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ICSE Class X Question Bank 2025 : Geography

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Sample Question Paper 12 (Detailed Solutions) 1. (i) (a) Given, number of elements = 34 CB = CD mn = 34 \ Now, the possible ordered pairs whose elements product is 34 are (1, 34), (34, 1), (2, 17), (17, 2). Hence, all possible orders of matrix are 1 34, 34 1, 2 17, 17 2 (ii) (d) Given equation is x 2 + x - ( a + 2)( a + 1) = 0 ...(i) On comparing Eq. (i) with Ax 2 + Bx + C = 0, we get A = 1, B = 1, C = - ( a + 2)( a + 1) = - ( a 2 + 3a + 2) \ x= = - B B2 - 4 AC 2A - 1 1 + 4( a 2 + 3a + 2) 2 1 2 = \ - 1 4a + 12a + 9 = - 1 ( 2a + 3)2 2 - 1 ( 12a + 3) = 2 - 1+ 2a + 3 - 1 - 2a - 3 , x= 2 2 2a + 2 - ( 2a + 4) , x= 2 2 2 Also, CDB + CBD + BCD = 180 85 + 85 + BCD = 180 170 + BCD = 180 BCD = 180 - 170 = 10 Now, BAD = BCD [Q angles of same segment are equal] \ BAD = 10 (vii) (c) We know that sin 2 q + cos 2 q = 1 ...(i) 1 2 2 2 Given, a = sec a, b =cosec a = cos a, a 1 = sin 2 a b 1 1 Then, from Eq. (i), + = 1 a b 1 1 ab and c = = = 1 1 1 1 a + b -1 1+ ab a b ab c( a + b - 1) = ab ac + bc - c = ab ac + bc - ab = c (viii) (a) Both the statements are true. x = a + 1 and - ( a + 2) (ix) (b) Given, money deposited per month ( P) = ` 200 Number of months, n = 36 months 5z = 7 20 7 20 z= = 28 5 and rate of interest, r = 11% per annum r( n + 1) \Maturity amount = Pn 1 + 2400 (iv) (a) We have, 11( 36 + 1) = 200 36 1 + 2400 2400 + 407 = 7200 2400 20 - 4x 16 - 20 + 20 - 4x 16 - 20 [add - 20 both sides] - 4x - 4 4x - 4 [divide by - 4 both sides] -4 -4 Hence, the smallest value of x is 1. (v) (a) The equation of a line parallel to X-axis is Since, it is passing through a point ( 3, - 5). \ - 5= a a = - 5 On putting a = - 5 in Eq. (i), we get y=-5 y+ 5=0 (vi) (a) Given, CB = CD and CDB = 85 Now, in DCBD, we have = 3( 2807) = ` 8421 x 1 y=a ...(i) [Q angles opposite to equal sides of a triangle are equal] (iii) (b) We have, 7: z :: 5 : 20 CBD = CDB = 85 \ All factors of 34 are 1, 2, 17, 34 [given] ...(i) Hence, Ritu will get the amount ` 8421 at the time of maturity. (x) (b) Let the radius of original right circular cylinder be r cm and height be h cm. Then, volume of original cylinder V1 = p r2 h cm3 Volume of circular cylinder, when radius is halved pr2 h r 2 V2 = p h cm3 = cm3 2 4 2 Thus, V2 p r h 1 1 = 2 = V1 4 pr h 4 V1 = (xi) (b) Let f ( x ) = x 3 + ax + 4 Since, ( x + 2) is a factor of f ( x ), Therefore, f ( - 2) = 0 ( - 2)3 + a ( - 2) + 4 = 0 - 8 - 2a + 4 = 0 a = - 2 (xii) (d) Given, BAD = 70 , DBC = 30 BCD = 180 - BAD [sum of opposite angles of a cyclic quadrilateral = 180 ] = 180 - 70 = 110 \ BDC = 180 - ( BCD + CBD) = 180 - ( 110 + 30 ) = 40 Marks and volume of a cone having metal B, 1 1 V2 = pr22 h2 = p ( 4)2 3 3 3 \ Ratio of the volume of metal A to the volume 1 p 72 12 V1 49 of the metal B = =3 = = 49 : 4 1 V2 4 2 p 4 3 3 Common Mistake Use different symbols for cone having metal A and cone having metal B. As it may result in substituting the wrong data and hence outcome is incorrect. (ii) Given, 4x - 3y + 12 = 0 meets X-axis at A, (xiii) (c) The frequency distribution table is given below. Number of students 1 2 1 pr 1 h1 = p ( 7)2 12 3 3 Cumulative frequency 0-10 3=3 3 10-20 12 - 3 = 9 12 20-30 27 - 12 = 15 27 30-40 57 - 27 = 30 57 40-50 75 - 57 = 18 75 50-60 80 - 75 = 5 80 Here, we see that the highest frequency is 30, which lies in the interval 30-40. So, it is the modal class. (xiv) (d) Here, S = {( 3,1),( 2,2),( 1,3),( 6,2),( 5,3), ( 4,4),( 3, 5), ( 2, 6),( 6, 6)}. Total number of possible outcomes = 36 Number of favourable outcomes = 9 \ P (sum of two numbers will be multiple of 4) 9 1 = = 36 4 (xv) (c) Dividend on one share of ` 100 = ` 9 Let the market value of one share be ` x. 1 The profit on one share = 7 % of ` x 2 1 3x 15 =` x = ` 2 100 40 Since, the dividend paid on one share = ` 9 3x = 9 x = 120 \ 40 \ The market value of each share = ` 120 i.e. Point A will satisfy the given equation. \ \ 4a - 0 + 12 = 0 4a = - 12 -12 a= 4 a=-3 Given, equation of line is 4x - 3y + 12 = 0. 3y = 4x + 12 4x 12 4x y= + +4 y= 3 3 3 On comparing it with slope intercept form of a line, y = m1 x + c1, we get 4 m1 = 3 Let slope of perpendicular line be m2 . We know that if lines are perpendicular to each other, then m1 m2 = - 1 4 m2 = - 1 3 m2 = \ Equation of a line with slope -3 4 -3 and passing 4 through point A ( -3, 0 ) is metal A are r1 = 7 cm and h1 = 12 cm, respectively. Now, volume of a cone having metal A, 4a - 3 0 + 12 = 0 Hence, coordinates of A are ( -3, 0 ). 2. (i) Given, radius and height of the cone of heavy Also, radius and height (depth) of the cone of lighter metal B are r2 = 4 cm and h2 = 3 cm. y = 0. So, let the coordinates of A be ( a , 0). \ y - y1 = m( x - x1 ) -3 y - 0 = [x - ( -3)] 4 [Q y1 = 0, x1 = - 3 and m = m2 ] -3 y = ( x + 3) 4y = - 3x - 9 4 4y + 3x + 9 = 0 which is the required equation of line. Value Points Mention the equation of the line with slope m and going through a point, i.e. y - y1 = m (x - x1 ). (iii) Given, shopkeeper buys a purifier from wholesaler = ` 22000 and rate of GST = 12% Here, transaction is from one state to another state, so IGST is levied on sale. \Wholesaler collects the IGST amount from shopkeeper = 12% of ` 22000 12 = 22000 = ` 2640 100 Price of an article inclusive tax (under GST) in which the shopkeeper bought = Cost price of article for shopkeeper + Shopkeeper pays GST amount = ` 22000 + shopkeeper pays the IGST amount to the wholesaler = ` 22000 + ` 2640 = ` 24640 3. (i) (a) Firstly, plot the points A( 6, 4) and B( 0, 4) on a graph paper. The reflections of A and B in the origin is shown in the figure. Scale : X-axis 1 cm = 1 unit Y-axis 1 cm = 1 unit Y 4 Class interval d Frequency Midd = xi - A ui = i (f i ) value ( x i ) i h 50-55 5 52.5 - 15 -3 - 15 55-60 20 57.5 - 10 -2 - 40 60-65 10 62.5 -5 -1 - 10 65-70 10 67.5 = A 0 0 0 70-75 9 72.5 5 1 9 75-80 6 77.5 10 2 12 80-85 12 82.5 15 3 36 85-90 8 87.5 20 4 32 Total fi = 80 3 (a) We know that Mean ( X ) = A + (iii) In the given figure, AB is the diameter of a circle. 5 4 3 2 [Q AB is a diameter and angle in a semi-circle is 90 ] ABD = DC A [Q angles made by same segment] 1 2 3 4 5 6 X DAB + ABD + BDA = 180 2 3 4 B (0, 4) Y (b) The coordinates of the images of A and B in the origin are A ( - 6, - 4) and B ( 0, - 4), respectively. (c) The geometrical name of the figure ABA B is parallelogram. (d) Here, ( A B)2 = ( 6)2 + ( 8)2 = 36 + 64 = 100 = ( 10)2 A B = 10 cm and A B = 6 cm \ Perimeter of parallelogram ABA B = 2( A B) + 2( A B ) = 2 10 + 2 6 = 20 + 12 = 32 cm (ii) Let assumed mean, A = 67.5 and class interval, h = 5. To make a table for the product of frequency and step-deviation method. [Q DC A = 40 ] ABD = 40 Now, in DABD, 1 A ( 6, 4) ADB = 90 \ Also, 1 X 6 Sfi ui 24 h = 67.5 + 5 Sf i 80 = 67.5 + 0. 3 5 = 67.5 + 1.5 = 69 (b) Here, we see that highest frequency is 20, whose corresponding class is 55-60. Hence, modal class is 55-60. 2 1 0 fi ui = 24 A (6,4) B (0,4) f iu i [Q sum of all angles in a triangle is 180 ] x + 40 + 90 = 180 x + 130 = 180 x = 180 - 130 \ x = 50 4. (i) Let the age of the son 5 yr ago be x yr, then the age of the woman 5 yr ago = x 2 yr \The present age of the son = ( x + 5) yr and the present age of the woman = ( x 2 + 5) yr 10 yr hence i.e. after 10 yr from now, the age of the son = (( x + 5) + 10) yr = ( x + 15) yr and the age of the woman = (( x 2 + 5) + 10) yr = ( x 2 + 15) yr According to the question, x 2 + 15 = 2( x + 15) x 2 + 15 = 2x + 30 2 x - 2x - 15 = 0 ( x - 5)( x + 3) = 0 x - 5 = 0 or x + 3 = 0 x = 5 or x = - 3 but x being age cannot be negative. x=5 \ 5. (i) Given, \ (a) The age of the son 5 yr ago = 5 yr (b) The present age of the woman = ( x 2 + 5) = ( 52 + 5) = 30 yr x y z (ii) Given, = = = k (say) a b c x = ka, y = kb, z = kc 2 2 2 2 (i) 2 2 a x + b y + c z LHS = 3 3 3 a x + b y+ c z 3/ 2 1 2 A= 3 4 A2 = A A 1 2 1 2 1 + 6 2 + 8 A2 = = 3 4 3 4 3 + 12 6 + 16 7 = 15 1 3A = 3 3 and Now, A2 + 3 A + I a 2 k 2 a 2 + b 2 k 2 b 2 + c2 k 2 c2 = a 3 ka + b 3 kb + c3 kc 3/ 2 7 10 3 6 1 0 = + + 15 22 9 12 0 1 1 0 Q I = 0 1 7 + 3 + 1 10 + 6 + 0 = 15 + 9 + 0 22 + 12 + 1 [using Eq. (i)] k 2 ( a 4 + b 4 + c4 ) = 4 4 4 k( a + b + c ) 3/ 2 = k3/ 2 xyz ka kb kc = = k3/ 2 abc abc and RHS = (ii) (iii) From Eqs. (ii) and Eqs. (iii), we get LHS = RHS Hence proved. (iii) Let f ( x ) = 2x 3 + ax 2 + bx - 14 Since, ( x - 2) is a factor of f ( x ). 3 2 8 + 4a + 2b - 14 = 0 4a + 2b + 2 = 0 4a + 2b = - 2 2a + b = - 1 (i) 54 + 9a + 3b - 14 = 52 9a + 3b = 12 3a + b = 4 \ 0= (ii) 2a + b = - 1 3a + b = 4 -a=-5 a=5 Now, putting the value of a = 5 in Eq. (ii), we get 3 5 + b = 4 15 + b = 4 b = 4 - 15 b = - 11 Hence, a = 5 and b = - 11 m1 ( -5) + m2 3 m1 + m2 [Q y2 = - 5, y1 = 3 and y = 0] On subtracting Eq. (ii) from Eq. (i), we get \ Since, y-coordinate of any point on X-axis is 0 m y + m2 y1 and y= 1 2 m+ n [by remainder theorem] B (6, 5) Let the point K divides AB in the ratio m1 : m2 . 2 ( 3 )3 + a ( 3 )2 + b ( 3 ) - 14 = 52 K A (2,3) Again, when f ( x ) is divided by ( x - 3), it leaves the remainder 52. f ( 3 ) = 52 16 35 m1 : m2 2 16 35 (ii) Given, points of a line segment AB are A(2, 3) and B ( 6, - 5). 2 ( 2 ) + a ( 2 ) + b ( 2 ) - 14 = 0 \ 11 = 24 11 A2 + 3 A + I = 24 [by factor theorem] f ( 2) = 0 \ [multiply row by column] 10 22 2 3 6 = 4 9 12 0 = - 5m1 + 3m2 5m1 = 3m2 m1 3 = m2 5 m1 : m2 = 3 : 5 m x + m2 x1 Now, x= 1 2 m1 + m2 \ 3 6 + 5 2 3+ 5 18 + 10 28 = = 8 8 7 x= 2 x= (a) Point K divides AB in the ratio 3 : 5. 7 (b) Thus, the coordinates of point K are , 0 . 2 (iii) Given, a sin q + b cos q = c \ Sum of first 20 terms = On squaring both sides, we get = 10 ( 2 1 + 19 5) ( a sin q + b cos q)2 = c2 a 2 sin 2 q + b 2 cos 2 q + 2 ab sin q cos q = c2 a 2 ( 1 - cos 2 q) + b 2 ( 1 - sin 2 q) +2ab sin q cos q = c2 [Qsin 2 A + cos 2 A = 1] a 2 + b 2 - ( a 2 cos 2 q + b 2 sin 2 q - 2ab sin q cos q) = c2 a 2 cos 2 q + b 2 sin 2 q - 2ab sin q cos q = a 2 + b 2 - c2 ( a cos q - b sin q)2 = a 2 + b 2 - c2 = 10 97 = 970 3x 2 (iii) Given, 4x - 19 < - 2 - + x, x R 5 5 3x Consider, 4x - 19 < -2 5 3x 4x < 19 - 2 5 20x - 3x < 17 5 17x < 17 5 a cos q - b sin q = a 2 + b 2 - c2 [multiply both sides by 5] Hence proved. (a) Let E1 be the event of drawing odd numbered balls. Then, n( E1 ) = 10 i.e. 1, 3, 5, 7, 9, 11, 13, 15, 17, 19 \ P (drawing an odd numbered ball) n( E1 ) 10 1 P( E1 ) = = = n( S) 20 2 (b) Let E2 be the event of drawing a prime numbered ball. Then, n( E2 ) = 8 i.e. 2, 3, 5, 7, 11, 13, 17, 19 \ P (a prime numbered ball), n( E2 ) 8 2 P( E2 ) = = = n( S) 20 5 (ii) Let a be the first term and d be the common difference of the AP. According to the question, 5 7 ( 2a + 4d ) + ( 2a + 6d ) = 167 2 2 ...(i) 3x 10 2 + 5x [multiply both sides by 5] x<5 3x -2 Again, consider -2 +x 5 5 3x - 10 -2 + 5x 5 5 2 - 10 5x - 3x [subtract (- 2 + 3x) from both sides] -8 2x - 4 x [divide both sides by 2] (ii) From Eqs. (i) and (ii), we get -4 x < 5 Hence, solution set is {x : - 4 x < 5, x R} + 5 4 3 2 1 0 1 2 3 4 5 6 5 ( a + 2d ) + 7( a + 3d ) = 167 (i) 12a + 31d = 167 10 and S10 = 235 ( 2a + 9d ) = 235 2 7. (i) To make a table for the cumulative frequency Marks Number of students ( f ) Cumulative frequency (cf ) 0-10 5 5 10-20 10 5+10 = 15 20-30 11 15 + 11 = 26 30-40 20 26+ 20 = 46 40-50 27 46+ 27 = 73 23d = 115 50-60 38 73+38 = 111 d=5 60-70 40 111+40 = 151 70-80 29 151+ 29 = 180 80-90 14 180 + 14 = 194 194+6 = 200 (ii) 2a + 9d = 47 On multiplying Eq. (ii) by 6, we get 12a + 54d = 282 (iii) On subtracting Eq. (i) from Eq. (iii), we get 17x < 85 85 [divide both sides by 17] x< 17 The graph of the solution set on number line is represented by dark line. S5 + S7 = 167 6. (i) Total number of balls in a bag, n ( S) = 20 20 ( 2a + 19d ) 2 On substituting d = 5 in Eq. (ii), we get 2a + 9 5 = 47 2a = 2 90-100 6 a=1 Total N = 200 On the graph paper, we plot the following points A( 10, 5), B( 20, 15), C( 30, 26), D( 40, 46), E( 50, 73), F( 60, 111), G( 70, 151), H( 80, 180), I( 90, 194) and J( 100, 200). Join all these points by a free hand drawing. (ii) Let AD be the height of the tree and C be the position of man. Let B be the another position of a man when he moves 40 m away from the bank of river. A The required ogive is shown on the graph paper given below Y 200 180 Scale : On X-axis 1 cm = 10 marks On Y-axis 1 cm = 20 students L M 160 J (100,200) I(90,194) H(80,180) B 40 m G(70,151) Number of students 140 80 D(40,46) 40 O [Q exterior angle is equal to the sum of interior opposite angles] T 50 20 Also, ABC + BAC = ACD E(50, 73) 60 S (20,15)B C(30,26) A(10,5) R U K 10 20 30 40 50 60 70 80 90 100 Marks D ABD = 30 and ACD = 60 Q F(60,111) P C BC = 40 m \ 120 100 60 30 X (a) Here, number of students ( N ) = 200, which is even. Let P be the point on Y-axis, N 200 representing frequency = = = 100. 2 2 Through P, draw a horizontal line to meet the ogive at point Q. Through Q, draw a vertical line to meet the X-axis at point R. The abscissa of the point represents 57 marks. Hence, median marks is 57. (b) Let S be the point on Y-axis representing N 200 frequency = = = 50. 4 4 Through S, draw a horizontal line to meet the ogive at point T. Through T, draw a vertical line to meet the X-axis at point U. The abscissa of the point represents 41 marks. Hence, lower quartile is 41 marks. (c) 80% marks = 80% of 100 = 80 Let the point K on X-axis represents 80 marks. Through K, draw a vertical line to meet the ogive at the point L. Through L, draw a horizontal line to meet the Y-axis at point M. The ordinate of point M represents 180 students. Hence, the number of students scoring more than 80% marks = 200 - 180 = 20 students. \ 30 + BAC = 60 BAC = 60 - 30 BAC = 30 \ BAC = B = 30 So, BC = AC [Q BC = 40 m] AC = 40 m (a) In right angled DADC, CD base cos 60 = Qcos q = hypotenuse AC 1 CD = [Q AC = 40 m] 2 40 40 CD = = 20 2 Hence, the width of the river is 20 m. (b) Again, in right angled DADC, AD perpendicular sin 60 = Qsin q = hypotenuse AC 3 AD [Q AC = 40 m] = 2 40 40 3 = 20 3 AD = 2 = 20 1.73 = 34.6 Hence, the height of the tree is 34.6 m. 8. (i) Given, dimensions of cuboidal block are l = 15 cm, 7 cm 2 Total surface area of cubodial block = 2( lb + bh + hl ) = 2 ( 15 10 + 10 5 + 5 15) = 550 cm 2 Curved surface area of cylinder = 2prh 22 7 = 2 5 = 110 cm 2 7 2 b = 10 cm, h = 5 cm and r = Area of two circular bases = 2pr2 22 7 7 = 2 = 77 cm 2 7 2 2 \ Surface area of the remaining block = Total surface area of cuboidal block + Curved surface area of cylinder - Area of two circular bases = 550 + 110 - 77 = 583 cm 2 Number of Mid-value d = x - A i i students ( fi ) ( xi ) 11-20 ...(ii) [reciprocal the terms] From Eqs. (i) and (ii), we get AE BF = ED FC Hence proved. 9. (i) Let the cost price of the toy be ` x. fi di Then, gain per cent = x % x x2 Gain = x \ =` 100 100 Now, (ii) The table for the given data is Marks CG CF AG BF or = = AG BF GC CF SP = CP + Gain = x + x2 100 2 15.5 -30 - 60 21-30 6 25.5 -20 - 120 31-40 10 35.5 -10 - 100 41-50 12 45.5 = A 0 0 \ 51-60 9 55.5 10 90 x 2 + 100 x - 2400 = 0 61-70 7 65.5 20 140 x 2 + 120 x - 20 x - 2400 = 0 71-80 4 75.5 30 120 Total fi = 50 fi d i = 70 But SP = ` 24 x2 x+ = 24 100 x + x 2 = 2400 100 [splitting the middle term] x ( x + 120) - 20( x + 120) = 0 ( x + 120)( x - 20) = 0 Here, fi = 50, fi di = 70 and A = 45.5 Using shortcut method, x + 120 = 0 or x - 20 = 0 x = 20, - 120 \ x = 20 Mean marks ( x ) = A + fi di 70 = 45. 5 + fi 50 (ii) Given, each side of a regular hexagon is 5 cm. Hence, the mean marks of the given distribution is 46.9. (iii) Given In trapezium ABCD, AB||DC and EF ||AB AE BF = ED FC Construction Join AC to intersect EF at G. We know that each interior angle of regular hexagon = 120 Steps of construction 1. First, draw a line segment AB = 5 cm. 2. Taking A and B as centres, draw two rays AX and BY, making an angle of 120 with AB. E B A E [Q cost cannot be negative] Hence, the cost price of a toy is ` 20. = 45. 5 + 14 . = 46.9 To prove G [given] F D X Y O F C C D 120 Proof Since, AB || DC and EF||AB A EF||DC \ [Q lines parallel to the same line are also parallel to each other] In DADC, EG||DC [Q EF|| DC] Using basic proportionality theorem, AE AG = ED GC In DABC, GF||AB ...(i) [Q EF ||AB] Using basic proportionality theorem, 120 B 5 cm 3. Cut AF = 5 cm from AX and BC = 5 cm from BY. 4. Now, take F, E, D and C respectively, as centres to repeat steps 2 and 3. Then, we get a regular hexagon ABCDEF with each side equal to 5 cm and each angle equal to 120 . 5. Draw the perpendicular bisectors of sides AB and BC, which intersect each other at point O. 6. Draw a circle with O as centre and OA as radius, which passes through all the vertices of the regular hexagon ABCDEF. Hence, ABCDEF is the required circumcircle of the regular hexagon. (iii) Let a be the first term and d be the common 1 1 difference of the AP, then ap = and aq = q p 1 ...(i) a + ( p - 1)d = q and 1 a + ( q - 1)d = p ...(ii) On subtracting Eq. (i) from Eq. (ii), we get 1 d= pq On putting d = 1 in Eq. (i), we get pq a + ( p - 1) a- (ii) Given, nominal value of one share = ` 50 and rate of dividend = 15% Now, dividend on one share = Total dividend that Salman got from the company = `600 (a) \Number of shares Salman bought Annual income 600 = = = 80 Income on one share 7.50 (b) Premium on one share = 20% of ` 50 20 50 = ` 10 = 100 \Market value of one share = 50 + 10 = 60 Total investment for 80 shares = 80 60 600 = 100 % = 12. 5% 4800 1 1 + 1=1 pq pq (iii) Let the GP be a, ar, ar2 , ar3 , ... According to the given condition, Sum of first three terms = a + ar3 + ar2 = 16 ...(i) 10. (i) In DABC, B = 90 [by Pythagoras theorem] BC2 = 52 - 32 = 25 - 9 = 16 and sum of next three terms = ar3 + ar4 + ar5 = 128 A E B C = D 1 1 BC AB = 4 3 = 6 cm 2 2 2 In DADE and DABC, 3 a( 1 + r + r2 ) 1 1 1 = = ar ( 1 + r + r ) 8 r 2 3 and A is common to both. On comparing the base of power 3 from both sides, we get 1 1 = r =2 r 2 On putting r = 2 in Eq. (i), we get a + 2a + 4a = 16 7a = 16 a = Now, sum of n terms, Sn = DADE ~ DABC [Q by AA similarity criterion] area of DADE DE2 = area of DABC BC2 3 2 ABC = AED = 90 \ ...(ii) On dividing Eq. (i) by Eq. (ii), we get a + ar + ar2 16 = ar3 + ar4 + ar5 128 BC = 4 cm 1 \Area of DABC = base height 2 A 15 50 = ` 7.50 100 (c) Rate of return on his investment Total dividend amount = 100 % Total investment Hence, ( pq ) th term is 1. 4 6 = 15 . cm 2 16 = ` 4800 1 1 = 0 a = pq pq AB2 + BC2 = AC2 area of DADE ( 2)2 = 2 6 ( 4) area of DADE= 1 1 1 1 1 = a+ = pq q q pq q Now, ( pq ) th term = a + ( pq - 1)d 1 1 = + ( pq - 1) pq pq = a( rn - 1) r- 1 16 7 [Q r = 2 > 1] 16 n ( 2 - 1) 16 = 7 = ( 2n - 1) 2-1 7 Hence, a = 16 16 , r = 2 and Sn = ( 2n - 1). 7 7

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