# Question 1.Use the given letters to write the answer.i. There are ‘a’ trees in the village Lat. If the number of trees increases every year by ‘b’, then how many trees will there be after ‘x’ years?ii. For the parade there are y students in each row and x such row is formed. Then, how many students are there for the parade in all?iii. The tens and units place of a two-digit number is m and n respectively. Write the polynomial which represents the two-digit number.Answer:i. Given:Total number of trees in the village = aIncrease in the number of trees by = bTo find: Number of trees after x yearsNumber of trees in x years = (a + b) xii. Given:Number of students in each row = yNumber of rows of Students = xSo total number of students = x × y= xyiii. For a number in tens place should be multiplied with ten and for the number in units place should be multiplied by one. So the number that can be formed as follows:Since m is in tens place and n is in units place,10 × m + 1 × n10m + nSo the polynomial representing the two digit number is = 10m + nQuestion 2.Add the given polynomials.i. x3 – 2x2 – 9; 5x3 + 2x +9ii. iii. 2y2 + 5; 3y +9; 3y2 – 4y – 3Answer:i. The result = 6x3 – 2x2 + 2xii. The result = 2m4 + 2m3 + 2m2 + 3m – 6 +√2iii. The result = 5y2 + 6y + 11Question 3.Subtract the second polynomial from the first.i. ii. 2ab2 + 3a2b – 4ab; 3ab – 8ab2 +2a2bAnswer:In the subtraction process, the sign of the subtrahend that is the second polynomial is inverted and then the operation is carried out. The result = 6x2 + 10xii. In the subtraction process, the sign of the subtrahend that is the second polynomial is inverted and then the operation is carried out. The result = 10ab2 – a2b – 7abQuestion 4.Multiply the given polynomials.i. 2x; x2 – 2xii. x5 – 1; x3 + 2x3 + 2iii. 2y + 1; y2 – 2y3 + 3yAnswer:The multiplication is as follows:⇒ 2x × (x2 – 2x – 1)⇒ 2x.x2 – 2x.2x – 2x.1⇒ 2x2+1 – 4x1+1 – 2x⇒ 2x3 – 4x2 – 2xTherefore, the product = 2x3 – 4x2 – 2x“.” represents multiplicationii. The multiplication is as follows:⇒ (x5 – 1) × (x3 + 2x3 + 2)⇒ x5.x3 + 2x3.x5 + 2.x5 – 1.x3 – 1.2x3 – 1.2⇒ x5+3 + 2x3 + 5 +2x5 – x3 – 2x3 – 2⇒ x8 + 2x8 + 5 + 2x5 – x3 – 2x3 – 2⇒ 3x8 + 2x5 + 3x3 + 3The product is = 3x8 + 2x5 + 3x3 + 3“.” represents multiplicationiii. The multiplication is as follows:⇒ (2y + 1) × (y2 – 2y3 + 3y)⇒ 2y.y2 – 2y3.2y + 3y.2y + 1.y2 – 1.2y3 + 1.3y⇒ 2y2+1 – 4y3+1 + 6y1+1 + y2 – 2y3 + 3y⇒ 2y3 – 4y4 + 6y2 + y2 – 2y3 + 3y⇒ -4y4 + 7y2 + 3yThe product is = -4y4 + 7y2 + 3y“.” represents multiplicationQuestion 5.Divide first polynomial by second polynomial and write the answer in the form ‘Dividend = Divisor × Quotient + Remainder’.i. x3 – 64; x – 4ii. 5x5 + 4x4 – 3x3 + 2x2 + 2; x2 – xAnswer:i. The division is as follows: x3-64 = [(x2 + 4x + 16) × (x-4)] + 0x3-64 = (x2 + 4x + 16) × (x-4)ii. The division is as follows: 5x5 + 4x4 – 3x3 + 2x2 + 2 = [(5x3 + 9x2 + 6x + 8) × (x2-x)] + 8x +2Quotient = 5x3 + 9x2 + 6x + 8Remainder = 8x + 2Question 6.Write down the information in the form of algebraic expression and simplify.There is a rectangular farm with length (2a2 + 3b2) meter and breadth (a2 + b2) meter. The farmer used a square shaped plot of the farm to build a house. The side of the plot was (a2 - b2) meter. What is the area of the remaining part of the farm?Answer:Given:Length of the farm, l = (2a2 + 3b2) mBreadth of the farm, b = (a2 + b2) mSide of the square plot, s = (a2 - b2) mTo find: Area of remaining farmExplain:Area of Remaining Farm = Area of Rectangle – Area of SquareArea of Rectangle = l × b= (2a2 + 3b2) × (a2 + b2)= 2a2.a2 + 2a2.b2 + 3b2.a2 + 3b2.b2= 2a4 + 5a2b2 + 3b4Area of Square = s × s= (a2 - b2) × (a2 - b2)= a2.a2 – a2.b2 – a2.b2 + b2.b2= a4 – 2a2b2 + b4Area of Remaining Part = (2a4 + 5a2b2 + 3b4) – (a4 – 2a2b2 + b4)= 2a4 + 5a2b2 + 3b4 - a4 + 2a2b2 - b4= a4 + 7a2b2 + 2b4Therefore, the area of remaining portion is = a4 + 7a2b2 + 2b4

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2019 Board Paper Solution

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