CBSE Class 8 > Class 8 Notes > ML Aggarwal: Percentage - 3
ML Aggarwal: Percentage - 3
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Page 1 1. Find the discount and the selling price, when: (i) the marked price = ? 575, discount = 12% (ii) the printed price = ? 12750, discount = % Solution: (i) the marked price = ? 575, discount = 12% Here Amount of discount = 12 % of ? 575 It can be written as = ) × 575* By further calculation = ) × 23* So we get = 3 × 23 = ? 69 We know that Net sale price = M.P. – discount Substituting the values = 575 – 69 = ? 506 (ii) the printed price = ? 12750, discount = 8 % = % Page 2 1. Find the discount and the selling price, when: (i) the marked price = ? 575, discount = 12% (ii) the printed price = ? 12750, discount = % Solution: (i) the marked price = ? 575, discount = 12% Here Amount of discount = 12 % of ? 575 It can be written as = ) × 575* By further calculation = ) × 23* So we get = 3 × 23 = ? 69 We know that Net sale price = M.P. – discount Substituting the values = 575 – 69 = ? 506 (ii) the printed price = ? 12750, discount = 8 % = % Here Amount of discount = % of ? 12750 It can be written as = ! × × 12750( By further calculation = ) × 1275* So we get = ) × 1275* = ? 1062.50 We know that Net sale price = M.P. – discount Substituting the values = 12750 – 1062.50 = ? 11687.50 2. Find the discount and the discount percentage, when: (i) marked price = ? 780, selling price = ? 721.50 (ii) advertised price = ? 28500, selling price = ? 24510 Solution: (i) marked price = ? 780, selling price = ? 721.50 We know that Discount = M.P. – Selling price Page 3 1. Find the discount and the selling price, when: (i) the marked price = ? 575, discount = 12% (ii) the printed price = ? 12750, discount = % Solution: (i) the marked price = ? 575, discount = 12% Here Amount of discount = 12 % of ? 575 It can be written as = ) × 575* By further calculation = ) × 23* So we get = 3 × 23 = ? 69 We know that Net sale price = M.P. – discount Substituting the values = 575 – 69 = ? 506 (ii) the printed price = ? 12750, discount = 8 % = % Here Amount of discount = % of ? 12750 It can be written as = ! × × 12750( By further calculation = ) × 1275* So we get = ) × 1275* = ? 1062.50 We know that Net sale price = M.P. – discount Substituting the values = 12750 – 1062.50 = ? 11687.50 2. Find the discount and the discount percentage, when: (i) marked price = ? 780, selling price = ? 721.50 (ii) advertised price = ? 28500, selling price = ? 24510 Solution: (i) marked price = ? 780, selling price = ? 721.50 We know that Discount = M.P. – Selling price Substituting the values = 780 – 721.50 = ? 58.50 Here Discount % = [Discount/M.P. × 100] % Substituting the values = ! . × 100( % By further calculation = % So we get = % = 7.5 % (ii) advertised price = ? 28500, selling price = ? 24510 We know that Discount = Advertised price – Selling Price Substituting the values = 28500 – 24510 = ? 3990 Here Discount % = ! QAG?B>C@ $=R<#@AG<= S#A?< × 100( % Substituting the values Page 4 1. Find the discount and the selling price, when: (i) the marked price = ? 575, discount = 12% (ii) the printed price = ? 12750, discount = % Solution: (i) the marked price = ? 575, discount = 12% Here Amount of discount = 12 % of ? 575 It can be written as = ) × 575* By further calculation = ) × 23* So we get = 3 × 23 = ? 69 We know that Net sale price = M.P. – discount Substituting the values = 575 – 69 = ? 506 (ii) the printed price = ? 12750, discount = 8 % = % Here Amount of discount = % of ? 12750 It can be written as = ! × × 12750( By further calculation = ) × 1275* So we get = ) × 1275* = ? 1062.50 We know that Net sale price = M.P. – discount Substituting the values = 12750 – 1062.50 = ? 11687.50 2. Find the discount and the discount percentage, when: (i) marked price = ? 780, selling price = ? 721.50 (ii) advertised price = ? 28500, selling price = ? 24510 Solution: (i) marked price = ? 780, selling price = ? 721.50 We know that Discount = M.P. – Selling price Substituting the values = 780 – 721.50 = ? 58.50 Here Discount % = [Discount/M.P. × 100] % Substituting the values = ! . × 100( % By further calculation = % So we get = % = 7.5 % (ii) advertised price = ? 28500, selling price = ? 24510 We know that Discount = Advertised price – Selling Price Substituting the values = 28500 – 24510 = ? 3990 Here Discount % = ! QAG?B>C@ $=R<#@AG<= S#A?< × 100( % Substituting the values = ! × 100( % So we get = % = 14 % 3. A notebook is marks at ? 30. Find the price a student pays for a dozen notebooks if he gets 15% discount. Solution: It is given that M.P. of one notebook = ? 30 M.P. of one dozen notebooks = 30 × 12 = ? 360 Discount = 15% We know that Amount of discount = 15% of M.P. It can be written as = 15% of ? 360 By further calculation = ) × 360* So we get = ) × 36* = ) × 36* On further simplification Page 5 1. Find the discount and the selling price, when: (i) the marked price = ? 575, discount = 12% (ii) the printed price = ? 12750, discount = % Solution: (i) the marked price = ? 575, discount = 12% Here Amount of discount = 12 % of ? 575 It can be written as = ) × 575* By further calculation = ) × 23* So we get = 3 × 23 = ? 69 We know that Net sale price = M.P. – discount Substituting the values = 575 – 69 = ? 506 (ii) the printed price = ? 12750, discount = 8 % = % Here Amount of discount = % of ? 12750 It can be written as = ! × × 12750( By further calculation = ) × 1275* So we get = ) × 1275* = ? 1062.50 We know that Net sale price = M.P. – discount Substituting the values = 12750 – 1062.50 = ? 11687.50 2. Find the discount and the discount percentage, when: (i) marked price = ? 780, selling price = ? 721.50 (ii) advertised price = ? 28500, selling price = ? 24510 Solution: (i) marked price = ? 780, selling price = ? 721.50 We know that Discount = M.P. – Selling price Substituting the values = 780 – 721.50 = ? 58.50 Here Discount % = [Discount/M.P. × 100] % Substituting the values = ! . × 100( % By further calculation = % So we get = % = 7.5 % (ii) advertised price = ? 28500, selling price = ? 24510 We know that Discount = Advertised price – Selling Price Substituting the values = 28500 – 24510 = ? 3990 Here Discount % = ! QAG?B>C@ $=R<#@AG<= S#A?< × 100( % Substituting the values = ! × 100( % So we get = % = 14 % 3. A notebook is marks at ? 30. Find the price a student pays for a dozen notebooks if he gets 15% discount. Solution: It is given that M.P. of one notebook = ? 30 M.P. of one dozen notebooks = 30 × 12 = ? 360 Discount = 15% We know that Amount of discount = 15% of M.P. It can be written as = 15% of ? 360 By further calculation = ) × 360* So we get = ) × 36* = ) × 36* On further simplification = 3 × 18 = ? 54 Price a student pays for a dozen notebooks = 360 – 54 = ? 306 4. A dealer gave 9% discount on an electric fan and charges ? 728 from the customer. Find the marked price of the fan. Solution: Consider ? x as the M.P. of the fan Discount = 9% We know that Amount of discount = 9% of ? x It can be written as = × 9 = ? / Here Charges for customer = ? x – ? / Substituting the values 728 = / – / By further calculation 728 = / So we getRead More
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Apr 25, 2026 Last updated
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