{"id":26713,"date":"2022-06-05T10:00:55","date_gmt":"2022-06-05T04:30:55","guid":{"rendered":"https:\/\/mcq-questions.com\/?p=26713"},"modified":"2022-05-23T15:48:49","modified_gmt":"2022-05-23T10:18:49","slug":"ncert-solutions-for-class-10-maths-chapter-3-ex-3-4","status":"publish","type":"post","link":"https:\/\/mcq-questions.com\/ncert-solutions-for-class-10-maths-chapter-3-ex-3-4\/","title":{"rendered":"NCERT Solutions for Class 10 Maths Chapter 3 Pair of Linear Equations in Two Variables Ex 3.4"},"content":{"rendered":"

These NCERT Solutions for Class 10 Maths<\/a> Chapter 3 Pair of Linear Equations in Two Variables Ex 3.4 Questions and Answers are prepared by our highly skilled subject experts.<\/p>\n

NCERT Solutions for Class 10 Maths Chapter 3 Pair of Linear Equations in Two Variables Exercise 3.4<\/h2>\n

\"NCERT<\/p>\n

Question 1.
\nSolve the following pair of linear equations by the elimination method and the substitution
\n(i) x + y = 5 and 2x – 3y = 4
\n(ii) 3x + 4y = 10 and 2x – 2y = 2
\n(iii) 3x – 5y – 4 = 0 and 9x = 2y + 7
\n(iv) x\/2 + 2y\/3 = -1 and x – y\/3 = 3
\nSolution:
\n(i) By Elimination Method:
\nFquations are x + y = 5
\nand 2x – 3y = 4
\nMultiply equation (i) by 2 and subtract equation (ii) from it, we have
\n\"NCERT<\/p>\n

(ii) By Elimination method:
\nEquations are 3x + 4y = 10
\nand 2x – 2y = 2
\nMultiplying equation (ii) by 2 and adding to equation (i), we
\n\"NCERT<\/p>\n

(iii) By Elimination Method:
\n\"NCERT<\/p>\n

(iv) By Elimination Method:
\n\"NCERT
\nSubstracting equation (i) from (iii) we get
\n5y = – 15
\n\u21d2 y = – 3
\nSubtracting the value of y = – 3 in (ii), we get
\n3x – (- 3) = 9
\n3x + 3 = 9
\n3x = 6 x = 2
\n\u2234 x = 2 and y = 3
\nSubstituting method
\nFrom equation (ii) we have
\n3x – y = 9
\n\u21d2 3x = 9 + y
\n\u21d2 x = \\(\\frac { 9 + y }{ 3 }\\)
\nSubstituting the value of x = equation, (i) we get
\n3(\\(\\frac { 9 + y }{ 3 }\\)) + 4y = – 6
\n\u21d2 9 + y + 4y = – 6
\n\u21d2 5y = – 15
\n\u21d2 y = – 3
\nAgain, substituting the value of y = -3 in equation (i) we get
\n3x – (-3) = 9
\n\u21d2 3x + 3 = 9
\n\u21d2 3x = 6
\n\u21d2 x = 2
\n\u2234 x = 2 and y = – 3<\/p>\n

\"NCERT<\/p>\n

Question 2.
\nForm the pair of linear equations in the following problems, and find their solutions (if they exist) by the elimination method:
\n(i) If we add 1 to the numerator and subtract 1 from the denominator, a fraction reduces to 1. It becomes – if we only add 1 to the denominator. What is the fraction\u20b9<\/p>\n

(ii) 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) 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 digits. Find the number.<\/p>\n

(iv) Meena went to a bank to withdraw \u20b9 2000. She asked the cashier to give her \u20b9 50 and \u20b9 100 notes only. Meena got 25 notes in all. Find how many notes of \u20b9 50 and \u20b9 100 she received.<\/p>\n

(v) A lending library has a fixed charge for the first three days and an additional charge for each day thereafter. Saritha paid \u20b9 27 for a book kept for seven days, while Susy paid \u20b9 21 for the book she kept for five days. Find the fixed charge and the charge for each extra day.
\nSolution:
\n(i) Let numerator be x and denominator be y.
\n\\(\\frac { x + 1 }{ y – 1 }\\) = 1
\nx – y = – 2 … (i)
\nAccording to question’s second condition
\n\\(\\frac { x }{ y + 1 }\\) = \\(\\frac { 1 }{ 2 }\\)
\n2x – y = 1 … (i)
\nSubstract equation (i) from equation we eliminate y
\nx = 3
\nSubstitute this value in equation (i) we get
\n3 – y = – 2
\ny = 5
\nHence the fraction is \\(\\frac { 3 }{ 5 }\\)<\/p>\n

(ii) Let the present age of Nuri be x
\nand let the present age of Sonu be y.
\nAccording to questions first condition
\n(x – 5) = 3 (y – 5)
\nx – 3y = – 10 … (i)
\nAccording to eq. question’s second condition
\n(x + 10) = 2 (y+ 10)
\nx – 2y = 10 … (ii)
\nSubstracting equation (i) from equation (ii)
\nWe eliminate for x.
\ny = 20
\nPutting this value in equation (i)
\nx – 60 = – 10
\nx = 50
\nPresent age of Nuri is 50 years old and Sonu is 20 years old.<\/p>\n

(iii) Let the ten’s digit of the number be x
\nand the unit’s digit of the number by y.
\nAccording to question’s first condition
\nx + y = 9 … (i)
\nAccording to question’s second condition
\n9(10 x + y) = 2(10y + x)
\n90x + 9y – 20y + 2x
\n88x = 11y
\n8x – y = 0 (dividing by 11)
\n8x – y = 0 … (ii)
\nAdding equation (i) and (ii) we eleminate for y, we get
\n9x = 9
\nx = 1
\nPutting the value in equation (i)
\n1 + y = 9
\ny = 8
\nHence the units digit is 8 and the ten’s digit is 1. Then the number is 18.<\/p>\n

(iv) Let the notes of \u20b9 50 be x
\nand let the notes of \u20b9 100 be y.
\nAccording to question’s first condition
\n50x + 100y = 2000
\nx + 2 y= 40 … (i)
\nAccording to question second condition
\nx + y = 25
\nSubstract equation (i) from equation (ii) we eleminate for x.
\n– y = – 15
\ny = 15
\nPutting this value in equation (i)
\nx + 30 = 40
\nx = 40
\nHence \u20b9 50 notes is 10 and the notes of \u20b9 100 is 15.<\/p>\n

(v) Let the fixed charge be \u20b9 x and let the additional charge per day be \u20b9 y.
\nAccording to questions first condition
\nx + 4y = 27 … (i)
\nAccording to questions second condition
\nx + 2y = 21 … (ii)
\nEliminating x for equation (i) and (ii)
\nWe subtract equation (ii) from equation (1)
\n2y = 6
\ny = 3
\nPutting this value in equation (ii)
\nx + 6 = 21
\nx = 21 – 6
\nx = 15
\nHence the fixed charge is \u20b9 15 and the additional charge per day is \u20b9 3.<\/p>\n","protected":false},"excerpt":{"rendered":"

These NCERT Solutions for Class 10 Maths Chapter 3 Pair of Linear Equations in Two Variables Ex 3.4 Questions and Answers are prepared by our highly skilled subject experts. NCERT Solutions for Class 10 Maths Chapter 3 Pair of Linear Equations in Two Variables Exercise 3.4 Question 1. Solve the following pair of linear equations …<\/p>\n

NCERT Solutions for Class 10 Maths Chapter 3 Pair of Linear Equations in Two Variables Ex 3.4<\/span> Read More »<\/a><\/p>\n","protected":false},"author":9,"featured_media":0,"comment_status":"closed","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"site-sidebar-layout":"default","site-content-layout":"default","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":"default","adv-header-id-meta":"","stick-header-meta":"default","header-above-stick-meta":"","header-main-stick-meta":"","header-below-stick-meta":"","spay_email":""},"categories":[2],"tags":[],"yoast_head":"\nNCERT Solutions for Class 10 Maths Chapter 3 Pair of Linear Equations in Two Variables Ex 3.4 - MCQ Questions<\/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:\/\/mcq-questions.com\/ncert-solutions-for-class-10-maths-chapter-3-ex-3-4\/\" \/>\n<meta property=\"og:locale\" content=\"en_US\" \/>\n<meta property=\"og:type\" content=\"article\" \/>\n<meta property=\"og:title\" content=\"NCERT Solutions for Class 10 Maths Chapter 3 Pair of Linear Equations in Two Variables Ex 3.4 - MCQ Questions\" \/>\n<meta property=\"og:description\" content=\"These NCERT Solutions for Class 10 Maths Chapter 3 Pair of Linear Equations in Two Variables Ex 3.4 Questions and Answers are prepared by our highly skilled subject experts. 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