{"id":26822,"date":"2021-06-28T14:50:41","date_gmt":"2021-06-28T09:20:41","guid":{"rendered":"https:\/\/ncertsolutions.guru\/?p=26822"},"modified":"2021-06-28T14:51:54","modified_gmt":"2021-06-28T09:21:54","slug":"ncert-solutions-for-class-8-maths-chapter-2-ex-2-4","status":"publish","type":"post","link":"https:\/\/ncertsolutions.guru\/ncert-solutions-for-class-8-maths-chapter-2-ex-2-4\/","title":{"rendered":"NCERT Solutions for Class 8 Maths Chapter 2 Linear Equations in One Variable Ex 2.4"},"content":{"rendered":"

These NCERT Solutions for Class 8 Maths<\/a> Chapter 2 Linear Equations in One Variable Ex 2.4 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 2.4<\/h2>\n

Question 1.
\nAmina thinks of numbers and subtracts \\(\\frac{5}{2}\\) from it. She multiplies the results by 8. The result now obtained is 3 times the same number she thought of. What is the number?
\nSolution:
\nLet the number be x
\nBy subtracting \\(\\frac{5}{2}\\), we get x – \\(\\frac{5}{2}\\)
\nAccording to the given question
\n\\(8\\left(x-\\frac{5}{2}\\right)=3 x\\)
\n8x – \\(\\frac{8 \\times 5}{2}\\) = 3x
\n8x – 20 = 3x
\nBy transposing 3x to L.H.S. and -20 to R.H.S., we get
\n8x – 3x = 20
\n5x = 20
\nDividing both sides by 5, we get
\n\\(\\frac{5 \\mathrm{x}}{5}=\\frac{20}{5}\\)
\nx = 4
\n\u2234 The required number is 4.<\/p>\n

\"NCERT<\/p>\n

Question 2.
\nA positive number is 5 times another number. If 21 is added to both the numbers, then one of the new numbers becomes twice the other new number. What are the numbers?
\nSolution:
\nLet the number be x
\nThe other positive number is 5x
\non adding 21 to both numbers, we get (x + 21) and (5x + 21)
\nAccording to the question, we get
\n2(x + 21) = 5x + 21
\n2x + 42 = 5x + 21
\nTransposing 42 to R.H.S and 5x to L.H.S., we get
\n2x – 5x = 21 – 42
\n-3x = -21
\nDividing both sides by -3, we get
\n\\(\\frac{-3 x}{-3}=\\frac{-21}{-3}\\)
\nx = 7
\nthe other number 5x = 5 \u00d7 7 = 35
\nThus, the required numbers are 7 and 35.<\/p>\n

Question 3.
\nSum of the digits of a two-digit number is 9. When we interchange the digits, it is found that the resulting new number is greater than the original number by 27. What is the two-digit number.
\nSolution:
\nLet the units digit be ‘x’.
\nthe tens digits = 9 – x(sum of the digits is 9)
\nThe original two digit number = 10(9 – x) + x
\n= 90 – 10x + x
\n= 90 – 9x
\nOn interchanging the digits,
\nthe new number = 10x + 9 – x = 9x + 9
\nAccording to the given question, we get
\nNew number = Original number + 27
\n9x + 9 = 90 – 9x + 27
\n9x + 9 = 117 – 9x
\nTransposing 9 to R.H.S. and -9x to L.H.S., we get
\n9x + 9x = 117 – 9
\n18x = 108
\nDividing both sides by 18, we get
\n\\(\\frac{18 \\mathrm{x}}{18}=\\frac{108}{18}\\)
\nx = 6
\n\u2234 The original number = 90 – 9x = 90 – 54 = 36.<\/p>\n

\"NCERT<\/p>\n

Question 4.
\nOne of the two digits of a two digit number is three times the other digit. If you interchange the digits of this two- digit number and add the resulting number to the original number, you get 88. What is the original number?
\nSolution:
\nLet the digit in the unit place be ‘x’
\nThen, the digit at tens place = 3x
\nThe number = 10(3x) + x
\n= 30x + x
\n= 31x
\nOn interchanging the digits,
\nThe new number = 10x + 3x = 13x
\nAccording to the question
\n31x + 13x = 88
\n44x = 88
\nDividing both sides by 44
\n\\(\\frac{44 x}{44}=\\frac{88}{44}\\)
\nx = 2
\n\u2234 The number = 31x = 31 \u00d7 2 = 62.<\/p>\n

Question 5.
\nShobo\u2019s mother\u2019s present age is six times Shobo\u2019s present age. Shobo\u2019s age five years from now will be one-third of his mother\u2019s present age. What are their present ages?
\nSolution:
\nLet Shobos present age be ‘x’ years
\nMother\u2019s present age = 6x years
\nAfter 5 years
\nShobos age = (x + 5) years
\nShobos mothers age = (6x + 5) years
\nAccording to the question, we get
\n\\(\\frac{1}{3}\\) (mothers present age) = Shobos age after 5 years
\n\\(\\frac{1}{3}\\) \u00d7 6x = x + 5
\n2x = x + 5
\nTransposing x to LHS
\n2x – x = 5
\nx = 5
\n\u2234 Shobo\u2019s present age = 5 years
\nMothers present age = (6 \u00d7 5) = 30 years<\/p>\n

\"NCERT<\/p>\n

Question 6.
\nThere is a narrow rectangular plot, reserved for a school, in Mahuli village. The length and breadth of the plot are in the ratio 11 : 4. At the rate of \u20b9 100 per metre it will cost the village panchayat \u20b9 75000 to fence the plot. What are the dimensions of the plot?
\nSolution:
\nLet the length of the rectangular plot be 11x metres
\nand the breadth of the rectangular plot be 4x metres
\nPerimeter of the plot = 2(l + b)
\n= 2(11x + 4x)
\n= 2 \u00d7 15x
\n= 30x
\nCost of fencing = \u20b9 75000
\n100 \u00d7 30x = 75000
\n3000x = 75000
\nDividing both sides by 3000, we get
\n\\(\\frac{3000 \\mathrm{x}}{3000}=\\frac{75000}{3000}\\)
\nx = 25
\nLength of the plot = 11 \u00d7 25 = 275 metres
\nBreadth of the plot = 4 \u00d7 25 = 100 metres<\/p>\n

Question 7.
\nHasan buys two kinds of cloth materials for school uniforms, shirt material that costs him \u20b9 50 per metre and trouser material that costs him \u20b9 90 per metre. For every 3 metres of the shirt material he buys 2 metres of the trouser material. He sells the materials at 12% and 10% profit respectively. His total sale of \u20b9 36,600. How much trouser material did he buy?
\nSolution:
\nLet the length of cloth for shirts be \u20183x\u2019 metres
\nand the length of cloth for trousers be \u20182x\u2019 metres
\nCost of shirts cloth = 3x \u00d7 50 = \u20b9 150x
\nCost of trouser cloth = 2x \u00d7 90 = \u20b9 180x
\nS.P of shirts cloth at 12% profit
\n= \u20b9 \\(\\frac{112}{100}\\) \u00d7 150x
\n= \u20b9 168x
\nS.P. of trousers cloth at 10% profit
\n= \\(\\frac{110}{100}\\) \u00d7 180x
\n= \u20b9 198x
\nTotal S.P = \u20b9 36,600
\n168x + 198x = \u20b9 36,600
\n366x = 36600
\nDividing both sides by 366, we get
\n\\(\\frac{366 \\mathrm{x}}{366}=\\frac{36600}{366}\\)
\nx = 100
\nMaterial bought for trousers (2 \u00d7 100) = 200 metres<\/p>\n

\"NCERT<\/p>\n

Question 8.
\nHalf of a herd of deer are grazing in the field and three-fourths of the remaining are playing nearby. The rest 9 are drinking water from the pond. Find the number of deer in the herd.
\nSolution:
\nLet the total number of deer be \u2018x\u2019
\nNumber of deer, grazing in the field = \\(\\frac{\\mathrm{x}}{2}\\)
\nNumber of deer playing near by = \\(\\frac{3}{4}\\left(x-\\frac{x}{2}\\right)=\\frac{3 x}{4}-\\frac{3 x}{8}=\\frac{6 x-3 x}{8}=\\frac{3 x}{8}\\)
\nNumber of deer drinking water = 9
\n\\(\\frac{x}{2}+\\frac{3 x}{8}\\) + 9 = x
\nTransposing 9 to RHS and x to LHS we get
\n\\(\\frac{x}{2}+\\frac{3 x}{8}-x=-9\\)
\n\\(\\frac{4 x+3 x-8 x}{8}=-9\\)
\n\\(\\frac{-x}{8}=-9\\)
\nMultiplying both sides by -8
\n\\(\\frac{-\\mathrm{x}}{8}\\) \u00d7 (-8) = -9 \u00d7 (-8)
\nx = 72
\n\u2234 Number of deer in herd = 72<\/p>\n

Question 9.
\nA grandfather is ten times older than his granddaughter. He is also 54 years older than her. Find their present ages.
\nSolution:
\nLet the present age of granddaughter be \u2018x\u2019 years
\nPresent age of grandfather = 10x years
\nAccording to the given question, we get
\n10x – x = 54
\n9x = 54
\nDividing both sides by 9, we get
\n\\(\\frac{9 \\mathrm{x}}{9}=\\frac{54}{9}\\)
\nx = 6
\nPresent age of granddaughter = 6 years.
\nPresent age of grandfather = 10 \u00d7 6 = 60 years.<\/p>\n

\"NCERT<\/p>\n

Question 10.
\nAman\u2019s age is three times his son\u2019s age. Ten years ago he was five times his son\u2019s age. Find their present ages.
\nSolution:
\nLet the present age of son be \u2018x\u2019 years
\nPresent age of Aman = 3x
\nTen years ago Son\u2019s age = (x – 10) years
\nAman\u2019s age = (3x – 10) years
\nAccording to the given question, we get
\n3x – 10 = 5(x – 10)
\n3x – 10 = 5x – 50
\nTransposing -10 to R.H.S. and 5x to L.H.S
\n3x – 5x = 10 – 50
\n-2x = -40
\nDividing both sides by -2
\n\\(\\frac{-2 x}{2}=\\frac{-40}{-2}\\)
\nx = 20
\n\u2234 Sons present age = 20 years
\nAman\u2019s present age = (3 \u00d7 20) = 60 years.<\/p>\n","protected":false},"excerpt":{"rendered":"

These NCERT Solutions for Class 8 Maths Chapter 2 Linear Equations in One Variable Ex 2.4 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 2.4 Question 1. Amina thinks of numbers and subtracts from it. She multiplies the …<\/p>\n

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NCERT Solutions for Class 8 Maths Chapter 2 Linear Equations in One Variable Exercise 2.4 Question 1. Amina thinks of numbers and subtracts from it. 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