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Word Problems Using Percents and Perimeter By Dr. Carol A. Marinas

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Is’s and Of’s n What % is a of b? x = a 100 b The Pattern is: % = “is” 100 “of”

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Examples of Is/Of % Questions n What number is 16% of 70? n 16 = x n Cross multiply n 16(70) = 100x n 1120 = 100x n 11.2 = x n The number 45 is 25% of what number? n 25 = x n Cross multiply n 25x = 45(100) n 25x = 4500 n x = 180

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Examples of Is/Of % Questions n Find 23% of 20. n 23 = x n Cross multiply n 23(20) = 100x n 460 = 100x n 4.6 = x n The number 42 is what percent of 35? n x = n Cross multiply n 35x = 42(100) n 35x = 4200 n x = 120 n 120 %

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Word Problem - Sale Price n Dillard’s advertised a 25% off sale. If a London Fog coat originally sold for $156, find the decrease and the sale price. n Decrease.25(156) = 39 dollars n Sale price = 117 dollars

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Word Problems - Increased Price n Time Saver increased the price of a $.75 cola by 15%. Find the increase and the new price. n Increase.15(.75) =.11 n New price =.86

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Word Problem - Interest #1 n Zoya invested part of her $25,000 advance at 8% annual simple interest and the rest at 9% annual simple interest. If her total yearly interest from both accounts was $2135, find the amount invested at each rate. n Explanation: $25,000 partly invested at 8% and the rest at 9%. The total interest from these investments is $2135. We are going to organize the information in a table. We will need the business formula: Principal*Rate=Interest

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Word Problem - Interest (cont.) n Put % rates as decimals in the Rate column. Put the total principal of $25,000 and total interest of $2135 in the bottom row. Put x in the principal row for 8% so then x must be the principal invested at 9%. Since Principal*rate=Interest multiply the top 2 rows across. Your equation comes from adding down the last column..08x +.09(25000-x) = x x = x = x = -115 So x = 8% = 9%

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Word Problem - Interest #2 n Bruce invested a sum of money at 10% annual simple interest and invested twice that amount at 12% annual simple interest. If his total yearly income from both investments was $2890, how much was invested at each rate? n Explanation: x amount invested at 10% and 2x amount invested at 12%. Complete the table.

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Word Problem - Interest #2 n Multiply across the top 2 rows. The equation is adding down the last column. n.10x +.12(2x) = 2890 n.10x +.24x = 2890 n.34x = 2890 n x = 10% and 12 %

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Word Problems - Perimeter #1 n If the length of a rectangular parking lot is 10 meters less than twice its width, and the perimeter is 400 meters, find the length of the parking lot. n If x is the width, then 2x - 10 is the length. Perimeter is distance around. n P= 2L + 2W n 400 = 2(2x - 10) + 2(x) n 400 = 6x - 20 n 420 = 6x So x = 70 meters (width) and n 2(70) - 10 = 130 meters (length)

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Word Problem - Perimeter #2 n A flower bed is in the shape of a triangle with one side twice the length of the shortest side, and the third side is 30 feet more than the length of the shortest side. Find the dimension if the perimeter is 102 feet. n If x is the shortest side, the second side is 2x, and the third side is x The perimeter is the distance around the triangular flower bed. n x + 2x + (x + 30) = 102 n 4x + 30 = 102 n 4x = 72 So x = 18 feet (shortest side); 2x = 36 feet (second side); x + 30 = 48 feet (third side)

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The End ! n Remember word problems are difficult. Some hints to help: – You must first understand the problem – Then assign x to your missing number – Come up with a strategy. Write an equation. – Solve the equation. – Check answer. – Write solution including UNITS.

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