{"id":25584,"date":"2022-06-05T02:30:30","date_gmt":"2022-06-04T21:00:30","guid":{"rendered":"https:\/\/mcq-questions.com\/?p=25584"},"modified":"2022-05-23T15:39:15","modified_gmt":"2022-05-23T10:09:15","slug":"ncert-solutions-for-class-10-maths-chapter-4-ex-4-1","status":"publish","type":"post","link":"https:\/\/mcq-questions.com\/ncert-solutions-for-class-10-maths-chapter-4-ex-4-1\/","title":{"rendered":"NCERT Solutions for Class 10 Maths Chapter 4 Quadratic Equations Ex 4.1"},"content":{"rendered":"

In this Page, you will learn how to solve questions of Ex 4.1 Class 10 Maths<\/a> NCERT Solutions recommended in CBSE 10th Board Maths Exam syllabus.<\/p>\n

These NCERT Solutions for Class 10 Maths<\/a> Chapter Ex 4.1 Questions and Answers are prepared by our highly skilled subject experts.<\/p>\n

NCERT Solutions for Class 10 Maths Chapter 4 Quadratic Equations Exercise 4.1<\/h2>\n

\"NCERT<\/p>\n

Question 1.
\nCheck whether the following are quadratic equations:
\n(i) (x+ 1)2<\/sup> = 2(x – 3)
\n(ii) x – 2x = (- 2) (3 – x)
\n(iii) (x – 2) (x + 1) = (x – 1) (x + 3)
\n(iv) (x – 3) (2x + 1) = x (x + 5)
\n(v) (2x – 1) (x – 3) = (x + 5) (x – 1)
\n(vi) x2<\/sup> + 3x + 1 = (x – 2)2<\/sup>
\n(vii) (x + 2)3<\/sup> = 2x(x2<\/sup> – 1)
\n(viii) x3<\/sup> -4x2<\/sup> – x + 1 = (x – 2)3<\/sup>
\nSolution:
\n(i) (x+ 1)2<\/sup> = 2(x – 3)
\n\u21d2 x2<\/sup> + 2x +1 = 2x – 6
\n\u21d2 x2<\/sup> + 2x – 2x+1 + 6 = 0
\n\u21d2 x2<\/sup> + 7 = 0
\nNow, it is in the form of ax\u00b2 + bx + c = 0, where b = 0.
\nTherefore, the given equation is a quadratic equation.<\/p>\n

(ii) x2<\/sup> – 2x = (- 2) (3 – x)
\n\u21d2 x2<\/sup> – 2x = – 6 + 2x
\n\u21d2 x2<\/sup> – 4x + 6 = 0
\nWhich is of the form
\nax2<\/sup> + bx + c = 0
\nNow, it is in the form of ax\u00b2 + bx + c = 0.
\nTherefore, the given equation is a quadratic equation.<\/p>\n

(iii) We have,
\n(x – 2) (x + 1) = (x – 1) (x + 3)
\n\u21d2 x2<\/sup> + x – 2x – 2 = x\u00b2 + 3x – x – 3
\n\u21d2 x2<\/sup> + x – 2x – 2 = x2<\/sup> – 3x + x + 3 = 0
\n\u21d2 – 3x + 1 = 0
\nIt is not in the form of ax\u00b2 + bx + c = 0.
\nTherefore, the given equation is a quadratic equation.<\/p>\n

(iv) We have
\n(x-3) (2x+ 1) = x (x + 5)
\n\u21d2 2x2<\/sup> + x – 6x – 3 = x2<\/sup> + 5x
\n\u21d2 2x2<\/sup> + x – 6x – 3 – x\u00b2 – 5x = 0
\n\u21d2 x2<\/sup> – 10x – 3 = 0
\nNow, it is in the form of ax\u00b2 + bx + c = 0.
\nTherefore, the given equation is a quadratic equation.<\/p>\n

(v) We have
\n(2x – 1)(x – 3) = (x + 5)(x – 1)
\n\u21d2 2x2<\/sup> – 6x-x + 3 = x2<\/sup> -x + 5x – 5
\n2x2<\/sup> – 6x-x + 3 = x2<\/sup> + x – 5x + 5 = 0
\n\u21d2 x2<\/sup> – 11x + 8 = 0
\nNow, it is in the form of ax\u00b2 + bx + c = 0.
\nTherefore, the given equation is a quadratic equation.<\/p>\n

(vi) We have
\nx2<\/sup> + 3x + 1 = (x – 2)2<\/sup>
\n\u21d2 x2<\/sup> + 3x + 1 = x2<\/sup> + 4 – 4x
\n\u21d2 x2<\/sup> + 3x + 1 = x2<\/sup>– 4 + 4c = 0
\n\u21d2 7x – 3 = 0
\nIt is not in the form of ax\u00b2 + bx + c = 0.
\nTherefore, the given equation is a quadratic equation.<\/p>\n

(vii) We have
\n(x + 2)3<\/sup> = 2x (x2<\/sup> – 1)
\n\u21d2 x3<\/sup> + 8 + 3.x.2 (x + 2) = 2x3<\/sup> – 2x
\n\u21d2 x3<\/sup> + 8 + 6x2<\/sup> + 12x = 2x3<\/sup> – 2x
\n\u21d2 x3<\/sup> – 6x2<\/sup> – 14x – 8 = 0
\nIt is not in the form of ax\u00b2 + bx + c = 0.
\nTherefore, the given equation is a quadratic equation.<\/p>\n

(viii) We have
\nx3<\/sup> – 4x2<\/sup> – x+1 = (x-2)3<\/sup>
\n\u21d2 x3<\/sup> – 4x2<\/sup> – x + 1 = x3<\/sup>-8 + 3x(-2)(x – 2)
\n\u21d2 x3<\/sup> – 4x2<\/sup> -x + 1 = x3<\/sup> – 6x2<\/sup> + 12x – 8
\n\u21d2 2x2<\/sup> – 13x + 9 = 0
\nNow, it is in the form of ax\u00b2 + bx + c = 0.
\nTherefore, the given equation is a quadratic equation.<\/p>\n

\"NCERT<\/p>\n

Question 2.
\nRepresent the following situations in the form of quadratic equations:
\n(i) The area of a rectangular plot is 528 m2<\/sup>. The length of the plot (in metres) is one more than twice its breadth. We need to find the length and breadth of the plot.
\n(ii) The product of two consecutive positive integers is 306. We need to find the integers.
\n(iii) Rohan\u2019s mother is 26 years older than him. The product of their ages (in years) 3 years from now will be 360. We would like to find Rohan\u2019s present age.
\n(iv) A train travels a distance of 480 km at a uniform speed. If the speed had been 8 km\/h less, then it would have taken 3 hours more to cover the same distance. We need to find the speed of the train.
\nSolution:
\n(i) Let breadth of the rectangular plot = x m
\nThen, length of the plot = (2x + 1)m
\nArea of a rectangular plot = l x b ,
\n\u21d2 528 (2x + 1)x
\n\u21d2 528 = 2x2<\/sup> +x
\n\u21d2 2x2<\/sup> + x – 528 = 0
\nWhich is the required quadratic equation.
\nTherefore, area of rectangle, satisfies the quadratic equation 2x\u00b2 + x – 528 = 0, where x is breadth (in metres) of the plot.<\/p>\n

(ii) Let first consecutive positive integer = x
\n\u2234 Second consecutive positive integer = x + 1
\nAccording to question,
\nx (x + 1) = 306
\nor x\u00b2 + x = 306
\nor x\u00b2 + x – 306 = 0
\nTherefore, the two consecutive positive integers whose product is 306, satisfies the quadratic equation.
\nx\u00b2 + x – 306 = 0, where x is the smallest integer.<\/p>\n

(iii) Let the present age of Rohan = x years
\n\u2234 Rohan\u2019s mother\u2019s present age = (x + 26) years
\nAfter 3 years, Rohan\u2019s age = (x + 3) years
\nAfter 3 years, Rohan\u2019s mother\u2019s age = (x + 26 + 3) years
\nAccording to question,
\n(x + 3) (x + 29) = 360
\n\u21d2 x2<\/sup> + 29x + 3x + 87 – 360 = 0
\n\u21d2 x2<\/sup> + 32x – 273 = 0
\nTherefore, product or Rohan’s and his mother’s age three years now satisfies the quadratic equation
\nx\u00b2 + 32x – 273 = 0, where x (in years) is the present age of Rohan.<\/p>\n

(iv) Case I :
\nLet the uniform speed of the train = u km\/h
\nTotal distance covered by the train = 480 km.
\ndistance
\nspeed
\nCase: II
\nSpeed of train = (u – 8) km\/h
\nand total distance covered by the train = 480 km
\n\u2234 Time taken = \\(\\frac { distance }{ speed }\\) = \\(\\frac { 480 }{ u }\\)h
\nCase: II
\nSpeed of train = (u – 8) km\/h
\nand total distance covered by the train = 480 km
\n\u2234 Time taken = \\(\\frac { distance }{ speed }\\) = \\(\\frac { 480 }{ u – 8 }\\)h
\nAccording to question,
\n\"NCERT
\nTherefore, speed of train satisfies the quadratic equation 3u2 – 24K – 1280, where K (in km\/h) is the speed of the train.<\/p>\n","protected":false},"excerpt":{"rendered":"

In this Page, you will learn how to solve questions of Ex 4.1 Class 10 Maths NCERT Solutions recommended in CBSE 10th Board Maths Exam syllabus. These NCERT Solutions for Class 10 Maths Chapter Ex 4.1 Questions and Answers are prepared by our highly skilled subject experts. NCERT Solutions for Class 10 Maths Chapter 4 …<\/p>\n

NCERT Solutions for Class 10 Maths Chapter 4 Quadratic Equations Ex 4.1<\/span> Read More »<\/a><\/p>\n","protected":false},"author":9,"featured_media":0,"comment_status":"closed","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"site-sidebar-layout":"default","site-content-layout":"default","ast-main-header-display":"","ast-hfb-above-header-display":"","ast-hfb-below-header-display":"","ast-hfb-mobile-header-display":"","site-post-title":"","ast-breadcrumbs-content":"","ast-featured-img":"","footer-sml-layout":"","theme-transparent-header-meta":"default","adv-header-id-meta":"","stick-header-meta":"default","header-above-stick-meta":"","header-main-stick-meta":"","header-below-stick-meta":"","spay_email":""},"categories":[2],"tags":[],"yoast_head":"\nNCERT Solutions for Class 10 Maths Chapter 4 Quadratic Equations Ex 4.1 - MCQ Questions<\/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:\/\/mcq-questions.com\/ncert-solutions-for-class-10-maths-chapter-4-ex-4-1\/\" \/>\n<meta property=\"og:locale\" content=\"en_US\" \/>\n<meta property=\"og:type\" content=\"article\" \/>\n<meta property=\"og:title\" content=\"NCERT Solutions for Class 10 Maths Chapter 4 Quadratic Equations Ex 4.1 - MCQ Questions\" \/>\n<meta property=\"og:description\" content=\"In this Page, you will learn how to solve questions of Ex 4.1 Class 10 Maths NCERT Solutions recommended in CBSE 10th Board Maths Exam syllabus. 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