{"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":"
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
\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