3-2-1 and Scheme Mathematics Total Essay

041/X/SA2/01/A1

Totally Confidential: (For Internal and Restricted Employ Only) Secondary School Evaluation [Class —X/2011] Summative Assessment — II

Marking Plan — MATHEMATICS

Total No . of Pages: 13 Basic Instructions: 1 ) The Marking Scheme delivers general suggestions to reduce subjectivity and maintain order, regularity. The answers given in the marking system are the best suggested answers. Tagging be done as per the instructions offered in the observing scheme. (It should not be completed according to one’s personal interpretation or any other consideration). Marking Structure be firmly adhered to and religiously followed. 3. 5. Alternative methods be accepted. Proportional markings be honored. If a problem is tried twice plus the candidate hasn't crossed virtually any answer, simply first make an effort be evaluated and ‘EXTRA’ written with second attempt. 5. An entire scale of marks 0 to 70 be used. Will not hesitate to award complete marks if the answer deserves it. In the event where no answers are offered or email address details are found wrong in this Observing Scheme, right answers may be found and used for valuation purpose.

2 .

6.

1

SECTION -- A

1 . 2 . several. 4. your five. 6. six. 8. 9. 10. (B) (D) (C) (C) (C) (D) (B) (B) (A) (C) you 1 1 1 one particular 1 1 1 you 1

SECTION - N

11. a54, b522 (k11), c5k11 D50 for real and equal roots

∴ 4(k11)2216 (k11)50

1 ½+½

(k11) (k1124)50 k521, a few

1 18, t85 a couple of 6

12.

a5

t 85a17d or

17 1 5 17d 6 a couple of 1, to 5a13d three or more 4

1

⇒ d5

5

one particular 1 3 13. 5 2 3 2

1 2

041/X/SA2/01/A1

13.

½

Let h be the height of the structure in DABC

h 5tan 60 60

h560. three or more

∴ The peak of the tower system is 70 3 m.

½ ½ ½

13.

Since DABC is equilateral

∴ � A5 � B5 � C5608

Area of sector AFEA5

u 3 p r 2 cm2 3608

 60  3 l (5)2  cm2 your five   360 

5

twenty-five p cm2 6

you

Area of all sectors will be equal

 25  ∴ Total area of shaded region53  p  cm2  6 

5

25 3 a few. 14 a couple of

539. 25 cm2. 041/X/SA2/01/A1 3

one particular

15.

Development: Through Um draw OR?? BA or perhaps OR?? COMPACT DISC as ABS and CD are seite an seite tangents. ½ Proof: � OPA5908 (radius is always perpendicular to tangent)

Since OR?? BA (By construction)

∴ � OPA1 � POR51808

⇒ � POR5180829085908

In the same way � QOR5908

∴ � POR1 � QOR51808

one particular

⇒ PQ is direct line through O. And so PQ is usually diameter.

OR

½

OA5OC (radii of same circle)

⇒ � OAC5 � OCA

∴ � OCD5908

------- (1)

½ ½

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

⇒ � ACD1 � OCA5908 ⇒ � ACD1 � OAC5908

[from (1)] ½ ½

⇒ � ACD1 � BAC5908

Hence turned out. 16. In the event that points happen to be collinear the area of the triangle formed simply by three items as vertices is actually zero So area5[24 (616)24(2622)24(226)] your five[248132116] 50 and so the points happen to be collinear. 041/X/SA2/01/A1 4

½ 1 ½

17.

Area point on x-axis end up being (a, o) since PA5PB ⇒ PA25PB2

∴ (a12)21(520)25(a22) 21(013)2

1

⇒ (a12) 22(a22) 2592255216

⇒ 8a5216 ⇒ a522.

∴ stage is (22, 0)

1

18.

Intended for given cone r five 7a, l 5 13a

∴ CSA5p (7a) (13a)

½

591pa 2 5286

286 six 3 fifty-one or a25 91 twenty two

⇒ a51

∴ radius of cone 57 cm.

1 ½

SECTION - C

nineteen. a2b 2x 21b 2x2a2x2150 or b2x(a2x11)21(a2x11)50 or (b2x21) (a2x11)50 1 1 or x5 2, x52 two b a

1 you

1 OR

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

5 six

1 ½ 1½ 5

041/X/SA2/01/A1

20.

a59, d58, sn5450 t n5

d [2a1(n21)d] 2 n [181(n21)(8)] 2

½

4505

45054n 2 15n or or perhaps 4n two 15n245050 4n 2 145n240n245050 4n 2 240n145n245050 4n(n210)145(n210)50 or n52 45 or n510 four

45 because number of conditions can not be negative. 4

1

1

Rejecting n52

∴ n510

Eight terms of the provided A. P. will make sum as 400.00. 21. Intended for correct and accurate building of 1st D Constructing second Deb 22.

½ 1 two

Let the the perfect time to fill the conical flask be times minutes. In 1 minute 20 meters water goes. So in x a few minutes 20x meters. will circulation. ∴

Volume of water flowed through cylindrical pipe 5p r2 (20 x) a few 100 cms 5p (2)2 (20 x) 3 100 cm one particular

Volume of water filled in cone-shaped tank five

1 p (40)2 3 72...

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