros: £364.70 × 1.15 = €419.40.

Top tip: Always convert through pounds when moving between two foreign currencies - it's less likely to cause errors than trying to calculate direct exchange rates.

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