{"id":26947,"date":"2021-06-28T18:20:18","date_gmt":"2021-06-28T12:50:18","guid":{"rendered":"https:\/\/mcq-questions.com\/?p=26947"},"modified":"2022-03-02T10:47:26","modified_gmt":"2022-03-02T05:17:26","slug":"ncert-solutions-for-class-7-maths-chapter-4-ex-4-1","status":"publish","type":"post","link":"https:\/\/mcq-questions.com\/ncert-solutions-for-class-7-maths-chapter-4-ex-4-1\/","title":{"rendered":"NCERT Solutions for Class 7 Maths Chapter 4 Simple Equations Ex 4.1"},"content":{"rendered":"

These NCERT Solutions for Class 7 Maths<\/a> Chapter 4 Simple Equations Ex 4.1 Questions and Answers are prepared by our highly skilled subject experts.<\/p>\n

NCERT Solutions for Class 7 Maths Chapter 4 Simple Equations Exercise 4.1<\/h2>\n

Question 1.
\nComplete the last column of the table.
\nAnswer:<\/p>\n\n\n\n\n\n\n\n\n\n\n\n\n\n
Equation<\/td>\nValue<\/td>\nSay, whether the Equation is Satisfied, (Yes\/No)<\/td>\n<\/tr>\n
(i) x + 3 = 0<\/td>\nx = 3<\/td>\nNo<\/td>\n<\/tr>\n
(ii) x + 3 = 0<\/td>\nx = 0<\/td>\nNo<\/td>\n<\/tr>\n
(iii) x + 3 = 0<\/td>\nx = -3<\/td>\nyes<\/td>\n<\/tr>\n
(iv) x – 7 = 1<\/td>\nx = 7<\/td>\nNo<\/td>\n<\/tr>\n
(v) x – 7 = 1<\/td>\nx = 8<\/td>\nyes<\/td>\n<\/tr>\n
(vi) 5x = 25<\/td>\nx = 0<\/td>\nNo<\/td>\n<\/tr>\n
(vii) 5x = 25<\/td>\nx = – 5<\/td>\nNo<\/td>\n<\/tr>\n
(viii) \\(\\frac{m}{3}\\) = 2<\/td>\nm = -6<\/td>\nNo<\/td>\n<\/tr>\n
(ix) \\(\\frac{m}{3}\\) = 2<\/td>\nm = 0<\/td>\nNo<\/td>\n<\/tr>\n
(x) \\(\\frac{m}{3}\\) = 2<\/td>\n\u00a0m = 6<\/td>\nyes<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n

\"NCERT<\/p>\n

Question 2.
\nCheck whether the value given in the brackets is a solution to the given equation or not:
\n(a) n + 5 = 19 (n = 1)
\n(b) 7n + 5 = 19 (n = – 2)
\n(c) 7n + 5 = 19 (n = 2)
\n(d) 4p – 3 = 13 (p = 1)
\n(e) 4p – 3 = 13 (p = – 4)
\n(f) 4p – 3 = 13 (p = 0)
\nAnswer:
\n(a) n + 5 = 19
\n1 + 5 =19 (Putting n = 1)
\n6 \u2260 19
\n\u2234 n = 1 is not a solution.<\/p>\n

(b) 7n + 5 =19
\n7(- 2) + 5 =19 (Put n = – 2)
\n-14 + 5 = 19
\n-9 \u2260 19
\nn = -2 is not a solution.<\/p>\n

(c) 7n + 5 =19
\n7(2) + 5=19 (Put n = 2)
\n14 + 5 = 19
\n19 = 19
\n\u2234 n = 2 is a solution.<\/p>\n

(d) 4p – 3 = 13
\n4(1) – 3 =13 (Put p = 1)
\n4 – 3 = 13
\n1 \u2260 13
\n\u2234 p = 1 is not a solution.<\/p>\n

(e) 4p – 3 = 13
\n4(- 4) – 3 = 13 (Put p = – 4)
\n-16 -3 = 13
\n-19 \u2260 13
\n\u2234 p = – 4 is not a solution.<\/p>\n

(f) 4p – 3 = 13
\n4(0) – 3 = 13 (Put p = 0)
\n0 – 3 = 13
\n-3 \u2260 13
\np = 0 is not a solution.<\/p>\n

\"NCERT<\/p>\n

Question 3.
\nSolve the following equations by trial and error method:
\n(i) 5p + 2 = 17
\n(ii) 3m – 14 = 4
\nAnswer:
\n(i) 5p + 2 = 17
\nPut p = 0
\nL.H.S = 5 (0) + 2
\n= 0 + 2
\n= 2 \u2260 RHS<\/p>\n

Put p = 1
\nL.H.S = 5(1)+ 2
\n= 5 + 2
\n= 7 \u2260 RHS<\/p>\n

Put p = -1
\nL.H.S = 5 (-1) + 2
\n= -5 + 2
\n= -3 \u2260 RHS<\/p>\n

Put p = 2
\nL.H.S = 5 (2) + 2
\n= 10 + 2
\n= 12 \u2260 RHS<\/p>\n

Put p = -2
\nL.H.S = 5 (-2) + 2
\n= -10 + 2
\n= -8 \u2260 17 RHS<\/p>\n

Put p = 3
\nL.H.S = 5 (3) + 2
\n= 15 + 2
\n= 17 = RHS
\np = 3 is the solution of 5P + 2 = 17<\/p>\n

(ii) 3m – 14 =4
\nPut m = 0
\nL.H.S = 3m – 14
\n= 3 (0) – 14
\n= – 14
\n– 14 \u2260 4
\nL.H.S \u2260 R.H.S<\/p>\n

Put m = 1
\nL.H.S = 3m -14
\n= 3(1) – 14
\n= 3 – 14
\n= – 11 \u2260 R.H.S
\nL.H.S \u2260 R.H.S<\/p>\n

Put m = 2
\nL.H.S = 3m – 14
\n= 3 (2) – 14
\n= 6 – 14
\n= – 8 \u2260 R.H.S
\nL.H.S \u2260 R.H.S<\/p>\n

Put m = 3
\nL.H.S = 3(3) – 14
\n= 9 – 14
\n= – 5 \u2260 R.H.S
\nL.H.S \u2260 R.H.S<\/p>\n

Put m = 4
\nL.H.S = 3(4) – 14
\n= 12 – 14
\n= – 2 \u2260 R.H.S
\nL.H.S \u2260 R.H.S<\/p>\n

Put m = 5
\nL.H.S = 3(5) – 14
\n= 15 – 14
\n= 1 \u2260 R.H.S
\nL.H.S \u2260 R.H.S<\/p>\n

Put m = 6
\nL.H.S = 3(6) – 14
\n= 18 – 14
\n= 4 = R.H.S
\nL.H.S = R.H.S
\nm = 6 is the solution to 3m – 14 = 4<\/p>\n

\"NCERT<\/p>\n

Question 4.
\nWrite equations for the following statements:
\n(i) The sum of numbers x and 4 is 9.
\n(ii) 2 subtracted from y is 8.
\n(iii) Ten times a is 70.
\n(iv) The number b divided by 5 gives 6.
\n(v) Three-fourth of t is 15.
\n(vi) Seven times m plus 7 gets you 77.
\n(vii) One-fourth of a number x minus 4 gives 4.
\n(viii) If you take away 6 from 6 times y, you get 60.
\n(ix) If you add 3 to one-third of z, you get 30.
\nAnswer:
\n(i) x + 4 = 9
\n(ii) y – 2 = 8
\n(iii) 10 a = 70
\n(iv) \\(\\frac { b }{ 5 }\\) = 6
\n(v) \\(\\frac { 3t }{ 4 }\\) = 15
\n(vi) 7m + 7 = 77
\n(vii) \\(\\frac { x }{ 4 }\\) x – 4 = 4
\n(viii) 6 y – 6 = 60
\n(ix) \\(\\frac { z }{ 3 }\\) + 3 = 30<\/p>\n

Question 5.
\nWrite the following equations in statement forms:
\n(i) p + 4 = 15
\n(ii) m – 7 = 3
\n(iii) 2m = 7
\n(iv) \\(\\frac { m }{ 5 }\\) = 3
\n(v) \\(\\frac { 3m }{ 5 }\\) = 6
\n(vi) 3p + 4 = 25
\n(vii) 4p – 2 = 18
\n(viii) \\(\\frac { p }{ 2 }\\) + 2 = 8
\nAnswer:
\n(i) The sum of p and 4 is 15.
\n(ii) 7 subtracted from m is 3.
\n(hi) Twice a number m is 7.
\n(iv) One-fifth of a number m is 3.
\n(v) Three-fifth of a number m is 6.
\n(vi) Three times a number p when added to 4 gives 25.
\n(vii) 2 subtracted from four times a number p is 18.
\n(viii) 2 added to half of a number p is 8.<\/p>\n

\"NCERT<\/p>\n

Question 6.
\nSet up an equation in the following cases:
\n(i) Irfan says that he has 7 marbles more than live times the marbles Parmit has. Irfan has 37 marbles. (Take m to be the number of Parmit\u2019s marbles.)<\/p>\n

(ii) Laxmi\u2019s father is 49 years old. He is 4 years older than three times Laxmi\u2019s age. (Take Laxmi\u2019s age to be y years.)<\/p>\n

(iii) The teacher tells the class that the highest marks obtained by a student in her class is twice the lowest marks plus 7. The highest score is 87. (Take the lowest score to be 1.)<\/p>\n

(iv) In an isosceles triangle, the verte x angle is twice either base angle. (Let the base angle be b in degrees. Remember that the sum of angles of a triangle is 180 degrees).
\nAnswer:
\n(i) Let Parmit has m marbles.
\nThen, five time the marbles Parmit has = 5 m.
\nIrfan has 7 marbles more than five times the marbles parmit has
\nSo, Irfan has (5m + 7) marbles
\nBut it is given that Irfan has 37 marbles.
\n5m + 7 = 37<\/p>\n

(ii) Let Laxmi\u2019s age = y years
\n3 times Laxmi\u2019s age = 3y years.
\nAge of Laxmi\u2019s father = 3 times
\nLaxmi\u2019s age + 4 years
\n= 3y + 4 years
\nBut Laxmi\u2019s father is 49 years old.
\n3y + 4 = 49<\/p>\n

(iii) Let the lowest score (marks) = 1
\nTwice the lowest marks = 2 l
\nSince highest marks = (twice the lowest marks) + 7 = 2l + 7
\nBut the highest marks = 87
\n2l + 7 = 87<\/p>\n

(iv) Let the base angle be b degrees
\nThe base angle of an isosceles triangle are equal.
\nThe other base angle = b degrees
\nSince the vertex angle = Twice either base angle = 2b degrees.
\nAlso, the sum of three angles of triangle = 180\u00b0
\nb + b + 2b = 180\u00b0
\n(OR)
\n4b = 180\u00b0<\/p>\n","protected":false},"excerpt":{"rendered":"

These NCERT Solutions for Class 7 Maths Chapter 4 Simple Equations Ex 4.1 Questions and Answers are prepared by our highly skilled subject experts. NCERT Solutions for Class 7 Maths Chapter 4 Simple Equations Exercise 4.1 Question 1. Complete the last column of the table. Answer: Equation Value Say, whether the Equation is Satisfied, (Yes\/No) …<\/p>\n

NCERT Solutions for Class 7 Maths Chapter 4 Simple Equations Ex 4.1<\/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 4.1 - 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-4-1\/\" \/>\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 4.1 - MCQ Questions\" \/>\n<meta property=\"og:description\" content=\"These NCERT Solutions for Class 7 Maths Chapter 4 Simple Equations Ex 4.1 Questions and Answers are prepared by our highly skilled subject experts. 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