MARKING SCHEME SQP
MATHEMATICS (STANDARD)
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2020-21
CLASS X
| S.NO. | ANSWER | MARKS | ||||||||||||||||||||||||||||||
|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|
| Part-A | ||||||||||||||||||||||||||||||||
| 1. | (LCM)(3) =180 ½ LCM=60 ½ OR --- Content provided by FirstRanker.com --- Four decimal places | 1 | ||||||||||||||||||||||||||||||
| 2. | a+B=k/3 ½ 3=k/3 K=9 ½ | |||||||||||||||||||||||||||||||
| 3. | 32143 ½ 6 k¢8 3.1 --- Content provided by FirstRanker.com --- k=2 ½ | |||||||||||||||||||||||||||||||
| 4. | Let the cost of 1 chair=Rs.x ½ And the cost of 1 table=Rs. y 3x+y=1500 ½ 6x+y=2400 | |||||||||||||||||||||||||||||||
| 5. | an=a+(n-1)d 0=27+(n-1)(-3) ½ --- Content provided by FirstRanker.com --- 30=3nn=10 ½ 10th OR an=a+(n-1)d --- Content provided by FirstRanker.com --- 4=a+6x(-4) ½a=-28 ½ | |||||||||||||||||||||||||||||||
| 6. | 9x2+6kx+4=0 (6k)2-4X9X4=0 ½ 36k2=144 K2=4 --- Content provided by FirstRanker.com --- K=+2 ½ | |||||||||||||||||||||||||||||||
| 7. | X2+7x+10=0 X2+5x+2x+10=0 ½ (x+5)(x+2)=0 X=-5,x=-2 ½ OR --- Content provided by FirstRanker.com --- 3ax2-6x+1=0 ½(-6)2>-4(3a) (1)<0 12a>36 =>a>3 ½ | |||||||||||||||||||||||||||||||
| 8. | PQ=PT PL+LQ=PM+MT PL+LN=PM+MN --- Content provided by FirstRanker.com --- Perimeter(APLM)=PL+LM+PM ½ =PL+LN+MN+PM =2(PL+LN) =2(PL+LQ) --- Content provided by FirstRanker.com --- =2X28=56cm ½ | |||||||||||||||||||||||||||||||
| 9. | In APAO ½ Tan30°=A0/PA 1/v3 =3/PA ½ PA=3v3 cm OR --- Content provided by FirstRanker.com --- In AOPQ?P+?Q+?0=180" 2?Q+?P=180° ½ 2?Q+90°=180° 2?Q=90° --- Content provided by FirstRanker.com --- ?Q= 45° ½ | |||||||||||||||||||||||||||||||
| 10. | AD/BD = AE/CE ½ 3/4 = 2/CE ½ CE=8/3cm | |||||||||||||||||||||||||||||||
| 11. | 8:5 | 1 | ||||||||||||||||||||||||||||||
| 12. | Sin30°+cosB=1 ½+cosB=1 ½ CosB=1/2 --- Content provided by FirstRanker.com --- B=60" ½ | |||||||||||||||||||||||||||||||
| 13. | X+y =2sin2? +2cos2?+1 ½ =2(sin2? +cos2?)+1 =3 ½ | |||||||||||||||||||||||||||||||
| 14. | length of arc=?/360°(2pr) ½ = 60/360(2X22/7X21) --- Content provided by FirstRanker.com --- =22 cm ½ | |||||||||||||||||||||||||||||||
| 15. | pR2H=12X4/3pr3 1X1x16=4/3Xr3 X12 ½ r3=1 r=1 d=2cm ½ | |||||||||||||||||||||||||||||||
| 16. | probability of getting a doublet=1/6 | 1 | ||||||||||||||||||||||||||||||
| OR --- Content provided by FirstRanker.com --- probability of getting a black queen=2/52=1/26 | ||||||||||||||||||||||||||||||||
| 17. | (a) ii)(15/2,33/2) (b) i)4 (c) iii)16 (d) iv)(2.0,8.5) (e) i) x-13=0 | 1x4=4 | ||||||||||||||||||||||||||||||
| 18. | (a) ii)15 cm --- Content provided by FirstRanker.com --- (b) iv)They are not the mirror image of one another(c) ii)Their altitudes have a ratio a:b (d) iv) 5m (e) iii)m | 1x4=4 | ||||||||||||||||||||||||||||||
| 19. | (a) ii) (4,-2) (b) i) Intersects x-axis --- Content provided by FirstRanker.com --- (c) iii) parabola | 1x4=4 | ||||||||||||||||||||||||||||||
| 20. | ii) x> — 36 i) 0 iii)43 iii)60 ii)Median --- Content provided by FirstRanker.com --- iii)80iii)31 | |||||||||||||||||||||||||||||||
| Part-B | ||||||||||||||||||||||||||||||||
| 21. | 4=2X2 ½ 7=7X1 ½ 14=2X7 ½ LCM=2X2X7=28 ½ --- Content provided by FirstRanker.com --- The three bells will ring together again at 6:28 am | 2 | ||||||||||||||||||||||||||||||
| 22. | Let P(x,0) be a point on X-axis PA=PB ½ PA2=pB2 ½ (x-2)2+(0+2)2=(x+4)2+(0-2)2 X2+4-4x+4=x2+16+8x+4 --- Content provided by FirstRanker.com --- -4x+4=8x+16X=-1 ½ P(-1,0) ½ OR PR:QR=2:1 ½ --- Content provided by FirstRanker.com --- R(1(-2)+2(3)/2+1 , 1(5)+2(2)/2+1 )R(4/3, 3) ½ | 1 | ||||||||||||||||||||||||||||||
| 23. | Sum of zeroes= 5-3v2+5+3v2=10 ½ Product of zeroes= (5-3v2)(5+3v2)= 7 ½ P(x)= X2-10x+7 ½ | |||||||||||||||||||||||||||||||
| 24. | Line seg=1/2 Circles=1 --- Content provided by FirstRanker.com --- Tangents=1/2 | 2 | ||||||||||||||||||||||||||||||
| 25. | tanA=3/4=3k/4k ½ sinA=3k/5k=3/5,cosA=4k/5k=4/5 ½ 1/sinA+1/cosA =5/3+5/4 ½ =(20+15)/12 ½ --- Content provided by FirstRanker.com --- =35/12OR v3 sin?=cos? ½ sin?/cos?=1/v3 ½ tan?=1/v3 ½ --- Content provided by FirstRanker.com --- ?=30° ½ | |||||||||||||||||||||||||||||||
| 26. | ?A = ?OPA =?OSA =90° ½ Hence, ?SOP=90° Also, AP=AS Hence, OSAP is a square AP=AS=10cm ½ --- Content provided by FirstRanker.com --- CR=CQ=27cmBQ=BC-CQ=38-27=11cm ½ BP=BQ=11 cm X=AB=AP+BP=10+11=21 cm ½ | |||||||||||||||||||||||||||||||
| 27. | Let 2-v3 be a rational number ½ We can find co-prime a and b (b?0) such that --- Content provided by FirstRanker.com --- 2-v3=a/b ½2-a/b=v3 ½ So we get,(2a-b)/b=v3 Since a and b are integers, we get (2a-b)/b is irrational and so v3 is rational. But v3 is an irrational number ½ --- Content provided by FirstRanker.com --- Which contradicts our statement ½Therefore 2-v3 is irrational ½ | |||||||||||||||||||||||||||||||
| 28. | 3x2+px+4=0 ½ 3(2/3)2+p(2/3)+4=0 4/3+2p/3+4=0 ½ P=-8 ½ --- Content provided by FirstRanker.com --- 3x2-8x+4=03x2-6x-2x+4=0 ½ X=2/3 or x=2 ½ Hence, x=2 ½ OR --- Content provided by FirstRanker.com --- a+ß=5 ----(1) ½a-ß=1 ----(2) ½ Solving (1) and (2), we get a=3 and ß=2 ½ also aß=6 ½ --- Content provided by FirstRanker.com --- or 3(k-1)=6 ½k-1=2 k=3 ½ | |||||||||||||||||||||||||||||||
| 29. | Area of 1 segment = area of sector —area of triangle ½ =(90°/360°)pr2 — ½ X7X7 =1/4x22/7x72 — ½ X7X7 ½ --- Content provided by FirstRanker.com --- = 14cm2 ½Area of 8 segments=8x14= 112 cm2 ½ Area of the shaded region = 14x14-112 ½ =196-112=84cm2 ½ (each petal is divided into 2 segments) | |||||||||||||||||||||||||||||||
| 30. | ?ABC~?DEF --- Content provided by FirstRanker.com --- Perimeter (?ABC) =AB+BC+CAPerimeter (?DEF) =DE+EF+FD 25/DE = 9/X ½ X=5.4cm ½ DE=5.4cm --- Content provided by FirstRanker.com --- ORConstruction-Draw AM ? BC ½ BD = 1/3BC,BM=1/2BC In ?ABM, AB2=AM2+BM2 ½ --- Content provided by FirstRanker.com --- =AM2+(BD+BM)2=AM2+DM2+BD2+2BD. DM ½ =AD2+BD2+2BD(BM-BD) =AD2+(BC/3)2+2. BC/3.(BC/2-BC/3) =AD2+2BC2/9 ½ --- Content provided by FirstRanker.com --- =AD2+2AB2/9Hence,7AB2=9AD2 | 1 | ||||||||||||||||||||||||||||||
| 31. |
a+b=15 median=l+(N/2 - cf)/f Xh ½ 16 =15+(35-12-a)/15 X5 --- Content provided by FirstRanker.com --- 1=(23-a)/3A=8 ½ 55+a+b=70 ½ 55+8+b=70 B=7 | |||||||||||||||||||||||||||||||
| 32. | Let AB=candle --- Content provided by FirstRanker.com --- C and D are coinsTan60°=AB/BC=h/b v3=h/b H=bv3 (1) ½ Tan30°=AB/BD=h/a --- Content provided by FirstRanker.com --- 1/v3=h/aH=a/v3 (2) ½ Multiplying (1) and (2), we get H2= bv3X a/v3 ½ H2=ba --- Content provided by FirstRanker.com --- H=vab ½ | |||||||||||||||||||||||||||||||
| 33. | Mode= l+(f1-f0)/(2f1-f2-f0) xh ½ 67 = 60+(15-12)/(30-x-12) x 10 7 = 30/(18-x) 7x(18-x)=10(15-x) 126-7x=150-10x --- Content provided by FirstRanker.com --- 3x=150-1263x=24 X=8 | |||||||||||||||||||||||||||||||
| 34. | Let BD=river AB=CD=palm trees=h BO=x --- Content provided by FirstRanker.com --- OD=80-xIn ?ABO, Tan60°=h/x v3=h/X (1) ½ H=v3x --- Content provided by FirstRanker.com --- In ?CDO,Tan 30°=h/(80-x) 1/v3=h/(80-X) (2) ½ Solving (1) and (2), we get X=20 --- Content provided by FirstRanker.com --- H=v3x=34.6the height of the trees=h=34.6m BO=x=20m DO=80-x=80-20=60m | |||||||||||||||||||||||||||||||
| OR Let AB=Building of height 50m --- Content provided by FirstRanker.com --- RT= tower of height=h m ½BT=AS=xm AB=ST=50 m ½ RS=TR-TS=(h-50)m In ?ARS, tan30°=RS/AS --- Content provided by FirstRanker.com --- 1/v3 = (h-50)/x (1)In ?RBT, tan60°=RT/BT ½ v3 =h/x (2) ½ Solving (1) and (2), we get ½ h=75 --- Content provided by FirstRanker.com --- from (2) ½x=h/v3 =75/v3 ½ =25v3 Hence, height of the tower=h=75m --- Content provided by FirstRanker.com --- Distance between the building and the tower=25v3=43.25m | 1 | |||||||||||||||||||||||||||||||
| 35. | For pipe, r=1cm ½ Length of water flowing in“1 sec, h=0.7m=7cm ½ Cylindrical Tank,R=40 cm , rise in water level=H ½ Volume of water flowing in 1 sec= pr2h=px1x1x70 =70p ½ --- Content provided by FirstRanker.com --- Volume of water flowing in 60 sec=70px60Volume of water flowing in 30 minutes=70px60x30 ½ Volume of water in Tank=pr2H=px40x40xH ½ Volume of water in Tank= Volume of water flowing in 30 minutes ½ px40x40xH = 70px60x30 --- Content provided by FirstRanker.com --- H=78.75cm | |||||||||||||||||||||||||||||||
| 36. | Let speed of the boat in still water =x km/hr, and Speed of the current =y km/hr Downstream speed =(x+y) km/hr Upstream speed =(x-y) km/hr 24/(x+y) + 16/(x-y) = 6 (1) ½ --- Content provided by FirstRanker.com --- 36/(x+y) + 12/(x-y) = 6 (2) ½Let u=1/(x+y) and v=1/(x-y) Put in the above equation we get, 24u+16v=6 Or, 12u+8v=3 .. (3) --- Content provided by FirstRanker.com --- 36u+12v=6Or, 6u+2v=1 .. (4) Multiplying (4) by 4, we get 24u+8v=4v ... (5) Subtracting (3) by (5), we get, --- Content provided by FirstRanker.com --- 12u=1u=1/12 Putting the value of u in (4), we get, v=1/4 1/(x+y) = 1/12 and 1/(x-y) = 1/4 x+y=12 and x-y=4 --- Content provided by FirstRanker.com --- Thus, speed of the boat in still water = 8 km/hr,Speed of the current = 4 km/hr, | |||||||||||||||||||||||||||||||
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