{"id":31825,"date":"2018-10-08T11:07:03","date_gmt":"2018-10-08T11:07:03","guid":{"rendered":"https:\/\/cbselibrary.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:\/\/cbselibrary.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":"
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> Question 2.<\/strong><\/span> Question 3.<\/strong><\/span> Question 4.<\/strong><\/span> Solve the following equations (5 to 8) by using formula :<\/strong><\/p>\n Question 5.<\/strong><\/span> Question 6.<\/strong><\/span> Question 7.<\/strong><\/span> Question 8.<\/strong><\/span> Question 9.<\/strong><\/span> Question 10.<\/strong><\/span> Question 11.<\/strong><\/span> Question 12.<\/strong><\/span> Question 13.<\/strong><\/span> Question 14.<\/strong><\/span> Question 15.<\/strong><\/span> Question 16.<\/strong><\/span> Question 17.<\/strong><\/span> Question 18.<\/strong><\/span> Question 19.<\/strong><\/span> Question 20.<\/strong><\/span> Question 21.<\/strong><\/span> Question 22.<\/strong><\/span> Question 23.<\/strong><\/span> Question 24.<\/strong><\/span> Question 25.<\/strong><\/span> Question 26.<\/strong><\/span> 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 … Read more<\/a><\/p>\n","protected":false},"author":8,"featured_media":0,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"spay_email":""},"categories":[6768],"tags":[6795,6794,6796,6775,6793,6799,6798,39620,6776,39619],"yoast_head":"\n
\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
\n<\/p>\n
\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
\n<\/p>\n
\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
\n<\/p>\n
\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
\n
\n<\/p>\n
\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
\n<\/p>\n
\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
\n
\n<\/p>\n
\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
\n
\n<\/p>\n
\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
\n
\n<\/p>\n
\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<\/p>\n
\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
\n<\/p>\n
\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
\nor k – 3 = 0, then k= 3
\nk = 0, 3 Ans.<\/p>\n
\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<\/p>\n
\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
\n<\/p>\n
\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<\/p>\n
\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<\/p>\n
\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<\/p>\n
\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
\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
\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<\/p>\n
\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<\/p>\n
\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<\/p>\n
\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<\/p>\n
\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<\/p>\n
\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<\/p>\n
\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<\/p>\n
\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<\/p>\n
\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<\/p>\n