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 Amsco Integrated Algebra I: Verbal Problems Involving Ratio
This is a free lesson from our course in Amsco's Integrated Algebra
This lesson explains the basics and approach to solve the real-world problems involving ratio. E.g. car engine is run on a petrol-ethanol mixture typically in a ratio of 9:1. The "9:1" means that for each nine units of petrol, 1 unit of ethanol is used. If you had 9 litres of petrol and 1 litre of ethanol, the mixture would measure 10 litres. It also means as fractions, the proportion of each liquid is 9/10 petrol and 1/10 ethanol. Remember: if the ratio of two or more numbers is known, you can use the terms of the ratio and a nonzero variable, x, to express the numbers. (More text below video...)
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(Continued from above) Any two numbers in the ratio a: b can be written as ax and bx where x is a nonzero real number. For example: The perimeter of the triangle is 72 feet and the sides are in the ratio 3:4:5. You are required to calculate the length of each side of the triangle. The solution can be worked following the steps below:
Let 3x = the length of the first side,
4x = the length of the second side,
5x = the length of the third side.
Given: The perimeter of the triangle is 72 feet.
3x+ 4x+ 5x = 72
12x = 72
x = 6
3x = 3(6) = 18
4x = 4(6) = 24
5x = 5(6) = 30
The final answer is - the lengths of the sides are 18 feet, 24 feet, and 30 feet. You can check the answer: 18: 24:30 = 3: 4: 5 and 18 + 24 + 30 =72
Take a look at the video above for detailed explanations of Ratios.

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