krit.club logo

Number - Ratio, proportion, and rate

Grade 11IGCSE

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

🔑Concepts

•

Simplifying Ratios: Expressing ratios in their lowest terms and dividing quantities into a given ratio.

•

Direct Proportion: A relationship where as one variable increases, the other increases at a constant rate (y=kxy = kx or y=kxny = kx^n).

•

Inverse Proportion: A relationship where as one variable increases, the other decreases (y=kxy = \frac{k}{x} or y=kxny = \frac{k}{x^n}).

•

Rates: Comparison of two different quantities (e.g., Speed, Density, Fuel consumption).

•

Map Scales: Understanding linear scale factors (1:n) and their relationship to area and volume scales.

•

Currency Exchange: Converting between different currencies using a given exchange rate.

📐Formulae

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

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

Average Speed=Total DistanceTotal Time\text{Average Speed} = \frac{\text{Total Distance}}{\text{Total Time}}

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

Area Scale Factor=(Linear Scale Factor)2\text{Area Scale Factor} = (\text{Linear Scale Factor})^2

Volume Scale Factor=(Linear Scale Factor)3\text{Volume Scale Factor} = (\text{Linear Scale Factor})^3

💡Examples

Problem 1:

Divide $540 in the ratio 2:3:4.

Solution:

  1. Total parts = 2+3+4=92 + 3 + 4 = 9.
  2. Value of one part = 540/9=60540 / 9 = 60.
  3. Shares: 2×60=1202 \times 60 = 120, 3×60=1803 \times 60 = 180, 4×60=2404 \times 60 = 240. Result: 120,120, 180, $240.

Explanation:

To divide a quantity in a ratio, find the total number of parts first, determine the value of a single part, and then multiply by the specific parts of the ratio.

Problem 2:

yy is inversely proportional to the square of xx. When x=3,y=2x = 3, y = 2. Find yy when x=6x = 6.

Solution:

  1. Formula: y=kx2y = \frac{k}{x^2}.
  2. Find kk: 2=k32⇒2=k9⇒k=182 = \frac{k}{3^2} \Rightarrow 2 = \frac{k}{9} \Rightarrow k = 18.
  3. New equation: y=18x2y = \frac{18}{x^2}.
  4. Substitute x=6x = 6: y=1862=1836=0.5y = \frac{18}{6^2} = \frac{18}{36} = 0.5. Result: y=0.5y = 0.5.

Explanation:

Establish the general equation with the constant kk first, use given values to solve for kk, and then use the completed equation to find the unknown value.

Problem 3:

A map has a scale of 1:20,000. If a forest has an actual area of 4 km², find the area of the forest on the map in cm².

Solution:

  1. Linear Scale Factor (LSF) = 20,000.
  2. Area Scale Factor (ASF) = 20,0002=400,000,00020,000^2 = 400,000,000.
  3. Convert actual area to cm²: 4 km2=4×106 m2=4×1010 cm24 \text{ km}^2 = 4 \times 10^6 \text{ m}^2 = 4 \times 10^{10} \text{ cm}^2.
  4. Map Area = 40,000,000,000/400,000,000=100 cm240,000,000,000 / 400,000,000 = 100 \text{ cm}^2. Result: 100 cm².

Explanation:

When dealing with area on maps, you must square the linear scale factor before performing the conversion.