{"id":26800,"date":"2021-06-28T11:46:10","date_gmt":"2021-06-28T06:16:10","guid":{"rendered":"https:\/\/mcq-questions.com\/?p=26800"},"modified":"2022-03-02T10:47:29","modified_gmt":"2022-03-02T05:17:29","slug":"ncert-solutions-for-class-8-maths-chapter-2-ex-2-2","status":"publish","type":"post","link":"https:\/\/mcq-questions.com\/ncert-solutions-for-class-8-maths-chapter-2-ex-2-2\/","title":{"rendered":"NCERT Solutions for Class 8 Maths Chapter 2 Linear Equations in One Variable Ex 2.2"},"content":{"rendered":"

These NCERT Solutions for Class 8 Maths<\/a> Chapter 2 Linear Equations in One Variable Ex 2.2 Questions and Answers are prepared by our highly skilled subject experts.<\/p>\n

NCERT Solutions for Class 8 Maths Chapter 2 Linear Equations in One Variable Exercise 2.2<\/h2>\n

Question 1.
\nIf you subtract \\(\\frac{1}{2}\\) from a number and multiply the result by \\(\\frac{1}{2}\\), you get \\(\\frac{1}{8}\\). What is the number.
\nSolution:
\nLet the number be \u2018x\u2019
\nAccording to the given condition, we get
\n\\(\\left(x-\\frac{1}{2}\\right) \\frac{1}{2}=\\frac{1}{8}\\)
\nMultiplying both sides by 2
\n\\(\\mathrm{x}-\\frac{1}{2}=\\frac{1}{8} \\times 2=\\frac{1}{4}\\)
\nTransposing \\(\\left(-\\frac{1}{2}\\right)\\) to R.H.S. we get
\nx = \\(\\frac{1}{4}+\\frac{1}{2}\\)
\nx = \\(\\frac{1+2}{4}=\\frac{3}{4}\\)
\n\u2234 The required number is \\(\\frac{3}{4}\\).<\/p>\n

\"NCERT<\/p>\n

Question 2.
\nThe perimeter of a rectangular swimming pool is 154 m. Its length is 2 m more than twice its breadth. What are the length and the breadth of the pool?
\nSolution:
\nLet the breadth of the pool be x m
\n\u2234 Length of the pool = (2 + 2x) m = (2x + 2) m
\nPerimeter of the pool = 154 m
\n2(2x + 2 + x) = 154 m
\n[\u2235 Perimeter of the rectangle = 2(l + b)]
\n2(3x + 2) = 154
\n6x + 4 = 154
\nTransposing 4 to the R.H.S.
\n6x = 154 – 4
\n6x = 150
\nDividing both sides by 6,
\n\\(\\frac{6 x}{6}=\\frac{150}{6}\\)
\n\u2234 x = 25
\n\u2234 Breadth of the pool = 25m
\nLength of the pool = 2(25) + 2 = 50 + 2 = 52 m<\/p>\n

Question 3.
\nThe base of an isosceles triangle is \\(\\frac{4}{3}\\) cm and the perimeter of the triangle is 4\\(\\frac{2}{15}\\) cm. What is the length of the remaining equal sides?
\nSolution:
\nLet the length of the equal sides of a triangle be ‘x’
\nBase of an isosceles triangle = \\(\\frac{4}{3}\\) cm
\nPerimeter of the triangle = 4\\(\\frac{2}{15}\\) cm
\n\"NCERT
\n\u2234 Length of the equal sides is 1\\(\\frac{2}{5}\\) cm.<\/p>\n

\"NCERT<\/p>\n

Question 4.
\nSum of two numbers is 95. If one exceeds the other by 15, find the numbers.
\nSolution:
\nLet the smaller number be x
\nthen the greater number = x + 15
\nSum of two numbers = 95
\nx + (x + 15) = 95
\n2x + 15 = 95
\nTransposing 15 to R.H.S, we get
\n2x = 95 – 15 = 80
\nDividing both sides by 2
\nx = \\(\\frac{80}{2}\\) = 40
\n\u2234 The smaller number = 40
\nThe greater number = (40 + 15) = 55<\/p>\n

Question 5.
\nTwo numbers are in the ratio 5 : 3. If they differ by 18, what are the numbers?
\nSolution:
\nLet the two number be 5x and 3x
\nDifference of two numbers = 18
\n5x – 3x = 18
\n2x = 18
\nDividing both sides by 2, we get
\n\\(\\frac{2 \\mathrm{x}}{2}=\\frac{18}{2}\\)
\n\u2234 x = 9
\nThe two numbers are (5 \u00d7 9) and (3 \u00d7 9) i.e. 45 and 27.
\nHence, the required numbers are 45 and 27.<\/p>\n

\"NCERT<\/p>\n

Question 6.
\nThree consecutive integers add up to 51. What are these integers?
\nSolution:
\nLet the three consecutive integers be x, x + 1 and x + 2
\nSum of three integers = 51
\nx + x + 1 + x + 2 = 51
\n3x + 3 = 51
\nTransposing 3 to RHS, we get
\n3x = 51 – 3 = 48
\nDividing both sides by 3, we get
\n\\(\\frac{3 \\mathrm{x}}{3}=\\frac{48}{3}\\)
\nx = 16
\nNow x = 16,
\nx + 1 = 16 + 1 = 17
\nx + 2 = 16 + 2 = 18
\n\u2234 The required three consecutive integers are 16, 17 and 18.<\/p>\n

Question 7.
\nThe sum of three consecutive multiple of 8 is 888. Find the multiples.
\nSolution:
\nLet the three consecutive multiples of 8 be x, x + 8, x + 16
\nSum of three consecutive multiples = 888
\nx + x + 8 + x + 16 = 888
\n3x + 24 = 888
\nTransposing 24 to R.H.S., we get
\n3x = 888 – 24 = 864
\nDividing both sides by 3, we get
\n\\(\\frac{3 x}{3}=\\frac{864}{3}\\)
\nx = 288
\nx + 8 = 288 + 8 = 296
\nx + 16 = 288 + 16 = 304
\n\u2234 The required multiples of 8 are 288, 296 and 304.<\/p>\n

\"NCERT<\/p>\n

Question 8.
\nThree consecutive integers are such that when they are taken in increasing order and multiplied by 2, 3 and 4 respectively, they add up to 74. Find these numbers.
\nSolution:
\nLet the three consecutive integers be x, x + 1 and x + 2
\nAccording to the given condition
\n2(x) + 3(x + 1) + 4(x + 2) = 74
\n2x + 3x + 3 + 4x + 8 = 74
\n9x + 11 = 74
\nTransposing 11 to R.H.S., we get
\n9x = 74 – 11
\n9x = 63
\nDividing both sides by 9, we get
\n\\(\\frac{9 x}{9}=\\frac{63}{9}\\)
\nx = 7
\nx + 1 = 7 + 1 = 8
\nx + 2 = 7 + 2 = 9
\n\u2234 The consecutive integers are 7, 8 and 9.<\/p>\n

Question 9.
\nThe ages of Rahul and Haroon are in the ratio 5 : 7. Four years later, the sum of their ages will be 56 years. What are their present ages?
\nSolution:
\nLet the present age of Rahul be \u20185x\u2019
\nand the present age of Harron be 7x;
\n4 years later
\nRahul\u2019s age = (5x + 4) years
\nHarron\u2019s age = (7x + 4) years
\nSum of their ages = 56 years
\n5x + 4 + 7x + 4 = 56
\n12x + 8 = 56
\nTransposing 8 to R.H.S. we get
\n12x = 56 – 8 = 48
\nDividing both sides by 12
\n\\(\\frac{12 \\mathrm{x}}{12}=\\frac{48}{12}\\)
\nx = 4
\nPresent age of Rahul = 5 \u00d7 4 = 20 years
\nPresent age of Haroon = 7 \u00d7 4 = 28 years<\/p>\n

\"NCERT<\/p>\n

Question 10.
\nThe number of boys and girls in a class are in the ratio 7 : 5. The number of boys is 8 more than the number of girls. What is the total class strength.
\nSolution:
\nLet the number of boys in the class be 7x
\nand the number of girls in the class be 5x
\nAs number of boys is 8 more than the number of girls.
\n7x = 5x + 8
\nTransposing 5x to L.H.S.
\n7x – 5x = 8
\n2x = 8
\nDividing both sides by 2
\n\\(\\frac{2 \\mathrm{x}}{2}=\\frac{8}{2}\\)
\nx = 4
\n\u2234 Number of boys = 7 \u00d7 4 = 28
\nNumber of girls = 5 \u00d7 4 = 20
\nTotal class strength = 28 + 20 = 48 students.<\/p>\n

Question 11.
\nBaichung\u2019s father is 26 years younger than Baichung\u2019s grandfather and 29 years older than Baichung. The sum of the ages of all the three is 135 years. What is the age of each one of them?
\nSolution:
\nLet Baichung\u2019s present age be x years
\nBaichung father\u2019s age = (x + 29) years
\nBaichung grandfather\u2019s age = (x + 29 +26) years = (x + 55) years
\nSum of the ages of all the three =135 years
\nx + x + 29 + x + 55 = 135
\n3x + 84 = 135
\nTransposing 84 to RHS, we have
\n3x = 135 – 84 = 51
\nDividing both sides by 3, we get
\n\\(\\frac{3 x}{3}=\\frac{51}{3}\\)
\nx = 17
\nBaichung\u2019s age =17 years
\nBaichung fathers age = 17 + 29 = 46 years
\nBaichung grandfather\u2019s age = 17 + 55 = 72 years.<\/p>\n

\"NCERT<\/p>\n

Question 12.
\nFifteen years from now, Ravi\u2019s age will be four times his present age. What is Ravi\u2019s present age?
\nSolution:
\nLet Ravi\u2019s present age be \u2018x\u2019 years
\n4 times his present age = 4x years
\n15 years from now, Ravi\u2019s age = (x + 15) years
\nAccording to the given condition
\nx + 15 = 4x
\nTransposing 15 and 4x, we get
\n-4x + x = -15
\n-3x = -15
\nDividing both sides by (-3) we get
\n\\(\\frac{-3 x}{-3}=\\frac{-15}{-3}\\)
\nx = 5
\n\u2234 Ravi\u2019s present age = 5 years.<\/p>\n

Question 13.
\nA rational number is such that when you multiply it by \\(\\frac{5}{2}\\) and add \\(\\frac{2}{3}\\) to the product, you get \\(\\frac{7}{12}\\). What is the number?
\nSolution:
\nLet the required rational number be \u2018x\u2019
\nAccording to the given condition
\n\"NCERT
\nx = \\(\\frac{-1}{2}\\)
\n\u2234 The rational number is \\(\\frac{-1}{2}\\).<\/p>\n

\"NCERT<\/p>\n

Question 14.
\nLakshmi is a cashier in a bank. She has currency notes of denominations \u20b9 100, \u20b9 50 and \u20b9 10 respectively. The ratio of the number of these notes is 2 : 3 : 5. The total cash with Lakshmi is \u20b9 4,00, 000. How many notes of each denomination does she have?
\nSolution:
\nLet the number of \u20b9 100 notes be 2x
\nthe number of \u20b9 50 notes be 3x
\nand the number of \u20b9 10 notes be 5x
\nValue of \u20b9 100 notes = 2x \u00d7 100 = \u20b9 200x
\nValue of \u20b9 50 notes = 3x \u00d7 50 = \u20b9 150x
\nValue of \u20b9 10 notes = 5x \u00d7 10 = \u20b9 50x
\nAccording to the given condition,
\n\u20b9 200x + \u20b9 150x + \u20b9 50x = \u20b9 4,00000
\n400x = 4,00,000
\nDividing both sides by 400, we get
\n\\(\\frac{400 \\mathrm{x}}{400}=\\frac{4,00,000}{400}\\)
\nx = 1000
\n\u2234 Number of \u20b9 100 notes = 2 \u00d7 1000 = 2000
\nNumber of \u20b9 50 notes = 3 \u00d7 1000 = 3000
\nNumbers of \u20b9 10 notes = 5 \u00d7 1000 = 5000<\/p>\n

Question 15.
\nI have a total of \u20b9 300 in coins of denomination \u20b9 1, \u20b9 2 and \u20b9 5. The number of \u20b9 2 coins is 3 times the number of \u20b9 5 coins. The total number of coins is 160. How many coins of each denomination are with me?
\nSolution:
\nLet the number of \u20b9 5 coins be x
\nthen number of \u20b9 2 coins = 3x
\nand the number of coins of \u20b9 1 = 160 – (x + 3x) = 160 – 4x
\nNow Value of \u20b9 5 coins = \u20b9 5 \u00d7 x = 5x
\nValue of \u20b9 2 coins = \u20b9 2 \u00d7 3x = 6x
\nValue of \u20b9 1 coins = \u20b9 1 \u00d7 (160 – 4x) = (160 – 4x)
\nAccording to the given condition,
\n160 – 4x + 5x + 6x = 300
\n160 + 7x = 300
\nTransposing 160 to the R.H.S.
\n7x = 300 – 160
\n7x = 140
\nDividing both sides by 7
\n\\(\\frac{7 \\mathrm{x}}{7}=\\frac{140}{7}\\)
\nx = 20
\n\u2234 Number of \u20b9 5 coins = 20
\nNumber of \u20b9 2 coins = (3 \u00d7 20) = 60
\nNumber of \u20b9 1 coin = 160 – (4 \u00d7 20) = 80<\/p>\n

\"NCERT<\/p>\n

Question 16.
\nThe organisers of an essay competition decide that a winner in the competition gets a prize of \u20b9 100 and a participant who does not win gets a prize of \u20b9 25. The total prize money distributed is \u20b9 3000. Find the number of winners, if the total number of participants is 63.
\nSolution:
\nLet the number of winners be ‘x’
\nThe total number of participants = 63
\nThe number of non-winners = 63 – x
\nPrize money given to winners = \u20b9 100 \u00d7 x
\nPrize money given to non-winner participants = \u20b9 25(63 – x)
\n= \u20b9 25 \u00d7 63 – \u20b9 25x
\n= \u20b9 1575 – \u20b9 25x
\nAccording to the condition given in the question,
\n100x + 1575 – 25x = 3000
\n75x + 1575 = 3000
\nTransposing 1575 to R.H.S. we get
\n75x = 3000 – 1575
\n75x = 1425
\nDividing both sides by 75
\n\\(\\frac{75 \\mathrm{x}}{75}=\\frac{1425}{75}\\)
\nx = 19
\n\u2234 The number of winners = 19.<\/p>\n","protected":false},"excerpt":{"rendered":"

These NCERT Solutions for Class 8 Maths Chapter 2 Linear Equations in One Variable Ex 2.2 Questions and Answers are prepared by our highly skilled subject experts. NCERT Solutions for Class 8 Maths Chapter 2 Linear Equations in One Variable Exercise 2.2 Question 1. If you subtract from a number and multiply the result by …<\/p>\n

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