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Number - Ratio, Proportion, and Rate

Grade 9IGCSE

Review the key concepts, formulae, and examples before starting your quiz.

🔑Concepts

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Simplifying Ratios: Expressing a ratio in its simplest form by dividing all terms by their highest common factor.

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Dividing an Amount in a Given Ratio: Finding the value of one 'part' by dividing the total amount by the sum of the ratio parts.

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Map Scales: Understanding ratios in the form 1 : n, where 1 unit on the map represents n units in real life.

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Direct Proportion: Two quantities increase or decrease at the same rate (y=kxy = kx).

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Inverse Proportion: As one quantity increases, the other decreases (y=k/xy = k/x).

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Compound Measures: Understanding the relationship between different units, specifically Speed, Density, and Pressure.

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Unit Conversions: Converting between different rates (e.g., km/h to m/s).

📐Formulae

Share=Total AmountSum of Ratio Parts×Specific Part\text{Share} = \frac{\text{Total Amount}}{\text{Sum of Ratio Parts}} \times \text{Specific Part}

Direct Proportion: y=kx\text{Direct Proportion: } y = kx

Inverse Proportion: y=kx\text{Inverse Proportion: } y = \frac{k}{x}

Speed=DistanceTime\text{Speed} = \frac{\text{Distance}}{\text{Time}}

Density=MassVolume\text{Density} = \frac{\text{Mass}}{\text{Volume}}

Pressure=ForceArea\text{Pressure} = \frac{\text{Force}}{\text{Area}}

💡Examples

Problem 1:

Divide $240 in the ratio 3 : 5.

Solution:

90and90 and 150

Explanation:

First, find the total number of parts: 3+5=83 + 5 = 8. Find the value of one part: 240/8=30240 / 8 = 30. Multiply each ratio part by this value: 3×30=903 \times 30 = 90 and 5×30=1505 \times 30 = 150.

Problem 2:

y is directly proportional to the square of x. When x = 3, y = 18. Find y when x = 5.

Solution:

y = 50

Explanation:

Start with the equation y=kx2y = kx^2. Substitute the known values: 18=k(32)⇒18=9k⇒k=218 = k(3^2) \Rightarrow 18 = 9k \Rightarrow k = 2. Now use the formula y=2x2y = 2x^2 for x=5x = 5: y=2(52)=2×25=50y = 2(5^2) = 2 \times 25 = 50.

Problem 3:

A map has a scale of 1 : 50,000. If the distance between two towns on the map is 8 cm, calculate the real-world distance in kilometers.

Solution:

4 km

Explanation:

Actual distance = 8 cm×50,000=400,000 cm8 \text{ cm} \times 50,000 = 400,000 \text{ cm}. Convert cm to meters: 400,000/100=4,000 m400,000 / 100 = 4,000 \text{ m}. Convert meters to km: 4,000/1,000=4 km4,000 / 1,000 = 4 \text{ km}.

Problem 4:

A car travels 150 km in 2 hours and 30 minutes. Calculate its average speed in km/h.

Solution:

60 km/h

Explanation:

Convert time to decimal hours: 2 hours 30 mins = 2.5 hours. Use the formula Speed=Distance/Time=150/2.5=60 km/h\text{Speed} = \text{Distance} / \text{Time} = 150 / 2.5 = 60 \text{ km/h}.