{"id":31825,"date":"2018-10-08T11:07:03","date_gmt":"2018-10-08T11:07:03","guid":{"rendered":"https:\/\/www.aplustopper.com\/?p=31825"},"modified":"2020-11-25T13:02:16","modified_gmt":"2020-11-25T07:32:16","slug":"ml-aggarwal-class-10-solutions-for-icse-maths-chapter-6-chapter-test","status":"publish","type":"post","link":"https:\/\/www.aplustopper.com\/ml-aggarwal-class-10-solutions-for-icse-maths-chapter-6-chapter-test\/","title":{"rendered":"ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 6 Quadratic Equations in One Variable Chapter Test"},"content":{"rendered":"

ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 6 Quadratic Equations in One Variable Chapter Test<\/h2>\n

These Solutions are part of ML Aggarwal Class 10 Solutions for ICSE Maths<\/a>. Here we have given ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 6 Quadratic Equations in One Variable Chapter Test<\/p>\n

ML Aggarwal Solutions<\/a>ICSE Solutions<\/a>Selina ICSE Solutions<\/a><\/p>\n

Solve the following equations (1 to 4) by factorisation :<\/strong><\/p>\n

Question 1.<\/strong><\/span>
\n(i) x\u00b2 + 6x – 16 = 0<\/strong>
\n(ii) 3x\u00b2 + 11x + 10 = 0<\/strong>
\nSolution:<\/strong><\/span>
\nx\u00b2 + 6x – 16 = 0
\n=> x\u00b2 + 8x – 2x – 16 = 0
\nx (x + 8) – 2 (x + 8) = 0
\n\"ML
\n\"ML<\/p>\n

Question 2.<\/strong><\/span>
\n(i) 2x\u00b2 + ax – a\u00b2 = 0<\/strong>
\n(ii) \u221a3x\u00b2 + 10x + 7\u221a3 = 0<\/strong>
\nSolution:<\/strong><\/span>
\n(i) 2x\u00b2 + ax – a\u00b2 = 0
\n=> 2x\u00b2 + 2ax – ax – a\u00b2 = 0
\n\"ML
\n\"ML<\/p>\n

Question 3.<\/strong><\/span>
\n(i) x(x + 1) + (x + 2)(x + 3) = 42<\/strong>
\n(ii) \\(\\frac { 6 }{ x } -\\frac { 2 }{ x-1 } =\\frac { 1 }{ x-2 } \\)<\/strong>
\nSolution:<\/strong><\/span>
\n(i) x(x + 1) + (x + 2)(x + 3) = 42
\n2x\u00b2 + 6x + 6 – 42 = 0
\n\"ML
\n\"ML<\/p>\n

Question 4.<\/strong><\/span>
\n(i)\\(\\sqrt { x+15 } =x+3 \\)<\/strong>
\n(ii)\\(\\sqrt { { 3x }^{ 2 }-2x-1 } =2x-2\\)<\/strong>
\nSolution:<\/strong><\/span>
\n(i)\\(\\sqrt { x+15 } =x+3 \\)
\nSquaring on both sides
\nx + 15 = (x + 3)\u00b2
\n\"ML
\n\"ML
\n\"ML<\/p>\n

Solve the following equations (5 to 8) by using formula :<\/strong><\/p>\n

Question 5.<\/strong><\/span>
\n(i) 2x\u00b2 – 3x – 1 = 0<\/strong>
\n(ii) \\(x\\left( 3x+\\frac { 1 }{ 2 } \\right) =6\\)<\/strong>
\nSolution:<\/strong><\/span>
\n(i) 2x\u00b2 – 3x – 1 = 0
\nHere a = 2, b = – 3, c = – 1
\n\"ML
\n\"ML<\/p>\n

Question 6.<\/strong><\/span>
\n(i) \\(\\frac { 2x+5 }{ 3x+4 } =\\frac { x+1 }{ x+3 } \\)<\/strong>
\n(ii) \\(\\frac { 2 }{ x+2 } -\\frac { 1 }{ x+1 } =\\frac { 4 }{ x+4 } -\\frac { 3 }{ x+3 } \\)<\/strong>
\nSolution:<\/strong><\/span>
\n(i) \\(\\frac { 2x+5 }{ 3x+4 } =\\frac { x+1 }{ x+3 } \\)
\n(2x + 5)(x + 3) = (x + 1)(3x + 4)
\n\"ML
\n\"ML
\n\"ML<\/p>\n

Question 7.<\/strong><\/span>
\n(i) \\(\\frac { 3x-4 }{ 7 } +\\frac { 7 }{ 3x-4 } =\\frac { 5 }{ 2 } ,x\\neq \\frac { 4 }{ 3 } \\)<\/strong>
\n(ii) \\(\\frac { 4 }{ x } -3=\\frac { 5 }{ 2x+3 } ,x\\neq 0,-\\frac { 3 }{ 2 } \\)<\/strong>
\nSolution:<\/strong><\/span>
\n(i) \\(\\frac { 3x-4 }{ 7 } +\\frac { 7 }{ 3x-4 } =\\frac { 5 }{ 2 } ,x\\neq \\frac { 4 }{ 3 } \\)
\nlet \\(\\frac { 3x-4 }{ 7 } \\) = y,then
\n\"ML
\n\"ML
\n\"ML<\/p>\n

Question 8.<\/strong><\/span>
\n(i)x\u00b2 + (4 – 3a)x – 12a = 0<\/strong>
\n(ii)10ax\u00b2 – 6x + 15ax – 9 = 0,a\u22600<\/strong>
\nSolution:<\/strong><\/span>
\n(i)x\u00b2 + (4 – 3a)x – 12a = 0
\nHere a = 1,b = 4 – 3a,c = – 12a
\n\"ML
\n\"ML
\n\"ML<\/p>\n

Question 9.<\/strong><\/span>
\nSolve for x using the quadratic formula. Write your answer correct to two significant figures: (x – 1)\u00b2 – 3x + 4 = 0. (2014)<\/strong>
\nSolution:<\/strong><\/span>
\n(x – 1)\u00b2 – 3x + 4 = 0
\nx\u00b2 + 1 – 2x – 3x + 4 = 0
\n\"ML<\/p>\n

Question 10.<\/strong><\/span>
\nDiscuss the nature of the roots of the following equations:<\/strong>
\n(i) 3x\u00b2 – 7x + 8 = 0<\/strong>
\n(ii) x\u00b2 – \\(\\\\ \\frac { 1 }{ 2 } x\\) – 4 = 0<\/strong>
\n(iii) 5x\u00b2 – 6\u221a5x + 9 = 0<\/strong>
\n(iv) \u221a3x\u00b2 – 2x – \u221a3 = 0<\/strong>
\nSolution:<\/strong><\/span>
\n(i) 3x\u00b2 – 7x + 8 = 0
\nHere a = 3, b = – 7,c = 8
\n\"ML
\n\"ML<\/p>\n

Question 11.<\/strong><\/span>
\nFind the values of k so that the quadratic equation (4 – k) x\u00b2 + 2 (k + 2) x + (8k + 1) = 0 has equal roots.<\/strong>
\nSolution:<\/strong><\/span>
\n(4 – k) x\u00b2 + 2 (k + 2) x + (8k + 1) = 0
\nHere a = (4 – k), b = 2 (k + 2), c = 8k + 1
\n\"ML
\nor k – 3 = 0, then k= 3
\nk = 0, 3 Ans.<\/p>\n

Question 12.<\/strong><\/span>
\nFind the values of m so that the quadratic equation 3x\u00b2 – 5x – 2m = 0 has two distinct real roots.<\/strong>
\nSolution:<\/strong><\/span>
\n3x\u00b2 – 5x – 2m = 0
\nHere a = 3, b = – 5, c = – 2m
\n\"ML<\/p>\n

Question 13.<\/strong><\/span>
\nFind the value(s) of k for which each of the following quadratic equation has equal roots:<\/strong>
\n(i)3kx\u00b2 = 4(kx – 1)<\/strong>
\n(ii)(k + 4)x\u00b2 + (k + 1)x + 1 =0<\/strong>
\nAlso, find the roots for that value (s) of k in each case.<\/strong>
\nSolution:<\/strong><\/span>
\n(i)3kx\u00b2 = 4(kx – 1)
\n=> 3kx\u00b2 = 4kx – 4
\n=> 3kx\u00b2 – 4kx + 4 = 0
\n\"ML
\n\"ML<\/p>\n

Question 14.<\/strong><\/span>
\nFind two natural numbers which differ by 3 and whose squares have the sum 117.<\/strong>
\nSolution:<\/strong><\/span>
\nLet first natural number = x
\nthen second natural number = x + 3
\nAccording to the condition :
\nx\u00b2 + (x + 3)\u00b2 = 117
\n\"ML<\/p>\n

Question 15.<\/strong><\/span>
\nDivide 16 into two parts such that the twice the square of the larger part exceeds the square of the smaller part by 164.<\/strong>
\nSolution:<\/strong><\/span>
\nLet larger part = x
\nthen smaller part = 16 – x
\n(\u2235 sum = 16)
\nAccording to the condition
\n\"ML<\/p>\n

Question 16.<\/strong><\/span>
\nTwo natural numbers are in the ratio 3 : 4. Find the numbers if the difference between their squares is 175.<\/strong>
\nSolution:<\/strong><\/span>
\nRatio in two natural numbers = 3 : 4
\nLet the numbers be 3x and 4x
\nAccording to the condition,
\n\"ML<\/p>\n

Question 17.<\/strong><\/span>
\nTwo squares have sides A cm and (x + 4) cm. The sum of their areas is 656 sq. cm.Express this as an algebraic equation and solve it to find the sides of the squares.<\/strong>
\nSolution:<\/strong><\/span>
\nSide of first square = x cm .
\nand side of second square = (x + 4) cm
\nNow according to the condition,
\n\"ML
\nor x – 16 = 0 then x = 16
\nSide of first square = 16 cm
\nand side of second square = 16 + 4 – 4
\n= 20 cm Ans.<\/p>\n

Question 18.<\/strong><\/span>
\nThe length of a rectangular garden is 12 m more than its breadth. The numerical value of its area is equal to 4 times the numerical value of its perimeter. Find the dimensions of the garden.<\/strong>
\nSolution:<\/strong><\/span>
\nLet breadth = x m
\nthen length = (x + 12) m
\nArea = l \u00d7 b = x (x + 12) m\u00b2
\nand perimeter = 2 (l + b)
\n= 2(x + 12 + x) = 2 (2x + 12) m
\nAccording to the condition.
\n\"ML<\/p>\n

Question 19.<\/strong><\/span>
\nA farmer wishes to grow a 100 m\u00b2 rectangular vegetable garden. Since he has with him only 30 m barbed wire, he fences three sides of the rectangular garden letting compound wall of his house act as the fourth side fence. Find the dimensions of his garden.<\/strong>
\nSolution:<\/strong><\/span>
\nArea of rectangular garden = 100 cm\u00b2
\nLength of barbed wire = 30 m
\nLet the length of side opposite to wall = x
\n\"ML<\/p>\n

Question 20.<\/strong><\/span>
\nThe hypotenuse of a right angled triangle is 1 m less than twice the shortest side. If the third side is 1 m more than the shortest side, find the sides of the triangle.<\/strong>
\nSolution:<\/strong><\/span>
\nLet the length of shortest side = x m
\nLength of hypotenuse = 2x – 1
\nand third side = x + 1
\nNow according to the condition,
\n\"ML<\/p>\n

Question 21.<\/strong><\/span>
\nA wire ; 112 cm long is bent to form a right angled triangle. If the hypotenuse is 50 cm long, find the area of the triangle.<\/strong>
\nSolution:<\/strong><\/span>
\nPerimeter of a right angled triangle = 112 cm
\nHypotenuse = 50 cm
\n\u2234 Sum of other two sides = 112 – 50 = 62 cm
\nLet the length of first side = x
\nand length of other side = 62 – x
\n\"ML<\/p>\n

Question 22.<\/strong><\/span>
\nCar A travels x km for every litre of petrol, while car B travels (x + 5) km for every litre of petrol.<\/strong>
\n(i) Write down the number of litres of petrol used by car A and car B in covering a distance of 400 km.<\/strong>
\n(ii) If car A uses 4 litres of petrol more than car B in covering 400 km. write down an equation, in A and solve it to determine the number of litres of petrol used by car B for the journey.<\/strong>
\nSolution:<\/strong><\/span>
\nDistance travelled by car A in one litre = x km
\nand distance travelled by car B in one litre = (x + 5) km
\n(i) Consumption of car A in covering 400 km
\n\"ML<\/p>\n

Question 23.<\/strong><\/span>
\nThe speed of a boat in still water is 11 km\/ hr. It can go 12 km up-stream and return downstream to the original point in 2 hours 45 minutes. Find the speed of the stream<\/strong>
\nSolution:<\/strong><\/span>
\nSpeed of boat in still water =11 km\/hr
\nLet the speed of stream = x km\/hr.
\nDistance covered = 12 km.
\nTime taken = 2 hours 45 minutes .
\n\"ML<\/p>\n

Question 24.<\/strong><\/span>
\nBy selling an article for Rs. 21, a trader loses as much percent as the cost price of the article. Find the cost price.<\/strong>
\nSolution:<\/strong><\/span>
\nS.R of an article = Rs. 21
\nLet cost price = Rs. x
\nThen loss = x%
\n\"ML<\/p>\n

Question 25.<\/strong><\/span>
\nA man spent Rs. 2800 on buying a number of plants priced at Rs x each. Because of the number involved, the supplier reduced the price of each plant by Rupee 1.The man finally paid Rs. 2730 and received 10 more plants. Find x.<\/strong>
\nSolution:<\/strong><\/span>
\nAmount spent = Rs. 2800
\nPrice of each plant = Rs. x
\nReduced price = Rs. (x – 1)
\n\"ML<\/p>\n

Question 26.<\/strong><\/span>
\nForty years hence, Mr. Pratap’s age will be the square of what it was 32 years ago. Find his present age.<\/strong>
\nSolution:<\/strong><\/span>
\nLet Partap\u2019s present age = x years
\n40 years hence his age = x + 40
\nand 32 years ago his age = x – 32
\nAccording to the condition
\n\"ML<\/p>\n

Hope given ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 6 Quadratic Equations in One Variable Chapter Test\u00a0are helpful to complete your math homework.<\/p>\n

If you have any doubts, please comment below. APlusTopper<\/a> try to provide online math tutoring for you.<\/p>\n

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

ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 6 Quadratic Equations in One Variable Chapter Test These Solutions are part of ML Aggarwal Class 10 Solutions for ICSE Maths. Here we have given ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 6 Quadratic Equations in One Variable Chapter Test ML Aggarwal SolutionsICSE SolutionsSelina […]<\/p>\n","protected":false},"author":9,"featured_media":0,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"_genesis_hide_title":false,"_genesis_hide_breadcrumbs":false,"_genesis_hide_singular_image":false,"_genesis_hide_footer_widgets":false,"_genesis_custom_body_class":"","_genesis_custom_post_class":"","_genesis_layout":"","footnotes":""},"categories":[6768],"tags":[6795,6794,6796,6775,6793,6799,6798,39620,6776,39619],"yoast_head":"\nML Aggarwal Class 10 Solutions for ICSE Maths Chapter 6 Quadratic Equations in One Variable Chapter Test - A Plus Topper<\/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:\/\/www.aplustopper.com\/ml-aggarwal-class-10-solutions-for-icse-maths-chapter-6-chapter-test\/\" \/>\n<meta property=\"og:locale\" content=\"en_US\" \/>\n<meta property=\"og:type\" content=\"article\" \/>\n<meta property=\"og:title\" content=\"ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 6 Quadratic Equations in One Variable Chapter Test\" \/>\n<meta property=\"og:description\" content=\"ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 6 Quadratic Equations in One Variable Chapter Test These Solutions are part of ML Aggarwal Class 10 Solutions for ICSE Maths. 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