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>\nNCERT 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
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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
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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
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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
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(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 \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n\t \n\t \n\t \n