{"id":14700,"date":"2020-09-09T11:08:05","date_gmt":"2020-09-09T05:38:05","guid":{"rendered":"https:\/\/kseebsolutions.guru\/?p=14700"},"modified":"2021-07-02T15:03:35","modified_gmt":"2021-07-02T09:33:35","slug":"kseeb-solutions-for-class-7-maths-chapter-4-ex-4-1","status":"publish","type":"post","link":"https:\/\/kseebsolutions.guru\/kseeb-solutions-for-class-7-maths-chapter-4-ex-4-1\/","title":{"rendered":"KSEEB Solutions for Class 7 Maths Chapter 4 Simple Equations Ex 4.1"},"content":{"rendered":"

Students can Download Chapter 4 Simple Equations Ex 4.1, Question and Answers, Notes Pdf, KSEEB Solutions for Class 7 Maths<\/a>, Karnataka State Board Solutions help you to revise complete Syllabus and score more marks in your examinations.<\/p>\n

Karnataka State Syllabus Class 7 Maths Chapter 4 Simple Equations Ex 4.1<\/h2>\n

Question 1.
\nComplete the last column of the table.
\nSolution:
\n\"KSEEB<\/p>\n

Question 2.
\nCheck whether the value given in the brackets is a solution to the given equation or not :
\na) n + 5 = 19 (n = 1)
\nSolution:
\n1 + 5 \u2260 19
\n6 \u2260 19
\n\u2234 LHS \u2260 RHS
\n\u2234 n = 1 is not a solution to the given equation n + 5 = 19.<\/p>\n

\"KSEEB<\/p>\n

b) 7n + 5 = 19(n = -2)
\nSolution:
\n7 \u00d7 -2 + 5 = 19
\n-14 + 5 = 19
\n-9 = 19
\nLHS \u2260 RHS
\n\u2234 n = -2 is not a solutions the given equation 7n + 5 = 19 (n = 2)
\n7 \u00d7 2 + 5 = 19
\n14 + 5 = 19
\n19 = 19
\nLHS = RHS
\n\u2234 n = 2 is a solution to the given equation 7n + 5 = 19<\/p>\n

c) 7n + 5 = 19 (n = 2)
\nSolution:
\n7 \u00d7 2+5 = 19
\n14 + 5 = 19
\n19 = 19
\nLHS = RHS
\n\u2234 n = 2 is a solution to the given equation 7n + 5 = 19.<\/p>\n

d) 4p – 3= 13 (p = 1)
\n4(1) – 3 = 13
\n4 – 3 = 13
\n1 \u2260 13
\nLHS \u2260 RHS
\n\u2234 p = 1 is not a solution to the given equation 4p – 3 = 13.<\/p>\n

e) 4p – 3 = 13 (p = -4)
\n4(-4) – 3 = 13
\n– 16 – 3 = 13
\n-19 \u2260 13
\nLHS \u2260 RHS
\n\u2234 p = -4 is not a solution to the given equation 4p – 3 = 13.<\/p>\n

f) 4p – 3 = 13 (p = 0)
\n4(0) – 3 = 13
\n0 – 3 = 13
\n– 3 \u2260 13
\nLHS \u2260 RHS
\n\u2234 p = 0 is not a solution to the given equation 4p – 3 = 13.<\/p>\n

\"KSEEB<\/p>\n

Question 3.
\nSolve the following equations by trial and error method :
\ni) 5p + 2 = 17
\nSolution:
\nIf p = 0 then 5(0) + 2 = 17, 0 + 2 \u2260 17
\nIf p = 1 then 5(1) + 2 = 17, 5 + 2 \u2260 17
\nIf p = 2 then 5(2) + 2 = 17, 10 + 2 \u2260 17
\nIf p = 3 then 5(3) + 2 = 17, 15 + 2 \u2260 17
\n\u2234 p = 3 is the solution to the given equation 5p + 2 = 17.<\/p>\n

ii) 3m – 14 = 4
\nIf m = 0 then 3(0) – 14
\n0 – 14 = – 14 \u2260 4
\n3m – 14 = 4
\nIf m = 1 then 3(1) – 14
\n3 – 14 = -11
\nIf m = 2 then 3(2) – 14
\n6 – 14 = – 8
\nIf m = 3 then 3(3) – 14 = -5
\nIf m = 4 then 3(4) – 14 = -2
\nIf m = 5 then 3(5) – 14 = 1
\nIf m = 6 then 3(6) – 14 = 4
\nLHS = RHS
\n\u2234 m = 6 is the solution to the given equation 3m – 14 = 4.<\/p>\n

\"KSEEB<\/p>\n

Question 4.
\nWrite equations for the following statements :
\ni) The sum of numbers x and 4 is 9.
\nSolution:
\nx + 4 = 9<\/p>\n

ii) 2 subtracted from y is 8.
\nSolution:
\ny – 2 = 8<\/p>\n

iii) Ten times a is 70.
\nSolution:
\n10a = 70<\/p>\n

iv) The number b divided by 5 gives 6.
\nSolution:
\n\\(3 \/ 4 t=15\\)<\/p>\n

v) Three-fourth of t is 15.
\nSolution:
\nt = 15<\/p>\n

vi) Seven times m plus 7 gets you 77.
\nSolution:
\n7m + 1 = 11<\/p>\n

vii) One-fourth of a number x minus 4 gives 4.
\nSolution:
\n\"KSEEB<\/p>\n

viii) If you take away 6 from 6 times y, you get 60.
\nSolution:
\n6y – 6 = 60<\/p>\n

ix) If you add 3 to one-third of Z, you get 30.
\nSolution:
\n\"KSEEB<\/p>\n

Question 5.
\nWrite the following equations in statement forms :
\ni) p + 4 = 15
\nSolution:
\nThe sum of P and 4 is equal to 15 .<\/p>\n

ii) m – 7 = 3
\nSolution:
\nSeven subtracted from m is equal to 3.<\/p>\n

iii) 2m = 1
\nSolution:
\nTwo times a number m is equal to 7.<\/p>\n

iv) \\(\\frac{\\mathrm{m}}{5}\\) = 3
\nSolution:
\nOne – fifth of \u2018m\u2019 number is equal to 3 3m<\/p>\n

v) \\(\\frac{\\mathrm{3m}}{5}\\) = 6
\nSolution:
\nThree – fifth of \u2018m\u2019 number is equal to 6<\/p>\n

vi) 3p + 4 = 25
\nSolution:
\nThree times a number \u2018P\u2019 is added to 4 is equal to 25.<\/p>\n

\"KSEEB<\/p>\n

vii) 4p – 2 = 18
\nSolution:
\nTwo is subtracted from 4 times \u2018p\u2019 is equal to 18.<\/p>\n

viii)
\n\"KSEEB
\nSolution:
\nTwo is added to half number of p is equal to 8.<\/p>\n

Question 6.
\nSet up an equation in the following cases:
\ni) Irfan says that he has 7 marbles more than five times the marbles Parmit has. Irfan has 37 marbles. (Take m to be the number of Parmit\u2019s marbles.)
\nSolution:
\nLet the number of marbles Parmit had be \u2018m\u2019
\n\u2234 Number of marbles Irfan has = Five times of parmits marble 5m.
\nTo this 7 marbles more means = 5m + 7
\n\u2234 5m + 7 = 37.<\/p>\n

\"KSEEB<\/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 bey years)
\nSolution:
\nLet the age of Lakshmi be \u2018y\u2019 years
\nThree times of age = 3y
\nMore than 4 years = 3y + 4
\n\u2234 Lakshmi\u2019s father age be = 3y + 4
\n\u2234 3y + 4 = 49<\/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 l.)
\nSolution:
\nLet the lowest score be ‘l’
\nTwice the lowest score be = 2l
\nPlus seven = 2l + 1
\n\u2234 The highest score = 2l + 7
\n\u2234 2l + 7 = 87<\/p>\n

iv) In an isosceles triangle, die vertex angle is twice either base angle. (Let the base angle be A in degrees. Remember that the sum of angles of a triangle is 180 degrees).
\nSolution:
\nLet the base angle be b\u00b0
\nVertex angle = Twice the base angle
\n\u2234 Vertex angle = 2b
\nSum of the angles of an isosceles triangle is =180\u00b0
\n\u2234 2b + b + b = 180\u00b0 = 4b = 180\u00b0<\/p>\n","protected":false},"excerpt":{"rendered":"

Students can Download Chapter 4 Simple Equations Ex 4.1, Question and Answers, Notes Pdf, KSEEB Solutions for Class 7 Maths, Karnataka State Board Solutions help you to revise complete Syllabus and score more marks in your examinations. Karnataka State Syllabus Class 7 Maths Chapter 4 Simple Equations Ex 4.1 Question 1. Complete the last column …<\/p>\n","protected":false},"author":5,"featured_media":0,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"footnotes":""},"categories":[84],"tags":[],"jetpack_sharing_enabled":true,"jetpack_featured_media_url":"","_links":{"self":[{"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/posts\/14700"}],"collection":[{"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/users\/5"}],"replies":[{"embeddable":true,"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/comments?post=14700"}],"version-history":[{"count":0,"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/posts\/14700\/revisions"}],"wp:attachment":[{"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/media?parent=14700"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/categories?post=14700"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/tags?post=14700"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}