{"id":11104,"date":"2022-06-18T21:54:48","date_gmt":"2022-06-18T16:24:48","guid":{"rendered":"https:\/\/mathemerize.com\/?p=11104"},"modified":"2022-06-18T21:54:51","modified_gmt":"2022-06-18T16:24:51","slug":"from-the-pair-of-linear-equation-in-the-following-problems-and-find-their-solutions-if-they-exist-by-the-elimination-method","status":"publish","type":"post","link":"https:\/\/mathemerize.com\/from-the-pair-of-linear-equation-in-the-following-problems-and-find-their-solutions-if-they-exist-by-the-elimination-method\/","title":{"rendered":"From the pair of linear equation in the following problems, and find their solutions (if they exist) by the elimination method"},"content":{"rendered":"

Question<\/strong> : From the pair of linear equation in the following problems, and find their solutions (if they exist) by the elimination method :<\/p>\n

(i)<\/strong>\u00a0 If we add 1 to the numerator and subtract 1 from the denominator, a fraction reduces to 1. It becomes \\(1\\over 2\\) if we only add 1 to the denominator. What is the fraction ?<\/p>\n

(ii)<\/strong>\u00a0 Five years ago, Nuri was thrice as old as Sonu. Ten years later, Nuri will be twice as old as Sonu. How old are Nuri and Sonu ?<\/p>\n

(iii)<\/strong>\u00a0 The sum of the digits of a two digit number is 9. also, nine times this number is twice the number obtained by reversing the order of the number. Find the number.<\/p>\n

(iv)<\/strong>\u00a0 Meena went to a bank to withdraw Rs 2000. She asked the cashier to give Rs 50 and Rs 1000 notes only. Meena got 25 notes in all. Find how many notes of Rs 50 and 100 she recieved ?<\/p>\n

(v)<\/strong>\u00a0 A lending library has fixed charge for the first three days and an additional charge for each day thereafter. Saritha paid Rs 27 for a book she kept for seven days. Find the fixed charge and the charge for each extra day.<\/p>\n

Solution :<\/h2>\n

(i)<\/strong>\u00a0 Let the numerator be x and the denominator be y of the fraction. Then, fraction is \\(x\\over y\\).<\/p>\n

According to Question,<\/p>\n

\\(x + 1\\over y – 1\\) = 1\u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \\(\\implies\\)\u00a0 \u00a0 \u00a0 \u00a0x + 1 = y – 1<\/p>\n

\\(\\implies\\)\u00a0 \u00a0x – y = -2\u00a0 \u00a0 \u00a0 \u00a0 \u00a0 ……..(1)<\/p>\n

and\u00a0 \\(x\\over y + 1\\) = \\(1\\over 2\\)\u00a0 \u00a0\\(\\implies\\)\u00a0 \u00a0 2x = y + 1<\/p>\n

\\(\\implies\\)\u00a0 2x – y = 1\u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0………(2)<\/p>\n

Now, subtract equation (2) from equation (1), we get<\/p>\n

(x – y) – ( 2x – y) = -2 – 1<\/p>\n

\\(\\implies\\)\u00a0 \u00a0 x – y – 2x + y = -3<\/p>\n

\\(\\implies\\)\u00a0 \u00a0x = 3<\/p>\n

Put the value of x in equation (1), we get<\/p>\n

3 – y = – 2\u00a0 \u00a0 \u00a0 \\(\\implies\\)\u00a0 \u00a0y = 5<\/p>\n

Hence, the required fraction is \\(3\\over 5\\)<\/strong>.<\/p>\n

(ii)<\/strong>\u00a0 Let Nuri’s present age be x years and Sonu’s present age be y years.<\/p>\n

Five years ago,<\/p>\n

Nuri’s age = (x – 5)\u00a0 years<\/p>\n

and\u00a0 Sonu’s age = (y – 5) years<\/p>\n

According to Question,<\/p>\n

(x – 5) = 3(y – 5)\u00a0 \u00a0\\(\\implies\\)\u00a0 \u00a0x – 5 = 3y – 15<\/p>\n

\\(\\implies\\)\u00a0 \u00a0x – 3y = – 15 + 5\u00a0 \u00a0\\(\\implies\\)\u00a0 \u00a0x – 3y = – 10\u00a0 \u00a0 \u00a0 \u00a0 \u00a0……..(1)<\/p>\n

Ten years later,<\/p>\n

Nuri’s age = (x + 10) years<\/p>\n

and\u00a0 Sonu’s age = (y + 10) years<\/p>\n

According to Question,<\/p>\n

x + 10 = 2(y + 10)\u00a0 \u00a0\\(\\implies\\)\u00a0 \u00a0x + 10 = 2y + 20<\/p>\n

\\(\\implies\\)\u00a0 x – 2y = 10\u00a0 \u00a0 \u00a0 \u00a0 \u00a0……….(2)<\/p>\n

Now, Subtract equation (2) form equation (1), we get<\/p>\n

y = 20<\/p>\n

Put the value of y in equation (2), we get<\/p>\n

x – 2(20) = 10\u00a0 \u00a0 \u00a0 \u00a0\\(\\implies\\)\u00a0 \u00a0 x = 40<\/p>\n

\\(\\therefore\\)\u00a0 \u00a0 Present age of Nuri is 50 years<\/strong> and present age of sonu is 20 years.<\/strong><\/p>\n

(iii)<\/strong>\u00a0 Let x and y be the ten’s and unit’s digit in the number respectively. So, the number may be written as 10x + y.<\/p>\n

By reversing the order of the number, unit’s digit becomes x and ten’s digit becomes y.<\/p>\n

Then, the number can be written as 10y + x.<\/p>\n

According to Question,<\/p>\n

x + y = 9\u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0…….(1)<\/p>\n

We are given in the question that nine times the number i.e. 9(10x + y) is twice the number obtained by reversing the order of number i.e. 2(10y + x).<\/p>\n

9(10x + y) = 2(10y + x)<\/p>\n

\\(\\implies\\)\u00a0 90x + 9y = 20y + 2x<\/p>\n

\\(\\implies\\)\u00a0 8x – y = 0\u00a0 \u00a0 \u00a0 ……(2)<\/p>\n

Adding the equation (1) and (2), we get<\/p>\n

9x = 9\u00a0 \u00a0 \u00a0 \\(\\implies\\)\u00a0 \u00a0 x = 1<\/p>\n

Now, Put the value of x in equation (1), we get<\/p>\n

y = 9 – 1 = 8<\/p>\n

Thus, the number is \\(10 \\times 1) + 8 = 10 + 8 = 18<\/strong><\/p>\n

(iv)<\/strong>\u00a0 Let x be the number of notes of Rs 50 and y be the number of notes of Rs 100.<\/p>\n

According to Question(ATQ),<\/p>\n

x + y = 25\u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 …….(1)<\/p>\n

50x + 100y = 2000\u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 …….(2)<\/p>\n

Divide equation (2) by 50, we get<\/p>\n

x + 2y = 40\u00a0 \u00a0 \u00a0 \u00a0 ………(3)<\/p>\n

Subtract equation (1) from equation (3), we get<\/p>\n

y = 15<\/p>\n

Now, put the value of y in equation (1), we get<\/p>\n

x + 15 = 25\u00a0 \u00a0 \u00a0\\(\\implies\\)\u00a0 \u00a0x = 10<\/p>\n

Hence,\u00a0 x = 10\u00a0 and\u00a0 \u00a0y = 15.<\/strong><\/p>\n

(v)<\/strong>\u00a0 Let x\u00a0 be the fixed charges for 3 days and y be the charges per day.<\/p>\n

According to the Question,<\/p>\n

x + 4y = 27\u00a0 \u00a0 \u00a0 \u00a0 \u00a0 ……(1)<\/p>\n

and\u00a0 x + 2y = 21\u00a0 \u00a0 \u00a0 \u00a0……..(2)<\/p>\n

Subtracting the equation (2)\u00a0 from equation (1),\u00a0 we get<\/p>\n

x + 4(3)\u00a0 = 27\u00a0 \u00a0\\(\\implies\\)\u00a0 \u00a0 x = 27 – 12 = 15<\/p>\n

Put the value of x in equation (2), we get<\/p>\n

y = 3<\/p>\n

Hence, x = 15 and y = 3.<\/strong><\/p>\n","protected":false},"excerpt":{"rendered":"

Question : From the pair of linear equation in the following problems, and find their solutions (if they exist) by the elimination method : (i)\u00a0 If we add 1 to the numerator and subtract 1 from the denominator, a fraction reduces to 1. It becomes \\(1\\over 2\\) if we only add 1 to the denominator. …<\/p>\n

From the pair of linear equation in the following problems, and find their solutions (if they exist) by the elimination method<\/span> Read More »<\/a><\/p>\n","protected":false},"author":1,"featured_media":0,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"site-sidebar-layout":"default","site-content-layout":"default","ast-global-header-display":"","ast-main-header-display":"","ast-hfb-above-header-display":"","ast-hfb-below-header-display":"","ast-hfb-mobile-header-display":"","site-post-title":"","ast-breadcrumbs-content":"","ast-featured-img":"","footer-sml-layout":"","theme-transparent-header-meta":"","adv-header-id-meta":"","stick-header-meta":"","header-above-stick-meta":"","header-main-stick-meta":"","header-below-stick-meta":""},"categories":[911,43],"tags":[],"yoast_head":"\nFrom the pair of linear equation in the following problems, and find their solutions (if they exist) by the elimination method - Mathemerize<\/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:\/\/mathemerize.com\/from-the-pair-of-linear-equation-in-the-following-problems-and-find-their-solutions-if-they-exist-by-the-elimination-method\/\" \/>\n<meta property=\"og:locale\" content=\"en_US\" \/>\n<meta property=\"og:type\" content=\"article\" \/>\n<meta property=\"og:title\" content=\"From the pair of linear equation in the following problems, and find their solutions (if they exist) by the elimination method - Mathemerize\" \/>\n<meta property=\"og:description\" content=\"Question : From the pair of linear equation in the following problems, and find their solutions (if they exist) by the elimination method : (i)\u00a0 If we add 1 to the numerator and subtract 1 from the denominator, a fraction reduces to 1. 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