{"id":26893,"date":"2021-06-28T15:58:10","date_gmt":"2021-06-28T10:28:10","guid":{"rendered":"https:\/\/ncertsolutions.guru\/?p=26893"},"modified":"2021-06-28T15:58:10","modified_gmt":"2021-06-28T10:28:10","slug":"ncert-solutions-for-class-8-maths-chapter-2-ex-2-5","status":"publish","type":"post","link":"https:\/\/ncertsolutions.guru\/ncert-solutions-for-class-8-maths-chapter-2-ex-2-5\/","title":{"rendered":"NCERT Solutions for Class 8 Maths Chapter 2 Linear Equations in One Variable Ex 2.5"},"content":{"rendered":"

These NCERT Solutions for Class 8 Maths<\/a> Chapter 2 Linear Equations in One Variable Ex 2.5 Questions and Answers are prepared by our highly skilled subject experts.<\/p>\n

NCERT Solutions for Class 8 Maths Chapter 2 Linear Equations in One Variable Exercise 2.5<\/h2>\n

Solve the following linear equations.<\/span><\/p>\n

Question 1.
\n\\(\\frac{x}{2}-\\frac{1}{5}=\\frac{x}{3}+\\frac{1}{4}\\)
\nSolution:
\n\\(\\frac{x}{2}-\\frac{1}{5}=\\frac{x}{3}+\\frac{1}{4}\\)
\nTransposing \\(-\\frac{1}{5}\\) to R.H.S. and \\(\\frac{\\mathrm{x}}{3}\\) to L.H.S
\n\\(\\frac{x}{2}-\\frac{x}{3}=\\frac{1}{4}+\\frac{1}{5}\\)
\n\\(\\frac{3 x-2 x}{6}=\\frac{5+4}{20}\\)
\n\\(\\frac{x}{6}=\\frac{9}{20}\\)
\nBy cross-multiplication, we get
\n20(x) = 9 \u00d7 6
\n20x = 54
\nDividing both sides by 20
\n\\(\\frac{20 \\mathrm{x}}{20}=\\frac{54}{20}\\)
\nx = \\(\\frac{27}{10}\\)<\/p>\n

\"NCERT<\/p>\n

Question 2.
\n\\(\\frac{\\mathrm{n}}{2}-\\frac{3 n}{4}+\\frac{5 \\mathrm{n}}{6}=21\\)
\nSolution:
\n\\(\\frac{\\mathrm{n}}{2}-\\frac{3 n}{4}+\\frac{5 \\mathrm{n}}{6}=21\\)
\nL.C.M of 2, 4 and 6 = 12
\nMultiplying each term by 12, we get
\n\\(12 \\times \\frac{\\mathrm{n}}{2}-12 \\times \\frac{3 \\mathrm{n}}{4}+12 \\times \\frac{5 \\mathrm{n}}{6}=21 \\times 12\\)
\n6n – 9n + 10n = 252
\n7n = 252
\nDividing both sides by 7
\n\\(\\frac{7 \\mathrm{n}}{7}=\\frac{252}{7}\\)
\nn = 36<\/p>\n

Question 3.
\n\\(x+7-\\frac{8 x}{3}=\\frac{17}{6}-\\frac{5 x}{2}\\)
\nSolution:
\n\\(x+7-\\frac{8 x}{3}=\\frac{17}{6}-\\frac{5 x}{2}\\)
\nTransposing 7 to R.H.S. and \\(\\frac{-5 x}{2}\\) to L.H.S
\n\\(x-\\frac{8 x}{3}+\\frac{5 x}{2}=\\frac{17}{6}-7\\)
\nL.C.M of 3, 2 and 6 = 6
\nMultiplying each term by 6, we get
\n\\(6 \\times x-6 \\times \\frac{8 x}{3}+6 \\times \\frac{5 x}{2}=6 \\times \\frac{17}{6}-6 \\times 7\\)
\n6x – 16x + 15x = 17 – 42
\n5x = -25
\nDividing both sides by 5
\n\\(\\frac{5 x}{5}=-\\frac{25}{5}\\)
\nx = -5<\/p>\n

\"NCERT<\/p>\n

Question 4.
\n\\(\\frac{x-5}{3}=\\frac{x-3}{5}\\)
\nSolution:
\n\\(\\frac{x-5}{3}=\\frac{x-3}{5}\\)
\nBy cross-multiplication we get
\n5(x – 5) = 3(x – 3)
\n5x – 25 = 3x – 9
\nTransposing -25 to R.H.S and 3x to L.H.S.
\n5x – 3x = 25 – 9
\n2x = 16
\nDividing both sides by 2
\n\\(\\frac{2 \\mathrm{x}}{2}\\) = \\(\\frac{16}{2}\\)
\nx = 8<\/p>\n

Question 5.
\n\\(\\frac{3 \\mathrm{t}-2}{4}-\\frac{2 \\mathrm{t}+3}{3}=\\frac{2}{3}-\\mathrm{t}\\)
\nSolution:
\n\\(\\frac{3 \\mathrm{t}-2}{4}-\\frac{2 \\mathrm{t}+3}{3}=\\frac{2}{3}-\\mathrm{t}\\)
\nTransposing -t to L.H.S., we get
\n\\(\\frac{3 \\mathrm{t}-2}{4}-\\frac{2 \\mathrm{t}+3}{3}+\\mathrm{t}=\\frac{2}{3}\\)
\nL.C.M of 4 and 3 is 12
\nMultiplying each term by 12
\n\\(12 \\frac{(3 \\mathrm{t}-2)}{4}-12 \\frac{(2 \\mathrm{t}+3)}{3}+12 \\mathrm{t}=12 \\times \\frac{2}{3}\\)
\n3(3t – 2) – 4(2t + 3) + 12t = 8
\n9t – 6 – 8t – 12 + 12t = 8
\n13t – 18 = 8
\nTransposing -18 to R.H.S., we get
\n13t = 8 + 18
\n13t = 26
\nDividing both sides by 13
\n\\(\\frac{13 t}{13}=\\frac{26}{13}\\)
\nt = 2<\/p>\n

\"NCERT<\/p>\n

Question 6.
\n\\(m-\\frac{m-1}{2}=1-\\frac{m-2}{3}\\)
\nSolution:
\n\\(m-\\frac{m-1}{2}=1-\\frac{m-2}{3}\\)
\nTransposing \\(\\left(\\frac{m-2}{3}\\right)\\) to L.H.S. we get
\n\\(\\mathrm{m}-\\frac{\\mathrm{m}-1}{2}+\\frac{\\mathrm{m}-2}{3}=1\\)
\nL.C.M of 2 and 3 is 6
\nMultiplying each term by 6, we get
\n6m – \\(\\frac{6(m-1)}{2}+\\frac{6(m-2)}{3}\\) = 1 \u00d7 6
\n6m – 3(m – 1) + 2(m – 2) = 6
\n6m – 3m + 3 + 2m – 4 = 6
\n5m – 1 = 6
\nTransposing -1 to the R.H.S., we get
\n5m = 6 + 1
\n5m = 7
\nDividing both sides by 5, we get
\n\\(\\frac{5 m}{5}=\\frac{7}{5}\\)
\nm = \\(\\frac{7}{5}\\)<\/p>\n

Simplify and solve the following linear equations.<\/span><\/p>\n

Question 7.
\n3(t – 3) = 5 (2t + 1)
\nSolution:
\n3(t – 3) = 5 (2t + 1)
\nOpening the brackets, we get
\n3t – 9 = 10t + 5
\nTransposing -9 to R.H.S. and 10t to L.H.S.
\n3t – 10t = 5 + 9
\n-7t = 14
\nDividing both sides by -7, we get
\n\\(\\frac{-7 t}{-7}=\\frac{14}{-7}\\)
\nt = -2<\/p>\n

\"NCERT<\/p>\n

Question 8.
\n15(y – 4) – 2(y – 9) + 5 (y + 6) = 0
\nSolution:
\n15(y – 4) – 2(y – 9) + 5(y + 6) = 0
\nOpening the brackets, we get
\n15y – 60 – 2y + 18 + 5y + 30 = 0
\n15y – 2y + 5y – 60 + 18 + 30 = 0
\n18y – 12 = 0
\nTransposing -12 to R.H.S.
\n18y = 12
\nDividing both sides by 18, we get.
\n\\(\\frac{18 y}{18}=\\frac{12}{18}\\)
\ny = \\(\\frac{2}{3}\\)<\/p>\n

Question 9.
\n3(5z – 7) – 2(9z – 11) = 4(8z – 13) – 17
\nSolution:
\n3(5z – 7) – 2(9z – 11) = 4(8z – 13) – 17
\nOpening the brackets, we get
\n15z – 21 – 18z + 22 = 32z – 52 – 17
\n15z – 18z + 22 – 21 = 32z – 69
\n-3z + 1 = 32z – 69
\nTransposing 1 to R.H.S. and 32z to L.H.S.
\n-3z – 32z = – 69 – 1
\n-35z = -70
\nDividing both sides by -35, we get
\n\\(\\frac{-35 z}{-35}=\\frac{-70}{-35}\\)
\nz = 2<\/p>\n

\"NCERT<\/p>\n

Question 10.
\n0.25(4f – 3) = 0.05(10f – 9)
\nSolution:
\n0.25(4f – 3) = 0.05(10f – 9)
\nOpening the brackets, we get
\n0.25(4f) – 0.25 \u00d7 3 = 0.05 \u00d7 10f – 0.05 \u00d7 9
\nf – 0.75 = 0.5f – 0.45
\nf – 0.5f = 0.75 – 0.45
\n0.5f = 0.3
\nDividing both sides by 0.5, we get
\n\\(\\frac{0.5 \\mathrm{f}}{0.5}=\\frac{0.3}{0.5}\\)
\nf = \\(\\frac{3}{5}\\) = 0.6
\nf = 0.6<\/p>\n","protected":false},"excerpt":{"rendered":"

These NCERT Solutions for Class 8 Maths Chapter 2 Linear Equations in One Variable Ex 2.5 Questions and Answers are prepared by our highly skilled subject experts. NCERT Solutions for Class 8 Maths Chapter 2 Linear Equations in One Variable Exercise 2.5 Solve the following linear equations. Question 1. Solution: Transposing to R.H.S. and to …<\/p>\n

NCERT Solutions for Class 8 Maths Chapter 2 Linear Equations in One Variable Ex 2.5<\/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":{"spay_email":""},"categories":[7],"tags":[],"yoast_head":"\nNCERT Solutions for Class 8 Maths Chapter 2 Linear Equations in One Variable Ex 2.5 - NCERT Solutions<\/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:\/\/ncertsolutions.guru\/ncert-solutions-for-class-8-maths-chapter-2-ex-2-5\/\" \/>\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 8 Maths Chapter 2 Linear Equations in One Variable Ex 2.5 - NCERT Solutions\" \/>\n<meta property=\"og:description\" content=\"These NCERT Solutions for Class 8 Maths Chapter 2 Linear Equations in One Variable Ex 2.5 Questions and Answers are prepared by our highly skilled subject experts. 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NCERT Solutions for Class 8 Maths Chapter 2 Linear Equations in One Variable Exercise 2.5 Solve the following linear equations. Question 1. 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