\n7<\/td>\n | 10 \u00d7 7 – 20 = 50<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n Thus the condition 10y – 20 is true for y = 7<\/p>\n NCERT In-text Question Page No. 80<\/span> \nQuestion 1. \nWrite atleast one other form for each equation (ii), (iii) and (iv)<\/p>\n\n\n\nEquation<\/td>\n | Other form of the equation<\/td>\n<\/tr>\n | \n(ii) 5p = 20<\/td>\n | (a) 5p – 4 = 16<\/td>\n<\/tr>\n | \n(iii) 3n + 7 = 1<\/td>\n | (iii) 4n + 9 = 45<\/td>\n<\/tr>\n | \n(iv) \\(\\frac{\\mathrm{m}}{\\mathrm{s}}\\)<\/td>\n | (iv) \\(\\frac{\\mathrm{m}}{\\mathrm{s}}\\)<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n Statement of other form of the equation: \n(a) Taking away 4 from five times p gives 16. \n(b) Add 9 to four times n to get 45. \n(c) Add 8 to one-third of m to get 17.<\/p>\n <\/p>\n NCERT In-text Question Page No. 88<\/span> \nQuestion 1. \nStart with the same step x = 5 and make two different equations. Ask two of your classmates to solve the equations. Check whether they get the solution x = 5. \nAnswer: \n(i) x = 5 \nMultiplying both sides by 2, we have \n2x = 10 \nAdding 6 to both sides, we have \nor 2x + 6 = 10 + 6 \nor 2x + 6 = 16 is an equation.<\/p>\n(ii) x = 5 \nDividing both sides by 3, we have \n\\(\\frac{x}{3}=\\frac{5}{3}\\) \nSubtracting 2 from both sides, we have \n\\(\\frac{\\mathrm{x}}{3}\\) – 2 = \\(\\frac{5}{3}\\) – 2 \nor \n\\(\\frac{\\mathrm{x}}{3}\\) – 2 = \\(\\frac{5-6}{3}=\\frac{-1}{3}\\) \nThus, \n\\(\\frac{\\mathrm{x}}{3}\\) – 2 = \\(\\frac{-1}{3}\\) is an equation.<\/p>\n Solution to I \n2x + 6 = 16 \nSubtracting 6 from both sides, we have \n2x + 6- 6 = 16 – 6 \nor 2x = 10 \nDividing both sides by 2, we have \n\\(\\frac{2 \\mathrm{x}}{2}=\\frac{10}{2}\\) or x = 5<\/p>\n Solution to II \n\\(\\frac{\\mathrm{x}}{3}\\) – 2 = \\(\\frac{-1}{3}\\) \nAdding 2 to both sides, we have \n\\(\\frac{\\mathrm{x}}{3}\\) – 2 + 2 = \\(\\frac{-1}{3}\\) + 2 or \\(\\frac{x}{3}=\\frac{5}{3}\\) \nMultiplying both sides by 3, we have \n\\(\\frac{\\mathrm{x}}{3}\\) \u00d7 3 = \\(\\frac{5}{3}\\) \u00d7 3 or x = 5<\/p>\n <\/p>\n NCERT In-text Question Page No. 88<\/span> \nQuestion 1. \nTry to make two number puzzles, one with the solution 11 and another with 100. \nAnswer: \nI. A puzzle having the solution as 11: \nThink of a number. Multiply bit by 3 and add 7. Tell me the sum. \nIf the sum is 40, then the number is 11.<\/p>\nII. A puzzle having the solution as 100: \nThink of a number. Divide it by 4 and add 5. Tell me what you get. \nIf you get 30, then the number is 100.<\/p>\n Note: \nInstead of making the same operation on both sides, we can move a number from one side to another by changing its sign from (+) to (-) and (-) to (+). This is called \u2018transposing a number\u2019. Thus, transposing a number is the same as adding or subtracting the number from both sides. In doing so, the sign of the number has to be changed.<\/p>\n NCERT In-text Question Page No. 90<\/span> \nQuestion 1. \n(i) When you multiply a number by 6 and subtract 5 from the product, you get 7. Can you tell what the number is?<\/p>\n(ii) What is that number one-third of which added to 5 gives 8? \nAnswer: \n(i) Let the number be x. \n\u2234 Multiply the number by 6, we have 6x. \nNow, according to the condition, we have \n6x = 7 + 5 = 12 \nDividing both sides by 6, we have \n\\(\\frac{6 x}{6}=\\frac{12}{6}\\) \nor x = 2 \n\u2234 The required number = 2<\/p>\n (ii) Let the required number be x. \n\u2235 One-third of the number = \\(\\frac{1}{3}\\)x \n\u2234 According to the condition, we have \n5 + \\(\\frac{1}{3}\\)x = 8 \nTransposing 5 from L.H.S. to R.H.S., we have \n\\(\\frac{1}{3}\\)x = 8 – 5 = 3 \nMultiplying both sides by 3, we have \n3 \u00d7 \\(\\frac{1}{3}\\)x = 3 \u00d7 3 \nor x = 9 \nThus, the required number = 9.<\/p>\n <\/p>\n NCERT In-text Question Page No. 90<\/span> \nQuestion 1. \nThere are two types of boxes containing mangoes. Each box of the larger type contains 4 more mangoes than the number of mangoes contained in 8 boxes of he smaller type. Each larger box contains 100 mangoes. Find the number of mangoes contained in the smaller box? \nAnswer: \nLet the number of mangoes contained in the smaller box be x. \n\u2234 Number of mangoes in 8 smaller boxes = 8x \nNow, according to the condition, we have \n8x + 4 = 100 \nTransposing 4 to R.H.S., we have \n8x = 100 – 4 or 8x = 96 \nDividing both sides by 8, we have \n\\(\\frac{8 x}{8}=\\frac{96}{8}\\) \nor x = 12 \nThus, the number of mangoes in the smaller box = 12.<\/p>\n","protected":false},"excerpt":{"rendered":"These NCERT Solutions for Class 7 Maths Chapter 4 Simple Equations InText Questions and Answers are prepared by our highly skilled subject experts. NCERT Solutions for Class 7 Maths Chapter 4 Simple Equations InText Questions NCERT In-text Question Page No. 78 Question 1. The value of the expression (10y – 20) depends on the value …<\/p>\n NCERT Solutions for Class 7 Maths Chapter 4 Simple Equations InText Questions<\/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 InText Questions - 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 | |