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CBSE Class 10 Board Exam 2024 : Mathematics

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Govindu Srinivas
The Institution of Engineers (India) IEI, Kolkata

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Basic Mathematics (241) Marking Scheme 2023-24 Section A 1) (b) 1 2) (c) 20 1 3) (b) 1 4) (d) No Solution 1 5) (d) 0,8 1 6) (c) 5 Unit 1 7) (a) ~ 1 8) (d) RHS 1 9) (b) 70 1 10) (b) 1 11) (b) 45 1 12) ( ) sin 1 13) ( ) :2 1 14) (a) 7 1 15) (d) 1 16) (a) 15 1 17) (a) 3.5 cm 1 18) (b) 12-18 1 19) (a) Both assertion and reason are true and reason is the correct explanation of assertion. 1 20) (d) Assertion (A) is false but reason(R) is true. 1 1 SECTION B 21) 3 +2 = 8 6 - 4 = 9 =3, =2, =8 =6, =-4, = 9 = = = 1 = = 1/2 The given pair of linear equations are consistent. 1/2 22) Given:-AB II CD II EF To prove:- = Construction:- Join BD to 1/2 intersect EF at G. Proof:- in ABD EG II AB ( EF II AB ) = ( by BPT )___________(1) 1/2 In GF II CD ( EF II CD ) = ( by BPT )___________(2) 1/2 (1) & (2) = 1/2 OR Given AD=6cm, DB=9cm AE=8cm, EC=12cm, ADE=48 To find:- ABC=? Proof: In = = = (1) ..(2) = From (1) & (2) 1 = DE II BC (Converse of BPT) ADE= ABC (Corresponding angles) 1 ABC=48 2 23) In OTA, OTA = 90 By Pythagoras theorem OA2 = OT2 + AT2 1/2 (5)2 = OT2 + (4) 2 25-16= OT2 9 = OT2 1/2 OT=3cm radius of circle = 3cm. 1 24) Sin2 60 + 2 tan 45 cos2 30 = = + 2(1) - 1 +2 - = 2 1 25) Area of the circle= sum of areas of 2 circles = (40)2 + (9)2 1/2 = x (402 + 92) 1/2 = 1600 + 81 = 1681 1/2 = 41 . 1/2 = 41 2 = 82 OR radius of circle = 10cm, = 90 Area of minor segment = = = = x r2 - r2 - Area of xbxh x3.14 x 10 x 10 - 1/2 x 10 x 10 1/2 50 = 78.5-50 = 28.5 cm2 1/2 Area of minor segment = 28.5 cm2 1/2 3 Section C 26) Let us assume that 3 be a rational number 3 = where a and b are co-prime. 1 squaring both the sides 3= 1/2 = 3 =3 is divisible by 3 so a is also divisible by 3_________(1) a=3c for any integer c. (3 ) =3b2 1/2 9 =3 =3 is divisible by 3 so, b is also divisible by 3 _____(2) From (1) & (2) we can say that 3 in a factor of a and b 1/2 which is contradicting the fact that a and b are co- prime. Thus, our assumption that 3 is a rational number is wrong. Hence, 3 is an irrational number. 1/2 27) P(S)= 4S2 -4S+1 4S2 -2S-2S+1=0 2S(2S-1)-1(2S-1)=0 (2S-1) (2S-1)=0 S= a=4 S= b=-4 + = + = c=1 1 1 4 , 2 4 =1 1=1 = = = , , 1 1 1 1 2 2 4 1 1 4 4 1 1 28) Let cost of one bat be Rs Let cost of one ball be Rs 1/2 ATQ 4 + 1 = 2050___________(1) 3 + 2 = 1600___________(2) (1)4 + 1 = 2050 1/2 = 2050 4 1/2 4 (2) 3 + 2(2050 4 ) = 1600 3x + 4100 8x =1600 -5x = 2500 = 500 (1) 4 + 1 = 2050 4(500) + = 2050 2000 + = 2050 = 50 Hence Cost of one bat = Rs. 500 Cost of one ball = Rs. 50 1/2 1/2 1/2 OR Let the fixed charge for first 3 days= Rs. And additional charge after 3 days= Rs. ATQ + 4 = 27---------------(1) + 2 = 21 --------------(2) Subtract eqn (2) from (1) 2 = 6 =3 Substitute value of in (2) + 2(3) = 21 = 21 6 = 15 Fixed charge= Rs. 15 Additional charge per day = Rs. 3 1/2 1/2 1 1 29) Given circle touching sides of ABCD at P,Q,R and S To prove- AB+CD=AD+BC ProofAP=AS-------(1) tangents from an external point PB=BQ-------(2) to a circle are equal in length DR=DS-------(3) CR=CQ-------(4) Adding eqn (1),(2),(3) & (4) AP+BP+DR+CR=AS+DS+BQ+CQ AB+DC=AD+BC 1 1 1 30) ( cot ) = LHS=( )2 = 1/2 1/2 = 5 = = = (1 cos ) ( ) 1 (1 cos ) (1 )(1 + ) =RHS = 1 LHS = RHS, Hence Proved OR secA (1 )(sec + )=1 LHS= (1 ) = ( )( ) = ( )( ) = 1 + (1- = ) 1 = = 1 = RHS 1 LHS=RHS. Hence Proved 31) (i) Red balls= 6 , Black balls = 4 , White balls = x P(white ball) = 1 = 3x = 10 + x x= 5 white balls 1/2 (ii) Let y red balls be removed, black balls = 4, white balls = 5 P(white balls)= ( = ) 1 = 10 = 15 = 5 So 5 balls should be removed. Section D 32) Let the speed of train be / 1/2 distance= 360 km Speed = Time = 1/2 New speed = ( + 5) / Time = 1 +5= ( + 5) 360 1 = 360 6 ( + 5)(360 ) = 360 5 + 1800 = 0 1 + 5 1800 = 0 + 45 40 1800 = 0 ( + 45) 40( + 45) = 0 ( + 45)( 40) = 0 + 45 = 0 , 40 = 0 = 45 , = 40 Speed cannot be negative Speed of train =40km/hr 1 1 OR Let the speed of the stream= / Speed of boat= 18 / Upstream speed= (18 ) / Downstream speed=(18 + ) / Time taken (upstream)=( 1/2 1/2 ) Time taken (downstream)=( ) ATQ ( ( ) ( ) = ) ( ) +1 1 =1 24(18 + ) 24(18 ) = (18 )(18 + ) 24(18 + 18 + ) = (18) 24(2 ) = 324 48 324 + = 0 + 48 324 = 0 6 + 54 324 = 0 ( 6) + 54( 6) = 0 ( 6) ( + 54) = 0 6=0 , + 54 = 0 =6 , = 54 Speed cannot be negative Speed of stream=6 / 33) Given , DE || BC To prove 1 1 1 = Construction: join BE and CD 1/2 Draw DM AC and EN AB Proof: = x b x h = x AD x EN-----------------------(1) Area ( ) = x DB x EN--------(2) Divide eqn (1) by (2) = = -----------(3) 1 7 area = x AE x DM -------(4) area = x EC x DM -------(5) Divide eqn (4) by (5) = = -----------(6) 1 and are on the same base DE and between same parallel lines BC and DE ( ) = ( ) ( ( ) = ( ( = ) ) [LHS of (3) =RHS of (6)] [RHS of (3) = RHS of (6) Since = 1/2 ( ) PST = PQR (Corresponding angles) 1 But PST = PRQ (given) PQR = PRQ PR = PQ ( sides opposite to equal angles are equal Hence is isosceles. 1 34) Diameter of cylinder and hemisphere = 5mm radius, (r) = Total length = 14mm Height of cylinder = 14 - 5 = 9mm 1 CSA of cylinder = 2 rh =2x = x x9 mm2 1 CSA of hemispheres = 2 r2 = 2x = x mm2 1 CSA of 2 hemispheres = 2 x = mm2 Total area of capsule = 1 + = = 220 mm2 1 OR 8 Diameter of cylinder = 2.8 cm of cylinder = . = 1.4 cm of cylinder = of hemisphere = 1.4 cm Height of cylinder = 5-2.8 1 = 2.2 cm Volume of 1 Gulab jamun = vol. of cylinder + 2 x vol. of hemisphere = 2h + 2 x 3 1 x (1.4)2 x 2.2 + 2 x x x (1.4)3 = 13.55 + 11.50 = 25.05 3 1 45 = 45 x25.05 45 = 30% x 45 x 25.05 = x 45 x 25.05 1 = 338.175 cm3 338 cm3 35) Life time (in hours) Number of lamps(f) Mid x d fd 1500-2000 14 1750 -1500 -21000 2000-2500 56 2250 -1000 -56000 2500-3000 60 2750 -500 -30000 3000-3500 86 3250 0 0 3500-4000 74 3750 500 37000 4000-4500 62 4250 1000 62000 4500-5000 48 4750 1500 72000 400 64000 2 Mean = a + 1/2 a = 3250 1/2 9 Mean = 3250 + 1 = 3250 + 160 = 3410 Average life of lamp is 3410 hr 1 Section E 36) a6 =16000 a9 = 22600 a+5d=16000-------(1) a+8d=22600 --------(2) substitute a = 1600 -5d from (1) 16000-5d + 8d = 22600 3d = 22600-16000 3d=6600 d= = 2200 a = 16000-5(2200) a = 16000-11000 a = 5000 (i) an = 29200, a = 5000, d = 2200 an = a + (n-1)d 29200 = 5000 + (n 1)2200 1/2 29200-5000 = 2200n-2200 24200+2200=2200n 26400=2200n n= n=12 1/2 in 12th year the production was Rs 29200 (ii) n=8, a=5000, d=2200 an = a + (n-1)d 1/2 = 5000+(8-1)2200 1/2 = 5000+7 x 2200 = 5000+15400 1/2 = 20400 The production during 8th year is = 20400 1/2 OR n = 3, sn = a = 5000, d = 2200 [ 2a + (n-1)d] 1/2 10 = [2(5000) + (3-1) 2200] S3 = (10000 + 2 x 2200) 1/2 = (10000 + 4400) 1/2 = 3 x 7200 = 21600 1/2 The production during first 3 year is 21600 (iii) a4 = a+3d = 5000 + 3 (2200) = 5000 + 6600 = 11600 1/2 a7 = a+6d = 5000 + 6 x 2200 =5000 + 13200 = 18200 a7 - a4 = 18200-11600 = 6600 1/2 37) coordinates of A (2, 3) Alia s house coordinates of B (2, 1) Shagun s house coordinates of C (4,1) (i) AB = ( ) + ( ) 1/2 (2 2) + (1 3) = = Library (0 + ( 2) = 0 + 4 = 4 = 2 units 1/2 2 (ii) C(4,1), B (2,1) CB = = 1/2 (2 4) + (1 1) ( 2) + 0 = 4 + 0 = 4 = 2 unit 1/2 (iii) 0(0,0), B(2,1) OB = (2 0) + (1 0) = 2 + 1 = 4 + 1 = 5 units 1 Distance between Alia s house and Shagun s house, AB = 2 units Distance between Library and Shagun s house, CB = 2 units OB is greater than AB and CB, 1/2 1/2 For shagun, school [O] is farther than Alia s house [A] and Library [C] 11 OR C (4, 1), A(2, 3) CA = = (2 4) + (3 1) ( 2) + 2 + = 4 + 4 = 2 2 units = 8 AC2= 8 1 Distance between Alia s house and Shagun s house, AB = 2 units Distance between Library and Shagun s house, CB = 2 units 1/2 AB2 + BC2 = 22 + 22 = 4 + 4 = 8 = AC2 Therefore A, B and C form an isosceles right triangle. 38) (i) XY PQ and AP is transversal. APD = PAX (alternative interior angles) 1/2 1/2 APD=45 (ii) Since XY || PQ and AQ is a transversal so alternate interior angles are equal hence YAQ = AQD=30 (iii) 1/2 In , = 45 tan = tan 45 = 1/2 PD=100 m 1/2 Boat P is 100 m from the light house 1 OR In , = 30 tan = 1/2 tan 30 = = 1/2 DQ = 100 3 m 1 12

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