{"id":27036,"date":"2021-06-29T11:37:12","date_gmt":"2021-06-29T06:07:12","guid":{"rendered":"https:\/\/mcq-questions.com\/?p=27036"},"modified":"2022-03-02T10:47:25","modified_gmt":"2022-03-02T05:17:25","slug":"ncert-solutions-for-class-7-maths-chapter-4-ex-4-2","status":"publish","type":"post","link":"https:\/\/mcq-questions.com\/ncert-solutions-for-class-7-maths-chapter-4-ex-4-2\/","title":{"rendered":"NCERT Solutions for Class 7 Maths Chapter 4 Simple Equations Ex 4.2"},"content":{"rendered":"

These NCERT Solutions for Class 7 Maths<\/a> Chapter 4 Simple Equations Ex 4.2 Questions and Answers are prepared by our highly skilled subject experts.<\/p>\n

NCERT Solutions for Class 7 Maths Chapter 4 Simple Equations Exercise 4.2<\/h2>\n

Question 1.
\nGive first step you will use to separate the variable and then solve the equation:
\n(a) x – 1 = 0
\n(b) x + 1 = 0
\n(c) x – 1 = 5
\n(d) x + 6 = 2
\n(e) y – 4 = -7
\n(f) y – 4 = 4
\n(g) y + 4 = 4
\n(h) y + 4 = – 4
\nAnswer:
\n(a) x – 1 = 0
\nAdding 1 on both sides, we get.
\nx – 1 + 1 = 0 + 1
\nx = 1
\nThe solution is x = 1.<\/p>\n

(b) x + 1 = 0
\nSubtracting 1 from both sides, we get.
\nx + 1 – 1 = 0 – 1
\nx = -1.
\nThe solution is x = – 1<\/p>\n

(c) x – 1 = 5
\nAdding 1 to both sides, we get.
\nx – 1 + 1 = 5 + 1
\nx = 6
\nThe solution is x = 6.<\/p>\n

(d) x + 6 = 2
\nSubtracting 6 from both sides, we get
\nx + 6 – 6 = 2 – 6
\nx = – 4
\nThe solution is x = – 4.<\/p>\n

(e) y – 4 = – 7
\nAdding 4 on both sides, we get
\ny – 4 + 4 = -7 + 4
\ny = – 3
\nThe solution is y = – 3.<\/p>\n

(f) y – 4 = 4
\nAdding + 4 on both sides, we get
\ny – 4 + 4 = 4 + 4
\ny = 8
\nThe solution is y = 8.<\/p>\n

(g) y + 4 = 4
\nSubtracting 4 from both sides, we get
\ny + 4 – 4 = 4 – 4
\ny = 0
\nThe solution is y = 0.<\/p>\n

(h) y + 4 = – 4
\nAdding – 4 on both sides, we get
\ny + 4 – 4 = -4 -4
\ny = – 8
\nThe solution is y = – 8.<\/p>\n

\"NCERT<\/p>\n

Question 2.
\nGive first step you will use to separate the variable and then solve the equation:
\n(a) 31 = 42
\n(b) \\(\\frac{\\mathrm{b}}{2}\\) = 6
\n(c) \\(\\frac{\\mathrm{p}}{7}\\) = 4
\n(d) 4x = 25
\n(e) 8y = 36
\n(f) \\(\\frac{z}{3}=\\frac{5}{4}\\)
\n(g) \\(\\frac{a}{5}=\\frac{7}{15}\\)
\n(h) 20t = -10
\nAnswer:
\n(a) 31 = 42
\nDividing both sides by 3, we get
\n\\(\\frac{31}{3}=\\frac{42}{3}\\)
\n1 = 14
\nThe solution is 1 = 14.<\/p>\n

(b) \\(\\frac{\\mathrm{b}}{2}\\) = 6
\nMultiplying both sides by 2, we get. b
\n\\(\\frac{\\mathrm{b}}{2}\\) \u00d7 2 = 6 \u00d7 2
\nb = 12
\nThe solution is b = 12.<\/p>\n

(c) \\(\\frac{\\mathrm{p}}{7}\\) = 4
\nMultiplying both sides by 7, we get
\n\\(\\frac{\\mathrm{p}}{7}\\) \u00d7 7 = 4 \u00d7 7
\np = 28
\nThe solution is p = 28.<\/p>\n

(d) 4x = 25
\nDividing both sides by 4, we get
\n\\(\\frac{4 x}{4}=\\frac{25}{4}\\)
\nx = \\(\\frac{25}{4}\\)
\nThe solution is x = \\(\\frac{25}{4}\\) or x = 6 \\(\\frac{1}{4}\\) .<\/p>\n

(e) 8y = 36
\nDividing both sides by 8, we get
\n\\(\\frac{8 y}{8}=\\frac{36}{8}\\)
\ny = \\(\\frac{9}{2}\\)
\nThe solution is y = \\(\\frac{9}{2}\\) or y = 4 \\(\\frac{1}{2}\\)<\/p>\n

(f) (f) \\(\\frac{z}{3}=\\frac{5}{4}\\)
\nMultiplying both sides by 3, we get
\n\\(\\frac{z}{3}\\) \u00d7 3 = \\(\\frac{5}{4}\\) \u00d7 3
\nz = \\(\\frac{25}{4}\\)
\nThe solution is z = \\(\\frac{15}{4}\\) or y = 3 \\(\\frac{3}{4}\\)<\/p>\n

(g) \\(\\frac{a}{5}=\\frac{7}{15}\\)
\nMultiplying both sides by 5, we get
\n\\(\\frac{a}{5}\\) \u00d7 5 = \\(\\frac{7}{15}\\) \u00d7 5
\na = \\(\\frac{7}{3}\\)
\nThe solution is a = \\(\\frac{7}{3}\\) or a = 2 \\(\\frac{1}{3}\\).<\/p>\n

(h) 20t = – 10
\nDividing both sides by 20, we get
\n\\(\\frac{20 \\mathrm{t}}{20}=\\frac{-10}{20}\\)
\nt = \\(\\frac{-1}{2}\\)
\nThe solution is t = \\(\\frac{-1}{2}\\).<\/p>\n

\"NCERT<\/p>\n

Question 3.
\nGive the steps you will use to separate the variable and then solve the equation:
\n(a) 3n – 2 = 46
\n(b) 5m +7 = 17
\n(c) \\(\\frac{20 \\mathrm{p}}{3}\\) = 40
\n(d) \\(\\frac{3 \\mathrm{p}}{10}\\) = 6
\nAnswer:
\n(a) 3n – 2 = 46
\nAdding 2 on both sides, we get
\n3n – 2 + 2 =46 + 2
\n3n = 48
\nDividing both sides by 3, we get
\n\\(\\frac{3 n}{3}=\\frac{48}{3}\\)
\nn = 16
\nThe solution is n = 16.<\/p>\n

(b) 5m + 7 = 17
\nAdding – 7 on both sides, we get
\n5m + 7 – 7 = 17 – 7
\n5m = 10
\nDividing both sides by 5, we get
\n\\(\\frac{5 m}{5}=\\frac{10}{5}\\)
\nm = 2
\nThe solution is m = 2<\/p>\n

(c) \\(\\frac{20 \\mathrm{p}}{3}\\) = 40
\nMultiplying both sides by 3, we get.
\n\\(\\frac{20 \\mathrm{p}}{3}\\) \u00d7 3 = 40 \u00d7 3
\n20 p = 120
\nDividing both sides by 20 we get.
\n\\(\\frac{20 p}{20}=\\frac{120}{20}\\)
\np = 6
\nThe solution is p = 6.<\/p>\n

(d) \\(\\frac{3 \\mathrm{p}}{10}\\) = 6
\nMultiplying both sides by 10, we get 3p
\n\\(\\frac{3 \\mathrm{p}}{10}\\) \u00d7 10 = 6 \u00d7 10
\n3p = 60
\nDividing both sides by 3, we get
\n\\(\\frac{3 p}{3}=\\frac{60}{3}\\)
\np = 20
\nThe solution is p = 20.<\/p>\n

\"NCERT<\/p>\n

Question 4.
\nSolve the following equations:
\n(a) 10p = 100
\n(b) 10p + 10 = 100
\n(c) \\(\\frac{\\mathrm{p}}{4}\\)
\n(d) \\(\\frac{-p}{3}\\) = 5
\n(e) \\(\\frac{3 \\mathrm{p}}{4}\\) = 6
\n(f) 3s = – 9
\n(g) 3s + 12 = 0
\n(h) 3s = 0
\n(i) 2q = 6
\n(j) 2q – 6 = 0
\n(k) 2q + 6 = 0
\n(l) 2q + 6 = 12
\nAnswer:
\n(a) 10p = 100
\nDividing both sides by 10, we get
\n\\(\\frac{10 \\mathrm{p}}{10}=\\frac{100}{10}\\)
\np = 10 10p = 90
\nThe solution is p = 10.<\/p>\n

(b) 10p + 10 = 100
\n10p + 10 – 10= 100 – 10
\nDividing both sides by 10, we get
\n\\(\\frac{10 p}{10}=\\frac{90}{10}\\)
\np = 9
\nThe solution is p = 9.<\/p>\n

(c) \\(\\frac{\\mathrm{p}}{4}\\) = 5
\nMultiplying both sides by 4, we get
\n\\(\\frac{\\mathrm{p}}{4}\\) \u00d7 4 = 5 \u00d7 4
\np = 20
\nThe solution is p = 20.<\/p>\n

(d) \\(\\frac{-p}{3}\\) = 5
\nMultiplying both sides by 3, we get
\n-p = 15
\n\\(\\frac{-p}{3}\\) \u00d7 3 = 5 \u00d7 3
\nMultiplying both sides by (- 1)
\n– p \u00d7 (- 1)= 15 \u00d7 (- 1)
\np = – 15
\nThe solution is p = -15.<\/p>\n

(e) \\(\\frac{3 \\mathrm{p}}{4}\\) = 6
\nMultiplying both sides by 4, we get
\n\\(\\frac{3 \\mathrm{p}}{4}\\) \u00d7 4 = 6 \u00d7 4
\n3p = 24
\nDividing both sides by 3, we get
\n\\(\\frac{3 p}{3}=\\frac{24}{3}\\)
\np = 8
\nThe solution is p = 8.<\/p>\n

(f) 3s = – 9
\nDividing both sides by 3, we get
\n\\(\\frac{3 s}{3}=\\frac{-9}{3}\\)
\ns = – 3
\nThe solution is s = – 3.<\/p>\n

(g) 3s + 12 = 0
\nSubtracting 12 from both sides, we get
\n3s + 12- 12 = 0 – 12
\n3s = – 12
\nDividing both sides by 3, we get
\n\\(\\frac{3 s}{3}=\\frac{-12}{3}\\)
\ns = – 4
\nThe solution is s = – 4.<\/p>\n

\"NCERT<\/p>\n

(h) 3s =0
\nDividing both sides by 3, we get
\n\\(\\frac{3 s}{3}=\\frac{0}{3}\\)
\ns = 0
\nThe solution is s = 0.<\/p>\n

(i) 2q = 6
\nDividing both sides by 2, we get
\n\\(\\frac{2 \\mathrm{q}}{2}=\\frac{6}{2}\\)
\nq = 3
\nThe solution is q = 3.<\/p>\n

(j) 2q – 6 = 0
\nAdding 6 on both sides, we get
\n2q – 6 + 6= 0 + 6
\n2q = 6
\nDividing both sides by 2, we get
\n\\(\\frac{2 \\mathrm{q}}{2}=\\frac{6}{2}\\)
\nq = 3
\nThe solution is q = 3.<\/p>\n

(k) 2q + 6 =0
\nSubtracting 6 from both sides, we get
\n2q + 6 – 6 = 0 – 6
\n2q = – 6
\nDividing both sides by 2, we get
\n\\(\\frac{2 q}{2}=\\frac{-6}{2}\\)
\nq = -3
\nThe solution is q = – 3.<\/p>\n

(l) 2q + 6 =12
\nSubtracting 6 from both sides, we get
\n2q + 6 – 6 = 12 – 6
\n2q = 6
\nDividing both sides by 2, we get
\n\\(\\frac{2 \\mathrm{q}}{2}=\\frac{6}{2}\\)
\nq = 3
\nThe solution is q = 3.<\/p>\n","protected":false},"excerpt":{"rendered":"

These NCERT Solutions for Class 7 Maths Chapter 4 Simple Equations Ex 4.2 Questions and Answers are prepared by our highly skilled subject experts. NCERT Solutions for Class 7 Maths Chapter 4 Simple Equations Exercise 4.2 Question 1. Give first step you will use to separate the variable and then solve the equation: (a) x …<\/p>\n

NCERT Solutions for Class 7 Maths Chapter 4 Simple Equations Ex 4.2<\/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":[9],"tags":[],"yoast_head":"\nNCERT Solutions for Class 7 Maths Chapter 4 Simple Equations Ex 4.2 - 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-7-maths-chapter-4-ex-4-2\/\" \/>\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 7 Maths Chapter 4 Simple Equations Ex 4.2 - MCQ Questions\" \/>\n<meta property=\"og:description\" content=\"These NCERT Solutions for Class 7 Maths Chapter 4 Simple Equations Ex 4.2 Questions and Answers are prepared by our highly skilled subject experts. 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