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SAT Math - Ratios and Proportions ReviewClick the link immediately below to view the SAT Verbal diagnostic test.
RATIOS & PROPORTIONS Ratio Proportion where k is a constant. The above equation shows that as x increases (or decreases) so does y. This simple concept has numerous applications in mathematics. For example, in constant velocity problems, distance is directly proportional to time: d = vt, where v is a constant. Note, sometimes the word directly is suppressed. Example: If the ratio of y to x is equal to 3 and the sum of y and x is 80, what is the value of y? Translating "the ratio of y to x is equal to 3" into an equation yields: y/x = 3 Translating "the sum of y and x is 80" into an equation yields: y + x = 80 Solving the first equation for y gives: y = 3x. Substituting this into the second equation yields 4x = 80 x = 20 Hence, y = 3x = 3(20) = 60. The answer is (E). In many word problems, as one quantity increases (decreases), another quantity also increases (decreases). This type of problem can be solved by setting up a direct proportion. Example: If Biff can shape 3 surfboards in 50 minutes, how many surfboards can he shape in 5 hours? As time increases so does the number of shaped surfboards. Hence, we set up a direct proportion. First, convert 5 hours into minutes: 5 hours = 5 x 60 minutes = 300 minutes. Next, let x be the number of surfboards shaped in 5 hours. Finally, forming the proportion yields 3(300)/50 = x 18 = x The answer is (C). If one quantity increases (or decreases) while another quantity decreases (or increases), the quantities are said to be inversely proportional. The statement "y is inversely proportional to x" is written as where k is a constant. Multiplying both sides of y = k/x by x yields Hence, in an inverse proportion, the product of the two quantities is constant. Therefore, instead of setting ratios equal, we set products equal. In many word problems, as one quantity increases (decreases), another quantity decreases (increases). This type of problem can be solved by setting up a product of terms. Example: If 7 workers can assemble a car in 8 hours, how long would it take 12 workers to assemble the same car? As the number of workers increases, the amount time required to assemble the car decreases. Hence, we set the products of the terms equal. Let x be the time it takes the 12 workers to assemble the car. Forming the equation yields 56/12 = x 4 2/3 = x The answer is (C). To summarize: if one quantity increases (decreases) as another quantity also increases (decreases), set ratios equal. If one quantity increases (decreases) as another quantity decreases (increases), set products equal.
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