Paper 1: Mathematical and Scientific Content — Gear Ratios, Mechanical Advantage, Energy and Forces

AQA GCSE D&T Paper 1 requires a minimum of 15% mathematical content and 10% scientific content. This lesson covers the key calculations and scientific principles you need to know, with worked examples and exam-style practice. Specification reference: AQA 8552, Sections 3.1.4 and 3.1.5.

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Gear Ratios

What Is a Gear Ratio?

A gear ratio describes the relationship between two meshing gears. It tells you how many times the driven gear rotates for each rotation of the driver gear.

Formula:

$$\text{Gear ratio} = \frac{\text{Number of teeth on driven gear}}{\text{Number of teeth on driver gear}}$$Gear ratio=Number of teeth on driver gearNumber of teeth on driven gear​

Worked Example 1

A driver gear has 20 teeth and a driven gear has 60 teeth.

Gear ratio = 60 ÷ 20 = 3:1

This means the driven gear rotates once for every three rotations of the driver gear. The driven gear turns more slowly but with more torque (turning force).

Worked Example 2

A driver gear has 40 teeth and a driven gear has 10 teeth.

Gear ratio = 10 ÷ 40 = 1:4 (or 0.25:1)

The driven gear rotates four times for every one rotation of the driver gear. This is a speed increase but with less torque.

Output Speed Calculation

$$\text{Output speed (RPM)} = \frac{\text{Input speed (RPM)}}{\text{Gear ratio}}$$Output speed (RPM)=Gear ratioInput speed (RPM)​

Example: Input speed = 300 RPM, gear ratio = 3:1. Output speed = 300 ÷ 3 = 100 RPM.

Gear Trains

When multiple pairs of gears are used, the overall gear ratio is the product of the individual ratios.

Example: Stage 1 has a ratio of 3:1, Stage 2 has a ratio of 2:1. Overall ratio = 3 × 2 = 6:1. If the input speed is 600 RPM, the output speed = 600 ÷ 6 = 100 RPM.

Gear Type	Effect
Spur gears	Transfer motion between parallel shafts
Bevel gears	Transfer motion through 90°
Worm and worm wheel	Large speed reduction; cannot be reversed (self-locking)
Rack and pinion	Convert rotary motion to linear motion (e.g. steering)

AQA Exam Tip: Always show your working in gear ratio questions. If you make an arithmetic error, you can still earn marks for the correct method. Write the formula first, then substitute the values.

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Mechanical Advantage

What Is Mechanical Advantage?

Mechanical advantage (MA) measures how much a machine multiplies the input force. A lever, pulley or gear system can provide a mechanical advantage, allowing a small effort to move a large load.

Formula:

$$\text{Mechanical Advantage} = \frac{\text{Load}}{\text{Effort}}$$Mechanical Advantage=EffortLoad​

Worked Example

A lever is used to lift a rock weighing 600 N. The effort applied is 150 N.

MA = 600 ÷ 150 = 4

This means the lever multiplies the effort by a factor of 4.

Levers

Class	Fulcrum Position	Example
Class 1	Between effort and load	Seesaw, crowbar, scissors
Class 2	Load between effort and fulcrum	Wheelbarrow, nutcracker, bottle opener
Class 3	Effort between load and fulcrum	Tweezers, fishing rod, human forearm

Pulleys

System	Mechanical Advantage	Description
Single fixed pulley	MA = 1	Changes direction of force only
Single movable pulley	MA = 2	Halves the effort needed
Block and tackle (4 ropes)	MA = 4	Quarters the effort needed

Key trade-off: Increasing mechanical advantage means the effort moves through a greater distance to move the load a shorter distance. You gain force but lose distance (and speed).

AQA Exam Tip: In lever and pulley questions, draw a diagram and label the load, effort and fulcrum (or ropes). This helps you visualise the system and reduces errors.

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Velocity Ratio

The velocity ratio (VR) of a machine is calculated differently depending on the system:

$$\begin{aligned}\text{VR (lever)} &= \frac{\text{Distance from effort to fulcrum}}{\text{Distance from load to fulcrum}} \\[6pt] \text{VR (pulley)} &= \text{Number of ropes supporting the load} \\[6pt] \text{VR (gears)} &= \frac{\text{Teeth on driven gear}}{\text{Teeth on driver gear}}\end{aligned}$$VR (lever)VR (pulley)VR (gears)​=Distance from load to fulcrumDistance from effort to fulcrum​=Number of ropes supporting the load=Teeth on driver gearTeeth on driven gear​​

Efficiency

No machine is 100% efficient due to friction. Efficiency links MA and VR:

$$\text{Efficiency (\%)} = \left(\frac{\text{MA}}{\text{VR}}\right) \times 100$$Efficiency (%)=(VRMA​)×100

Example: A pulley system has MA = 3.6 and VR = 4. Efficiency = (3.6 ÷ 4) × 100 = 90%

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Energy Calculations

Electrical Energy

$$\text{Energy (J)} = \text{Power (W)} \times \text{Time (s)}$$Energy (J)=Power (W)×Time (s)

Example: A kettle has a power rating of 2,000 W. How much energy does it use in 3 minutes?

Energy = 2,000 × (3 × 60) = 2,000 × 180 = 360,000 J (or 360 kJ)

Electrical Power

$$\text{Power (W)} = \text{Voltage (V)} \times \text{Current (A)}$$Power (W)=Voltage (V)×Current (A)

Example: A motor operates at 12 V and draws 2.5 A. Power = 12 × 2.5 = 30 W

Cost of Energy

$$\begin{aligned}\text{Energy (kWh)} &= \text{Power (kW)} \times \text{Time (hours)} \\[6pt] \text{Cost} &= \text{Energy (kWh)} \times \text{Price per kWh}\end{aligned}$$Energy (kWh)Cost​=Power (kW)×Time (hours)=Energy (kWh)×Price per kWh​

Example: A 3 kW oven is used for 2 hours. Electricity costs 30p per kWh. Energy = 3 × 2 = 6 kWh Cost = 6 × 30p = 180p = £1.80

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Forces and Stress

Types of Force

Force	Description	Example
Tension	Pulling/stretching force	A rope in a tug-of-war
Compression	Pushing/squashing force	The legs of a table supporting weight
Shear	Opposing forces sliding past each other	Scissors cutting paper
Torsion	Twisting force	Turning a screwdriver
Bending	Combination of tension and compression	A shelf loaded with books

Stress, Strain and Young's Modulus (Higher Tier Context)

$$\begin{aligned}\text{Stress (Pa)} &= \frac{\text{Force (N)}}{\text{Cross-sectional area (m}^{2}\text{)}} \\[6pt] \text{Strain} &= \frac{\text{Extension (m)}}{\text{Original length (m)}} \\[6pt] \text{Young's Modulus (Pa)} &= \frac{\text{Stress}}{\text{Strain}}\end{aligned}$$Stress (Pa)StrainYoung’s Modulus (Pa)​=Cross-sectional area (m2)Force (N)​=Original length (m)Extension (m)​=StrainStress​​

Example: A steel wire has a cross-sectional area of 2 × 10⁻⁶ m². A force of 100 N is applied. Stress = 100 ÷ (2 × 10⁻⁶) = 50,000,000 Pa = 50 MPa

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Practice Calculations

Question 1: Gear Ratio

A driver gear has 15 teeth and meshes with a driven gear of 45 teeth. The driver rotates at 900 RPM. Calculate the gear ratio and the output speed.

Answer: Gear ratio = 45 ÷ 15 = 3:1 Output speed = 900 ÷ 3 = 300 RPM

Question 2: Mechanical Advantage

A wheelbarrow carries a load of 800 N. The effort applied is 200 N. Calculate the mechanical advantage.

Answer: MA = 800 ÷ 200 = 4

Question 3: Energy Cost

A 2.5 kW washing machine runs for 1.5 hours. Electricity costs 34p per kWh. Calculate the cost.

Answer: Energy = 2.5 × 1.5 = 3.75 kWh Cost = 3.75 × 34p = 127.5p = £1.28 (rounded to 2 decimal places)

Question 4: Electrical Power

An LED strip operates at 24 V and draws 1.2 A. Calculate the power.

Answer: Power = 24 × 1.2 = 28.8 W

AQA Exam Tip: In the exam, ALWAYS show your formula, substitution and answer. Even if your arithmetic is wrong, you will earn method marks. Write units clearly — J, W, N, Pa, RPM, kWh.

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Summary

Topic	Key Formula
Gear ratio	Driven teeth ÷ Driver teeth
Output speed	Input speed ÷ Gear ratio
Mechanical advantage	Load ÷ Effort
Efficiency	(MA ÷ VR) × 100
Energy (electrical)	Power × Time
Power	Voltage × Current
Energy cost	Power (kW) × Time (h) × Price per kWh
Stress	Force ÷ Area

Paper 1 includes at least 15 marks of mathematical content. Practise these calculations until they are automatic — they are guaranteed marks if you know the formulae and show your working.

AQA Exam Tip: Create a formula card with all the key formulae and practise one calculation from each topic every day in the lead-up to the exam.

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Worked Example: Past-Paper-Style Multi-Part Calculation