{"id":26076,"date":"2022-05-28T10:00:32","date_gmt":"2022-05-28T04:30:32","guid":{"rendered":"https:\/\/ncertsolutions.guru\/?p=26076"},"modified":"2022-05-28T10:25:42","modified_gmt":"2022-05-28T04:55:42","slug":"ncert-solutions-for-class-10-maths-chapter-4-ex-4-3","status":"publish","type":"post","link":"https:\/\/ncertsolutions.guru\/ncert-solutions-for-class-10-maths-chapter-4-ex-4-3\/","title":{"rendered":"NCERT Solutions for Class 10 Maths Chapter 4 Quadratic Equations Ex 4.3"},"content":{"rendered":"

These NCERT Solutions for Class 10 Maths<\/a> Chapter Ex 4.3 Questions and Answers are prepared by our highly skilled subject experts.<\/p>\n

NCERT Solutions for Class 10 Maths Chapter 4 Quadratic Equations Exercise 4.3<\/h2>\n

\"NCERT<\/p>\n

Question 1.
\nFind the roots of the following quadratic equations, if they exist, by the method of completing the square:
\n(i) 2x2<\/sup> – 7x + 3 = 0
\n(ii) 2x2<\/sup> + x – 4 = 0
\n(iii) 4x2<\/sup> + 4\u221a3x + 3 = 0
\n(iv) 2x2<\/sup> + x + 4 = 0
\nSolution:
\n(i) We have,
\n\"NCERT
\nAdding and subtracting (\\(\\frac { 7 }{ 4 }\\))\u00b2 in equation (i), we get,
\n\"NCERT<\/p>\n

(ii) We have,
\n\"NCERT
\nAdding and subtracting (\\(\\frac { 1 }{ 4 }\\))\u00b2 in equation (i), we get,
\n\"NCERT<\/p>\n

(iii) We have,
\n\"NCERT
\nAdding and subtracting \\(\\left(\\frac{\\sqrt{3}}{2}\\right)^{2}\\) in equation (i), we get,
\n\"NCERT<\/p>\n

(iv) We have,
\n\"NCERT
\nAdding and subtracting (\\(\\frac { 1 }{ 4 }\\))\u00b2 in equation (i), we get,
\n\"NCERT
\nSo, roots do not exist.<\/p>\n

\"NCERT<\/p>\n

Question 2.
\nFind the roots of the quadratic equations by applying the quadratic formula.
\n(i) 2x2<\/sup> – 7x + 3 = 0
\n(ii) 2x2<\/sup> + x – 4 = 0
\n(iii) 4x2<\/sup> + 4\u221a3x + 3 = 0
\n(iv) 2x2<\/sup> + x + 4 = 0
\nSolution:
\n(i) We have,
\n2x2<\/sup> – 7x + 3 = 0
\nHere a = 2, b = – 7, and c = 3
\n\u2234 By using quadratic formula, we get,
\n\"NCERT<\/p>\n

(ii) We have,
\n2x2<\/sup> + x – 4 = 0
\nHere a = 2, b = 1, and c = – 4
\n\u2234 By using quadratic formula, we get,
\n\"NCERT<\/p>\n

(iii) We have,
\n4x2<\/sup> + 4\u221a3x + 3 = 0
\nHere a = 4, b = 4\u221a3 and c = 3
\n\u2234 By using quadratic formula, we get,
\n\"NCERT<\/p>\n

(iv) We have,
\n2x2<\/sup> + x + 4 = 0
\nHere a = 2, b = 1, and c = 4
\n\u2234 By using quadratic formula, we get,
\n\"NCERT
\nHere, b\u00b2 – 4ac < 0
\nSo, the given equation has no any real roots.<\/p>\n

\"NCERT<\/p>\n

Question 3.
\nFind the roots of the following equations:
\n(i) x – \\(\\frac { 1 }{ x }\\) = 3 ; x \u2260 0
\n(ii) \\(\\frac { 1 }{ x+4 }\\) – \\(\\frac { 1 }{ x-7 }\\); x \u2260 – 4, 7
\nSolution:
\n\"NCERT<\/p>\n

\"NCERT<\/p>\n

Question 4.
\nThe sum of the reciprocals of Rehman\u2019s ages, (in years) 3 years ago and 5 years from now is \\(\\frac { 1 }{ 3 }\\) Find his present age.
\nSolution:
\nLet the present age of Rehman be x years
\n3 years ago Rehman\u2019s age was = (x – 3) years
\n5 years from now Rehman\u2019s age will be = (x + 5) years
\n\u2234 According to question
\n\"NCERT
\nUsing the quadratic formula
\n\"NCERT
\nSince, age cannot be in negative,
\nso we ignore the value of x = – 3
\nTherefore, x = 7
\nSo, present age of Rehman = x = 7 years.<\/p>\n

\"NCERT<\/p>\n

Question 5.
\nIn a class test, the sum of Shefali\u2019s marks in Mathematics and English is 30. Had she got 2 marks more in Mathematics and 3 marks less in English, the product of their marks would have been 210. Find her marks in the two subjects.
\nSolution:
\nLet the marks secured by Shefali in Mathematics = x Then,
\nAccording to question,
\n(x + 2) x (30 – x – 3) = 210
\nor (x + 2) x (27 – x) = 210
\nor 27x – x\u00b2 + 54 – 2x – 210
\nor 25x – x\u00b2 + 54 = 210
\nor x\u00b2 – 25x – 54 + 210 = 0
\nor x\u00b2 – 25x + 156 = 0
\nHere, a = 1, b = – 25 and c = 156
\nUsing the quadratic formula
\n\"NCERT
\nTherefore, if Shefali’s marks in Mathematics = x = 13
\nThen, Shefail’s marks in English = 30 – x = 30 = 17
\nand if Shefail’s marks in Mathematics = x = 12
\nThen, Shefail’s marks in Mathematics = 30 – x = 30 – 12 = 18<\/p>\n

Question 6.
\nThe diagonal of a rectangular field is 60 metres more than the shorter side. If the longer side is 30 metres more than the shorter side, find the sides of the field.
\n\"NCERT
\nSolution:
\nLet ABCD be a rectangular field.
\nLet AB, its shorter side, be x metres And BC, the longer side be (x + 30) metres
\nDiagonal AC of the rectangular field be x + 60 metres.
\nBy Pythagoras theorem, we have,
\n(AC)\u00b2 = (AB)\u00b2 + (BC)\u00b2
\n\u2234 (x – 60)\u00b2 = x\u00b2 + (x + 30)\u00b2
\nor x\u00b2 + 120x + 3600 = x\u00b2 + x\u00b2 + 60x + 900
\nor x\u00b2 + 120x + 3600 = 2x\u00b2 + 60x + 900
\n2x\u00b2 + 60x + 900 – x\u00b2 – 120x – 3600 = 0
\nor x\u00b2 – 60x – 2700 = 0
\nHere, a = 1, b = – 60 and c = – 2700
\nUsing the quadratic formula, we get
\n\"NCERT
\nSince, dimension cannot be in negative,
\nso we ignore the value of x = – 30
\nTherefore, x = 90
\n\u2234 Length of shorter side of rectangular field = x = 90 metres
\nand length of longer side of rectangular = x + 30 = 90 + 30 = 120 metres.<\/p>\n

\"NCERT<\/p>\n

Completing the square Calculator<\/a> online with solution and steps.<\/p>\n

Question 7.
\nThe difference of squares of two numbers is 180. The square of the smaller number is 8 times the larger number. Find the two numbers.
\nSolution:
\nLet x be the larger number.
\nSo, the square of the smaller number = 8x According to the second condition, the square of the larger number – the square of the smaller number = 180
\nx\u00b2 – 8x = 180
\n\u21d2 x\u00b2 – 8x – 180 = 0
\n\u21d2 x\u00b2 – 18x + 10x – 180 = 0
\n\u21d2 x (x – 18) + 10 (x – 18) = 0
\n\u21d2 (x – 18) (x + 10) = 0
\n\u21d2 x = 18 or x = – 10
\nx = – 10 does not exist, so it is neglected.
\nSo, x = 18 (the larger number)
\nSmaller number = \\(\\sqrt{8x}\\)
\n\u21d2 \\(\\sqrt{8\u00d718}\\) = \\(\\sqrt{144}\\) = \u00b1 12
\n\u2234 The smaller and larger numbers are \u00b112and 18 respectively.<\/p>\n

Question 8.
\nA train travels 360 km at a uniform speed. If the speed had been 5 km\/h more, it would have taken 1 hour less for the same journey. Find the speed of the train.
\nSolution:
\nTotal distance travelled = 360 km
\nLet uniform speed be x km\/h
\nThen, increased speed = (x + 5) km\/h
\nAccording to question,
\n\"NCERT
\nSince, speed cannot be in negative, so we reject the value of x = – 45
\n\u2234 x = 40
\nSo, the usual speed of train = x km\/h = 40 km\/h.<\/p>\n

\"NCERT<\/p>\n

Question 9.
\nTwo water taps together can fill a tank in 9\\(\\frac { 3 }{ 8 }\\) hours. The tap of larger diameter takes 10 hours less than the smaller one to fill the tank separately. Find the time in which each tap can separately fill the tank.
\nSolution:
\nLet larger pipe fills the tank in x hours and the smaller pipe fills the tank in y hours.
\nThe tank filled by the larger pipe in 1 hour = \\(\\frac { 1 }{ x }\\)
\nand part of tank filled by larger tap in 1 hour = \\(\\frac { 1 }{ x-10 }\\)
\nTotal = 9\\(\\frac { 3 }{ 8 }\\) hours = \\(\\frac { 75 }{ 8 }\\)
\n\u2234 According to questions,
\n\"NCERT
\nUsing the quadratic formula
\n\"NCERT
\nSo, if smaller tap take 3.75 hour to fill the tank then time taken by larger tap = 3.75 = 10 = – 6.25 hour
\nBut we know that time cannot be in negative The value of x = 3.75 is neglected.
\nTherefore, time taken by the smaller tap to fill the tank = x = 25 hours.
\nand time taken by the larger tap to fill the bank = x – 10 = 25 – 10 = 15 hours.<\/p>\n

Question 10.
\nAn express train takes 1 hour less than a passenger train to travel 132 km between Mysore and Bengaluru (without taking into consideration the time they stop at intermediate stations). If the average speed of the express train is 11 km\/h more than that of the passenger train, find the average speed of the two trains.
\nSolution:
\nLet the average speed of passenger train be x km\/h
\nThen, average speed of express train = x + 11 km\/h
\nTotal distance from Mysore to Bangalore is 132 km.
\nTherefore, time taken by the passenger train to cover the distance of 132 km = \\(\\frac { 132 }{ x }\\)h
\nAccording to question,
\n\"NCERT
\nUsing the quadratic formula
\n\"NCERT
\nBut we know that speed of train cannot be in negative.
\n\u2234 The value of x = – 44 is neglected.
\n\u2234 x = 33
\nTherefore, the speed of passenger train = x = 33 km\/h
\nand the speed of express train = x + 11 = 33+ 11 = 44 km\/h.<\/p>\n

\"NCERT<\/p>\n

Question 11.
\nSum of the areas of two squares is 468 m\u00b2. If the difference of their perimeters is 24 m, find the sides of the two squares.
\nSolution:
\nLet the sides of smaller square = x
\nThe sides of larger square = \\(\\sqrt{468-x^{2}}\\)
\n[\u2234 Area of square = (side)\u00b2]
\nNow, perimeter of smaller square = 4x
\nand perimeter of larger square = 4 x \\(\\sqrt{468-x^{2}}\\)
\nAccording to question,
\n4 x \\(\\sqrt{468-x^{2}}\\) = 24
\nor 4 x \\(\\sqrt{468-x^{2}}\\) = 24 + 4x
\nor \\(\\sqrt{468-x^{2}}\\) = \\(\\frac{4(6+x)}{4}\\)
\nor \\(\\sqrt{468-x^{2}}\\) = 6 + x
\nSquaring both sides, we get,
\n468 – x\u00b2 = (6 + x\u00b2)
\nor 468 – x\u00b2 = 36 + x\u00b2 + 12x
\n[\u2234 (a + b)\u00b2 = a\u00b2 + b\u00b2 – 2ab]
\nor 36 + x\u00b2 + 12x + x\u00b2 – 468 = 0
\nor 2x\u00b2 + 12x – 432 = 0
\nor x\u00b2 + 6x – 216 = 0
\nHere, a = 1, b = 6 and c = – 216
\nUsing the quadratic fomula,
\n\"NCERT
\nBut we know that length of the side of square cannot be in negative.
\nSo, the value of x = – 18 is neglected. Therefore, x = 12
\nSo, length of the sides of smaller square = x = 12 m
\nand length of the sides of larger square = \\(\\sqrt{468-x^{2}}\\)
\n= \\(\\sqrt{468-(12)^{2}}\\)
\n= \\(\\sqrt{468-144}\\)
\n= \\(\\sqrt{324}\\)
\n= 18 m<\/p>\n

\"NCERT<\/p>\n","protected":false},"excerpt":{"rendered":"

These NCERT Solutions for Class 10 Maths Chapter Ex 4.3 Questions and Answers are prepared by our highly skilled subject experts. NCERT Solutions for Class 10 Maths Chapter 4 Quadratic Equations Exercise 4.3 Question 1. Find the roots of the following quadratic equations, if they exist, by the method of completing the square: (i) 2×2 …<\/p>\n

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NCERT Solutions for Class 10 Maths Chapter 4 Quadratic Equations Exercise 4.3 Question 1. 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