\nQuestion 1.
\nFill in the blanks:
\n(i) An equation of the type ax + b = 0 where a \u2260 0 is called a …………. in variable x.
\n(ii) Any value of the variable which satisfies the equation is called a …………. of the equation.
\n(iii) The process of finding all the solutions of an equation is called ………….
\n(iv) We can add the …………. to both sides of an equation.
\n(v) We can divide both sides of an equation by the same …………. number.
\n(vi) The solution set of the inequality 3x \u2264 10, x \u03f5 N is ………….
\nSolution:
\n(i) An equation of the type ax + b = 0
\nwhere a \u2260 0 is called a linear equation in variable x.
\n(ii) Any value of the variable which satisfies
\nthe equation is called a solution of the equation.
\n(iii) The process of finding all the solutions of
\nan equation is called solving the equation.
\n(iv) We can add the same number to both sides of an equation.
\n(v) We can divide both sides of an equation
\nby the same non-zero number.
\n(vi) The solution set of the inequality 3x \u2264 10, x \u03f5 N is (1, 2, 3).<\/p>\n
Question 2.
\nState whether the following statements are true (T) or false (F):
\n(i) An equation is a statement that two expressions are equal.
\n(ii) A term may be transposed from-one side of the equation to the other side, but its sign will not change.
\n(iii) We cannot subtract the same number from both sides of an equation.
\n(iv) 3x + 2 = 4(x + 7) + 9 is a linear equation in variable x.
\n(v) x = 1 is the solution of equation 4(x + 5) = 24.
\nSolution:
\n(i) An equation is a statement that two expressions are equal. True
\n(ii) A term may be transposed from one side of
\nthe equation to the other side, but its sign will not change. False
\nCorrect:
\nThe sign will change.
\n(iii) We cannot subtract the same number from
\nboth sides of an equation. False
\nCorrect:
\nWe can subtract.
\n(iv) 3x + 2 = 4(x + 7) + 9 is a linear equation in variable x. True
\n(v) x = 1 is the solution of equation 4(x + 5) = 24. True
\n4(1 + 5) = 24 \u21d2 4 \u00d7 6 = 24<\/p>\n
Multiple Choice Questions<\/strong> Question 4. Question 5. Question 6. Question 7. Question 8. Question 9. Question 10. Question 11. Question 12. Question 13. Question 14. Question 15. Question 16. Value Based Questions<\/strong> Question 2. Higher Order Thinking Skills (Hots)<\/strong> Question 2. Question 3. Question 4. Question 5.
\nChoose the correct answer from the given four options (3 to 16):<\/strong>
\nQuestion 3.
\nWhich of the following is not a linear equation in one variable?
\n(a) 3x + 2 = 0
\n(b) 2y – 4 = y
\n(c) x + 2y = 7
\n(d) 2(x – 3) + 7 = 0
\nSolution:
\nx + 2y = 7 is not a linear equation in one variable
\nas there are two variables x and y. (c)<\/p>\n
\nThe solution of the equation \\(\\frac{2}{3} x+1=\\frac{15}{9}\\) is
\n(a) 1
\n(b) \\(\\frac{3}{2}\\)
\n(c) 2
\n(d) \\(\\frac{2}{3}\\)
\nSolution:
\nThe solution of the equation
\n![\"ML](\"https:\/\/cbselibrary.com\/wp-content\/uploads\/2019\/07\/ML-Aggarwal-Class-8-Solutions-for-ICSE-Maths-Chapter-12-Linear-Equations-and-Inequalities-in-one-Variable-Objective-Type-Questions-Q4.1.png\")
<\/p>\n
\nThe solution of the equation 4z + 3 = 6 + 2z is
\n(a) 1
\n(b) \\(\\frac{3}{2}\\)
\n(c) 2
\n(d) 3
\nSolution:
\nSolution of equation 4z + 3 = 6 + 2z
\n\u21d2 4z – 2z = 6 – 3 \u21d2 2z = 3 \u21d2 z = \\(\\frac{3}{2}\\) (b)<\/p>\n
\nThe solution of the equation \\(\\frac{3 x}{5}+1=\\frac{4 x}{15}\\) is +7 is
\n(a) 12
\n(b) 14
\n(c) 16
\n(d) 18
\nSolution:
\n![\"ML](\"https:\/\/cbselibrary.com\/wp-content\/uploads\/2019\/07\/ML-Aggarwal-Class-8-Solutions-for-ICSE-Maths-Chapter-12-Linear-Equations-and-Inequalities-in-one-Variable-Objective-Type-Questions-Q6.1.png\")
<\/p>\n
\nThe solution of the equation \\(\\frac{x}{2}-\\frac{1}{5}=\\frac{x}{3}+\\)\\(\\frac{1}{4}\\) is
\n(a) 2.7
\n(b) 1.8
\n(c) 2.9
\n(d) 1.7
\nSolution:
\nThe solution of the equation
\n![\"ML](\"https:\/\/cbselibrary.com\/wp-content\/uploads\/2019\/07\/ML-Aggarwal-Class-8-Solutions-for-ICSE-Maths-Chapter-12-Linear-Equations-and-Inequalities-in-one-Variable-Objective-Type-Questions-Q7.1.png\")
<\/p>\n
\nThe solution of the equation \\(\\frac{8 x-3}{3 x}=2\\) is
\n![\"ML](\"https:\/\/cbselibrary.com\/wp-content\/uploads\/2019\/07\/ML-Aggarwal-Class-8-Solutions-for-ICSE-Maths-Chapter-12-Linear-Equations-and-Inequalities-in-one-Variable-Objective-Type-Questions-Q8.1.png\")
\n![\"ML](\"https:\/\/cbselibrary.com\/wp-content\/uploads\/2019\/07\/ML-Aggarwal-Class-8-Solutions-for-ICSE-Maths-Chapter-12-Linear-Equations-and-Inequalities-in-one-Variable-Objective-Type-Questions-Q8.2.png\")
\nSolution:
\n![\"ML](\"https:\/\/cbselibrary.com\/wp-content\/uploads\/2019\/07\/ML-Aggarwal-Class-8-Solutions-for-ICSE-Maths-Chapter-12-Linear-Equations-and-Inequalities-in-one-Variable-Objective-Type-Questions-Q8.3.png\")
<\/p>\n
\nIf we subtract \\(\\frac{1}{2}\\) from a number and multiply the result by \\(\\frac{1}{2}\\), we get \\(\\frac{1}{8}\\), then the number is
\n![\"ML](\"https:\/\/cbselibrary.com\/wp-content\/uploads\/2019\/07\/ML-Aggarwal-Class-8-Solutions-for-ICSE-Maths-Chapter-12-Linear-Equations-and-Inequalities-in-one-Variable-Objective-Type-Questions-Q9.1.png\")
\nSolution:
\nLet number be x, then
\n![\"ML](\"https:\/\/cbselibrary.com\/wp-content\/uploads\/2019\/07\/ML-Aggarwal-Class-8-Solutions-for-ICSE-Maths-Chapter-12-Linear-Equations-and-Inequalities-in-one-Variable-Objective-Type-Questions-Q9.2.png\")
<\/p>\n
\nFifteen years from now Ravi’s age will be four times his present age. What is Ravi’s present age?
\n(a) 4 years
\n(b) 5 years
\n(c) 6 years
\n(d) 3 years
\nSolution:
\nLet present age of Ravi = x years
\nAfter 15 years, his age will be = (x + 5) years
\n\u2234 x + 15 = 4x
\n\u21d2 15 = 4x – x = 3x
\n\u21d2 x = \\(\\frac{15}{3}\\) = 5
\n\u2234 His present age = 5 years (b)<\/p>\n
\nIf the sum of three consecutive integers is 51, then the largest integer is
\n(a) 16
\n(b) 17
\n(c) 18
\n(d) 19
\nSolution:
\nLet first integers = x
\nThen next two integers = x + 1, x + 2
\n\u2234 x + x + 1 + x + 2 = 51
\n\u21d2 3x + 3 = 51
\n\u21d2 3x = 51 – 3 = 48
\n\u21d2 x = \\(\\frac{48}{3}\\) = 16
\n\u2234 First integer = 16
\nand other two integer = 17, 18
\nLargest integers =18 (c)<\/p>\n
\nIf the perimeter of a rectangle is 13 cm and its Width is \\(2 \\frac{3}{4}\\) cm, then its length is
\n![\"ML](\"https:\/\/cbselibrary.com\/wp-content\/uploads\/2019\/07\/ML-Aggarwal-Class-8-Solutions-for-ICSE-Maths-Chapter-12-Linear-Equations-and-Inequalities-in-one-Variable-Objective-Type-Questions-Q12.1.png\")
\nSolution:
\nPerimeter of a rectangle = 13 cm
\n![\"ML](\"https:\/\/cbselibrary.com\/wp-content\/uploads\/2019\/07\/ML-Aggarwal-Class-8-Solutions-for-ICSE-Maths-Chapter-12-Linear-Equations-and-Inequalities-in-one-Variable-Objective-Type-Questions-Q12.2.png\")
\n![\"ML](\"https:\/\/cbselibrary.com\/wp-content\/uploads\/2019\/07\/ML-Aggarwal-Class-8-Solutions-for-ICSE-Maths-Chapter-12-Linear-Equations-and-Inequalities-in-one-Variable-Objective-Type-Questions-Q12.3.png\")
<\/p>\n
\nWhat should be added to twice the rational number \\(\\frac{-7}{3}\\) to get \\(\\frac{3}{7}\\) ?
\n![\"ML](\"https:\/\/cbselibrary.com\/wp-content\/uploads\/2019\/07\/ML-Aggarwal-Class-8-Solutions-for-ICSE-Maths-Chapter-12-Linear-Equations-and-Inequalities-in-one-Variable-Objective-Type-Questions-Q13.1.png\")
\nSolution:
\nLet x be added
\nAccording to the condition,
\n![\"ML](\"https:\/\/cbselibrary.com\/wp-content\/uploads\/2019\/07\/ML-Aggarwal-Class-8-Solutions-for-ICSE-Maths-Chapter-12-Linear-Equations-and-Inequalities-in-one-Variable-Objective-Type-Questions-Q13.2.png\")
<\/p>\n
\nSum of digits of a two digit number is 8. If the number obtained by reversing the digits is 18 more than the original number, then the original number is
\n(a) 35
\n(b) 53
\n(c) 26
\n(d) 62
\nSolution:
\nSum of digits of a two digit number = 8
\nLet unit digit = x
\nThen tens digit = 8 – x
\n\u2234 Number = x + 10(8 – x) = x + 80 – 10x = 80 – 9x
\nBy reversing the digits,
\nUnit digit = 8 – x
\nand tens digit = x
\n\u2234 Number = 8 – x + 10x = 8 + 9x
\n\u2234 8 + 9x = 80 – 9x + 18
\n\u21d2 9x + 9x = 80 + 18 – 8
\n\u21d2 18x = 90
\n\u21d2 x = \\(\\frac{90}{18}\\) =5
\n\u2234 Number = 80 – 9x = 80 – 9 \u00d7 5 = 80 – 45 = 35 (a)<\/p>\n
\nArjun is twice as old as Shriya. If five years ago his age was three times Shriya’s age, then Arjun’s present age is
\n(a) 10 years
\n(b) 15 years
\n(c) 20 years
\n(d) 25 years
\nSolution:
\nLet Shriya’s age = x years
\nThen Arjun’s age = 2x
\n5 years ago,
\nAge of Shriya was = (x – 5) years
\nand age of Arjun’s = (2x – 5) years
\n\u2234 2x – 5 = 3(x – 5)
\n\u21d2 2x – 5 = 3x – 15
\n\u21d2 3x – 2x = 15 – 5 = 10
\n\u21d2 x = 10
\n\u2234 Arjun’s present age = 2x = 2 \u00d7 10 = 20 years (c)<\/p>\n
\nIf the replacement set is {-5, -3, -1,0, 1, 3}, then the solution set of the inequation -3 < x < 3 is
\n(a) {-2,-1, 0, 1, 2}
\n(b) {-1, 0, 1, 2}
\n(c) {-3,-1, 0, 1, 3}
\n(d) {-1,0, 1}
\nSolution:
\nReplacement set = {-5, -3, -1, 0, 1,3}
\n-3 < x < 3
\n\u2234 x = {-2, -1, 0, 1, 2} from the replacement set,
\nSolution set x = {-1, 0, 1} (d)<\/p>\n
\nQuestion 1.
\nSeema is habitual of saving her pocket money. She collected some 50 paise and 25 paise coins in her piggy bank. If she collected \u20b925 and number of 50 paise coins is double the number of 25 paise coins. How many coins of each type did she collect? What values are being promoted? Is saving a good habit?
\nSolution:
\nSeema has 50 paise and 25 paise coins in his piggy bank.
\nTotal amount = \u20b925
\nLet 25 paise coins = x
\nThen 50 paise coins = 2x
\n![\"ML](\"https:\/\/cbselibrary.com\/wp-content\/uploads\/2019\/07\/ML-Aggarwal-Class-8-Solutions-for-ICSE-Maths-Chapter-12-Linear-Equations-and-Inequalities-in-one-Variable-Objective-Type-Questions-value-Q1.1.png\")
\n\u2234 25-paise coins = 20
\nand 50 paise coins = 20 \u00d7 2 = 40
\nThis is a good habit to save some money,
\nwe can solve any financial problem with its help at any time.<\/p>\n
\nRamesh gave one-fourth of his property to his two sons in equal shares and rest to his wife Sunita. Sunita gave one-third of her share to an orphanage. If the amount given by Sunita to the orphanage was \u20b920000, find the total value of the Ramesh’s property and the amount each person got? What value is shown by the Sunita?
\nSolution:
\nLet Ramesh’s property = x
\n\\(\\frac{1}{4}\\)th part of property was given to two sons equally = \\(\\frac{x}{4}\\)
\nSo, each son’s share = \\(\\frac{x}{4} \\times \\frac{1}{2}=\\frac{x}{8}\\)
\nRest to his wife Sunita = \\(x-\\frac{1}{4} x=\\frac{3}{4} x\\)
\nSunita gave one third of her share to orphanages
\nProperty given to orphanage
\n\\(\\frac{1}{3} \\text { of } \\frac{3}{4} x=\\frac{1}{4} x\\)
\n\u2234 \\(\\frac{1}{4}\\)x = \u20b920000
\n\u2234 Total value of Ramesh property = \u20b920000 \u00d7 \\(\\frac{4}{1}\\) = \u20b980000
\nEach son will get = \u20b9\\(\\frac{x}{8}\\) \u00d7 80000 = \u20b9 10000
\nand wife will get = \u20b980000 \u00d7 \\(\\frac{3}{4}\\) = \u20b960000
\nSunita done a good deed to help the orphanage
\nwhere needy person are living and they need your help and support.<\/p>\n
\nQuestion 1.
\nA man covers a distance of 24 km in \\(3 \\frac{1}{2}\\) hours partly on foot at the speed of 4.5 km\/h and partly on bicycle at the speed of 10 km\/h. Find the distance covered on foot.
\nSolution:
\nTotal distance = 24 km
\nTime taken = \\(3 \\frac{1}{2}\\) hours = \\(\\frac{7}{2}\\) hours
\nLet a man travels x km on foot at the speed of 4.5 km
\nand (24 – x) km on bicycle at the speed of 10 km\/hr
\n![\"ML](\"https:\/\/cbselibrary.com\/wp-content\/uploads\/2019\/07\/ML-Aggarwal-Class-8-Solutions-for-ICSE-Maths-Chapter-12-Linear-Equations-and-Inequalities-in-one-Variable-Objective-Type-Questions-hots-Q1.1.png\")
\n\u2234 He travelled 9 km on foot.<\/p>\n
\nThe perimeter of a rectangle is 240 cm. If its length is decreased by 10% and breadth is increased by 20% we get the same perimeter. Find the original length and breadth of the rectangle.
\nSolution:
\nPerimeter of a rectangle = 240 cm
\n\u2234 Length + breadth = \\(\\frac{240}{2}\\) = 120 cm
\nLet length = x cm
\nThen breadth = (120 – x) cm
\nBy decreasing length by 10%
\nand increasing breadth by 20%, we get
\n![\"ML](\"https:\/\/cbselibrary.com\/wp-content\/uploads\/2019\/07\/ML-Aggarwal-Class-8-Solutions-for-ICSE-Maths-Chapter-12-Linear-Equations-and-Inequalities-in-one-Variable-Objective-Type-Questions-hots-Q2.1.png\")
\nAccording to the condition,
\nLength + Breadth =120
\n![\"ML](\"https:\/\/cbselibrary.com\/wp-content\/uploads\/2019\/07\/ML-Aggarwal-Class-8-Solutions-for-ICSE-Maths-Chapter-12-Linear-Equations-and-Inequalities-in-one-Variable-Objective-Type-Questions-hots-Q2.2.png\")
\n\u2234 Length = 80 cm
\nand breadth = 120 – 80 = 40 cm<\/p>\n
\nA person preparing a medicine wants to convert 15% alcohol solution into 32% alcohol solution. Find how much pure alcohol he should mix in 400 mL of 15% alcohol solution to obtain required solution?
\nSolution:
\n15% of alcohol mixture = 400 mL
\n\u2234 Alcohol = \\(\\frac{15}{100}\\) \u00d7 400 = 60 mL
\nand other solution = 400 – 60 = 340 mL
\nIn new mixture alcohol = 32%
\nOther solution = 100 – 32 = 68%
\nIn 86 mL, alcohol = 32
\nand in 340 mL, alcohol will be = \\(\\frac{32 \\times 340}{68}\\) = 160 mL
\nAlready alcohol = 60 mL
\n\u2234 More alcohol required = 160 – 60 = 100 mL<\/p>\n
\nRahul covers a distance from P to Q on bicycle at 10 km\/h and returns back at 9 km\/h. Anuj covers the distance from P to Q and Q to P both at 12 km\/h. On calculating we find that Anuj took 10 minutes less than Rahul. Find the distance between P and Q.
\nSolution:
\nLet distance between P and Q = x km
\nSpeed of Rahul from P to Q = 10 km\/h
\nand back Q to P = 9 km\/h
\n![\"ML](\"https:\/\/cbselibrary.com\/wp-content\/uploads\/2019\/07\/ML-Aggarwal-Class-8-Solutions-for-ICSE-Maths-Chapter-12-Linear-Equations-and-Inequalities-in-one-Variable-Objective-Type-Questions-hots-Q4.1.png\")
<\/p>\n
\nSolve:
\n![\"ML](\"https:\/\/cbselibrary.com\/wp-content\/uploads\/2019\/07\/ML-Aggarwal-Class-8-Solutions-for-ICSE-Maths-Chapter-12-Linear-Equations-and-Inequalities-in-one-Variable-Objective-Type-Questions-hots-Q5.1.png\")
\nSolution:
\n![\"ML](\"https:\/\/cbselibrary.com\/wp-content\/uploads\/2019\/07\/ML-Aggarwal-Class-8-Solutions-for-ICSE-Maths-Chapter-12-Linear-Equations-and-Inequalities-in-one-Variable-Objective-Type-Questions-hots-Q5.2.png\")
\n![\"ML](\"https:\/\/cbselibrary.com\/wp-content\/uploads\/2019\/07\/ML-Aggarwal-Class-8-Solutions-for-ICSE-Maths-Chapter-12-Linear-Equations-and-Inequalities-in-one-Variable-Objective-Type-Questions-hots-Q5.3.png\")
<\/p>\nML Aggarwal Class 8 Solutions for ICSE Maths<\/a><\/h4>\n","protected":false},"excerpt":{"rendered":"