{"id":27141,"date":"2021-06-29T16:15:58","date_gmt":"2021-06-29T10:45:58","guid":{"rendered":"https:\/\/ncertsolutions.guru\/?p=27141"},"modified":"2021-06-29T16:20:52","modified_gmt":"2021-06-29T10:50:52","slug":"ncert-solutions-for-class-7-maths-chapter-4-intext-questions","status":"publish","type":"post","link":"https:\/\/ncertsolutions.guru\/ncert-solutions-for-class-7-maths-chapter-4-intext-questions\/","title":{"rendered":"NCERT Solutions for Class 7 Maths Chapter 4 Simple Equations InText Questions"},"content":{"rendered":"

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

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

NCERT In-text Question Page No. 78<\/span>
\nQuestion 1.
\nThe value of the expression (10y – 20) depends on the value of y. Verify this by giving five different values to y and finding for each y the value of (10y – 20). From the different values of (10y – 20) you obtain, do you see a solution to 10y – 20 = 50? If there is no solution, try giving more values to y and find whether the condition 10y – 20 = 50 is met.
\nAnswer:<\/p>\n\n\n\n\n\n\n\n\n\n\n\n
Value of y<\/td>\nValue of expression (10y – 20)<\/td>\n<\/tr>\n
0<\/td>\n10 \u00d7 0 – 20 = -20<\/td>\n<\/tr>\n
1<\/td>\n10 \u00d7 1 – = – 10<\/td>\n<\/tr>\n
2<\/td>\n10 \u00d7 2 – 20 = 0<\/td>\n<\/tr>\n
3<\/td>\n10 \u00d7 3 – 20 = 10<\/td>\n<\/tr>\n
4<\/td>\n10 \u00d7 4 – 20 = 20<\/td>\n<\/tr>\n
5<\/td>\n10 \u00d7 5 – 20 = 30<\/td>\n<\/tr>\n
6<\/td>\n10 \u00d7 6 – 20 = 40<\/td>\n<\/tr>\n
7<\/td>\n10 \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\n\n\n\n
Equation<\/td>\nOther form of the equation<\/td>\n<\/tr>\n
(ii) 5p = 20<\/td>\n(a) 5p – 4 = 16<\/td>\n<\/tr>\n
(iii) 3n + 7 = 1<\/td>\n(iii) 4n + 9 = 45<\/td>\n<\/tr>\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

\"NCERT<\/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

\"NCERT<\/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>\n

II. 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

\"NCERT<\/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":{"spay_email":""},"categories":[9],"tags":[],"yoast_head":"\nNCERT Solutions for Class 7 Maths Chapter 4 Simple Equations InText Questions - 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-7-maths-chapter-4-intext-questions\/\" \/>\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 InText Questions - NCERT Solutions\" \/>\n<meta property=\"og:description\" content=\"These NCERT Solutions for Class 7 Maths Chapter 4 Simple Equations InText Questions and Answers are prepared by our highly skilled subject experts. 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