Page 1 of 12

Strictly Confidential : (For Internal and Restricted Use only)

Secondary School Examination

MARKING SCHEME

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.

2. 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. Alternative methods be accepted. Proportional marks to be awarded.

4. 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. 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.

MARKING SCHEME

CLASS – IX

MATHEMATICS

Term – I (2012-2013)

SECTION – A ( 1 mark each)

1. ( iii)

2. (iii)

3. (iv)

4. (iii)

5. (iv)

6. (iii)

7. (i)

8. (i)

SECTION – B ( 2 mark each)

9. =

X ------

7-5 2____ --------------------

(7) 2 –(5√2) 2

Page 2 of 12

7-5 2____ --------------------

49 – 50

7-5 2____ = -7 +5 √2 --------------------

-1

10.

( 43)3 + (-27) 3 + (-16) 3

Let a = 43 , b = -27 c = -16

Then a+b+c = 43 + (-27) +(-16) = 0 -----------

a3+b3+c3 = 3 abc ---------------

(43) 3 + (-27) 3 + (-16) 3 = 3 x 43 x (-27)x (-16) --------

= 55728. -------

11. 2x2 + 3 + 7x

= 2x2 + 7x+3 -----------------

= 2x2+6x+1x+3 -----------------

= 2x(x+3) +1(x+3)----------------

=(2x+1)(x+3)----------------------

12.

Sol :- AB = 2 AC ( C is the midpoint of AB) ---------------------

XY = 2XD (D is the midpoint of XY) ---------------------

Ac = XD (given) => 2AC = 2 XD

AB = XY ----------------------------

Things which are double of the same things are equal to one another. ----------------

Page 3 of 12

13.

Sol :- GEF = GED - FED = 126 – 90 = 36o ------------------ 1 mark

FGE = 180- (36+90) = 180-126 = 54 ( Angle sum property) ------------ 1 mark

OR

Sol :

In ∆PQR , Q > R (given)

Q > R ------------------

SQR SRQ ( QS and RS bisects Q , R respectively) --------

SR > SQ --------- (side opposite to greater angle is longer ) -----------

14.

Sol. (-4,2) – II Quadrant

(4,0) - X – axis

(0,-3) - Y –axis

(3,3) - I quadrant -------------------------------------- ½ X 4 = 2

SECTION – C ( 3 mark each)

15.

Sol :- P(x) = Kx3 + 9x2 + 4x -8 , Divisor = x+3

x+3 =0 => x = -3

P(-3) = 7 ---------------------------- 1 mark

K(-3)3 + 9 (-3)2 + (4 (-3) - 8 =7

-27 K + 81 -12 -8 = 7 -------------------------- 1 mark

Page 4 of 12

-27k = -54

K = = 2 ----------------- 1 mark

16.

Sol : -2 and 5

-2/4 X 4 and 5/4 X 4 --------------------1 mark

-8/4 and 20/4 -----------------------1/2 mark

Any three rational numbers between -8/4 and 20/4 -----------1 ½ mark

OR

Sol :- x= 0. 3232..........

100x= 32.3232............(ii)-------------------------

(ii) – (i) => 100x –x = 32.3232..... – 0.3232..... ---------------

99x = 32 ----------------

X = 32/99------------- 1 mark

17.

Sol :- Correct no line -------------- 1 mark

Correct measurement (4,1) ------------1 mark

And correct construction ------------- 1 mark

18.

Sol :- x – 4 =4

(x – y) 3 = 4 3 = 64 ---------------1/2 mark

X3 – y3 – 3xy (x-y) = 64 --------------- 1 mark

X3 – y3 – 3 X 21 X 4 = 64 --------------1/2 mark

X3 – y3 = 64 + 252 -----------------1/2 mark

= 316 ----------------------1/2 mark

OR

Sol :-

Let p(x) = ax3 + 3x2 -3

q(x) = 2x3 – 5x + a

Page 5 of 12

The remainder when p(x) and q(x) are divided by x – 4 are p(4) and q(4) respectively

p(4) = a (4)3 + 3 (4)2 – 3 -----------------

q(4) = 2(4)3 – 5(4) +a ---------------------

P(4) = q(4) -------------------

64 a + 48 – 3 = 128 – 20 +a --------------------------

64 a – a = 63

63 a = 63

a = 1 -------------------------------- 1 mark

19.

Sol :-

XZY = 180 – (62 +54) (Angle sum property)

= 180 – 116

= 64 ----------------------- 1 mark

OYZ = (54) ( YO bisects Y)

= 27 ------------------

OZY = (64 ) ( ZO bisects Z)

= 32 ------------------

YOZ = 180 – ( OYZ + OZY) (Angle sum property)

= 180 – (27 +32)

= 180 – 59

= 121 -------------------------- 1 mark

OR

Page 6 of 12

Sol :

In the given figure AB CD. Through O , draw a line l parallel to both AB and CD.

------------------------

1 = ABO = 30o ( alternate interior angle ) --------------------

2 = DCO = 45o ( alternate interior angle ) ---------------------

BOC = 1 + 2 = 30 + 45 = 75 ----------------------------

x = 360 – BOC ----------------------

= 360 – 75

= 285o -----------------------------

20.

Sol :-

Y + 55 = 180 ( Interior angles on the same side of the transversal ED)

Y = 180 – 55 = 125 ---------------------------- 1 mark

X = y ( AB CD , corresponding angles )

= 125 ---------------------- 1 mark

AB CD , CD EF AB EF

EAB + FEA = 180o (Interior angles on the same side of the transversal EA)

= 90 + z + 55 =180o

z = 35o -------------------------------- 1 mark

21.

Page 7 of 12

Sol:

Since OA stands on the line CD .

AOC + AOD = 180 ( l.p) ----------------------

Also OD stands on the given line AB

AOD + BOD = 180 (l.p) -------------------- 1 mark

AOC + AOD = AOD + BOD

AOC = BOD

22.

Sol :

Perimeter = 96 m a = 40 m b = 24 m

Third side C = 96 – (40 +24) = 32 m --------------- ½ mark

2S = 96 S = 96 / 2 = 48 m ------------------------- ½ mark

Area of the triangle = -------------------

= --------------------

= ------------------------- 1/2

= 384 m2 ------------------------- ½ mark

23.

Sol:

Correct figure, Given , to prove----------1 Mark

Correct proof ----------------------2 Mark

.

24.

Page 8 of 12

Sol :

BE = CD

BE – DE = CD – DE

BD = CE ------------------------- ½ mark

In ∆ ABD and ∆ ACE , AB=AC (given)

B = C (isosceles triangle property)

BD = CE (proved) 2 marks

∆ABD ∆ ACE ( SAS)

AD = AE (cpct) ------------------ ½ marks

SECTION – D ( 4 mark each)

25.

Sol :

= x ---------------1/2 mark

= (3 + ) 2

(3)2- )2 ------------------------------ ½ mark

= 9 + 8 + 6 ----------------------- 1 mark

9 – 8

= 17 + 6 -------------------- 1 mark

a+ b 17 + 6

a = 17 , b= 6 --------------------------- 1 mark

OR

Sol :

x= = √7 + √ 6

√7 - √6

1/x = √7 - √ 6 --------------------------------- ½ mark

√7 + √ 6

x + 1/x = √7 + √ 6 + √7 - √ 6 --------------------- ½ mark

√7 - √6 √7 + √ 6

Page 9 of 12

= (√7 + √6)2 + (√7 – √6)2 ------------------ ½ mark

7 – 6

= 7 + 6+ 2 √42 + 7 + 6 - 2√42 ------------------- 1 mark

= 26

(x+1/x)2 = (26)2 = 676 -------------------------- 1 mark

26.

Sol :

(i)

( 3 + 3) ( 2 + √2) = 3 X 2 + 3√2 + 2 √3 + √3 X √2 ------------------------- 1 mark

= 6 + 3 √2 + 2 √3 + √6 ------------------------ 1mark

(ii). = 5 X √5 X √4 / 6 x √5 x √25 ----------------- 1 mark

= 1/3 ------------------------- 1 mark

27.

Sol :

p (x) = x3 - 3x2 - 9x -5.

p(-1) = (-1) 3 – 3 (-1) 2– 9( -1) -5 = 0

x+1 is a factor of p(x) ------------------------- 1 mark

Divide p(x) by (x+1)

p(x) = (x+1) ( x2 -4x -5) ------------------- 1 ½ mark

X2 – 4x – 5 = (x-5)(x+1) ---------------------------- 1 mark

p(x) = (x+1) (x-5) (x+1) ------------------------ ½ mark

OR

p (x) = X3 – 23 x2 +142 x -120

p(1) = 1- 23+142-120 = 0

(x-1) is a factor of p(x) ---------------------1 mark

Divide p(x) by (x-1)

p(x) = (x-1)(x2 – 22x +120) ---------------------- 1 ½ mark

x2 – 22x +120 = (x-12)(x-10) --------------------- 1 mark

p(x) = (x-1)(x-12)(x-10) ----------------------- ½ mark

28.

Sol : Fig. , given , to prove , construction ------------------ ½ mark each.

Page 10 of 12

Correct proof ----------------------- 2 marks.

29.

Sol :

For plotting the points correctly --------------------- 3 marks

For finding the co-ordinate of D as (2,3) ---------------- 1 mark

30.

Sol :

In ∆ABC , BC > AB ( given)

1 > 3 ( angle opposite to longer side is larger ) ----- ½ mark

In ∆ADC , CD > AD (given)

2 > 4 --------------------- ½ mark

1 + 2 > 3 + 4 --------------------- ½ mark

A > C ---------------------- ½ mark

Similarly from fig. 2 , for proving B > D ----------------- 2 marks

Page 11 of 12

Fig.2

31.

Sol:

i) In ∆ABD and ∆ ACD

AB = AC ( given)

BD = CD ( given) 2 mark

AD = AD (common)

ABD ACD ( SSS)

BAD = CAD (cpct) ------------------- ½ mark

ii) In ∆ ABP and ∆ACP

AB = AC (given)

BAD = CAD (proved) 1 ½ mark

AP = AP (common)

∆ABP ∆ACP (SAS)

32.

Sol :

EPA = DPB (given)

EPA + EPD = EPD + DPB

APD = BPE ----------------- 1 mark

I n ∆DAP and ∆EBP

AP = BP ( p midpoint)

PAD = PBE (given)

APD = BPE ( proved) 2 mark

∆DAP ∆EBP ( ASA)

AD = BE (cpct) -------------------- 1 mark

33.

Sol :

p(x) = 2x3+ax2+11x+a+3

p(x) is exactly divisible by 2x-1.

p(1/2) = 0 --------------------- 1 mark

2 (1/2)3 + a (1/2) 2 +11 X ½ + a + 3 =0

2 x 1/8 + a x ¼ + 11/2 + a +3 = 0 --------------------- 1 mark

¼ + a/4 + 11/2 + a +3 = 0

(1 + a + 22 + 4a + 12) / 4 = 0 --------------------- 1 mark

(5a +35) / 4 =0

5a + 35 = 0 1 mark

a = - 7

34.

sol :

p(x) = 2x4 – 6x3 + 3x2+3x-2

Page 12 of 12

g(x) = x2 – 3x +2.

= ( x-2) (x-1) ----------------------- 1 mark

(x-2) and (x-1) are factors of g (x)

p(2) = 2 (2) 4 – 6(2) 3 + 3(2) 2 +3 X 2 – 2 = 0 ----------------1 mark

p(1)= 2 (1) 4 – 6(1) 3 + 3(1) 2 +3 X 1 – 2 =0 ----------------------- 1 mark

p(x) is divisible by (x-2) and (x-1)

Hence p(x) is exactly divisible by g(x) ----------------1 mark