Mathematics Class X Board Paper Solution Section - A (4 Marks) Soln.. (a). Here, p(x) = x + x kx + For (x-) to be the factor of p(x) = x + x kx + P () = Thus, () + () k() + = 8 + 8 - k + = k = Thus p(x) becomes x + x x + Now, (x+5) would be the factor of p(x) if p(-5) = p(-5) = (-5) + (-5) (-5) + = -5 + 5 + 65 + = So, (x+5) is also a factor of p(x) = x + x x +. (b) Yes, product AB is possible since the number of columns of matrix A is equal to the number of rows of matrix B. (Matrix A is of the order x and B is of the order of x ) The required product AB = 6 = 8 8 6 = (c). Here Principal, P = Rs. 5 Rate of interest, R = 8% for first year and % for second year P R T 58 Interest for st year = - Rs Amount at the end of first year = Rs. 5 + = Rs. 6 Kumar repays Rs. 6 Principal for second year = Rs. 6 Rs. 6 = Rs.
PR T Interest for second year = Rs. Amount at the end of second year = Rs. + = Rs. Soln.. (a) In a deck of cards, for each suit we have three cards with number, 6, 9 which are multiples of. Thus for four different suits Spade, Heart, Diamond, Club, x 4 = such cards will be removed. Total number of possible outcomes = 5 = 4 (i) Each suit has face cards. Four suits (Spade, Heart, Diamond, Club) will have x 4 = face cards. So, required probability will be given by P (getting a face card) = 4 (ii) Each suit has 4 (cards with number, 4, 8, ) even numbered cards. Suits Heart and Diamond are of red colour. 8 (b) X 6 X X Thus, two suits will have x 4 = 8 even numbered cards. So, required probability would be given by 8 P (getting an even numbered red card) = 4 5 8 6 X X 6x 8 = Here a =,b = -6 and c = 8 Thus the roots of the equation will be b x b a 4ac ( 6) x ( 6) () 4()( 8) 6 x 8
6 6 x X X.7 using,.7 X 8.9and.9 (C) IN ACO, ACO = (Given) OAC = 9 o [radius is perpendicular to the tangent at the point of contact] By angle sum property, ACO+OAC+AOC = 8 AOC = 8 o (9 o + o ) = 6 Consider AOC and BOC AO = BO (radii) AC = BC (tangents to a circle from an external point are equal in length) OC = OC (Common) AOC is congruent to BOC. (i) BCO =ACO = o (ii) AOC =BOC = 6 o AOB =AOC + BOC = o (iii) We know that, If two angles stand on the same chord, then the angle at the centre is twice the angle at the circumference. AOB and APB stand on the same chord AB. AOB = APB So, APB = AOB =6 o
Soln.. (a) P = Rs. 5, n = years = ( ) months= 4 months Total principal = Rs.,5 X 4 = Rs. 6, Amount = Rs. 66,5 Interest = Amount Principal = Rs. 66,5 Rs. 6,= Rs. 6,5 Thus, the interest paid by the bank is Rs. 6,5 Let r be the rate of interest. n(n ) 4 5 N 5 yrs This is equivalent to depositing Rs.,5 for 5 yrs. P N R Interest = 6,5 = R =,5 5 R Thus, the rate of interest is %. (b) Diameter of the semi circle is 4 cm. ED = AC = 4cm Therefore, AB = BC = AE = CD = 7cm Area of the shaded region = Area of semi circle EFD + Area of rectangle AEDC Area of quadrant ABE Area of quadrant BCD Area of semi circle EFD = πr 7 7 7 77cm Area of rectangle AEDC = AC X AE = 4 cm X 7cm = 98cm Area of quadrant ABE = Area of quadrant BCD = 9 6 πr = 7 4 7 77 cm 8
77 Area of the shaded region = 77 cm + 98 cm x cm 8 55.75cm (c) The coordinates of the vertices of ABC are A (, ), B (4, b) and C (a, ). It is known that A(x, y ), B(x, y ) and C(x, y ) are vertices of a triangle, then the coordinates of centroid are X x x y y y G =,, Thus, the coordinates of the centroid of ABC are 4 a, b 5 a 4 b =, It is given that the coordinates of the centroid are G (4, ). Therefore, we have: 5 a 4 5+a = a = 7 4 b 4+b=9 b =5 Thus, the coordinates of B and C are (4, 5) and (7, ) respectively. Using distance formula, we have: BC = 9 6 (7 4) = 5 5units ( 5) Soln.4. (a) The given inequation is x 5 5x + 4 <, where x I x 55x + 4 x 5x4 + 5 x 9 X 5x + 4 < 5x < -4 5X < 7 X <.4 Since X I, the solution set is {-, -, -,, }. The solution set can be represented on a number line as follows:
(b) tan5 cot55 sec4 cot55 tan5 cosec5 tan(9 55 ) cot(9 5 ) sec(9 5 ) = cot55 tan5 cosec5 tan(9 ) cot cot55 tan5 cosec5 cot(9 ) tan cot55 tan5 cosc5 sec(9 ) cosec =() + () -() =+-= (c) The histogram for the given data can be drawn by taking the marks on the x - axis and the number of students on the y-axis. To locate the mode from the histogram, we proceed as follows: i Find the modal class. Rectangle ABCD is the largest rectangle. It represents the modal class, that is, the mode lies in this rectangle. The modal class is 8 9 ii Draw two lines diagonally from the vertices C and D to the upper corners of the two adjacent rectangles. Let these rectangles intersect at point H. iii The x-value of the point H is the mode. Thus, mode of the given data is approximately 8.
Section B (4 marks) Soln.5. (a) Given cost of washing machine = Rs 5 (i) Amount of tax collected by manufacturer = 8% of Rs 5 8 5 Rs As profit of wholesaler is Rs, VAT to be payed by wholesaler = 8% of Rs 8 Rs96 As trader earns a profit of RS 8 VAT to be payed by trader = 8% of Rs 8 8 = 8 Rs44 Amount of tax received by government = Rs ( + 96 +44) = Rs 44 (i) Value of machine paid by the consumer = Price charged by manufacturer + Profit of wholesaler + Profit of trader = Rs (5++8) = Rs 8 Tax paid by consumer = 8% of Rs 8 8 8 Rs44 Therefore, amount paid by customer = Rs (8+44) = Rs 944 (b) Volume of solid cone = r h 5 8 58 7 7
Volume of a small sphere = 4 r 4 7 5 4 7 5 58 volume of cone Number of spheres formed = 7 4 volume of sphere 4 5 7 Thus 4 spheres are obtained by melting the solid cone. (c) We know that the diagonals of a parallelogram bisect each other So, coordinates of mid point of BD = coordinateof Bcoordinateof D Ycoordinateof B Ycoordinateof D, 5 8 4 7,,..() X 4 Y 7 Now the midpoint of diagonal AC =,..() X 4 Y 7 7, =, Comparing we get, X+4 = 7 and y+7 = 4 Thus X = and Y = - So, the coordinates of point A are (, -) Y Y (ii) Equation of a line is given by y y = (X X) X X Coordinates of point B and D are (5, 8) and (,-4) respectively. 4 8 Equation of a diagonal BD, y 8 = (X5) 5 y - 8 = 4(x-5) Or 4X y =
Ans.6. (a) (i) The points A(4,4), B(4,-6) and C(8,) are plotted on the coordinate plane as follows: (ii) To reflect the points ABC across Y axis we will keep y coordinate as it is and negate the x coordinate.
The reflected image of the triangle is shown in blue. (iii) (iv) The coordinates of point A, B, C are (-4, 4), (-4,-6) and (-8, ) respectively. The figure AA C B BC obtained is a polygon with six sides. Thus such a figure would be called a hexagon.
(v) The line of symmetry of AA C B BC would be the y axis. (b) Here rate of interest = 5% Principal for April, 7 = Rs 75 Principal for May, 7 = Rs 9 Principal for June, 7 = Rs 9 Principal for July, 7 = Rs 4 Principal for August, 7 = Rs 4 Principal for September, 7 = Rs 44 Principal for October, 7 = Rs 44
Principal for November, 7 = Rs 5 Principal for December, 7 = Rs 5 Principal for January, 7 = Rs Principal for Feb, 7 = Rs Principal for March, 7 = Rs 45 Total principal for April 7 to April 8 = Rs 577 577 5 principal RateTime Interest paid = Rs 657.8 Rs 657 Soln.7. x 4 x 5 (a) 9 x 4 x 5 Using componendo and dividendo, x 4 x 4 x 5 x 5 x 4 x 4 x 5 x 5 9 9 x 4 x 5 8 x 4 x 5 5 4 Squaring both sides, x 4 x 5 5 6 6(x+4) = 5(x -5) 48x + 64 = 75x 5 75x - 48x = 64 + 5 7x = 89 x = 7
(b) Given, A = 4, 5 B A t = 5 A t.b+bi = 5. 4 + 4. 4 9 4 8 7 7 4 7 4 7 6 (c) (i) InADB and CAB, ADB =CAB ( both 9 o ) ABD =CBA (common angle) ABC ~ DBA (AA similarity criterion)
In ADC and BAC, ADC =BAC (both 9 o ) ACD= ACB (common angle) DAC ~ ABC (AA similarity criterion) If two triangles are similar to one triangle, then the two triangles are similar to each other. DAC ~ DBA or CDA ~ ADB (ii) Since the corresponding sides of similar triangles are proportional. CD DA AD DB AD = DB X CD AD =8 X 8 AD = cm (iii) The ratio of the areas of two similar triangles is equal to the ratio of the squares of their corresponding sides. So Ar( ADB ) Ar( CDA) AD CD 44 64 9 4 Thus, the required ration is 9:4. Soln.8. (a) Class interval Frequency (f) X d = x A= x 67.5 t = i d i=5 f t 5-55 5 5.5-5 - -5 55-6 57.5 - - -4 6-65 6.5-5 - - 65-7 67.5 7-75 9 7.5 5 9 75-8 6 77.5 8-85 8.5 5 6 85-9 8 87.5 4 Total 8 4
Assumed mean (A) = 67.5 Class size, i = 5 ft Mean = A + i f 4 = 67.5 + 5 X 8 = 67.5 +.5 = 69 Thus, the mean of the given data is 69. Modal class is the class corresponding to the greatest frequency. So, the modal class is 55 6. (b) Marks No. of Students (f) cf - 5 5-6 - 6-4 46 4-5 8 74 5-6 7 6-7 4 5 7-8 9 8 8-9 4 94 9-6
Median marks will be 57.5 as the x coordinate corresponding to n/ i.e., is 57.5. (ii) The number of students who failed is 46, which is the y coordinate corresponding to 4 marks. (iii) Number of students who secured more than 85 marks (grade one) = Total number of students 84 = 84 = 6 Soln.9. (a) (i) Amount invested = Rs 5, Face value of share = Rs Discount = Rs Market price = Rs Rs = Rs 8 Number of shares = Rs 5,/ Rs 8=65 Dividend % = 8% Total FV = FV of each share X Number of shares = Rs X 65 = Rs 65, 8 D = D% X Total FV = 65, 5, Thus, the annual dividend is Rs 5, (ii) Amount at which the shares were sold = Rs X 65 = Rs 78, Profit earned including his dividend = Rs (78, 5,) + Rs 5, = Rs, (b) For constructing the pair of tangents to the given circle following steps will be followed. Taking any point O of the given plane as centre draw a circle of.5 cm radius. Locate a point P, 6 cm away from O. Join OP.. Bisect OP. Let M be the midpoint of PO.. Taking M as centre and MO as radius draw a circle. 4. Let this circle intersect our circle at point Q and R. 5. Join PQ and PR. PQ and PR are the required tangents. We may find that length of tangents PQ and PR are 8 cm each.
(C) Consider LHS LHS = (cosec A sin A) (sec A cos A)sec A sina sina cosa cosa cos A sin A cos A sina cosa cos A cos A sin A.. sin A cos A cos A sina cosa tana RHS Soln. (a) 6 is the mean proportion between two numbers x and y, i.e. 6 = xy So, 6 = xy.() It is given that 48 is the third proportional of x and y So, y = 48x () From () and (), we get 6 y = 48 y y = 78 Hence, y = (b) Here Principal = Rs,, Rate = %, Compound Interest = Rs 97
Compound Interest = P N R N 97 = N 97 = 97 = N = years N N (c) Let the height of the building be AB = h and BC = x In ABC, h Tan6 = x h () In ADB, h tan = 6 h 6
x+6 = h () From () and () x+6=x. x=6 x= Thus, h= =5.96m Soln.. (a) Let the radii of the circles with A, B and C as centres be r, r and r respectively. According to the given information, AB = cm = r + r () BC = 8 cm = r + r () CA = 6 cm = r + r () Adding equations (), () and (), (r + r + r ) = 4 r + r + r = (4) Subtracting () from (4), r = = Subtracting () from (4), r = 8 = 4 Subtracting () from (4), r = 6 = 6 Thus, the radii of the three circles are cm, 4 cm and 6 cm.
(b) Number of children = x. Share of each child = Rs 48 If number of children are x +, then share of each child = Rs According to the given information: 48 48 48 48 ( ) 48 ( ) 48 ( ) X + X 8 = X - X + 4X 8 = X(X - ) + 4(X - ) = (X - ) (X + 4) = X = or X = -4 But, the number of children cannot be negative. Therefore, x = 48
(c) The equation of the line L is y = 4. It is given that L is the bisector of angle O and O = 9. Thus, the line L makes an angle of 45 with the x-axis. Thus, slope of line L = tan 45 = The line L passes through (, ) and its slope is. So, its equation is given by y y = m(x x ) y = (x ) y = x Now, the point P is the point of intersection of the lines L and L. Solving the equations y = 4 and x = y, we get x = y = 4 Thus, the coordinates of the point P are (4, 4).