Trending ▼   ResFinder  

ICSE Class X Board Exam 2020 : Mathematics

11 pages, 23 questions, 0 questions with responses, 0 total responses,    0    0
shilpa P
   		+Fave Message
ProfileTimelineUploads
 Home > shilpaset >

Formatting page ...

ICSE Solved Paper 2020 Mathematics Class-X (Maximum Marks : 80) (Time allowed : Two hours and a half) Attempt all questions from Section A and any four questions from Section B. All working, including rough work, must be clearly shown and must be done on the same sheet as the rest of the answer. Omission of essential working will result in loss of marks. The intended marks for questions or parts of question are given in brackets [ ]. Mathematical tables are provided. SECTION-A (40 marks) Attempt all questions from this Section 1. (a) Solve the following Quadratic Equation : [3] x2 7x + 3 = 0 Give your answer correct to two decimal places. x 3 (b) Given A = y 3 If A 2 = 3I, where I is the identity matrix of order 2, find x and y. [3] (c) Using ruler and compass construct a triangle ABC where AB = 3 cm, BC = 4 cm and ABC = 90 . Hence construct a circle circumscribing triangle ABC. Measure and write down the radius of the circle. [4] Ans. (a) x2 7x + 3 = 0 Compare the equation by ax2 + bx + c = 0 then, a = 1, b = 7 and c = 3 = 7 49 4 1 3 2 1 = 7 37 2 7 37 7 6.08 = 2 2 7 + 6.08 13.08 \ x = = 2 2 = 6.54 7 6.08 0.92 and x = = 2 2 = 0.46 Hence, x = 6.54 (Approx.) and x = 0.46 (Approx). x 3 (b) A = y 3 = 2 = x + 3 y 3x + 9 xy + 3 y 3 y + 9 A2 = 3I (Given) x 2 + 3 y 3x + 9 1 0 3 0 = 3 = xy + 3 y 3 y + 9 0 1 0 3 Comparing the elements of matrices 3x + 9 = 0 x = 3 3y + 9 = 3 y = 2 (c) b b 4 ac 2a x 3 x 3 = y 3 y 3 2 x = A2 = A A Steps of construction (i) Draw BC = 4 cm (ii) Make PBC = 90 (iii) Taking centre B and radius 3 cm draw an arc which intersect BP at the point A. (iv) Join AC, DABC is required triangle. B = 90 \ AC is the diameter of circle. (v) Draw perpendicular bisector of AC. (vi) Taking centre O and radius equal to BO draw a circle which passes through vertex A, B and C. Radius of circumscribe circle = 2.5 cm. 2. (a) Use factor theorem to factorise 6x3 + 17x2 + 4x 12 completely. [3] (b) Solve the following inequation and represent the solution set on the number line. [3] 3x x + 2< x + 4 + 5 , x R 5 2 Oswaal ICSE 5 Previous Years Solved Papers, Class X (c) Draw a Histogram for the given data, using a graph paper : [4] Weekly Wages (in `) No. of People 3000 4000 4 4000 - 5000 9 5000 - 6000 18 6000 - 7000 6 7000 - 8000 7 8000 - 9000 2 9000 - 10000 4 (c) Estimate the mode from the graph. Ans. (a) p(x) = 6x3 + 17x2 + 4x 12 p( 2) = 6( 2)3 + 17 ( 2)2 + 4 ( 2) 12 = 48 + 68 8 12 = 68 68 = 0 \ (x + 2) is a factor of given polynomial p(x) ) x + 2 6 x 3 + 17 x 2 + 4 x 12 ( 6 x 2 + 5x 6 6 x 3 + 12 x 2 ( ) ( ) = Mode -5400 (approx) 3. (a) In the figure given below, O is the centre of the circle and AB is a diameter. [3] If AC = BD and AOC = 72 . Find: (i) ABC (ii) BAD (iii) ABD 5x 2 + 4 x 12 5x 2 + 10 x ( ) ( ) 6 x 12 6 x 12 (+) (+) 0 \ 6x3 + 17x2 + 4x 12 A 72 O = (x + 2) (6x2 + 5x 6) = (x + 2) {6x2 + 9x 4x 6} B = (x + 2) {3x(2x + 3) 2(2x + 3)} = (x + 2) (2x + 3) (3x 2) C D Prove that : sin A cos A = sin A cos A 1 + cot A 1 + tan A 3x x + 2 < x + 4 + 5, x R 5 2 (b) 3x + 2 < x + 14 5 x+4 3x x < 14 2 5 x 3x 5x < 12 5 Multiply by 5 2 x < 12 5 x> x +5 2 x 5 4 2 [3] (c) In what ratio is the line joining P(5, 3) and Q( 5, 3) divided by the y-axis ? Also find the coordinates of the point of intersection. [4] \ ADB = 90 (semicircle angle) x 1 2 1 (i) ABC = AOC 2 (same arc angles at circumference and centres) x 2 \ x > 30 < x 2 30 20 10 5 60 2 40 (b) Ans. (a) AB is the diameter of circle x > 30 0 1 2 \ ABC = 1 72 = 36 2 (ii) AC = BD \ ABC = BAD (equal arcs make equal angle) \ BAD = 36 (iii) In DABD ABD + BAD + ADB = 180 (sum of all angles of D) MATHEMATICS (SOLVED PAPER - 2020) ABD + 36 + 90 = 180 (ii) one of the first 9 letters of the English alphabet which appears in the given word ? (iii) one of the last 9 letters of the English alphabet ABD = 54 which appears in the given word ? [3] sin A cos A (c) Mr. Bedi visits the market and buys the (b) L.H.S. = following articles : 1 + cot A 1 + tan A Medicines costing ` 950, GST @ 5% sin A cos A = A Pair of shoes costing ` 3000, GST @ 18% cos A sin A 1+ 1+ A Laptop bag costing ` 1000 with a discount of sin A cos A 30% GST @ 18% 2 sin A cos2 A (i) Calculate the total amount of GST paid. = (ii) The total bill amount including GST paid by sin A + cos A cos A + sin A Mr. Bedi. [4] sin 2 A cos2 A 4 3 = Ans. (a) Volume of solid spherical ball = pr sin A + cos A 3 2 2 [a b = (a + b) (a b)] 4 3 (sin A + cos A )(sin A cos A ) = p( 6 ) 3 = (sin A + cos A ) 4 = 216 = sin A cos A 3 = R.H.S. Hence proved. 216 4 (c) Let the line joining the (5, 3) and ( 5, 3) divided = cm 3 3 by the y-axis in the ratio k : 1. Let the radius of spherical marble be R cm. Then 64 volume of spherical marbles = volume of spherical ball 4 3 4 216 cm 3 64 R = 3 3 ABD = 180 126 Let the coordinate of the point of intersection R be (0, y). m1x 2 + m2 x1 Rx = m1 + m2 1 5 + k ( 5) 1+ k y = 1 3 + k 3 1+ k y = 3+3 =3 1+1 0 = 5 5k = 0 k = 1 m1 y 2 + m2 y1 Ry = m1 + m2 [Put, k= 1] Ratio = k : 1 = 1 : 1 Point of intersection of the line (0, 3). 4. (a) A solid spherical ball of radius 6 cm is melted and recast into 64 identical spherical marbles. Find the radius of each marble. [3] (b) Each of the letters of the word 'AUTHORIZES' is written on identical circular discs and put in a bag. They are well shuffled. If a disc is drawn at random from the bag, what is the probability that the letter is : (i) a vowel ? R3 = R = 216 64 3 216 6 3 = = cm 64 4 2 = 1.5 cm \ Radius of spherical marble is 1.5 cm. (b) 'AUTHORIZES' Total number of disc n(S) = 10 (i) Number of discs a letter of vowel = 5 (A, U, O, I, E) \ n(E) = 5 Probability of a disc is vowel n( E) 5 1 = = = n(S ) 10 2 (ii) One of the first 9 letter in english alphabet = 4 (A, E, H, I) \ n(E) = 4 Probability of a disc is drawn is written one of first 9 letters of english alphabet n( E) 4 = = n(S ) 10 2 = 5 (iii) One of the last 9 (R, S, T, U, Z) letters of english alphabet n(E) = 5 Probability of a disc is drawn is written one of the last 9 letters of english alphabet n( E) 5 = = n(S ) 10 Oswaal ICSE 5 Previous Years Solved Papers, Class X 1 2 30 1000 100 \ Net cost = 1000 (c) Medicine costing = ` 950 5 950 = ` 47.50 GST @ 5% = 100 (i) Total amount of GST = ` 47.50 + 540 + 126 = ` 713.50 (ii) Total bill amount including GST = ` 950 + 3000 + 700+ 713.50 = ` 4650 + 713.50 = ` 5363.50 = A pair of shoes costing = ` 3000 18 GST @ 18% = 3000 = ` 540 100 Laptop bag costing ` 1000 with a discount of 30% = ` 700 18 GST @ 18% = 700 = ` 126 100 SECTION-B (40 marks) Attempt any four questions from this Section 5. (a) A company with 500 shares of nominal value Investment Market value of share = ` 120 declares an annual dividend of 15%. No. of share Calculate : 14400 (i) the total amount of dividend paid by the = 80 company. (ii) annual income of Mr. Sharma who holds 80 \ Market value of share = ` 180 shares of the company. [3] (b) If the return percent of Mr. Sharma from his Marks No. of Students shares is 10%. Find the market value of each share. (x) (f) f x (b) The mean of the following data is 16. Calculate 5 3 15 the value of f. [3] 10 7 70 Marks 5 10 15 20 25 15 f 15f No. of Students 3 7 f 9 6 (c) The 4th, 6th and the last term of a geometric progression are 10, 40 and 640 respectively. If the common ratio is positive, find the first term, common ratio and the number of terms of the series. [4] Ans. (a) Number of shares = 500 Nominal value = ` 120 Annual dividend = 15% (i) Total amount of dividend paid by the company = Rate of dividend Nominal value of share Number of share 15 120 500 = 100 = 15 600 =` 9000 \ Total amount paid by the company = ` 9000 (ii) Mr sharma's Annual Income = Rate of dividend Nominal value of share Number of share 15 = 120 80 100 = 15 96=` 1440 \ Mr sharma's Annual Income = ` 1440 Income Return of Investment = 100 Investment 1440 100 10 = Investment \ Investment = 144000 = ` 14400 10 20 9 180 25 6 150 25 +f 415 + 15 f Mean = 16 fx = 16 f (Given) 415 + 15 f = 16 25 + f 415 + 15f = 400 + 16 f 16f 15f = 415 400 \ f = 15 (c) Let the first term of G.P. be a, common ratio be r and number of terms be n. a4 = 10, a6 = 40 and an = 640 (given) a4 = 10 ar3 = 10 ...(i) Again, a6 = 40 ar5 = 40 ...(ii) From eqs. (i) & (ii) ar 5 40 = ar 10 r 2 = 4 \ r = 2 \ r = 2 3 (r > 0) MATHEMATICS (SOLVED PAPER - 2020) From eq. (i) ar3 = 10 3 a(2) = 10 10 5 a = = 8 4 Again, an = 640 arn 1 = 640 5 n-1 ( 2 ) = 640 4 2n 1 = 128 4 = 29 Comparing the power n 1 = 9 n = 10 5 \ First term of G.P. = , common ratio = 2 4 2n 1 = 640 4 5 and number of terms = 10 3 0 4 2 6. (a) If A= and B = 5 1 1 0 [3] Find A2 2AB + B2. (b) In the given figure AB = 9 cm, PA = 7.5 cm and PC = 5 cm. [3] Chords AD and BC intersect at P. B 9 cm D = 12 + 0 6 + 0 20 + 1 10 + 0 12 6 AB = 19 10 4 2 4 2 B2 = B B = 1 0 1 0 16 + 2 8 + 0 = 4 + 0 2 + 0 18 8 \ B2 = 4 2 \ A2 2AB + B2 12 6 18 8 9 0 2 + 20 1 19 10 4 2 9 0 24 12 18 8 + + 20 1 38 20 4 2 51 20 9 + 24 + 18 0 12 8 20 + 38 4 1 20 + 2 = 54 17 (b) (i) DPAB and DPCD APB = CPD (Vertically opposite angle) ABP = PDC (same arc AC angles) \ DPAB ~ DPCD (AA similarly test) B P P m 5c 7. A 5 cm C (i) Prove that DPAB ~ DPCD. (ii) Find the length of CD. (iii) Find area of DPAB : area of DPCD. (c) From the top of a cliff, the angle of depression of the top and bottom of a tower are observed to be 45 and 60 respectively. If the height of the tower is 20 m. [4] Find : (i) the height of the cliff. (ii) the distance between the cliff and the tower. 3 0 4 2 Ans. (a) A = and B = 5 1 1 0 A2 = A A = 3 0 3 0 5 1 5 1 9 + 0 0 + 0 = 15 + 5 0 + 1 9 0 2 \ A = 20 1 AB = 3 0 4 2 5 1 1 0 D 9 cm 7. A 5 cm m 5c (ii) C PA AB PB = = PC CD PD (Property of similar triangle) PA AB = PC CD 7.5 9 = 5 CD \ CD = \ CD = 6 cm (iii) 7.5 = 5 = 9 5 9 5 10 = 7.5 75 2 AP 2 AP arDPAB = = 2 CP CP arDPCD 9 4 \ arDPAB : arDPCD = 9 : 4 2 Oswaal ICSE 5 Previous Years Solved Papers, Class X = 20 + 27.32 \ Height of the cliff = 47.32 m (ii) Distance between the cliff and the tower OA = BC = CP OA = 27.32 m 7. (a) Find the value of 'p' if the lines, 5x 3y + 2 = 0 and 6x py + 7 = 0 are perpendicular to each other. Hence, find the equation of a line passing through ( 2, 1) and parallel to 6x py + 7 = 0. [3] (b) Using properties of proportion find x : y, given : [3] x 2 + 2 x y 2 +3 y = 2 x + 4 3 y +9 (c) In the given figure TP and TQ are two tangents to the circle with centre O, touching at A and C respectively. If BCQ = 55 and BAP = 60 , find : [4] (i) OBA and OBC (ii) AOC (iii) ATC Q (c) OP is the cliff and AB is the tower. CP In DPBC, tan B = BC tan 45 = CP BC 1 = CP BC (Q QPA = PAO = 60 ) OC + CP tan 60 = OA 3 = AB + CP OA (Q OC = AB = 20 m) 3 = 20 + CP (OA = BC) BC 3 = ( 20 + CP CP (BC = CP) 3 CP = 20 +CP ) 3 1 CP = 20 CP = CP = CP = 20 3 -1 20 3 1 m 3 +1 3 +1 20 ( 3 + 1) 3 1 = 10 ( 3 + 1) m CP = 10 (1.732 + 1)m CP = 27.32 m (i) height of the cliff OP = OC + CP C B \ CP = BC ...(i) In DOAP, OP tan A = OA O P A T Ans. (a) Slope of line 5x 3y + 2 = 0 is coefficent of x = coefficent of y = 5 5 = 3 3 Slope of line 6x py + 7 = 0 is = 6 6 = p p Lines are perpendicular each other \ Product of slopes = 1 5 6 = 1 3 p 3p = 30 p = 10 \ p = 10 Given line 6x py + 7 = 0 is 6x + 10y + 7 = 0 Equation of the line, parallel to 6x + 10y + 7 = 0 is 6x + 10y + k = 0 Since, line passes through ( 2, 1), \ 6 ( 2) + 10 ( 1) + k = 0 12 10 + k = 0 k = 22 Equation of the line parallel to 6x + 10y + 7 is 6x + 10y + 22 = 0 or 3x + 5y + 11 = 0 MATHEMATICS (SOLVED PAPER - 2020) (b) x 2 + 2x y 2 + 3y = 3y + 9 2x + 4 (using compodendo dividendo a+b c+d = ) a b c d x 2 + 2x + 2x + 4 x 2 + 2x 2x 4 x2 + 4x + 4 x2 4 = y2 + 3y + 3y + 9 y2 + 3y 3y 9 y2 + 6y + 9 y2 9 x+2 y+3 = x 2 y 3 (using compodendo - dividendo) = ( x + 2 )2 ( y + 3 )2 = ( x + 2 )( x 2 ) ( y + 3)( y 3) a c = b d y+3+ y 3 x+2+x 2 = y+3 y+3 x+2 x+2 2x 2y = 4 6 x y = 2 3 x 2 = y 3 x : y = 2 : 3 BCQ = 55 and BAP = 60 \ (c) Given, Q C B O P A T PAO = 90 (As PAT is the tangent) \ OAB = 90 60 = 30 In D OAB, OA = OB (radii of circle) OBA = OAB = 30 Similarly \ QCO = 90 (As QCT is the tangent) \ OCB = 90 55 = 35 In D OCB, OC = OB (radii of circle) OBC = OCB = 35 (ii) Now, ABC = OBA + OBC = 30 + 35 = 65 Again, AOC = 2 ABC = 2 65 = 130 ( Angle subtended at the center of the circle by an arc is twice the angle subtended at the circle) (i) (iii) AOC + ATC = 180 130 + ATC = 180 \ ATC = 180 130 = 50 \ ATC = 50 8. (a) What must be added to the polynomial 2x3 3x2 8x, so that it leaves a remainder 10 when divided by 2x + 1 ? [3] (b) Mr. Sona has a recurring deposit account and deposits ` 750 per month for 2 years. If he gets ` 19125 at the time of maturity, find the rate of interest. [3] (c) Use graph paper for this question. Take 1 cm = 1 unit on both x and y axes. [4] (i) Plot the following points on your graph sheets. A ( 4, 0), B ( 3, 2), C (0, 4), D (4, 1) and E (7, 3) (ii) Reflect the points B, C, D and E on the x-axis and name them as B', C', D' and E' respectively. (iii) Join the points A, B, C, D, E, E', D', C', B' and A in order. (iv) Name the closed figure formed. Ans. (a) p(x)= 2x3 3x2 8x q(x)= 2x + 1 Remainder= 10 Let k be added to get remainder 10 when divided by (2x + 1). \ p (x)= 2x3 3x2 8x + k Remainder= 10 1 p = 10 2 3 2 1 1 1 2 3 8 + k = 10 2 2 2 1 1 2 3 + 4 + k = 10 8 4 1 + 4 + k = 10 k=7 Hence, 7 be added to get remainder 10 when divided by 2x + 1 to given polynomial. (b) Maturity amount (M.A.) = ` 19125 Monthly deposit (P) = ` 750 Time (n) = 2 years = 24 months M.A. = P n + P 750 24 + 18000 + r n( n + 1) 1 100 2 12 750 r ( 24 ) ( 24 + 1) = 19125 2400 750 r = 19125 4 750r = 19125 18000 4 r = \ 1125 4 4500 = 750 750 r=6 Rate of interest = 6% Oswaal ICSE 5 Previous Years Solved Papers, Class X (c) (i) (ii) B'(-3, -2), C'(0, -4), D'(4, -1) and E'(7, -3) respectively are the reflected points through the x-axis. (iii) On joining the points A, B, C, D, E, E', D', C', B' and A in order, we get a closed figure. (iv) Closed figure formed is nine sided polygon or nonagon, polygon fish or kite. 9. (a) 40 Students enter for a game of shot-put competition. The distance thrown (in metres) is recorded below : Distance in m 12 13 13 14 14 15 15 16 16 17 17 18 18 19 Number of Students 3 9 12 9 4 2 1 Use a graph paper to draw an ogive for the above distribution. Use a scale of 2 cm = 1 m on one axis and 2 cm = 5 students on the other axis. Hence, using your graph find : (i) the median. (ii) upper qartile. Ans. (a) (iii) number of students who cover a distance 1 which is above 16 m. 2 [6] (b) If x = = 0. 2 a + 1 + 2 a 1 2 a + 1 2 a 1 , prove that x2 4ax + 1 [4] MATHEMATICS (SOLVED PAPER - 2020) Distance in (m) Number of students Less than Cumulative frequency 12 -13 13 -14 14 - 15 15 - 16 16 - 17 17 - 18 18 - 19 3 9 12 9 4 2 1 13 14 15 16 17 18 19 3 12 24 33 37 39 40 (i) Median = 14.6 0.2 (ii) Upper quartile = 15.75 0.2 (iii) Number of students above 16 m = 40 35= 5 2a + 1 + 2a 1 (b) x = x+1 = x 1 2a + 1 + 2a 1 2a + 1 + 2a 1 A C = B D A+B C+D = ) A B C D 2a + 1 x + 1 2 2a + 1 = = x 1 2 2a 1 2a + 1 + 2a 1 + 2a + 1 2a 1 2a + 1 2a 1 (Apply compodendo-dividendo 2a 1 Squaring both sides ( x + 1)2 = 2 a + 1 ( x 1)2 2 a 1 2 x 2 + 2x + 1 2a + 1 = 2 x 2x + 1 2a 1 Apply compodendo - dividendo x 2 + 2x + 1 + x 2 2x + 1 2a + 1 + 2a 1 = x 2 + 2x + 1 x 2 + 2x 1 2a + 1 2a + 1 4a 2( x + 1) = 2 4x x2 + 1 2a = 2x 1 x2 + 1 = 4ax (c) Use ruler and compass for this question. Construct a circle of radius 4.5 cm. Draw a chord AB = 6 cm. (i) Find the locus of points equidistant from A and B. Mark the point where it meets the circle as D. (ii) Join AD and find the locus of points which are equidistant from AD and AB. Mark the point where it meets the circle as C. (iii) Join BC and CD, Mesuare and write down the length of side CD of the quadrilateral ABCD. [4] Ans. (a) Let the first term of an A.P. be a and common difference be d respectively 2 x2 4ax + 1= 0 a6 a6 a + 5d 3a = 4 a (given) = 4a ...(i) = 4a {an = a + (n 1)d} = 5d 5 a = d ...(ii) 3 S6 = 75 (given) n [ a + a6 ] = 75 2 6 ( a + a6 ) = 75 2 a + a6 = a + 4a = 25 25 a = =5 5 \ a = 5 From equation (i) a = 5 d 3 5 = 5 d 3 d = 5 3 3 5 \ 75 3 \ d = 3 First term of an A.P. = 5 and common difference =3 (b) Let the numbers be x and y. Hence proved. 10. (a) If the 6th term of an A.P. is equal to four times its first term and the sum of fist six terms is 75, find the first term and the common difference. [3] (b) The difference of two natural numbers is 7 and their product is 450. [3] Find the numbers. x y = 7 y = x 7 [ x>y] ...(i) xy = 450 ...(ii) From (i) & (ii) x (x 7) = 450 x2 7x 450 = 0 x2 25x + 18x 450 = 0 According the given condition Oswaal ICSE 5 Previous Years Solved Papers, Class X x(x 25) + 18(x 25)= 0 (x 25) (x + 18) = 0 If x + 18 = 0 x = 18 it is not a natural number. If x 25 = 0 x = 25, y = 25 7 = 18 Thus, numbers 25 and 18. (c) BC = CD = 5.1 cm (Approx.) 11. (a) A model of a high rise building is made to a scale of 1 : 50. [3] (i) If the height of the model is 0.8 m, find the height of the actual building. (ii) If the floor area of a flat in the building is 2 20 m , find the floor area of that in the model. (b) From a solid wooden cylinder of height 28 cm and diameter 6 cm, two conical cavities are hollowed out. The diameters of the cones are also of 6 cm and height 10.5 cm. 22 Taking = find the volume of the remaining 7 solid. [3] (c) Prove the identity 2 1 tan 2 1 cot = tan [4] Height of the model 0.8 m \ Height of the building = 0.8 50 m = 40 m Floor area of building = 20 m2 20 Floor area of the model = m2 50 50 1 2 m 125 = 0.008 m2 \ Floor area of the model = 0.008 10000 cm2 = 80 cm2 (b) Dimension of cylinder Height (h) = 28 cm Diameter (2r) = 6 cm Dimension of conical cavities Height (H) = 10.5 cm diameter (2r) = 6 cm Volume of remaining solid = Volume of cylinder 2 Volume of cavity 1 = r 2 h 2 r 2 H 3 2 2 = r h H 3 = 2 22 3 3( 28 10.5) 7 3 = 22 3 3( 28 7 ) 7 = 22 3 3 21 7 = 22 3 3 3 =594 cm3 Hence, volume of remaining solid = 594 cm3 1 - tan q 2 (c) = tan 2 q 1 - cot q 1 tan L.H.S. = 1 cot 2 1 tan = 1 1 tan Ans. (a) Scale = 1 : 50 = 2 MATHEMATICS (SOLVED PAPER - 2020) 1 tan = tan 1 tan 2 2 tan (1 tan ) = (1 tan ) tan =tan2 q = 1 = R.H.S. Hence proved. 2

Formatting page ...

Related ResPapers
ICSE Class X Board Specimen 2024 : Mathematics
by abhosjwme 
ICSE Class X Board Exam 2023 : Mathematics
by theakshatgypta 
ICSE Class X Board Exam 2024 : Mathematics
by pranihita 
ICSE Class X Board Exam 2024 : Mathematics
by qwerty1272768 

Formatting page ...

Formatting page ...

Formatting page ...

Formatting page ...

Formatting page ...

Formatting page ...

Formatting page ...

Formatting page ...

Formatting page ...

 

  Print intermediate debugging step

Show debugging info


 

 


© 2010 - 2026 ResPaper. Terms of ServiceContact Us Advertise with us

 

shilpaset chat