3 90

Time allowed : 3 hours Maximum Marks : 90

(i)

(ii) 34 8

1 6 2 10

3 10 4

(iii) 1 8

(iv) 2 3 3 4 2

(v)

General Instructions:

(i) All questions are compulsory.

(ii) The question paper consists of 34 questions divided into four sections A, B, C and D.

Section-A comprises of 8 questions of 1 mark each; Section-B comprises of 6 questions of 2

marks each; Section-C comprises of 10 questions of 3 marks each and Section-D comprises

of 10 questions of 4 marks each.

(iii) Question numbers 1 to 8 in Section-A are multiple choice questions where you are required

to select one correct option out of the given four.

(iv) There is no overall choice. However, internal choices have been provided in 1 question of

two marks, 3 questions of three marks each and 2 questions of four marks each. You have to

attempt only one of the alternatives in all such questions.

(v) Use of calculator is not permitted.

MA1-014 SUMMATIVE ASSESSMENT – I, 2012

MATHEMATICS

Class – IX

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SECTION–A

1 8 1

Question numbers 1 to 8 carry one mark each. For each question, four

alternative choices have been provided of which only one is correct. You have

to select the correct choice.

1. 4 28 3 7

(a) 8

3 (b)

16

3 (c)

24

3 (d)

18

3

The value of 4 28 3 7 is :

(a) 8

3 (b)

16

3 (c)

24

3 (d)

18

3

1

2. p(x)x2 11x k 4 k

(a) 40 (b) 28 (c) 28 (d) 5If 4 is the zero of the polynomial p(x) x2 11x k, then value of k is : (a) 40 (b) 28 (c) 28 (d) 5

1

3.

(a) 0 (b) 1 (c) 2 (d) 3Maximum number of zeroes in a cubic polynomial are : (a) 0 (b) 1 (c) 2 (d) 3

1

4. x2 8x 15 x2 3x 10

(a) x 3 (b) x 5 (c) x 5 (d) x3

Common factor in quadratic polynomials x2 8x 15 and x2 3x 10 is :

(a) x 3 (b) x 5 (c) x 5 (d) x3

1

5. ABC AB 105, B C 120 B

(a) 65 (b) 80 (c) 35 (d) 45

If in a triangle ABC, AB 105, B C 120 then B is :

(a) 65 (b) 80 (c) 35 (d) 45

1

6. ABC A B

(a) 45 (b) 60 (c) 30 (d) 90

A rt. angled isosceles triangle ABC is right angled at A. Then B is :

(a) 45 (b) 60 (c) 30 (d) 90

1

7. (2, 5)

(a) I (b) II (c) III (d) IV

Point (2, 5) lies in the quadrant :

(a) I (b) II (c) III (d) IV

1

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8. x≠ y, (x, y) ≠ (y, x), x y

(a) (x, y) (y, x) (b) (x, y) ≠ ( y, x) (c) (x, y ) (x, y) (d) (x, y) (x, y) If x≠ y, then (x, y) ≠ (y, x), But if x y, then (a) (x, y) (y, x) (b) (x, y) ≠ ( y, x) (c) (x, y ) (x, y) (d) (x, y) (x, y)

1

/ SECTION-B

9 14 2

Question numbers 9 to 14 carry two marks each.

9. 0.5 0.55

Find the two irrational numbers between 0.5 and 0.55

2

10. 2y3 y2 2y 1

Factorize : 2y3 y2 2y 1

2

11. (103)3

Using suitable identity evaluate (103)3

2

12. ACBD ABCD

In figure, if AC BD, then prove that AB CD

2

13. AB CD, APQ40 PRD118 x, y

In figure, if ABCD, APQ 40 and PRD 118, find x and y.

/ OR

AB CD EF x : y3 : 2 z

2

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In figure if ABCDEF and x : y 3 :2, find z.

14. 62 cm 24 cm 10 cm

Find the area of a triangle when two sides are 24cm and 10 cm and the perimeter of the triangle is 62 cm.

2

/ SECTION-C

15 24 3

Question numbers 15 to 24 carry three marks each.

15. 1

6(729)

Find the value of

1

6(729)

/ OR

0.235 p

q p, q q ≠ 0

Show that 0.235 can be expressed in the form p

q, where p and q are integers and

q≠0.

3

16.

5 3a b 15

5 3

a b

Find the value of a and b, when 5 3

a b 15 5 3

3

17. 3 21 9 127 p p p

216 2 4

Factorize 3 21 9 127 p p p

216 2 4

/ OR

3

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a6 b6

Factorize : a6 b6

18. x2 x0 2x35x2pxb p b

If x 2 and x 0 are zeroes of the polynomial 2x3 5x2 px b, then find the value of p and b

3

19.

If the bisectors of a pair of alternate angles formed by a transversal with two given lines are parallel, prove that the given lines are parallel.

/ OR

Prove that if two lines intersect, then the vertically opposite angles are equal.

3

20. ACBC, DCAECB DBCEAC

DCEC

In the given figure AC BC, DCA ECB and DBC EAC Prove that DC EC

3

21.

Prove that “Two triangles are congruent if two angles and the included side of one triangle are equal to two angles and the included side of other triangle”.

3

22. ABC B C AC AB BE CF

BECF (i) ABE ACF (ii) ABAC

3

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ABC is triangle in which altitudes BE and CF are equal. Then show that :

(i) ABE ACF and (ii) AB AC

23. PQ RS AB,

PQ B BC RS C

CD AB CD

In figure PQ and RS are two mirrors placed parallel to each other. An incident ray AB strikes the mirror PQ at B, the reflected ray moves along the path BC and strikes the mirror RS at C and again reflects back along CD. Prove that ABCD.

3

24. 25 m 10 m

14 m 13 m

A field is in the shape of a trapezium whose parallel sides are 25 m and 10 m, The non – parallel sides are 14 m and 13 m. Find the area of the field.

3

/ SECTION-D

25 34 4

Question numbers 25 to 34 carry four marks each.

25.

3 1 3 1

3 1 3 1x , y

x2 y2 xy

If3 1 3 1

3 1 3 1

x , y

, then find the value of x2 y2 xy.

/ OR

4

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3 2 2x , 33

1 x

x

If 3 2 2x , find 33

1 x

x

26. 1

4

1 1 1 1 1

5 5 6 6 7 7 8 8 9

Prove that : 1

4

1 1 1 1 1

5 5 6 6 7 7 8 8 9

4

27. (a b c)2 (a b c) 2 4b2 4c2

Simplify and factorise (a b c)2 (a b c) 2 4b2 4c2

4

28. abc6 abbcca11 a3b3

c33abc

If a b c6 and ab bc ca 11, find the value of a3 b3 c3 3abc

4

29. bx33x2

3 2x35xb x4 R1 R2

2R1R20 b

The polynomial bx3 3x2 3 and 2x3 5x b when divided by x 4 leave the remainders R1 and R2 respectively. Find the value of b if 2R1 R2 0

4

30. A(2, 2), B(6, 0) C(0, 4) D (3, 2) ABCD

A D

Plot the points A(2, 2), B(6, 0), C(0, 4) and D (3, 2) on the graph paper.

Draw figure ABCD and write in which quadrant A and D lie.

4

31.

If two parallel lines are intersected by a transversal , then prove that bisectors of

the interior angles from a rectangle.

4

32. ABC ABAC BA D

BAAD BCD

4

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ABC is an isosceles triangle in which AB AC. Side BA is produced to D such

that BA AD. Show that BCD is a right angle.

/ OR

ACAE ABAD BADEAC BCDE

In the given figure ACAE, ABAD and BAD EAC Show that

BC DE

33. ABC BC D E BDEC ADAE

ABAC

In ABC points D and E are on BC such that BD EC and AD AE, Prove that

AB AC

4

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34.

Show that the sum of the three altitudes of a triangle is less than the sum of the three sides of a triangle.

4

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Page 1 of 7

MARKING SCHEME

MATHEMATICS

Class – IX

SECTION – A

1. (A) 1

2. (C) 1

3. (D) 1

4. (B) 1

5. (D) 1

6. (A) 1

7. (B) 1

8. (A) 1

SECTION – B

9. 0.5101001000100001……… and 0.502002000200002…….. 2

10. 2y3 y2 2y 1 2y3 2 y2 2y 1

2(y 1) (y2 y 1) (y 1 ) 2

(y 1 ) [2y2 3y 1] (y 1) (y1) (2y 1)

1

1

11. (100 3)3 1003 33 3.100.3 (100 3) 1092709

1 1

MA1-014

SUMMATIVE ASSESSMENT – I, 2012

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Page 2 of 7

12. AC BD (Given ) -------- (1) AC AB BC and ---------(2) BD BC CD --------- (3)

(1), (2) and (3) AB CD

1

1

13. BPR 180 118 62

APB is a line 40 y 62 180

y 78 x 40 (alt. angle )

OR Let x 3k, y 2k

x y 180 5k 180 k 36 x 108, y 72

z x 108 (alternate angles)

½

1 ½

1 ½ ½

14. Let third side x 24 10 x 60 x 26

60S 30

2

Area 30 6 20 4

120 cm2

½

1 ½

SECTION-C

15. 1 6 1 66 1(729) (3 ) 3

1

3

OR

Let x 235. .2353535 ………….

10x 2 35. , 1000x 235 35. .

990 x 233 233

990

x

2

1

1½

1½

16. 5 3 5 3 5 3

5 3 5 3 5 3

8 2 15 4 15

2

a b 15 4 15 a 4, b 1

1

1

1

17. 3 21 9 127 p p p

216 2 4

3 23 21 1 1

(3p) 3.(3p) 3(3p) 6 6 6

31 1 1 1

3p 3p 3p 3p 6 6 6 6

2

1

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Page 3 of 7

OR

a6 b6 (a3) 2 (b3)2

(a3 b3) (a3 b3)

(ab) (a2ab b2) (a b) (a2 ab b2)

1 1

1

18. Let P(x) 2x3 5x2 px b P(2) 0 2p b 4 ---------(1)

P(0) 0 b 0 --------- (2)

(1) and (2) p 2

1 1 1

19. Given, to prove, figure Proof

Given, to prove, figure Proof : GM HL 2 3 ---------(1)

Also 1 2 and 3 4 --------(2)

(1) and (2) 1 4

1 2 3 4

AGH DHG ABCD

1½

½

½

½

20. DCAECB

DCADCEECBDCE

ACEBCD

ACE BCD (ASA)

ECDC (Cpct)

1

2

1

21. Given , to prove, figure Proof

1 2

22. In ABE and ACF A A

E F 90

BE CF ABE ACF

So AB AC (Cpct)

2 1

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23.

Draw BL PQ and CM RS

1 2 and 3 4

BL CM 2 3 alternate angles

1 4

1 2 3 4

ABC BCD and they are alternate angles

ABCD

½

½

1

1

24.

Draw DE BC and DF AB

AE 25 10 15m

For AED, 13 14 15

S 21 m2

Area AED 2 21 8 7 6 84 m

Also area of AED 1

15 DF 842

2 84 56

DF 15 5

Area of parallelogram EBCD 256 10 112m

5

Area of trapezium 84 112 196 m2.

1

½

1 ½

SECTION – D

25. 3 1 3 1 3 1 4 2 3 2 3

23 1 3 1 3 1x

1

1

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3 1 3 1 3 1 4 2 3 2 3

23 1 3 1 3 1y

3 1 3 1 1

3 1 3 1xy

2 2 2 2 (2 3) (2 3) 1 15x y xy

OR

3 2 2 x

1 1 1 3 2 2 3 2 2

13 2 2 3 2 2 3 2 2x

1 3x

x 2 2 3 2 2

4 2

3

3

3

1 1 1 3 x x x

x x x

3 3

3 3

1 1128 2 3( 4 2) 140 2x x

x x

½

1½

1

1

1

26. LHS

1 1 1 1 1

4 5 5 6 6 7 7 8 8 9

4 5 5 6 6 7 7 8 8 9

4 5 5 6 6 7 7 8 8 9

4 5 5 6 6 7 7 8 8 9

2 3 1

2½

1 ½

27. (abc)2 (abc) 24b2

4c2

(a b c) 2 {(a (b) (c)} 2 4(b2 c2)

(a2b2

c22ab2bc2ca){a2

(b)2(c)22a(b)2(b)(c)

2(c)a)}4(b2c2)

a2b2

c22ab2bc2caa2 b2

c22ab2bc2ac4(bc) (bc)

4ab 4ac 4(b c) (b c)

4(b c) (ab c)

1 1 1 1

28. a3 b3c3

3abc (abc) [a2b2

c2 (ab bc ca)]--------(1)

(abc) 2 (6) 2 (a2 b2 c2) 2(ab bc ca)

a2 b2 c2 36 211 14

(1) a3 b3 c3 3abc 6(14 11) 18

1

1 1 1

29. p(x)bx3 3x2 3 R1 p(4)64b48364b45

q(x)2x3 5x b q(4) R2 2(64) 5(4) b108 b 2R1 R2 0 2(45 64b) (b 108) 0

18

b 127

1½

1½

½

½

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30. Plotting of points A, B, C and D

Location of A and D is IIIrd and IInd quadrant. Figure ABCD

2 1 1

31.

AGH GHD

1

2AGH

1

2GHD 1 2

GM LH Similarly GLMH GMHL is a parallelogram

BGH GHD 180 1

2BGH

1

2 (GHD) 90

3 2 90

In GLH 2 3 GLH 180 90 GLH GLH 90

GMH 90

So MGL GLH 180 MGL 90

MGL 90 MHL 90

½

1

½

1

1

32. In ABC, AB AC ABC ACB

In ACD, AC AD ACD ADC

Also BAC CAD 180

CAD ABC BCA 2ACB

and BAC ACD ADC 2ACD

BAC CAD 2(ACB ACD)

180 2BCD BCD 90

OR

BAD EAC

Adding CAD both sides

1

1 1 1

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BAD CAD EAC CAD

BAC DAE

BAC DAE ------ (SAS)

BC ED (cpct)

2

1 1

33.

In ADE, AD AE 1 2

Also 1 ADB 180 2 AEC

Hence ADB AEC

ADB AEC (SAS)

AB AC

1

1 1 1

34.

AD BC AB > AD and AC > AD

AB AC > 2AD ---------(1)

Similarly BE AC BC BA > 2BE--------- (2)

CF AB AC BC > 2CF-------- (3)

Adding (1), (2) and (3) AB AC BC BA AC BC > 2AD 2BE 2CF 2(AB BC CA) > 2(AD BE CF) AD BE CF < AB BC CA.

1

1

1

1

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