{"id":4763,"date":"2020-09-09T12:24:09","date_gmt":"2020-09-09T06:54:09","guid":{"rendered":"https:\/\/kseebsolutions.guru\/?p=4763"},"modified":"2021-07-02T15:03:34","modified_gmt":"2021-07-02T09:33:34","slug":"kseeb-solutions-class-8-maths-chapter-8-ex-8-2","status":"publish","type":"post","link":"https:\/\/kseebsolutions.guru\/kseeb-solutions-class-8-maths-chapter-8-ex-8-2\/","title":{"rendered":"KSEEB Solutions for Class 8 Maths Chapter 8 Linear Equations in One Variable Ex 8.2"},"content":{"rendered":"

Students can Download Maths Chapter 8 Linear Equations in One Variable Ex 8.2 Questions and Answers, Notes Pdf, KSEEB Solutions for Class 8 Maths<\/a> helps you to revise the complete Karnataka State Board Syllabus and score more marks in your examinations.<\/p>\n

Karnataka Board Class 8 Maths Chapter 8 Linear Equations in One Variable Ex 8.2<\/h2>\n

Question 1.
\nIf 4 is added to a number and the sum is multiplied 3, the result is 30. Find the number.
\nAnswer:
\nLet the number be \u2018x\u2019.
\nIf 4 is added to it will be x + 4.
\nSum is multiplied by 3 the result is 30.
\n\u2234 (x + 4)3 = 30
\n3x +12 = 30
\n3x = 30 – 12
\n3x = 18
\nx = \\(\\frac { 18 }{ 3 }\\)
\nx = 6
\n\u2234The number is 6<\/p>\n

Question 2.
\nFind three consecutive odd numbers whose sum is 219.
\nAnswer:
\nLet the odd number be \u2018x\u2019.
\nThe next two consecutive numbers are x + 2 and x + 4
\nx + (x + 2) + (x + 4) = 219
\n3x + 6 = 219
\n3x = 219 – 6
\n3x = 213
\nx = \\(\\frac { 213 }{ 3 }\\)
\nx = 71
\nx + 2 = 71 + 2 = 73
\nx + 4 = 71 + 4 = 75
\nThree consecutive odd numbers are 71, 73, 75<\/p>\n

Question 3.
\nA number subtracted by 30 gives 14 subtracted by 3 times the number. Find the number.
\nAnswer:
\nLet the number be x
\nNumber subtracted by 30 = 30 – x
\n14 subtracted by 3 times the number 3x – 14
\n\u223430 – x = 3x – 14
\n30+ 14 = 3x + x
\n44 = 4x
\nx = \\(\\frac { 44 }{ 4 }\\)
\nx =11
\n\u2234 The number is 11<\/p>\n

Question 4.
\nIf 5 is subtracted from three times a number the result is 16. Find the number.
\nAnswer:
\nLet the number be x, 5 is subtracted from 3 times the number the result is 16.
\n3x – 5 = 16
\n3x= 16 + 5
\n3x = 21
\nx = \\(\\frac { 21 }{ 3 }\\)
\nx = 7
\n\u2234 The number is 7<\/p>\n

Question 5.
\nFind two numbers such that one of them exceeds the other by 9 and their sum is 81.
\nAnswer:
\nLet the number is x. The other number is x + 9.
\nTheir sum is 81
\n\u2234 x + (x + 9) = 81
\n2x = 81 \u2013 9
\n2x = 72
\nx = \\(\\frac{72}{2}\\)
\nx = 36
\nx + 9 = 36 + 9 = 45
\n\u2234 The number are 36 and 45<\/p>\n

Question 6.
\nPrakruthi\u2019s age is 6 time Sahil\u2019s age. After 15 years prakruthi will be 3 times as old as Sahil. Find their age.
\nAnswer:
\nLet Sahil\u2019s present age be x. Prakruthi\u2019s present age is 6x, 15 years later Sahil age will be (x + 15) years and Prakruthi age will be (6x + 15) years.
\nGiven that the Prakruthi age will be 3 times as old as Sahil.
\n\u22346x +15 = 3(x + 15)
\n6x + 15 = 3x + 45
\n6x \u2013 3x = 45 \u2013 15
\n3x = 30
\nx = \\(\\frac{30}{3}\\)\u00a0= 10
\nSahils age = x = 10 years Prakruthis age = 6x = 6 x 10 = 60 years<\/p>\n

Question 7.
\nAhmed\u2019s father is thrice as old as Ahmed. After 12 years, his age will be twice that of his son. Find their present age.
\nAnswer:
\nAhmed\u2019s presents age be x
\nFathers present age = 3x
\n12 years later Ahmed\u2019s age = x + 12
\nand father\u2019s age = 3x + 12
\nGiven 3x + 12 = 2(x + 12)
\n3x+ 12 = 2x + 24
\n3x – 2x = 24 – 12
\nx = 12 years
\n\u2234 Ahmed\u2019s age = 12 years
\nFathers age = 3x = 3 \u00d7 12 = 36 years.<\/p>\n

Question 8.
\nSanju is 6 years older than his brother Nishu. If the sum of their ages is 28 years what are their present ages.
\nAnswer:
\nLet Nishu\u2019s age be \u2018x\u2019 Sanju\u2019s age = x + 6
\nSum of their ages = 28 x + (x + 6) = 28
\n2x + 6 = 28
\n2x = 28 – 6
\n2x = 22
\nx = \\(\\frac{22}{2}\\)
\nx = 11
\nNishu\u2019sage = x = 11 years
\nSanju\u2019 s age = x + 6 = 11 + 6 = 17 years<\/p>\n

Question 9.
\nViji is twice as old as his brother Deepu. If the difference of their ages is 11 years, find their present age.
\nAnswer:
\nLet Deep\u2019s age be \u2018x\u2019, Viji\u2019s age is 2x
\nDifference of their age = 11
\n2x – x = 11
\nx = 11
\n\u2234 Deepu\u2019s age = x = 11 years
\nViji\u2019s age = 2x = 2 x 11 = 22 years<\/p>\n

Question 10.
\nMrs. Joseph is 27 years older than her daughter Bindu. After 8 years she will be twice as old as Bindu. Find their present ages.
\nAnswer:
\nLet Bindu\u2019s present age be \u2018x\u2019 years.
\nMrs. Joseph\u2019s present age = x + 27 years.
\nAfter 8 years Bindu\u2019s age = x + 8 and Mrs.
\nJosephs age = x + 27 + 8
\n= x + 35 years.
\nGiven that x + 35 = 2(x + 8)
\nx + 35 = 2x + 16
\n35 – 16 = 2x – x
\n19 = x
\nx= 19
\nBindu\u2019s present age = 19 years
\nMrs.Joseph\u2019s age = x + 27 = 19 + 27 = 46 years<\/p>\n

Question 11.
\nAfter 16 years Leena will be three times as old as she is now. Find her present age.
\nAnswer:
\nLet Leena\u2019s present age be \u2018x\u2019 years
\n16 years later she will be (x + 16) years
\nGiven x + 16 = 3x
\n16 = 2x
\nx = \\(\\frac { 16 }{ 2 }\\)
\nx = 8 years
\nLeena’ s present age = 8 years<\/p>\n

Question 12.
\nA rectangle has a length which is 5 cm less than twice its breadth. If the length is decreased by 5 cm and breadth is increased by 2 cm the perimeter of the resulting rectangle will be 74 cm. Find the length and breadth of the original rectangle.
\nAnswer:
\nLet the breadth of the original rectangle be \u2018b\u2019 twice the breadth is 2b.
\nThe length of the rectangle is 5 cm less than twice the breadth.
\n\u2234 Length = 2b \u2013 5
\nIf the length is decreased by 5 then length
\nis 2b \u2013 5 \u2013 5 = 2b \u2013 10
\nIf the breadth is increased by 2cm then breadth is b + 2 cm.
\nPerimeter of new rectangle = 2(length + breadth)
\n74 = 2(2b \u2013 10 + b + 2)
\n74 = 2(3b \u2013 8)
\n74 + 16 = 6b 90 = 6b
\nb = \\(\\frac{90}{6}\\)
\nb = 15cm.
\nBreadth of the original rectangle = 15 cm
\nLength of the original rectangle = 2b \u2013 5
\n= 2 \u00d7 15 \u2013 5
\n= 30 \u2013 5
\n= 25 cm<\/p>\n

Question 13.
\nThe length of a rectangular field is twice its breadth. If the perimeter of the field is 288m. Find the dimensions of the field.
\nAnswer:
\nLet breadth of the rectangular field bee \u2018x\u2019m
\n\u2234 Its length = 2x
\nIts perimeter = 288
\n2(1 + b) = 288
\n2(2x + x) = 288
\n2(3x) = 288
\n6x = 288
\nx = \\(\\frac { 288 }{ 6 }\\) = 48
\nIts length = 2x = 2 \u00d7 48 = 96m
\nbreadth = x = 48m<\/p>\n

Question 14.
\nSrishti’s salary is the same as 4 times Azar\u2019s salary. If together they earn Rs 3750 a month find their individual salaries.
\nAnswer:
\nLet Axar\u2019s salary be x. Sristi\u2019s salary is 4x together they earn Rs. 3750
\n\u2234 x + 4x = 3750
\n5x = 3750
\nx = \\(\\frac{3750}{5}\\)
\nx = 750
\n4x = 4 \u00d7 750 = 3000
\n\u2234 Azar\u2019s salary is Rs.750 and Sristi\u2019s salary is Rs.3000<\/p>\n","protected":false},"excerpt":{"rendered":"

Students can Download Maths Chapter 8 Linear Equations in One Variable Ex 8.2 Questions and Answers, Notes Pdf, KSEEB Solutions for Class 8 Maths helps you to revise the complete Karnataka State Board Syllabus and score more marks in your examinations. Karnataka Board Class 8 Maths Chapter 8 Linear Equations in One Variable Ex 8.2 …<\/p>\n","protected":false},"author":5,"featured_media":0,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"footnotes":""},"categories":[6],"tags":[],"jetpack_sharing_enabled":true,"jetpack_featured_media_url":"","_links":{"self":[{"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/posts\/4763"}],"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=4763"}],"version-history":[{"count":0,"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/posts\/4763\/revisions"}],"wp:attachment":[{"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/media?parent=4763"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/categories?post=4763"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/tags?post=4763"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}