3-2-1 and Scheme Mathematics Total

Topics: Circle, Area Pages: 16 (1776 words) Published: July 2, 2013
041/X/SA2/01/A1

Strictly Confidential : (For Internal and Restricted Use Only) Secondary School Examination [Class —X/2011] Summative Assessment — II

Marking Scheme — MATHEMATICS
Total No. of Pages : 13 General Instructions : 1. The Marking Scheme provides general guidelines to reduce subjectivity and maintain uniformity. The answers given in the marking scheme are the best suggested answers. Marking be done as per the instructions provided in the marking scheme. (It should not be done according to one’s own interpretation or any other consideration). Marking Scheme be strictly adhered to and religiously followed. 3. 4. Alternative methods be accepted. Proportional marks be awarded. If a question is attempted twice and the candidate has not crossed any answer, only first attempt be evaluated and ‘EXTRA’ written with second attempt. 5. A full scale of marks 0 to 80 be used. Do not hesitate to award full marks if the answer deserves it. In case where no answers are given or answers are found wrong in this Marking Scheme, correct answers may be found and used for valuation purpose.

2.

6.

1

SECTION - A
1. 2. 3. 4. 5. 6. 7. 8. 9. 10. (B) (D) (C) (C) (C) (D) (B) (B) (A) (C) 1 1 1 1 1 1 1 1 1 1

SECTION - B
11. a54, b522 (k11), c5k11 D50 for real and equal roots
∴ 4(k11)2216 (k11)50

1 ½+½

(k11) (k1124)50 k521, 3
1 17 , t85 2 6

12.

a5

t 85a17d or
17 1 5 17d 6 2 1 , t 5a13d 3 4

1

⇒ d5
5

1 1 3 13. 5 2 3 2

1 2

041/X/SA2/01/A1

13.

½

Let h be the height of the tower in DABC

h 5tan 60 60
h560. 3
∴ The height of the tower is 60 3 m.

½ ½ ½

14.

Since DABC is equilateral
∴ ∠ A5 ∠ B5 ∠ C5608

Area of sector AFEA5
u 3 p r 2 cm2 3608

 60  3 p (5)2  cm2 5   360 

5

25 p cm2 6

1

Area of all three sectors are equal
 25  ∴ Total area of shaded region53  p  cm2  6 

5

25 3 3.14 2

539.25 cm2. 041/X/SA2/01/A1 3

1

15.

Construction : Through O draw OR??BA or OR?? CD as AB and CD are parallel tangents. ½ Proof : ∠ OPA5908 (radius is always perpendicular to tangent)

Since OR??BA (By construction)
∴ ∠ OPA1 ∠ POR51808

⇒ ∠ POR5180829085908
Similarly ∠ QOR5908
∴ ∠ POR1 ∠ QOR51808

1

⇒ PQ is straight line through O. So PQ is diameter.
OR

½

OA5OC (radii of same circle)

⇒ ∠ OAC5 ∠ OCA
∴ ∠ OCD5908

------- (1)

½ ½

Since the tangents at any point of a circle is perpendicular to radius.

⇒ ∠ ACD1 ∠ OCA5908 ⇒ ∠ ACD1 ∠ OAC5908
[from (1)] ½ ½

⇒ ∠ ACD1 ∠ BAC5908
Hence proved. 16. If points are collinear the area of the triangle formed by three points as vertices is zero So area5[24 (616)24(2622)24(226)] 5[248132116] 50 so the points are collinear. 041/X/SA2/01/A1 4

½ 1 ½

17.

Let the point on x-axis be (a, o) since PA5PB ⇒ PA25PB2
∴ (a12)21(520)25(a22) 21(013)2

1

⇒ (a12) 22(a22) 2592255216
⇒ 8a5216 ⇒ a522.
∴ point is (22, 0)

1

18.

For given cone r 5 7a, l 5 13a
∴ CSA5p (7a) (13a)

½

591pa 2 5286
286 7 3 51 or a25 91 22

⇒ a51
∴ radius of cone 57 cm.

1 ½

SECTION - C
19. a2b 2x 21b 2x2a2x2150 or b2x(a2x11)21(a2x11)50 or (b2x21) (a2x11)50 1 1 or x5 2 , x52 2 b a

1 1

1 OR

6x 2 112x25x21050 6x(x12)25(x12)50 or (6x25) (x12)50 x522,
5 6

1 ½ 1½ 5

041/X/SA2/01/A1

20.

a59, d58, sn5450 s n5
n [2a1(n21)d] 2 n [181(n21)(8)] 2

½

4505

45054n 2 15n or or 4n 2 15n245050 4n 2 145n240n245050 4n 2 240n145n245050 4n(n210)145(n210)50 or n52 45 or n510 4
45 as number of terms can not be negative. 4

1

1

Rejecting n52
∴ n510

Ten terms of the given A.P. will make sum as 450. 21. For correct and accurate construction of first D Constructing second D 22.

½ 1 2

Let the time to fill the conical flask be x minutes. In 1 minute 20 m water flows. So in x minutes 20x m. will flow. ∴

Volume of water flowed through cylindrical pipe 5p r2 (20 x) 3 100 cms 5p (2)2 (20 x) 3 100 cm 1

Volume of water filled in conical tank 5
1 p (40)2 3 72...
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