{"id":44678,"date":"2022-05-30T08:30:15","date_gmt":"2022-05-30T03:00:15","guid":{"rendered":"https:\/\/cbselibrary.com\/?p=44678"},"modified":"2023-01-25T12:16:37","modified_gmt":"2023-01-25T06:46:37","slug":"ml-aggarwal-class-8-solutions-for-icse-maths-chapter-7-ex-7-3","status":"publish","type":"post","link":"https:\/\/cbselibrary.com\/ml-aggarwal-class-8-solutions-for-icse-maths-chapter-7-ex-7-3\/","title":{"rendered":"ML Aggarwal Class 8 Solutions for ICSE Maths Chapter 7 Percentage Ex 7.3"},"content":{"rendered":"

ML Aggarwal Class 8 Solutions for ICSE Maths Chapter 7 Percentage Ex 7.3<\/h2>\n

Question 1.
\nFind the discount and the selling price, when:
\n(i) the marked price = \u20b9575, discount = 12%
\n(ii) the printed price = \u20b912750, discount = \\(8 \\frac{1}{3}\\)%
\nSolution:
\n(i) Marked price = \u20b9575,
\ndiscount = 12%
\n\u2234 Amount of discount = 12% of Rs. 575
\n\"ML
\nNet Sale Price = M.P. – discount
\n= \u20b9575 – \u20b969 = \u20b9506
\n(ii) Marked price = Rs. 12750,
\n\"ML
\nNet sale price = M.P. – discount
\n= \u20b912750 – \u20b91062\u00b750
\n= \u20b911687\u00b750<\/p>\n

Question 2.
\nFind the discount and the discount percentage, when:
\n(i) marked price = \u20b9780, selling price = \u20b9721\u00b750
\n(ii) advertised price = \u20b928500, selling price = \u20b924510
\nSolution:
\n(i) Marked price = \u20b9780,
\nSelling price = \u20b9721\u00b750
\nDiscount = M.P. – selling price
\n= \u20b9780 – \u20b9721\u00b750 = \u20b958\u00b750
\n\"ML
\n(ii) Advertised price = \u20b928500,
\nselling price = \u20b924510
\nDiscount = advertised price – selling price
\n= \u20b928500 – \u20b924510 = \u20b93990.
\n\"ML<\/p>\n

Question 3.
\nA notebook is marked at \u20b930. Find the price a student pays for a dozen notebooks if he gets 15% discount.
\nSolution:
\nMarked price of one notebook = \u20b930
\nMarked price for one dozen notebook
\n= \u20b930 \u00d7 12 = \u20b9360
\nDiscount = 15%
\nAmount of discount = 15% of M.P.
\n\"ML
\nHence the student pays = \u20b9360 – \u20b954 = \u20b9306<\/p>\n

Question 4.
\nA dealer gave 9% discount on an electric fan and charges \u20b9 728 from the customer. Find the marked price of the fan.
\nSolution:
\nLet the marked price of the fan = \u20b9 x
\nDiscount = 9%
\namount of discount = 9% of \u20b9 x
\n\"ML
\nHence marked the price of the fan = \u20b9800<\/p>\n

Question 5.
\nThe list price of an article is \u20b9800 and a dealer is selling it at a discount of 20%. Find:
\n(i) the selling price of the article.
\n(ii) the cost price of the article if he makes 25% profit on selling it.
\nSolution:
\n(i) M.P. = \u20b9800, discount = 20%
\n\"ML<\/p>\n

Question 6.
\nA shopkeeper marks his goods at such a price that would give him a profit of 10% after allowing a discount of 12%. If an article is marked at \u20b92250, find its:
\n(i) selling price
\n(ii) cost price.
\nSolution:
\n(i) Marked price of an article = \u20b92250
\ndiscount = 12%, S.P. = ?
\n\"ML
\n\"ML
\nHence the cost price = \u20b91800<\/p>\n

Question 7.
\nA shopkeeper purchased a calculator for \u20b9650. He sells it at a discount of 20% and still makes a profit of 20%. Find :
\n(i) the selling price
\n(ii) marked price
\nSolution:
\n(i) C.P. = \u20b9650, Profit = 20%, S.P. = ?
\n\"ML
\nHence the seeling price of the calculator = \u20b9780.
\n(ii) S.P. = \u20b9780, discount = 20%, M.P. = ?
\n\"ML
\nHence the marked price of calculator = \u20b9975<\/p>\n

Question 8.
\nA shopkeeper buys a dinner set for \u20b9 1200 and marks it 80% above the cost price. If he gives 15% discount on it, find :
\n(i) the marked price
\n(ii) the selling price
\n(iii) his profit percentage.
\nSolution:
\n(i) C.P. of a dinner set = \u20b91200
\nMarked price
\n= \u20b91200 + 80% of (\u20b91200)
\n= \u20b9[1200 + \\(\\frac{80}{100}\\) \u00d7 12]
\n= \u20b9[1200 + 80 \u00d7 12]
\n= \u20b9[1200 + 960] = \u20b92160<\/p>\n

(ii) M.P. = \u20b92160, discount = 15%,
\nS.P. = ?
\nS.P. = \\(\\left(1-\\frac{d}{100}\\right)\\) of M.P.
\n\"ML<\/p>\n

Question 9.
\nThe cost price of an article is \u20b91600, which is 20% below the marked price. If the article is sold at a discount of 16%, find :
\n(i) the marked price
\n(ii) the selling price
\n(iii) profit percentage.
\nSolution:
\n(i) C.P. = \u20b91600
\nSince the cost price of an article is 20% below the marked price
\nLet the marked price of an article = \u20b9x
\n\u2234 Cost price = Marked price – 20% of Marked price
\n\u21d2 \u20b9 1600 = x – 20% of x
\n\u21d2 \u20b91600 = x – \\(\\frac{20}{100}\\) \u00d7 x
\n\u21d2 \u20b91600 = \\(\\frac{80x}{100}\\)
\n\u21d2 x = \u20b91600 \u00d7 \\(\\frac{100}{80}\\)
\n= \u20b920 \u00d7 100 = \u20b92000
\nHence the marked price of an article = \u20b92000<\/p>\n

(ii) M.P. of \u20b92000, discount = 16%,
\nS.P. = ?
\n\"ML<\/p>\n

Question 10.
\nA shopkeeper allows 20% discount on his goods and still earns a profit of 20%. If an article is sold for \u20b9360, find:
\n(i) the marked price
\n(ii) tine cost price.
\nSolution:
\n(i) Since the dealer allows a discount of 20%
\n\"ML
\n\"ML
\n(ii) Let the cost price of the article be \u20b9 x
\nand Profit is 20%
\n\"ML
\nHence, the cost price of the article = \u20b9300<\/p>\n

Question 11.
\nThe printed price of a refrigerator is \u20b928600. A dealer allows two successive discount of 10% and 5%. Find the price which a customer has to pay for the refrigerator.
\nSolution:
\nThe printed price of refrigerator = \u20b928600
\n\"ML
\nHence the customer has to pay for the refrigerator = \u20b924453.<\/p>\n

Question 12.
\nTwo dealers have marked an article at the same price. The first dealer allows two successive discounts of 15% and 5%. The other allows for a discount of 20% which is the better offer?
\nSolution:
\nLet the marked price of an article = \u20b9x
\nFor the first dealer
\n\"ML
\n\"ML
\nHence a 20% discount is better to offer.<\/p>\n

Question 13.
\nFind a single discount equivalent to two successive discounts of 30% and 10%.
\nSolution:
\nLet the market price of an article be \u20b9 x
\nand a single discount of d% be equivalent to giving
\nsuccessive discounts of 30% and 10%, then
\n\"ML
\nHence the single discount of 37% is
\nequivalent to two given successive discounts.<\/p>\n

ML Aggarwal Class 8 Solutions for ICSE Maths<\/a><\/h4>\n","protected":false},"excerpt":{"rendered":"

ML Aggarwal Class 8 Solutions for ICSE Maths Chapter 7 Percentage Ex 7.3 Question 1. Find the discount and the selling price, when: (i) the marked price = \u20b9575, discount = 12% (ii) the printed price = \u20b912750, discount = % Solution: (i) Marked price = \u20b9575, discount = 12% \u2234 Amount of discount = … Read more<\/a><\/p>\n","protected":false},"author":5,"featured_media":0,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"spay_email":""},"categories":[3034],"tags":[],"yoast_head":"\nML Aggarwal Class 8 Solutions for ICSE Maths Chapter 7 Percentage Ex 7.3 - CBSE Library<\/title>\n<meta name=\"robots\" content=\"index, follow, max-snippet:-1, max-image-preview:large, max-video-preview:-1\" \/>\n<link rel=\"canonical\" href=\"https:\/\/cbselibrary.com\/ml-aggarwal-class-8-solutions-for-icse-maths-chapter-7-ex-7-3\/\" \/>\n<meta property=\"og:locale\" content=\"en_US\" \/>\n<meta property=\"og:type\" content=\"article\" \/>\n<meta property=\"og:title\" content=\"ML Aggarwal Class 8 Solutions for ICSE Maths Chapter 7 Percentage Ex 7.3\" \/>\n<meta property=\"og:description\" content=\"ML Aggarwal Class 8 Solutions for ICSE Maths Chapter 7 Percentage Ex 7.3 Question 1. Find the discount and the selling price, when: (i) the marked price = \u20b9575, discount = 12% (ii) the printed price = \u20b912750, discount = % Solution: (i) Marked price = \u20b9575, discount = 12% \u2234 Amount of discount = ... Read more\" \/>\n<meta property=\"og:url\" content=\"https:\/\/cbselibrary.com\/ml-aggarwal-class-8-solutions-for-icse-maths-chapter-7-ex-7-3\/\" \/>\n<meta property=\"og:site_name\" content=\"CBSE Library\" \/>\n<meta property=\"article:publisher\" content=\"https:\/\/www.facebook.com\/aplustopper\/\" \/>\n<meta property=\"article:published_time\" content=\"2022-05-30T03:00:15+00:00\" \/>\n<meta property=\"article:modified_time\" content=\"2023-01-25T06:46:37+00:00\" \/>\n<meta property=\"og:image\" content=\"https:\/\/cbselibrary.com\/wp-content\/uploads\/2019\/07\/ML-Aggarwal-Class-8-Solutions-for-ICSE-Maths-Chapter-7-Percentage-Ex-7.3-Q1.1.png\" \/>\n<meta name=\"twitter:card\" content=\"summary\" \/>\n<meta name=\"twitter:label1\" content=\"Written by\" \/>\n\t<meta name=\"twitter:data1\" content=\"Prasanna\" \/>\n\t<meta name=\"twitter:label2\" content=\"Est. reading time\" \/>\n\t<meta name=\"twitter:data2\" content=\"4 minutes\" \/>\n<script type=\"application\/ld+json\" class=\"yoast-schema-graph\">{\"@context\":\"https:\/\/schema.org\",\"@graph\":[{\"@type\":\"Organization\",\"@id\":\"https:\/\/cbselibrary.com\/#organization\",\"name\":\"Aplus Topper\",\"url\":\"https:\/\/cbselibrary.com\/\",\"sameAs\":[\"https:\/\/www.facebook.com\/aplustopper\/\"],\"logo\":{\"@type\":\"ImageObject\",\"@id\":\"https:\/\/cbselibrary.com\/#logo\",\"inLanguage\":\"en-US\",\"url\":\"https:\/\/cbselibrary.com\/wp-content\/uploads\/2018\/12\/Aplus_380x90-logo.jpg\",\"contentUrl\":\"https:\/\/cbselibrary.com\/wp-content\/uploads\/2018\/12\/Aplus_380x90-logo.jpg\",\"width\":1585,\"height\":375,\"caption\":\"Aplus Topper\"},\"image\":{\"@id\":\"https:\/\/cbselibrary.com\/#logo\"}},{\"@type\":\"WebSite\",\"@id\":\"https:\/\/cbselibrary.com\/#website\",\"url\":\"https:\/\/cbselibrary.com\/\",\"name\":\"CBSE Library\",\"description\":\"Improve your Grades\",\"publisher\":{\"@id\":\"https:\/\/cbselibrary.com\/#organization\"},\"potentialAction\":[{\"@type\":\"SearchAction\",\"target\":{\"@type\":\"EntryPoint\",\"urlTemplate\":\"https:\/\/cbselibrary.com\/?s={search_term_string}\"},\"query-input\":\"required name=search_term_string\"}],\"inLanguage\":\"en-US\"},{\"@type\":\"ImageObject\",\"@id\":\"https:\/\/cbselibrary.com\/ml-aggarwal-class-8-solutions-for-icse-maths-chapter-7-ex-7-3\/#primaryimage\",\"inLanguage\":\"en-US\",\"url\":\"https:\/\/cbselibrary.com\/wp-content\/uploads\/2019\/07\/ML-Aggarwal-Class-8-Solutions-for-ICSE-Maths-Chapter-7-Percentage-Ex-7.3-Q1.1.png\",\"contentUrl\":\"https:\/\/cbselibrary.com\/wp-content\/uploads\/2019\/07\/ML-Aggarwal-Class-8-Solutions-for-ICSE-Maths-Chapter-7-Percentage-Ex-7.3-Q1.1.png\",\"width\":137,\"height\":159,\"caption\":\"ML Aggarwal Class 8 Solutions for ICSE Maths Chapter 7 Percentage Ex 7.3 Q1.1\"},{\"@type\":\"WebPage\",\"@id\":\"https:\/\/cbselibrary.com\/ml-aggarwal-class-8-solutions-for-icse-maths-chapter-7-ex-7-3\/#webpage\",\"url\":\"https:\/\/cbselibrary.com\/ml-aggarwal-class-8-solutions-for-icse-maths-chapter-7-ex-7-3\/\",\"name\":\"ML Aggarwal Class 8 Solutions for ICSE Maths Chapter 7 Percentage Ex 7.3 - 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Find the discount and the selling price, when: (i) the marked price = \u20b9575, discount = 12% (ii) the printed price = \u20b912750, discount = % Solution: (i) Marked price = \u20b9575, discount = 12% \u2234 Amount of discount = ... 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