{"id":42716,"date":"2022-05-25T16:30:04","date_gmt":"2022-05-25T11:00:04","guid":{"rendered":"https:\/\/cbselibrary.com\/?p=42716"},"modified":"2023-11-10T09:51:51","modified_gmt":"2023-11-10T04:21:51","slug":"ml-aggarwal-class-7-solutions-for-icse-maths-chapter-9-ex-9-2","status":"publish","type":"post","link":"https:\/\/cbselibrary.com\/ml-aggarwal-class-7-solutions-for-icse-maths-chapter-9-ex-9-2\/","title":{"rendered":"ML Aggarwal Class 7 Solutions for ICSE Maths Chapter 9 Linear Equations and Inequalities Ex 9.2"},"content":{"rendered":"

ML Aggarwal Class 7 Solutions for ICSE Maths Chapter 9 Linear Equations and Inequalities Ex 9.2<\/h2>\n

Question 1.
\nIf 7 is added to five times a number, the result is 57. Find the number.
\nSolution:
\nLet the number be x
\nFive times a number is = 5x
\nIf 7 is added, it becomes 7 + 5x
\nAccording to given condition,
\n7 + 5x = 57
\n\u21d2 5x = 57 – 7
\n\u21d2 5x = 50
\n\u21d2 x = 10<\/p>\n

Question 2.
\nFind a number, such that one-fourth of the number is 3 more than 7.
\nSolution:
\nLet number = x
\nAccording to the condition,
\n\\(\\frac { 1 }{ 4 }\\) x – 3 = 7
\n\\(\\frac { 1 }{ 4 }\\) x = 7 + 3 = 10
\nx = 10 \u00d7 4 = 40
\nNumber = 40<\/p>\n

Question 3.
\nA number is as much greater than 15 as it is less than 51. Find the number.
\nSolution:
\nLet the number be x
\nIf it is greater than 15, it becomes x – 15.
\nIf it is less than 51, it becomes 51 – x
\nAccording to statement,
\nx – 15 = 51 – x
\n\u21d2 x + x = 51 + 15
\n\u21d2 2x = 66
\n\u21d2 x = 33<\/p>\n

Question 4.
\nIf \\(\\frac { 1 }{ 2 }\\) is subtracted from a number and the difference is multiplied by 4, the result is 5. What is the number?
\nSolution:
\nLet the number = x
\nAccording to the condition,
\n(x – \\(\\frac { 1 }{ 2 }\\)) \u00d7 4 = 5
\n\u21d2 4x – 2 = 5
\n\u21d2 4x = 5 + 2
\n\u21d2 4x = 7
\n\u21d2 x = \\(\\frac { 7 }{ 4 }\\)
\nNumber = \\(\\frac { 7 }{ 4 }\\)<\/p>\n

Question 5.
\nThe sum of two numbers is 80 and the greater number exceeds twice the smaller by 11. Find the numbers.
\nSolution:
\nLet the numbers be x and y
\nSmaller number = x
\nGreater number = y
\nIf greater number exceeds twice the smaller by 11,
\nIt becomes y = 2x + 11
\nAccording to statement,
\nx + 2x + 11 = 80
\n\u21d2 3x + 11 = 80
\n\u21d2 3x = 80 – 11
\n\u21d2 3x = 69
\n\u21d2 x = 23
\nSmaller number = 23
\nGreater number = 2x + 11 = 2 \u00d7 23 + 11 = 46 + 11 = 57<\/p>\n

Question 6.
\nFind three consecutive odd natural numbers whose sum is 87.
\nSolution:
\nLet the three consecutive odd natural numbers be
\nx, x + 2, x + 4.
\nAccording to statement,
\nx + x + 2 + x + 4 = 87
\n\u21d2 3x + 6 = 87
\n\u21d2 3x = 81
\n\u21d2 x = 27
\nx + 2 = 27 + 2 = 29 and x + 4 = 27 + 4 = 31
\nThree consecutive odd natural numbers are 27, 29 and 31.<\/p>\n

Question 7.
\nIn a class of 35 students, the number of girls is two-fifths of the number of boys. Find the number of girls in the class.
\nSolution:
\nLet the number of boys = x
\nThe number of girls = \\(\\frac { 2x }{ 5 }\\)
\nAccording to statement,
\n\"ML<\/p>\n

Question 8.
\nA chair costs \u20b9 250 and the table costs \u20b9 400. If a housewife purchased a certain number of chairs and two tables for \u20b9 2800, find the number of chairs she purchased.
\nSolution:
\nCost of a chair = \u20b9 250
\nand cost of a table = \u20b9 400
\nLet number of chairs = x
\nand number of tables = 2
\nTotal cost = \u20b9 2800
\nx \u00d7 250 + 2 \u00d7 400 = 2800
\n\u21d2 250x + 800 = 2800
\n\u21d2 250 x = \u20b9 2800 – \u20b9 800 = \u20b9 2000
\n\u21d2 x = 8
\nNumber of chairs = 8<\/p>\n

Question 9.
\nAparna got \u20b9 27840 as her monthly salary and over-time. Her salary exceeds the overtime by \u20b9 16560. What is her monthly salary?
\nSolution:
\nLet Apama’s monthly salary = \u20b9 x
\nThen over-time payment = \u20b9 (27840 – x)
\nAccording to the condition,
\nx – (27840 – x) = 16560
\n\u21d2 x – 27840 + x = 16560
\n\u21d2 2x = 16560 + 27840 = \u20b9 44400
\n\u21d2 x = 22200
\nMonthly salary = \u20b9 22200<\/p>\n

Question 10.
\nHeena has only \u20b9 2 and \u20b9 5 coins in her purse. If in all she has 80 coins in her purse amounting to \u20b9 232, find the number of \u20b9 5 coins.
\nSolution:
\nTotal number of coins = 80
\nLet the number of \u20b9 2 coins = x
\nThe number of \u20b9 5 coins = 80 – x
\nAccording to given statement,
\n2x + 5 (80 – x) = 232
\n\u21d2 2x + 400 – 5x = 232
\n\u21d2 -3x = 232 – 400
\n\u21d2 -3x = -168
\n\u21d2 x = 56
\nNumber of \u20b9 5 coins = 80 – x = 80 – 56 = 24
\nNumber of \u20b9 5 coins = 24<\/p>\n

Question 11.
\nA purse contains \u20b9 550 in notes of denominations of \u20b9 10 and \u20b9 50. If the number of \u20b9 50 notes is one less than that of \u20b9 10 notes, then find the number of \u20b9 50 notes.
\nSolution:
\nTotal amount in a purse = \u20b9 550
\nLet number of notes of \u20b9 10 = x
\nThe number of notes of \u20b9 50 = x – 1
\nAccording to the condition,
\nx \u00d7 10 + (x – 1) \u00d7 50 = 550
\n\u21d2 10x + 50x – 50 = 550
\n\u21d2 10x = 550 + 50 = 600
\n\u21d2 x = 10
\n50 rupees notes = 10 – 1 = 9<\/p>\n

Question 12.
\nAfter 12 years, 1 shall be 3 times as old as I was 4 years ago. Find my present age.
\nSolution:
\nLet my present age = x years
\nAfter 12 years, I will be = (x + 12) years
\nand 4 years ago, I was = (x – 4) years
\nAccording to the condition,
\n(x – 4) \u00d7 3 = x + 12
\n\u21d2 3x – 12 = x + 12
\n\u21d2 3x – x = 12 + 12
\n\u21d2 2x = 24
\n\u21d2 x = 12
\nMy present age = 12 years<\/p>\n

Question 13.
\nTwo equal sides of an isosceles triangle are 3x – 1 and 2x + 2. The third side is 2x units. Find x and the perimeter of the triangle.
\nSolution:
\nTwo equal sides of an isosceles \u0394 are 3x – 1 and 2x + 2
\n3x – 1 = 2x + 2
\n3x – 2x = 2 + 1
\nx = 3
\nWe know that
\nPerimeter of a \u0394 = (3x – 1) + (2x + 2) + (2x)
\n= (3 \u00d7 3 – 1) + (2 \u00d7 3 + 2) + (2 \u00d7 3)
\n= (9 – 1) + (6 + 2) + (6)
\n= 8 + 8 + 6
\n= 22 units<\/p>\n

Question 14.
\nThe length of a rectangle plot is 6 m less than thrice its breadth. Find the dimensions of the plot if its perimeter is 148 m.
\nSolution:
\nLet the breadth of a rectangle = x m.
\nThrice its breadth = 3x m
\nLength of a rectangle = 3x – 6 m
\nPerimeter of a rectangle = 2 (l + b)
\n= 2 (3x – 6 + x)
\n= 2 (4x – 6)
\n= 8x – 12
\nBut we are given, perimeter = 148 m
\n8x – 12= 148
\n8x = 148 + 12
\n8x = 160
\nx = 20 metres
\nBreadth = x = 20 metres
\nand Length = 3x – 6 = 3 \u00d7 20 – 6 = 60 – 6 = 54 metres.<\/p>\n

Question 15.
\nTwo complementary angles differ by 20\u00b0. Find the measure of each angle.
\nSolution:
\nWe know that
\nSum of measures of two complementary angles = 90\u00b0
\n\u21d2 x + y = 90\u00b0 ……. (i)
\nBut we are given x – y = 20\u00b0 …… (ii)
\n2x = 110\u00b0 [On comparing (i) and (ii)]
\n\u21d2 x = 55\u00b0
\nNow, x + y = 90\u00b0
\n\u21d2 y = 90\u00b0 – x
\n\u21d2 y = 90\u00b0 – 55\u00b0 = 35\u00b0<\/p>\n

ML Aggarwal Class 7 Solutions for ICSE Maths<\/a><\/h4>\n","protected":false},"excerpt":{"rendered":"

ML Aggarwal Class 7 Solutions for ICSE Maths Chapter 9 Linear Equations and Inequalities Ex 9.2 Question 1. If 7 is added to five times a number, the result is 57. Find the number. Solution: Let the number be x Five times a number is = 5x If 7 is added, it becomes 7 + … Read more<\/a><\/p>\n","protected":false},"author":5,"featured_media":0,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"spay_email":""},"categories":[3034],"tags":[],"yoast_head":"\nML Aggarwal Class 7 Solutions for ICSE Maths Chapter 9 Linear Equations and Inequalities Ex 9.2 - CBSE Library<\/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:\/\/cbselibrary.com\/ml-aggarwal-class-7-solutions-for-icse-maths-chapter-9-ex-9-2\/\" \/>\n<meta property=\"og:locale\" content=\"en_US\" \/>\n<meta property=\"og:type\" content=\"article\" \/>\n<meta property=\"og:title\" content=\"ML Aggarwal Class 7 Solutions for ICSE Maths Chapter 9 Linear Equations and Inequalities Ex 9.2\" \/>\n<meta property=\"og:description\" content=\"ML Aggarwal Class 7 Solutions for ICSE Maths Chapter 9 Linear Equations and Inequalities Ex 9.2 Question 1. 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If 7 is added to five times a number, the result is 57. Find the number. Solution: Let the number be x Five times a number is = 5x If 7 is added, it becomes 7 + ... 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