S O L U T I O N S

SAMPLE
QUESTION PAPER - 6

MATHEMATICS Oswaal CBSE (CCE) Class -9, SA-1 Examination Sample Question Paper
Self Assessment_________________________________

Time : 3 Hours Maximum Marks : 90

SECTION ‘A’
— —
1. (B) 0. 3 + 0. 4 = 0.333..... + 0.444......
= 0.777....
Let x = 0.777....
10x – x = 7.777.... – 0.777.....
9x = 7.0
7
x = .
9 1
3
83 + 17 3
(83 + 17)(832 − 83 × 17 + 17 2 )
2. (C) =

832 − 83 × 17 + 17 2

( 832 − 83 × 17 + 17 2 )
= 83 + 17 = 100. 1
3. (C) Degree of x3 + 5 = 3
Degree of 4 – x5 = 5
∴ Degree of (x3 + 5) (4 – x5) = 3 + 5 = 8. 1
4. (B) a = 3 + b
a – b = 3
(a – b)3 = (3)3
3 3
a – b – 3ab(a – b) = 27
a3 – b3 – 3ab × 3 = 27
a3 – b3 – 9ab = 27. 1

SECTION ‘B’

5.

(
) = 5 2 (15 + 3 2 + 5
5 2 3+ 2 5+ 2 )( 2 +2 ) 1

= 5 2 ( 17 + 8 2 ) ½

= 85 2 + 40 × 2
= 85 2 + 80 ½
[CBSE Marking Scheme, 2012]

2 | OSWAAL CBSE (CCE) Term-1, Mathematics, Class – 9

x4y4 – 256z4 = ( x y ) − ( 16 z )
2 2 2
2 2
6. ½

= ( x y − 16 z )( x y + 16 z )
2 2 2 2 2 2
½


= ( xy ) − ( 4 z )   x 2 y 2 + 16 z 2 
2 2

  ½
= ( xy − 4 z )( xy + 4 z ) ( x y + 16 z 2 ) .
2 2

½

( )
3
64x3 +
7. 125 y3 =(4x)3 + 5y
½
( )
= 4 x + 5 y ( 4 x ) + 5 y – 4 x × 5 y  ( )
2 2
1
 

( 2
)
= 4 x + 5 y 16 x + 5 y − 4 5 xy  .
2

½
8.
In a circle, having centre at P, we have
PR = PQ = radius ½

In a circle having centre at Q, we have
QR = QP = radius ½
Euclid’s first axiom : Things which are equal to the same thing are equal to one another. ½
∴ PR = PQ = QR ½
9.
∠M = ∠N (Given)

Ext. ∠NLQ = ∠M + ∠N = 2∠N 1
2∠NLP = 2∠N
∠NLP = ∠N (Alternate interior angles)
∴ LP || MN. 1
[CBSE Marking Scheme, 2012]
OR

AF || BE

Solutions | 3

70°
∠GAF = ∠DAF = = 35° (AF bisects ∠GAD) ½
2
∠DAF = ∠ADB = 35° (Alternate angles)
∴ ∠ADE = 180° – 35° = 145° ½
Ext. ∠GAD = ∠ABC + ∠ADB
⇒ 70° = ∠ABC + 35°
∴ ∠ABC = 35° 1
[CBSE Marking Scheme, 2012]
10. Drawing the points, A(3, 2), B(2, 3), C(–4, 5) and D(5, –3). Joining AB, BC, CA, AD, we get a
quadrilateral ABCD.
y

7
(–4,5) 6
C 5
4 B (2,3)
3 A
2 (3,2)
1
1 2 3 4 5 6 7 8
x' x
–6 –5 –4 –3 –2 –1
–1
–2
–3 (5, –3)
D
–4
–5
–6
–7
1
y'
No, because both require different position in plane. 1

SECTION ‘C’
( ) +( ) = ( 2) ( 3) ( 5) +( 2)
2 2 2 2 2 2
11. 2+ 3 5− 2 + +2 2 × 3 + –2 5× 2 1

= 2 + 3 + 2 6 + 5 + 2 –2 10
= 12 + 2 6 – 2 10 1

(
= 2 6 + 6 − 10 ) 1

OR
x = 2 + 3

1 = 1 2− 3
× =
2− 3
x 2 + 3 2 − 3
1
x– =2+ 3 – (2 – 3)=2 3)
x
2
1 1
=  x −  = (2 3 ) =
2
x + 2 2 +2 +2 14
x  x
4+ 5 4+ 5 4– 5 4– 5 16 + 8 5 + 5 16 – 8 5 + 5
12. × + × = + 1+1
4 – 5 4 + 5 4 + 5 4 – 5 16 – 5 16 – 5
21 + 8 5 + 21 – 8 5 42 ½+½
= =
11 11

4 | OSWAAL CBSE (CCE) Term-1, Mathematics, Class – 9

13. f(x) = px2 + 5x + r
1
f(2) = 0 and f   = 0
2
f(2) = 0 ⇒ 4p + 10 + r = 0
4p + r = – 10 ...(1) ½
1 p 5
f  = 0 ⇒ + + r = 0
2 4 2
p + 10 + 4r = 0 ½
p + 4r = – 10 ...(2)
(1) = (2) as both are equal to – 10 1
∴ p + 4r = 4p + r
4r – r = 4p – p
3r = 3p ½
⇒ r = p [CBSE Marking Scheme, 2012] ½

OR
2 2
2 2  3+ 2  3− 2  3+ 2  3− 2
x – y + xy =   –  + × ½
 3− 2  3+ 2  3− 2  3+ 2
5+2 6 5−2 6
= − +1 ½
5 − 2 6 5 + 2 6

(5 + 2 6 ) − (5 − 2 6 )
2 2

= +1 ½

( 5 − 2 6 )( 5 + 2 6 )
25 + 24 + 20 6 − 25 − 24 + 20 6
= +1 ½
25 − 24
= 40 6 + 1
= 40 × 2.4 + 1 = 96 + 1= 97. 1
14. p(x) = 3x3 + 2mx2 + 3x + 6
If p(x) is exactly divisible by x + 2, then p (– 2) = 0 ½
p(– 2) = 3(–2)3 + 2m(– 2)2 + 3(–2) + 6 = 0 1
–24 + 8m – 6 + 6 = 0
8m = 24
m = 3
3x3 + 6x2 + 3x + 6 1
= 3x2(x + 2) + 3(x + 2)
= 3(x2 + 1) (x + 2) ½
15. 23° + 40° + 35° + ∠D = 360° (Angle sum property of quadrilateral) 1
D = 262°

1

Solutions | 5

x° = Reflex ∠D = 360° – 262°

= 98° [CBSE Marking Scheme, 2012] 1
OR
Sol. Given,
∠POR : ∠ROQ = 5 : 7
Let, ∠POR = 5x
and ∠ROQ = 7x
5x + 7x = 180°
⇒ 12x = 180°
180
⇒ x = = 15° 2
12
∠POR = d = b = 5x = 5 × 15° = 75°
∠ROQ = c = a = 7x = 7 × 15°
= 105°. 1
[C.B.S.E. Marking Scheme, 2012]

16. D A

M

B C
(i) In ∆s AMC and ∆BMD, we have
AM = BM
(Q M is the mid-point of AB)
∠AMC = ∠BMD
(Vertically opp. angles)
and CM = MD (Given)
∴ By SAS criterion of congruence, we have

∆AMC ≅ ∆BMD 1
(ii) Now, ∆AMC ≅ ∆BMD

BD = CA and ∠BDM = ∠ACM 1

(Q Corresponding parts of congruent triangles are equal)
Thus, transversal CD cuts CA and BD at C and D respectively such that the alternate angle ∠BDM
and ∠ACM are equal. Therefore BD || CA.

∠CBD + ∠BCA = 180° ½

(Q Sum of the interior angles on the same side of transversal = 180°)
∠CBD + 90° = 180°
∠DBC = 90°. ½

17.
∠1 + ∠5 = 180° = ∠2 + ∠6 ⇒ ∠1 = ∠2, ( ∠5 = ∠6) 1
In ∆CAP and ∆BAP,
∠1 = ∠2 (Proved)

c. 18.p. Mathematics. (By c.) 1 [CBSE Marking Scheme.6 | OSWAAL CBSE (CCE) Term-1.t. a + b + c 6 + 6 + 4 16 s = = = = 8 cm ½ 2 2 2 A = s( s − a)( s − b)( s − c ) 1 . QR = BC (Proved above) AB = PQ (Given) 1 ∠BAC = ∠QPR = 90° (Given) ∴ ∆ABC ≅ ∆PQR (RHS) 1 ⇒ AC = PR.t.c.t. Sides of a triangle are a = 6 cm.p. Class – 9 ∠3 = ∠4 (AD is the bisector of ∠BAC) AP = AP ∴ ∆CAP ≅ ∆BAP (By AAS) 1 ⇒ CP = BP (By c. 2012] 19.) ½ [CBSE Marking Scheme.c.p. 2012] 20. A R C B Q P Given. 2012] In ∆COD and ∆BOA. b = 6 cm. c = 4 cm.) (Proved) 1 [CBSE Marking Scheme. BR = CQ BR + BQ = CQ + BQ QR = BC ½ In ∆ABC and ∆PQR. OD = OA (Given) OC = OB (Given) 1 ∠DOA = ∠BOC = 90° Given ⇒ ∠DOA + ∠AOC = ∠BOC + ∠AOC (Adding ∠AOC to both sides) ⇒ ∠DOC = ∠BOA ∴ ∆COD ≅ ∆BOA (By SAS) 1 ⇒ CD = AB (By c.

2012] SECTION ‘D’ 2− 5 2+ 5 21. we get 1 2  1  x +  = 52  x 2 1 x2 +   + 2 × x × 1 = 25 x x [ (a + b)2 = a2 + b2 + 2ab)] 1 1 x2 + 2 + 2 = 25 1 x 2 1 x + 2 = 25 – 2 x 1 x2 + 2 = 23 1 x . 2012] OR 1 x+ =5 x On squaring both sides. Solutions | 7 = 8×2×2×4 = 8 2 cm2 ½ Area of two black triangles = 8 2 × 2 = 16 2 cm2 ½ 2 Area of two white triangles = 16 2 cm ½ [CBSE Marking Scheme. a + b = + 1 2+ 5 2− 5 (2 − 5 ) + (2 + 5 ) 2 2 = 1 ( 2 + 5 )( 2 − 5 ) 4+5−4 5 +4+5+4 5 = 1 4−5 18 = = −18 ½ −1 \ (a + b)3 = (–18)3 = –5832 ½ [CBSE Marking Scheme.

(i) 2½ (ii) Graph is a straight line. 1½ 27. x8 – y8 = (x4)2 – (y4)2 = (x4 + y4) (x4 – y4) 1 = (x4 + y4) [(x2)2 – (y2)2] 1 4 4 2 2 2 2 = (x + y ) (x + y ) (x – y ) 1 = (x4 + y4) (x2 + y2) (x + y) (x – y) 1 2 2 2 2 24.S. ½ 26. ½ 3 3 3 3 3 25. Marking Scheme.8 | OSWAAL CBSE (CCE) Term-1. ∠PYZ + ∠ZYX = 180° ⇒ 2x + 64° = 180° ⇒ x = 58° 1 1 ∠XYQ = 64° + 58° = 122° 1 Reflex ∠QYP = 360° – x = 360° – 58° = 302° 1 .E. Class – 9 22. 2012] 23. Mathematics.B. RHS = (x – y) (x2 + y2 + xy) = x3 + xy2 + x2y – x2y – y3 – xy2 = x3 – y3 = LHS 2 Now. (a + b + c) – (a – b – c) + 4b – 4c = (a + b + c + a – b – c)(a + b + c – a + b + c) + (2b)2 – (2c)2 1 = 2a × (2b + 2c) + (2b – 2c)(2b + 2c) 1 = (2b + 2c)(2a + 2b – 2c) 1 = 2(b + c) × 2(a + b – c) ½ = 4(b + c)(a + b – c). 216x3 – 125y3 = (6x)3 – (5y)3 = (6x – 5y) [(6x)2 + (5y)2 + 6x × 5y] = (6x – 5y) (36x2 + 25y2 + 30xy). 2 [C. x – 8y – 36xy – 216 = (x) + (–2y) + (–6) – 3(x)(–2y)(–6) 1 = [x + (– 2y) + (– 6)][x2 + (–2y)2 + (– 6)2 –(x)(– 2y) – (– 2y)(– 6) – (x)(– 6)] 1 2 2 = (x – 2y – 6) (x + 4y + 36 + 2xy – 12y + 6x) ½ = 0 × [x2 + 4y2 + 36 + 2xy – 12y + 6x] ( x = 2y + 6 or x – 2y – 6 = 0) 1 ⇒ x3 – 8y3 – 36xy – 216 = 0.

(ii) 1 On adding (i) and (ii)... 1 T A P B l O S m C R D (ii) Lines and angles. (i) ∠TOP = ∠TOB 1 2 1 ∠ORS = ∠ORD 2 But ∠TOB = ∠ORD (l || m and corresponding angles) ∴ ∠TOP = ∠ORS 1 But they are corresponding angles w. OP || RS..(i) 1 AB > AC ⇒ ∠C > ∠B . we get 2∠C > ∠A + ∠B (by adding ∠C both the sides) 2∠C + ∠C > ∠A + ∠B + ∠C 1 3∠C > 180° ∠C > 60° 1 OR Given.t. AD = BC 1 Then. Hence. transversal TR and lines OP and RS. from the ASA Congruence Rule ∆ABC ≅ ∆ABD 1 ∴ BD = AC Proved 1 29.. ½ (iii) Bisectoring among human beings gives rise to deterioration in the society. Solutions | 9 28. ∠1 + ∠3 = ∠2 + ∠4 1 ∠DAB = ∠CBA and. ½ .r. AB > BC ⇒ ∠C > ∠A . AD = BC ∠1 = ∠2 ∠3 = ∠4 Let us consider ∆ABC and ∆ABD AB = AB (Common side) 1 ∠1 = ∠2 ∠3 = ∠4 So. In ∆ABC.

Ext. c = 1 cm a+b+ c 5+5+1 s = = cm 2 2 = 5. 6.5 − 5)(5. ∠AMC + ∠ACM + ∠2 + ∠3 = 180° 90° + 20° + ∠2 + 20° + ∠2 = 180° ½ 2∠2 = 50° ∠2 = 25°. Area = s( s − a)( s − b)( s − c ) = 5.5 − 1) cm2 .5(5. Class – 9 30. ∠M = ∠1 + 70° 1 90° = ∠1 + 70° ∠1 = 20° ∠1 + ∠2 = ∠3 (AN is angle bisector ) 1 20° + ∠2 = ∠3 In ∆AMC. Area (Shade IV) = Area (Shade V) 5 cm 6 cm (IV) 1·5 cm(V) 6·5 cm 1 cm 1 cm 2 cm 1 = × 6 × 1. ∠MAN = 25° 1½ 31.5 cm2 1 For (Shade II).5 cm.5 cm2 For (Shade 1).5 cm2 2 = 4. a = 5 cm.5 − 5)(5. Mathematics. In ∆ABM. b = 5 cm.5 × 1 = 6.10 | OSWAAL CBSE (CCE) Term-1.

75 × 3. consider rt. 2 A 1 cm B 1 cm 1 cm h D M N C 0·5 cm 1 cm 0·5 cm ∴ Area (Shade III) 1 = (AD + BC) × h 2 1 3 = (2 + 1) × cm2 2 2 1.5 × 0.5 × 0.5 × 0.5 × 0.5 × 3 11 cm2 = 0. ∆ADM In rt.49 cm2 1 2 = 2.5 cm For area (Shade III).5 + 6.3 cm2 Area of total paper used = (2. 2 1 3 h2 = 12 –   = 2 4 3 ∴ h = cm.5 × 9 cm = 0.3 + 4.5 ) cm2 = 19. ∆ADM.5 × 0.5 × 4.5 + 1.5 + 4.3 cm2 1 qqq .732 = 1.32 cm2 = 2.5 × cm2 1 2 1. Solutions | 11 2 = 5.866 cm2 = 1.5 cm 2 = 11 × 0.5 × 0.5 × 0.

L. SA-1 Examination Sample Question Paper Self Assessment_________________________________ Time : 3 Hours Maximum Marks : 90 SECTION ‘A’ ( ) ( 2) +( 5) ( 2 )×( 5 ) 2 2 2 1. 1 2. S O L U T I O N S SAMPLE QUESTION PAPER-7 MATHEMATICS Oswaal CBSE (CCE) Class -9. (C) p(x) = 2x + 3x – p If x = 2 is a zero of p(x) then p(2) = 0 ⇒ 2(2)2+ 3(2) – p = 0 ⇒ 8 + 6 – p = 0 p = 14 1 4. coefficient of x2 is –7. < < < < ½ 35 35 35 35 35 . 1 2 3.C. 1 SECTION ‘B’ 5. (D) (9x2 – 1) – ( 1 + 3x)2 = [(3x)2 – 12] – (1 + 3x)2 = (3x + 1) (3x – 1) – (3x + 1)2 = (3x + 1) [(3x – 1) – (3x + 1)] = – 2(3x + 1) Hence one of the factor is (3x + 1). (D) (2x2 – 5) (4 + 3x2) = 8x2 + 6x4 – 20 – 15x2 = 6x4 –7x2 – 20 Hence. 3 3 7 21 ∴ = × = ½ 5 5 7 35 5 5 5 25 and = × = ½ 7 7 5 35 21 22 23 24 25 So. of 5 and 7 is 35. (D) 2+ 5 = +2 = 2 + 5 + 2 10 = 7 + 2 10 = Irrational number.M.

. AC + CB = DE + CE ½ AB = DE ½ If equals are added to equals. 2012] 9. AC = DC (Given) CB = CE Adding. ½ 35 35 35. p(x) = x3 + x2 + x + 1 1 1 Put x– =0⇒x= in p(x) ½ 2 2 1 Remainder = p   ½ 2 3 2 1 1 1 =   +  + +1 ½  2   2   2  1 1 1 + + +1 = 8 4 2 1 + 2 + 4 + 8 15 = = .. ½ 8. the wholes are equal. 1 [CBSE Marking Scheme. 2012] 6. ½ 8 8 [CBSE Marking Scheme. 2012] 3 3 3  8  1 1 x 7.(ii) ½ Using (i) and (ii). ½ 2 ..   −  –  = . we get 8 −1 −1 x 3× × × = ½ 15 3 5 75 x = 8. x = ∠1 (Corresponding angles) ½ ∠2 = ∠1 (Corresponding angles) ½ ∠2 + 2y = 180° ⇒ x + 2y = 180° ½ 2y = 180° – x 1 y = 90° – x. . Solutions | 13 22 23 24 The required three rational numbers are .(i)  15   3   5  75 8 −1 −1 Let a= = .. [CBSE Marking Scheme. b = .c 15 3 5 8 1 1 8−5−3 a + b + c = −= − = 0 ½ 15 3 5 15 ∴ a3 + b3 + c3 = 3abc. and .

. ∠A + 140° = 180° ∠A = 40° ½ Again by (i)... Mathematics. Given. 2012] 10.(i) 65° + ∠C = 180° 1 ∠C = 115° Again by (i). Class – 9 [CBSE Marking Scheme. 2012] OR We know that ∠A + ∠B + ∠C = 180° . A 90 cm B 54 cm C Let the longest side be 90 cm and other side BC be 54 cm AB = ( 90 )2 − ( 54 )2 = 72 cm ½ 1 ∴ ×b×h Area of the triangle = 2 1 = × 54 × 72 ½ 2 = 27 × 72 = 1944 cm2 1 SECTION ‘C’ 11. 40° + ∠B + 115° = 180° ∠B = 25° ½ [CBSE Marking Scheme. x = 1 + 2 1 = 1 ( × 2 −1 ) x ( 2 + 1) ( 2 − 1) = ( 2 −1 = ) 2 −1 1 2 −1 1 x − = 1+ 2 − x ( ) ( 2 −1 ) ½ .14 | OSWAAL CBSE (CCE) Term-1.

Solutions | 15 = 1 + 2 – 2 + 1 = 2 ½ 3  1  x −  = 23 = 8 1  x [CBSE Marking Scheme. we get 2x – 2 = 4 ½ 2x = 6 ⇒ x = 3 ½ [CBSE Marking Scheme. ½ a2 – b2 [CBSE Marking Scheme. p3 – q3 = (p – q) (p2 + q2 + pq) = (p – q)[(p – q)2 + 3pq] 1 10  10  5 2 =  10    + 3 ×  Q p – q = 9  1 9  9  3   10  100  = + 5 9  81  . 2012] OR 1 1 a −1 a −1 a a −1 −1 + −1 −1 = 1 1 + 1 1 ½ a +b a −b + – a b a b   1 1 1  =  +  ½ a b+a b–a  ab ab  1  ab ab  = – ½ a  a + b a – b  1  1 1  = ( ab)  – ½ a  a + b a – b  a – b – a – b = b 2 2  ½  a –b  = b  –2 b   a2 – b2    –2 b 2 = Proved. 2012] 13. Given. 52x–1 – (25)x–1 = 2500 ⇒ 52x–1 – [(52)]x–1 = 2500 ½ ⇒ 52x–1 – 52x–2 = 2500 ⇒ 52x–2 (5 – 1) = 2500 ½ 2500 ⇒ 52x–2 = = 625 ½ 4 ⇒ 52x–2 =54 ½ comparing the powers of both sides. 2012] 12.

2012] OR 3 1  3 1  p(x) =  x  − ( 2 y ) + xy  x − 2 y  3 ½  4  4  4  1  1 2 1  =  x − 2 y   x  + ( 2 y ) + x × 2 y  + 3 xy  1 x − 2 y  2 4   4  4 4  4   1  1 2 2 1 3  =  x − 2 y   x + 4 y + xy + xy  ½  4   16 2 4   x 5  2 1 =  x − 2 y   + 4 y 2 + xy  ½  4   16 4   x 2  2 1 1 =  x − 2 y   + xy + xy + 4 y  ½ 4   16 4  1 x  x  x  =  x − 2 y    + y  + 4 y  + y  ½ 4 4 4 4       1  x  x  =  x − 2 y   + y   + 4 y  ½ 4  4  4  14. x4 + 2x3y – 2xy3 – y4 = (x2)2 – (y2)2 + 2x3y – 2xy3 ½ = (x2 – y2) (x2 + y2) + 2xy(x2 – y2) 1 2 2 2 2 = (x – y ) (x + y + 2xy) = (x – y) (x + y) (x + y)2 1 = (x – y) (x + y)3. ∠1 = ∠3 = 2x + 90° (V.A) ∠3 + ∠6 = 180° ⇒ 2x + 90° + 70° – x = 180° ½ ⇒ x = 20° ½ ∠1 = 130°. ∠6 = 50° ½ ∠4 = ∠6 = 50° ½ ∠8 = ∠6 = 50° 1 OR In ∆APD. ½ 15.O. Class – 9 10  100 + 405  = 9  81  5050 = = 6. ∠APD + ∠DAP + ∠ADP = 180° ∠A ∠D ∠APD + + = 180° 1 2 2 .927 1 729 [CBSE Marking Scheme.16 | OSWAAL CBSE (CCE) Term-1. Mathematics.

... ∠B + ∠C = 360° – (∠A + ∠D) . 2010. AC > AB (Given) ∠ABC >∠ACB (Angle opposite to larger side is greater) ½ A 1 2 B D C ∴ ∠ABC + ∠1 > ∠ACB + ∠1 (Adding ∠1 on both sides) ½ ∴ ∠ABC + ∠1 > ∠ACB + ∠2 (AD bisects angle A. ∠DAP = ∠A 2 1 and ∠ADP = ∠D ½ 2 1 alow ∠APD = 180° – (∠A + ∠D) (mul by 2 both the sides) 2 ⇒ 2∠APD = 360° – (∠A + ∠D) . ∠1 = ∠2) ½ ∴ ∠ADC > ∠ADB (Exterior angle property of triangle) 1½ . A 1 2 B D C E Let ∠DAC be ∠3 ∠1 = ∠2 (Given) ∠1 + ∠3 = ∠2 + ∠3 ∠BAC = ∠EAD ... 2∠APD = ∠B + ∠C.. 2011.(iii) From (i).(ii) ½ On comparing (i) and (ii). Proved.(i) ½ By ASP. we get ∆ABC ≅ ∆AED (By AAS rule) 1 17. BD = CE BD + DC = CE + DC BC = DE . (ii) and (iii).. Solutions | 17 1 Given.. 2012] 16..(ii) 1 ∠B = ∠E (Given) .. ½ [CBSE Marking Scheme.(i) 1 Given that.

(iv) i..(ii) Thus.. AC = AD] ... Class – 9 18.18 | OSWAAL CBSE (CCE) Term-1. y = 3a.. ∠CAD = ∠CAB + ∠BAD = x + x = 2x [From (i)] . ∆ACD is an equilateral triangle. AC = 2BC (Since BC = BD)..e. [From (iii) and (iv)] or AC = CD. A x D B C In ∆ABC and ∆ABD..c.t.(iii) 1 and ∠ACD = ∠ADB = 2x [From (ii). ∠CAB = ∠DAB (c. i.e.) . 1 19.(i) and AC = AD .p. ∆ABC ≅ ∆ABD (SAS rule) So. Produce CB to a point D such that BC = BD and join AD. OP ⊥ AB ⇒ ∠POA = 90° ½ Let x = a. Mathematics. we have BC = BD (By construction) AB = AB (Same side) ∠ABC = ∠ABD (Each of 90°) 1 Therefore.... z = 5a ½ P Q R y x z A O B ⇒ x + y + z = 90° ⇒ a + 3a + 5a = 90° ½ 9a = 90° a = 10° ½ ∴ x = 10° y = 3 × 10 = 30° z = 5 × 10 = 50° 1 .

142857. Solutions | 19 20. 2012] SECTION ‘D’ 21. Marking Scheme. ½ [C.S.B. b = 170 m. c = 250 m ½ a + b + c 120 + 170 + 250 540 s= = = = 270 ½ 2 2 2 Area of ∆ = s( s – a )( s – b )( s – c ) = 270 × (270 – 120)(270 – 170)(270 – 250) = 270 × 150 × 100 × 20 = 81000000 = 9000 m2 1 1 Area = × base × height ½ 2 1 ⇒ 9000 = × 250 × h 2 \ h = 72 m.E. b = –5 1 22. 2 6 + 6 2 + 8 = 2 6 ( 2− 3 )+6 2( 6− 3 ) + 8( 6− 2 ) 1½ 2+ 3 6+ 3 6+ 2 2−3 6−3 6−2 4 3 − 6 2 12 3 − 6 6 = −1 + 3 +2 6 − 2 ( ) 1½ = 6 2 −2 6 +2 6 −2 2 = 4 2 1 OR 5 + 3 7 5 −2 3 + ( ) LHS = 5− 3 5 +2 3 ( )( ) + 7( ) 2 5+ 3 5+ 3 5 −2 3 = 1 5−3 5 − 12 8 + 2 15 = 2 − 17 − 4 15 ( ) 1 = – 13 + 5 15 = −13 + 5 15 =a − 15 b 1 a = –13. a = 120 m. 0. We have.00000000 7 30 ... ½ 7 ) 1.

.428571 ½ 4 1 = 4 × = 4 × 0. (i) Plot the points A(1. we get the line segment AB. we can predict the decimal expansions of . Class – 9 28 20 14 60 56 1½ 40 35 50 49 1 2 3 4 5 6 Yes.571428 ½ 5 1 = 5 × = 5 × 0. .20 | OSWAAL CBSE (CCE) Term-1. Joining these points. (z + x) – (z + x). y + z = q. 1 .142857 7 7 = 0. 5). x + y = p. –1) and B(4. We can take the point C(2. (x + y) 1 = x2 + y2 + z2 – xy – yz – zx ½ 2 2 2 2 2 2 (p + q + r)(p + q + r – pq – qr – rp) = 2(x + y + z)(x + y + z – xy – yz – zx) 1 3 3 3 = 2(x + y + z – 3xyz) 25.142857 7 7 = 0. . . Mathematics. 1) between the points A and B.142857 7 7 = 0. without actually doing the long division 7 7 7 7 7 as follows : 2 1 = 2 × = 2 × 0. z + x = r ½ p + q + r = 2(x + y + z) ½ 3 3 3 2 2 2 p + q + r = (p + q + r) [q + q + r – pq – qr – rp] ½ 2 2 2 2 2 2 Now p + q + r – pq – qr – rp = (x + y) + (y + z) + (z + x) – (x + y) (y + z) – (y + z).285714 ½ 3 1 = 3 × = 3 × 0. 4x3 + 12x2 – x – 3 = 4x2 (x + 3) – (x + 3) 1 = (4x2 – 1)(x + 3) 1 2 2 = {(2x) – 1 )} (x + 3) 1 = (2x – 1)(2x + 1) (x + 3) 1 26.714285 ½ 3 2 3 2 23. Let P(x) = x + 2x – 5ax – 7 and Q(x) = x + ax – 12x + 6 P(–1) = p ⇒ p = 5a – 6 1 Q(2) = q ⇒ q = 4a – 10 1 2p + q = 6 (Given) ⇒ 2(5a – 6) + (4a – 10) = 6 1 a = 2 1 24.142857 7 7 = 0. Let.

1 27.. ∴ ∠FED = ∠PDQ (Corresponding angles) ⇒ ∠PDQ = 68° [From (iii)] and ∠AED = 34° 1 .7) 6 5 B(4. ∴ ∠AEF + ∠BEF = 180° 1 ⇒ ∠AED + ∠PEF + ∠BEF = 180° ⇒ 34° + ∠PEF + 78° = 180° ⇒ ∠PEF = 180° – 112° = 68° .–1) –2 –3 –4 –5 –6 y' 2 (ii) We can take the point D(5.(iii) 1 Now...(ii) P Q 34º D C F 78º A E B Now.1) O 1 2 3 4 5 6 7 8 x' x –4 –3 –2 –1–1O A(1. EF || DQ and transversal DE intersect them at E and D respectively.. Since AB || CD and transversal DE intersect them at E and D respectively. ∴ ∠AED = ∠CDP (Corresponding angles) 1 ⇒ ∠AED = 34° .. 7) lying on the AB produced.5) 4 3 2 1 C(2..(i) and ∠BEF = 78° . ray EF stands on AB at E. Solutions | 21 y 8 7 D(5.

Since.(ii) 1 Adding (i) and (ii). ∠GHE = ∠HFA = 95° ⇒ x° = ∠GHE + ∠EGH = 95° + 30° = 125° Thus.. ∠D + ∠1 = 180° ⇒ ∠1 = 180° – ∠D = 180° – 95° = 85° But. 1 OR Draw. ∠BGE + ∠GEH = 180° 1 (AB || EH.e. x = 100° 1 . ∠DFE + ∠FEH = 180° (CD || EH.. Class – 9 28. Therefore..... DE || AF and AD || FG ∴ AD || FG So. EH || AB A G B 135° E x H 125° C F D Now. ∠1 = ∠2 = 85° (Vetically opposite angles) 1 Also in ∆ABC.22 | OSWAAL CBSE (CCE) Term-1. ∴ ∠D = ∠F = 95° 1 Now. x° = 125° and y° = 35°.(i) 1 Again. we get ∠GEH + ∠FEH = 45° + 55° i. ∠2 + 60° + y° = 180° ⇒ 85° + 60° + y° = 180° ⇒ y° = 180° – 145° = 35° 1 Again. DAFH is a parallelogram. Mathematics. Co-interior angles) 125° + ∠FEH = 180° ∠FEH = 180° – 125° = 55° . G 30º H F xº E D 95º 1 A 2 yº B 60º C Produce DE to meet FG at H. EH || AB and AB || CD. EH || CD Now. Co-interior angles) 135° + ∠GEH = 180° ∠GEH = 180° – 135° = 45° .

c = 6 m a + b + c 15 + 11 + 6 s == = 16 m 1 2 2 . CD > BC ∠7 > ∠8 ∠5 + ∠7 > ∠6 + ∠8 ∠B > ∠D. CD > AD 1 ∠3 > ∠4 In ∆ABC. (SAS) ½ 30. 1 31.) ½ Now. b = 11 m.c. BC > AB 1 ∠1 > ∠2 ∠1 + ∠3 > ∠2 + ∠4 ∠A > ∠C 1 In ∆ABD. AB be smallest and CD be longest.t. D 6 8 A 3 1 5 4 7 2 B C Join AC and BD. In ∆ALB and ∆NCM. Solutions | 23 29. AL = CN (by adding ∠C both sides) ⇒ AL + LC = LC + CN AC = LN ½ In ∆ABC and ∆NML. Given. In ∆ADC.p. Let a = 15 m. AL = CN (Given) BL = CM (Given) ∠BLA = ∠MCN = 90° (Given) 1 ∴ ∆ALB ≅ ∆NCM (SAS) ½ ⇒ AB = NM and ∠A = ∠N (c. AD > AB ∠5 > ∠6 In ∆BCD. AB = NM (Proved) AC = LN (Proved) ∠A = ∠N (Proved) 1 ∴ ∆ABC ≅ ∆NML.

Class – 9 Using Heron’s formula. required area = s( s – a)( s − b)( s − c ) 1 = 16(16 − 15)(16 − 11)(16 − 6) m 2 16 × 1 × 5 × 10 m 2 = 2 = 4 × 5 × 2 m 2 = 20 2 m 1 Hence area painted in blue colour = 20 2 m2. Mathematics. 1 qqq .24 | OSWAAL CBSE (CCE) Term-1.

1 SECTION ‘B’ 2157 5. (a2 + a + 1) 1 3. 10x – x = 7. – 0. (A) x = 360 – (120 + 140) = 360 – 260 = 100 4. 10x = 7...777......777. (D) a3 – 1 = (a – 1).S O L U T I O N S SAMPLE QUESTION PAPER .. (A) The degree of the polynomial (5 – x3) (x2 + 3x + 2) is 5..4152 1 ⇒ 625 [CBSE Marking Scheme.. (C) Let x = 0. 2012] . = 625 2157 × 16 = 625 × 16 34512 = 10000 2157 = 3.777..8 MATHEMATICS Oswaal CBSE (CCE) Class -9. 9x = 7 7 x = 1 9 2.777. SA-1 Examination Sample Question Paper Self Assessment__________________________________ Time : 3 Hours Maximum Marks : 90 SECTION ‘A’ 1.

Given. AE = DF (Given) ∴ AB = CD (Things which are double of the same thing are equal to one another) 1 9. c = 13 cm a+b+c s = 2 5 + 12 + 13 = = 15 cm 2 Area of triangle = s( s − a)( s − b)( s − c ) = 15(15 − 5)(15 − 12)(15 − 13) = 15 × 10 × 3 × 2 = 30 cm2 1 Let length of perpendicular is h. ½ 3 1 ∠2 = ∠1 + of right angle 3 [CBSE Marking Scheme. Area = 30 1  Area = × 13 × h = 30 2 2 × 30 ∴ h = 13 60 = cm. Mathematics. b = 12 cm. 2012] 10. 1 13 . m 1 x 2 n ∠1 = ∠x = 75° (Alternate angles) ½ ∠2 = 180° – x = 180 – 75° = 105° 1 ∠2 = 105° = 75° + 30° 1 = 75° + × 90°. Class – 9 6.26 | OSWAAL CBSE (CCE) Term-1. 249 × 251 = (250 – 1) × (250 + 1) 1 2 2 = (250) – (1) = 62500 – 1 = 62499 1 [CBSE Marking Scheme. a = 5 cm. (7p)3 – (11b)3 = (7p – 11b)[(7p)2 + (7p)(11b) + (11b)2] 1½ 2 2 = (7p – 11b)(49p + 77pb + 121b ) ½ 8. 2012] 7. AB = 2AE (E is the mid point of AB) ½ CD = 2DF (F is the mid point of CD) ½ Also.

½ 990 198 [CBSE Marking Scheme.3282828 ½ 10x = 3. ½ 1000x – 10x = 328..000 ½ 325 65 x = = .....  x +  = (3)  x  1 (3)3 = x3 + +3×3 ½ x3 1 27 = x3 + + 9 1 x3 1 x3 + = 27 – 9 = 18 1 x3 [CBSE Marking Scheme... Let x = 0.282828. Given. 2012] 1 1 −   1 1   4 2    1 2  1    4    1  – 3  1  – 3       1 3× − 3   1 3× − 3    12.282828.2828.. – 3.. x + =3 x 3  1 3 1  1  x +  = x + 3 + 3 x +  ½  x x  x 3  1 3 But. ½ 990x = 325... ½ 1000x = 328.. Solutions | 27 SECTION ‘C’ — 11... 2012] OR p(x) = x3 + 3x2 – 2x – 4 p(–2) = (– 2)3 + 3(– 2)2 – 2(– 2) – 4 = – 8 + 12 + 4 – 4 = 4 1 p(1) = (1)3 + 3(1)2 – 2(1) – 4 .3 28 = 0.  5   +      =  5   +       8   27     2 3                  1 2 – = 5 {2 + 3}  2 4 1 1 2 2 – = 5 × 5  4 1  1 – = 4 4 [5 ] 1 = 1 5 1 13.2828.

28 | OSWAAL CBSE (CCE) Term-1. Angle : A figure formed by two rays with a common initial point. 1 2 14. ½ Point : Undefined term. ½ Right angle : Angle whose measure is 90°. In a square. Line segment : Part of a line with two end points. all angles are right angles. Euclid's fourth postulate says that “all right angles are equal to one another”. Three line segments are equal to fourth line segment. Undefined terms used are : line & point. ∠PQS + ∠SQR = ∠QRT (Alternate angle) ⇒ x + 28° = 65° ⇒ x = 65 – 28° = 37° 1 and ∠SPQ + ∠PQS + ∠QSP = 180° 90° + 37° + y = 180° y = 180° – (127°) = 53°. all the four sides of a square are equal. P Q x 28º y z 65º S T R  ∠SRQ + ∠QRT = 180° z + 65° = 180° z = 180° – 65° = 115° 1 Again. Ray : Part of a line with one end point. Class – 9 = 1 + 3 – 2 – 4 = – 2 1 p(0) = 0 + 0 – 0 – 4 = – 4 Now. 7a – 25a + 12 = 7a2 – 21a – 4a +12 2 = 7a(a – 3) – 4(a – 3) = (a – 3) (7a – 4) 1 = (2x – y – 3) [7 (2x – y) – 4) = (2x – y – 3) (14x – 7y – 4) 1 15. therefore. 7(2x – y) – 25 (2x – y) + 12 1 Let. 1 OR Sol. all angles are equal. . (Given) Therefore. (2x – y) = a then. Mathematics. Line : Undefined term. (From Euclid’ s fourth postulate). p(–2) + p(1) + p(0) = 4 – 2 – 4 = – 2. (i) The terms need to be defined are : Polygon : A simple closed figure made up of three or more line segments.

(i) 1 In DACD. AB = AC ∴ ∠ABC = ∠ACB 1 BP = PC ∴ ∠PBC = ∠PCB 1 ∴ ∠ABC – ∠PBC = ∠ACB – ∠PCB ½ ∠ABP = ∠ACP ½ 18. Let. s = = 15 m 2 . 1 [CBSE Marking Scheme. ∴ CD + DA > AC .. ½ 16. as sum of two sides of a triangle is greater than the 3rd side. 2012] 1 20. we get CD + DA + AB + BC > 2AC.. 1 [CBSE Marking Scheme. 2012] 19. ∠PBC < ∠QCB 1 ∴ ∠A + ∠C < ∠A + ∠B B C ∴ ∠C < ∠B 1 P Q ∠B > ∠C AC > AB 1 17. ½ A (iii) Equality leads to democracy. In DABC. ∠a + ∠b = 90° 1 2x + 3x = 90° x = 18° a = 2 × 18 = 36° b = 3 × 18° = 54° 1 ∴ c = 180° – b = 180° – 54° = 126°. ar (∆BCD) = × 12 × 5 = 30 m2 ½ 2 BD = 12 2 + 52 = 13 m2 ½ 13 + 8 + 9 For ∆ABD..”) 1 (ii) Introduction to Euclid’ s Geometry.. ∴ AB + BC > AC .. ∠a = 2x and ∠b = 3x Then.(ii) 1 Adding (i) and (ii). as sum of two sides of a triangle is greater than the 3rd side. Solutions | 29 (By Euclid's firt axiom “things which are equal to the same thing are equal to one another.

x = × 2 − 3 (2 + 3) ½ 2+ 3 = = 2+ 3 ½ 4−3 ( ) ( ) ( ) 3 2 2x3 – 2x2 – 7x + 5 = 2 2 + 3 −2 2+ 3 −7 2 + 3 +5 ½ .30 | OSWAAL CBSE (CCE) Term-1. OR 1 1 1  xa  a+b  xb  b+c  xc  c+a 2 2 2  b  . Mathematics. xc–a ½ = xa–b+b–c+c–a 1 0 = x = 1 ½ 1 (2 + 3) 22.( b 2 – c2 b + c x ) c 2 – a2 c + a 1 1 1 1 ( a − b )( a + b )× ( b − c )( b + c )× ( c − a )( c + a )× ( a+b) (b+c) ( c + a) = x . Class – 9 Area of ∆ABD = 15 × 2 × 7 × 6 1½ = 6 35 m2 ar (quad ABCD) = (30 + 6 35 ) m2 ½ SECTION ‘D’ 3+ 2 21.  a  x 2  2  2    x  x  1 1 1 = ( x ) .x . xb–c.( a 2 – b2 a + b x ) .x 1 = xa–b. x = 3− 2 3+ 2 3+ 2 = × 3− 2 3+ 2 ( ) 2 3+ 2 = =3+2+2 6 =5+2 6 1 3–2 3− 2 3− 2 y = × 3+ 2 3− 2 ( )= 2 3− 2 3+2−2 6 = = 5−2 6 1 3−2 1 x2 + y2 = (5 + 2 6 )2 + (5 – 2 6 )2 1 = 25 + 24 + 20 6 + 25 + 24 – 20 6 1 = 98.  c  .

(i) Draw X′OX and Y′OY as the co-ordinate axes and mark their point of intersection O as the origin (0. (a + b + c)2 = a2 + b2 + c2 + 2(ab + bc + ca) (5)2 = a2 + b2 + c2 + 2(10) a2 + b2 + c2 = 25 – 20 = 5 ½ a3 + b3 + c3 – 3abc = (a + b + c) [(a2 + b2 + c2) – (ab + bc + ca)] 1 = 5[5 – 10] ½ = – 25. 2012] 23. x3 + x2 – 4x – 4 Now put x = – 1. 1 [CBSE Marking Scheme. Solutions | 31 ( 3)  3 3 = 2 2 + + 3 × 2 × 3(2 + 3)    –2(4 + 3 + 4 3 ) – 14 – 7 3 + 5 ½ = 2[8 + 3 3 + 12 3 + 18] – 14 –8 3 – 9 –7 3 ½ = 2(26 + 15 3 ) – 23 – 15 3 ½ = 52 + 30 3 – 23 – 15 3 = 29 + 15 3 . then p(–1) = (–1)3 + (–1)2 – 4(– 1) – 4 = – 1 + 1 + 4 – 4 = 0. (x + 1) is a factor of this polynomial. 0). x3 + x2 – 4x – 4 = (x + 1) (x2 – 4) 1 = (x + 1) (x + 2) (x – 2). 1 2 24. Hence. . g(x) = x – 3x + 2 = x2 – 2x –x + 2 = x (x– 2) –1(x –2) = (x – 2)(x –1) 1 Zero of (x – 2) is 2 Zero of (x – 1) is 1 f(x) = 2x4 – 6x3 + 3x2 + 3x – 2 f(2) = 2(24) – 6(23) + 3(22) + 3(2) – 2 1 = 32 – 48 + 12 + 6 – 2 = 0 f(1) = 2(1)4 – 6(1)3 + 3(1)2 + 3(1) – 2 1 = 2 – 6 + 3 + 3 – 2 = 0 ⇒ (x – 1) and (x – 2) are the factors of f(x) ∴ f(x) is exactly divisible by g(x). 1 25. 1 26. 2 Hence.

½ Similarly. 8). we take 2 units on OX′ and then 8 units parallel to OY to obtain the point A(– 2. 7).25) ½ In order to plot (1. –1. Mathematics. A 90° B C 1 We have. ½ (iii) Co-ordination among people is good for progress. ∠A = 90° AB = AC ∠B = ∠C ( Q Angles opp. ∠A + ∠B + ∠C = 180° (Angle sum property) 1 90° + 2∠B = 180° ⇒ 2∠B = 180° – 90° = 90° ( Q ∠C = ∠B) 90° ⇒ ∠B = =45° 2 ∴ ∠C = ∠B = 45° 1 . we take 1 unit on OX and then 3 units parallel to OY to obtain the point D(1. –1.3) 2 1 1 2 3 4 5 6 7 8 x' x –5 –4 –3 –2 –1–1O E(3.7) 6 5 4 3 D(1. to equal sides of a triangle are equal) Also.8) 8 B 7 (–1. Class – 9 y A(–2.–1.–1) C –2 (0.25). ½ 27. 8). we take 3 units on OX and then 1 unit below x-axis parallel to OY′ to obtain the point E(3. 3) ½ In order to plot (3. we plot the point B(–1.25) –3 –4 –5 y' ½ In order to plot the points (– 2. –1) ½ (ii) Co-ordinate geometry.32 | OSWAAL CBSE (CCE) Term-1. 3). – 1). ½ In order to plot (0.25 units below x-axis on the y-axis to obtain C(0. we take 1.

c. we have AB = CD ½ BC = AD ½ and AC = AC ∴ We have ∆ABC ≅ ∆CDA. i. ∠FEC = ∠ECD = 10°. ½ 30. 2012] .p. Let. AD || BC and AB || CD ½ ⇒ ABCD is a parallelogram. ∠B + ∠BED + ∠BDE = 180° 40° + 90° + ∠BDE = 180° ∠BDE = 50° 1 [CBSE Marking Scheme. (By SSS criterion of congruence) 1 (ii) Congruency of triangles. ½ (iii) Equility is the sign of democracy. 2012] OR (i) ∆ADB ≅ ∆ADC (RHS) ∴ BD = DC ½ ∴ BE = FC 1½ ∠B = ∠C (Since AB = AC) AB = AC (∴ ∠B = ∠C) 1½ ∆ABE ≅ DACF (By SAS) ½ 29.e. AB = CD and BC = AD ½ Now in ∆ABC and ∆CDA. Solutions | 33 28. (Alternate angles are equal) ½ ∠BCA = 90° ∠BCD + ∠DCA = 90° 10° + ∠DCA = 90° 1 ∠DCA = 80° AC = DC (Given) 180° − 80° ∠DAC = ∠ADC = = 50° ½ 2 In ∆BAC.t) [CBSE Marking Scheme. AC = AE. ∠BAC = ∠EAD ½ ∴ BC = DE (By c. ∠A + ∠B + ∠BCA = 180° 50° + ∠B + 90° = 180° ∠B = 40° 1 In ∆BDE. (i) l and m are two parallel lines intersected by another pair of parallel lines p and q. A E B D C ∠DAC = x ∠BAD + x = ∠EAC + x ∠BAC = ∠EAD 1 ∆ABC ≅ DADE (By SAS) 2½ AB = AD.

34 | OSWAAL CBSE (CCE) Term-1. b = 22 cm.000. 1 qqq . Let a = 122 cm. Mathematics.  3  2 Total rent for 3 months = `  5. Class – 9 31.50. c = 120 m a+b+c s = 2 122 + 22 + 120 = m = 132 m 1 2 Area of triangular side wall = s( s − a)( s − b)( s − c ) (Heron’s formula) 1 2 = 132(132 − 122)(132 − 22)(132 − 120) m = 2 132 × 10 × 110 × 12 m = 1320 m2 1 Rate of rent = ` 5000 per m2 per year. 000 × 1320 ×  m  12  = ` 16.

707 ½ 2 2 1 + π = 0. we get 2  1 ( 3) 2 x −  = ½  x 1 1 x2 + –2×x× = 3 1 x2 x 1 x2 + 2 = 3 + 2 = 5. (D) Put x + 1 = 0 ⇒ x = –1 in given expression. the remainder is f(– 1) = (–1)11 + 101 1 = –1 + 101 = 100. (D) Constant polynomal is 7. = × (Rationalising) ½ 2 2 2 2 1. (C) (x.S O L U T I O N S SAMPLE QUESTION PAPER . 4. 1 SECTION ‘B’ 1 1 2 5. ½ x . 1 2 2 3. SA-1 Examination Sample Question Paper Self Assessment___________________________________ Time : 3 Hours Maximum Marks : 90 SECTION ‘A’ 1.707 + 3. 2012] 1 6.141 = 3.848 1 2 [CBSE Marking Scheme. 0) 1 π π 2. x– = 3 x Squaring both sides.9 MATHEMATICS Oswaal CBSE (CCE) Class -9. (D) Coefficient of x in the expression x2 + x – 7 is .414 == = 0.

(Angles on the same side of transversal) ½ ⇒ EF || CD ½ ∠BCD = ∠BCE + ∠ECD = 35° + 25° = 60° = ∠ABC (Alternate angles) ½ ∴ AB || CD So. ½ 16 4 4 2 8. 2012] OR ∠1 = ∠2 = x. ∠ADC = ∠1 + ∠B = 40° + 45° = 85° ½ . ∠C = 180° – 108° = 72° ½ So.. from (i).. ∠C = 55° 2x + 45° + 55° = 180° . ∠B = 130° – 72° = 58° and ∠A = 180° – 72° – 58° = 50° 1 9. ∠B = 45° ∠A + ∠B + ∠C = 180°..(i) (Given) 1 Again ∠B + ∠C = 130° .  − + 1  =  4 +  − 2  + 1 ½ 4 2      2 2 a  b a b  b =   +  −  + (1)2 + 2 ×  −  + 2  −  × (1) 4    2  4  2   2 a + 2 ×   × (1) 1 4 a2 b2 ab a = + + 1− −b+ . Mathematics.. AB || EF. AB || CD || EF Thus. Class – 9 2 2 a b  a  b   7.(iii) (Property of triangle) ½ So.. (Exterior angles is the sum of the two interior opposite angles) 1 = 40° + 55° = 95° ½ Similarly... ∠FEC + ∠DCE = 155° + 25° = 180°.36 | OSWAAL CBSE (CCE) Term-1.(ii) (Given) and ∠A + ∠B + ∠C = 180° ..(i) 2x = 80° x = 40° ∠ADB = ∠2 + ∠C. ½ [CBSE Marking Scheme. ∠A + ∠B = 108° .

Area of an equilateral triangle = a ½ 4 3 2 \ a = 81 3 1 4 a2 = 81 × 4 a = 9 × 2 = 18 cm Perimeter of an equilateral triangle = 3a = 3 × 18 = 54 cm.  = 1 3 2 16 2 x 32 x 34 .   . Solutions | 37 3 2 10. 2012] SECTION ‘C’ x 2x 2 3 81 11. 2012] . ½ [CBSE Marking Scheme. we get 4q2x2 + p2 – 4pqx = p2 – 4q2 ½ 2 2 4q(qx – px) = –4q ½ 2 qx – px + q = 0. = 1 3x 2 2 x 2 4 2x–2x. ½ [CBSE Marking Scheme.32x–x = 2–4 34 x – 2x = –4 1 \ x = 4 OR p + 2q + p − 2q p + 2q + p − 2q x = × ½ p + 2q − p − 2q p + 2q + p − 2q ( p + 2q ) + ( p − 2q ) + 2 × p + 2q + p − 2q = ½ ( p + 2q ) − ( p − 2q ) 2 p + 2 p 2 − 4q 2 = ½ p + 2q − p + 2q = ( 2 p + p 2 − 4q 2 ) 4q 2 2 2qx = p + p − 4 q 2qx – p = p 2 − 4q 2 Squaring both sides.

. then from eqs. Let p(x) = 3x3 + x2 – 20x + 12 Given (3x – 2) is a factor of p(x). . Class – 9 12.. 2012] 13. (i) and (ii). 5 + 2 6 + 8 − 2 15 = 3 + 2 + 2 3×2 + 5 + 3 − 2 5×3 ½ ( 3) +( 2) ( 5) +( 3) 2 2 2 2 +2 3× 2 + −2 5× 3 1 = ( ) ( ) 2 2 3+ 2 + 5− 3 ½ = = 3+ 2+ 5− 3 ½ = 2+ 5 ½ [CBSE Marking Scheme. p(–2) = q(–2) – 46 – 2k = – 10 + k 3k = – 36 ½ k = –12 ½ OR 2 1 2 1 1 2 (i) x2 + 2 + 2 – 2x – = (x)2 +   + 2 × x × – 2x – x x x   x x 2  1  1 = x +  – 2 x +  ½  x  x  1  1  =  x +  x + − 2  1  x   x  (ii) x4 – y4 = (x2)2 – (y2)2 ½ = (x2 – y2) (x2 + y2) ½ 2 2 = (x – y) (x + y) (x + y ) ½ 14. p(x) = 3x3 – 5x2 + kx – 2 Remainder on dividing p(x) by (x + 2) is p(–2) = 3(–2)3 – 5(–2)2 + k(–2) –2 ½ = – 24 – 20 – 2k – 2 p(–2) = – 46 – 2k .(ii) ½ Since remainders are same.. 3x – 2 ) 3x3 + x2 – 20x + 12 ( x2 + x – 6 –3x3 + – 2x2 3x2 – 20x + 12 2 – 3x + – 2x – 18x + 12 1½ +–18x –+12 × x2 + x – 6 = x2 + 3x – 2x – 6 ½ = x(x + 3) – 2(x + 3) ½ = (x + 3) (x – 2) ½ The other factors are (x – 2) and (x + 3). Mathematics. we have.38 | OSWAAL CBSE (CCE) Term-1.(i) ½ q(x) = – x3 – x2 + 7x + k Remainder on dividing q(x) by (x + 2) is q(–2) = –(– 2)3 – (– 2)2 + 7(–2) + k = 8 – 4 – 14 + k ½ = – 10 + k ..

1 2 [CBSE Marking Scheme. ½ OR Proof : OR ⊥ PQ ∴ ∠POR = ∠ROQ = 90° ∴ ∠POS + ∠ROS = 90° ∠ROS = 90° – ∠POS 1 Add ∠ROS on both sides R S P O Q ∠ROS + ∠ROS = 90° – ∠POS + ∠ROS ⇒ 2∠ROS = 90° + ∠ROS – ∠POS 1 2∠ROS = ∠QOS – ∠POS ( 90° + ∠ROS = ∠QOS) 1 ∴ ∠ROS = (∠QOS – ∠POS). AB = AC ⇒ = ½ 2 2 ⇒ AE = AF. CD || EF. A E F B C . Solutions | 39 15. Given AB || CD. EA ⊥ AB ½ ∠BEF = 55° y = 180° – 55° = 125° (Co-interior angles) ½ x = y = 125° (Corresponding angles) ½ EA ⊥ AB ∴ z = 90° – 55° = 35°. 2012] AB AC 16.

to greater angle is larger) ½ [CBSE Marking Scheme. 2012] .. P S R Q Proof : In DPQR.) ½ Similarly. using (i) ½ ⇒ ∠PRS > ∠PSR ⇒ PS > PR ½ PS > PQ (\ PR = PQ) (Side opp. ∠BEC = ∠BFC = 90° (Given) ½ BE = CF (Given) BC = CB (Common) ∴ ∆BCE ≅ ∆CBF (RHS) ½ ⇒ ∠B = ∠C ½ ∴ AC = AB (By c.p. 2012] 17. AB = BC = AC ½ Thus.. to equal sides are equal) .t.) ½ \ BF = CE [CBSE Marking Scheme. Mathematics. angle) \ ∠PRQ > ∠PSQ. [CBSE Marking Scheme. ∠PQR > ∠PSQ ½ (Ext. AB = AC (Given) ½ ∠A = ∠A (Common) ½ AF = AE (Proved) ½ \ DABF ≅ DACE (By SAS cong. Class – 9 Since E and F are the mid-points of AB and AC. ∆ABD ≅ ∆BAE AC = BC ½ Therefore. In DABF and DACE.(i) 1 In DPQS. In ∆BCE and ∆CBF. ∆ABC is an equilateral triangle. PQ = PR ∠PQR = ∠PRQ (Angles opp.40 | OSWAAL CBSE (CCE) Term-1. angle of a D is greater than each of interior opp.c. 2012] 18.

=3+2 2 (rationalizing) ½ x  1 2 1  x + x  = 6 ⇒ x + 2 = 34 1   x 1 x– = –4 2 ½ x 1 x2 – = 6 × – 4 2 = –24 2 1 x2 1  1  1  x4 – 4 =  x2 − 2   x2 + 2  x  x  x  ( = 34 −24 2 = –816 ) 2 1 . Solutions | 41 19. 2012] 20.cm ½ 2 ∴ Area of unshaded portion = 43. 1 [CBSE Marking Scheme.3 sq. Area of an equilateral triangle 3 2 3 = a = × 10 × 10 ½ 4 4 = 25 3 cm = 25 × 1. DC = 10 2 − 8 2 = 6 cm ½ 1 ∴ Area of ∆BDC = × 6 × 8 = 24 sq.3 sq. cm In ∆BDC. y – x = 50° (½+ ½) ∴ (x + y) : (y – x) = 21 : 5. y = 180° – (30° + 20°) = 130° ½ l || m ⇒ x + 100° = 180° ⇒ x = 80° ½ A B l 100º 40º xº m C D E yº 30º 20º n E ∴ x + y = 210°.3 – 24 = 19.732 1 = 43. cm ½ SECTION ‘D’ 1 21.

2) = = 5. 1 1 1 = + + 1 1 1 1 1+ x + 1+ y + 1+ z + y z x 1 1 1 1 = + + ( xyz = 1 ⇒ = xy) 1 1 1 + y + xy 1 1 z 1+ x + 1+ + y xy x . 2x3 + x2 + 3x – 5 (Quotient) 3x + 4 ) 6x4 + 11x3 + 13x2 – 3x + 27 4 3 –6x –+ 8x 3 2 –3x + 13x 3 –3x + –4x2 9x2 – 3x 2 –9x + –12x –15x + 27 –+ 15x – + 20 47 Thus. 1 1 1 = + + y + xy + 1 y + xy + 1 xy + 1 + y 1 y xy y 1 xy = + ½ y + xy + 1 y + xy + 1 xy + 1 + y ( y + 1 + xy ) = = 1 ½ ( y + xy + 1) [CBSE Marking Scheme. 2012] (1 + x + y–1)–1 + (1 + y + z–1)–1 + (1 + z + x–1)–1 22. quotient = 2x3 + x2 + 3x – 5 and remainder = 47. Class – 9 OR 15 15 1 = 10 + 20 + 40 – 5 – 80 10 + 2 5 + 2 10 – 5 – 4 5 15 = ½ 3 10 − 3 5 15 ( 10 + 5) = × 1 3( 10 – 5) ( 10 + 5) 5 × ( 10 + 5) = ½ 10 – 5 5.4 1 5 [CBSE Marking Scheme.2 + 2. 2012] 23. Mathematics.(3.42 | OSWAAL CBSE (CCE) Term-1. 1 4 3 2 Verificaiton : Let p(x) = 6x + 11x + 13x – 3x + 27 Zero of 3x + 4 is – 4/3 .

Let f(x) = x3 + 2x2 – 5ax –8 and g(x) = x3 + ax2 – 12x – 6 When p(x) and q(x) are divided by (x – 2) and (x – 3) and leave p and q as remainders. 1 4 3 2 4  4 4 4  4 Remainder = p  −  = 6  −  + 11  −  + 13  −  – 3  −  + 27 1  3   3   3   3   3 512  −704  208 = + + + 4 + 27 ½ 27  27  9 512 − 704 + 624 = + 31 27 = 16 + 31 = 47. (i) Plot the point M(5. then f(2) = p and g(3) = q f(2) = 23 + 2 × 22 – 5a × 2 – 8 1½ = 8 + 8 – 10a –8 f(2) = 8 – 10a = p g(3) = 33 + a × 32 – 12 × 3 – 6 g(3) = 27 + 9a – 36 – 6 g(3) = – 15 + 9a = q 1½ If. – 3) and N(– 3. 1 19 [CBSE Marking Scheme. ½ 2 2 2 2 24. q – p = 10 \ – 15 + 9a – 8 + 10a = 10 19a – 23 = 10 19a = 33 33 a = . 2 . 1 [CBSE Marking Scheme. 2012] 26. (a) 4a – 9b – 2a – 3b = (2a) – (3b) – (2a + 3b) ½ = (2a – 3b) (2a + 3b) – (2a + 3b) 1 = (2a + 3b) (2a – 3b –1) ½ (b) a2 + b2 – 2(ab – ac + bc) = a2 + b2 – 2ab + 2ac – 2bc ½ = (a – b)2 + 2c(a – b) ½ = (a – b) [(a – b) + 2c] = (a – b) (a – b + 2c). 2012] 25. Solutions | 43 By remainder theorem. – 3) on the graph paper.

Given : A parallelogram ABCD in which AN and CP are perpendiculars to the diagonal BD. Mathematics. AN = CP (By c. they will not meet on the other side also. Again we know that the sum of the interior angles on the other side of line l will also be two right angles. So. – 3) ½ B(1. A(3. to equal angles are equal). 1 RS = QT (Given)  QS = RT (Given) S T QR = RQ (Common) ∴ ∆QRS ≅ ∆RQT (SSS rule) 1 ∴ ∠QRS = ∠RQT P (C.) Or ∠QRP = ∠RQP 1 Hence. ∠ANB = ∠CPD (Each 90°)  ∠1 = ∠2 (Alternate angles) 1 AB = CD (Opposite sides of a parallelogram) 1 ∴ ∆ANB ≅ ∆CPD (A.c.A. 1 Proved OR Given : PQ = PR (Given) ∠PRQ = ∠PQR (Angles opp. then by Euclid’s fifth postulate the lines will not meet on this side of l.P. 2 (ii) Introduction to Euclid’s Geometry. Given : RS = QT and QS = RT To prove : PQ = PR R Q Proof : In ∆QRS and ∆RQT. (i) If a straight line l falls on two straight lines m and n such that the sum of the interior angles on one side of l is two right angles. Therefore.) Proved 1 . the lines m and n never meet and are therefore parallel.C. To prove : AN = CP A D P 1 3 4 2 N B C Proof : In ∆ANB and ∆CPD.p.44 | OSWAAL CBSE (CCE) Term-1. – 3) ½ 27. – 3) ½ C(– 1. PQ = PR (In a triangle sides opp.T.S. congruence) 1 Hence. 1 28. 1 (iii) Universal truth.t. Proved 1 29. to equal sides of a ∆) 1 P S R Q Now. ∠SQR < ∠PQR 1 ∴ ∠SQR < ∠PRQ or ∠SRQ 1 ∴ RS < QS. Class – 9 (ii) Length of MN = 3 + 5 = 8 units ½ (iii) From figure.

we get DP + PQ = PQ + CQ 1 ⇒ DQ = CP. For one triangular piece. Solutions | 45 30.) 1 Proved. AQ = Hyp.t. c = 50 cm a+b+c s = 2 20 + 50 + 50 = = 60 cm 1 2 50 cm 30 cm 20 cm Area of one triangular piece = s( s − a)( s − b)( s − c ) = 60(60 − 20)(60 − 50)(60 − 50) cm2 2 = 60 × 40 × 10 × 10 cm = 6 × 10 × 10 × 4 × 10 × 10 cm2 1 = 200 6 cm2 Area of 5 triangular pieces of one colour = (5 × 200 6 ) cm2 = 1000 6 cm2 1 Area of 5 triangular pieces of other colour = 1000 6 cm2. 31. Given : AD ⊥ CD and BC ⊥ CD AQ = BP and DP = CQ To prove : ∠DAQ = ∠CBP A B D P Q C Proof : AD ⊥ CD and BC ⊥ CD 1 ∴ ∠D = ∠C (Each 90°)  DP = CQ (Given) Adding PQ to both sides.p. BT (Given) Side DQ = Side CP (Proved above) 1 ∴ ∆ADQ ≅ ∆BPC (RHS) ∴ ∠DAQ = ∠CBP. in right triangles ADQ and BPC. Now. 1 qqq . b = 50 cm. a = 20 cm.c. (c.  Hyp.

...237237.237237237237............ (B) = 50 5×5×2 1 2 2 = × = 1 5 2 2 10 So.. (C) Put x + 1 = 0 ⇒ x = – 1 in given expression.) ½ 999x = 237 ½ 237 79 x = = ½ 999 333 6. (B) The degree of the polynomial 2x3 + x2 – 2x + 34 is 3.237237...237237237. ½ .. x4 – 125xy3 = x(x3 – 125y3) ½ 3 3 = x[(x) – (5y) ] = x(x – 5y) [x2 + (5y)2 + x × 5y] 1 = x(x – 5y) (x2 + 25y2 + 5xy).. ½ 1000x – x = (237.. Let x = 0...237 = 0.... 2... 1 3. SA-1 Examination Sample Question Paper Self Assessment__________________________________ Time : 3 Hours Maximum Marks : 90 SECTION ‘A’ 1 1 1... (B)  x + 2   x + 2  = x2 + x+ x+     2 2 4 3 = x2 + 2x + 1 4 SECTION ‘B’ 5. 1000x = 237..) – (0.... The remainder is : p(–1) = x31 + 31 = (– 1)31 + 31 = – 1+ 31 = 30 1  1  3 3 1 3 4.. rationalising factor is 2.. S O L U T I O N S SAMPLE QUESTION PAPER-10 MATHEMATICS Oswaal CBSE (CCE) Class -9.

then AC = AB. 2012] 9. ½ 8. ½ 4 [CBSE Marking Scheme.O. c = 12 cm a + b + c 15 + 15 + 12 s = = = 21 cm ½ 2 2 Area = s( s − a)( s − b)( s − c ) 21 × (21 – 15)(21 − 15)(21 − 12) = 1 = 21 × 6 × 6 × 9 2 = 18 21 cm . AC || DE. then given expression = 12p2 – 8pq – 15q2 = 12p2 – 18pq + 10pq –15q2 ½ = 6p(2p – 3q) + 5q(2p – 3q) ½ = (2p – 3q) (6p + 5q) = [2(x2 + 7) – 3(2x – 1)] [6(x2 + 7)+ 5(2x – 1)] ½ 2 2 = (2x + 14 – 6x + 3) (6x + 42 + 10x – 5) = (2x2 – 6x + 17) (6x2 + 10x + 37). AB || DE \ ∠ACB = ∠DEC (Corresponding angles) A x D Thus. a = 15 cm. (Corresponding angles) 1 y 55° B C OR E ∠DEC = ∠GEF = 80° (V.) ½ B D A ∠ACH = 80° 70° \ ∠DEC = ∠ACH = 80° ½ G E 80° But they are corresponding angles. ½ [CBSE Marking Scheme. y = 55° 1 70° ∠ABC = 180° – (70° + 55°) = 55° \ x = ∠ABC = 55°. b = 15 cm. Solutions | 47 Let x2 + 7 = p and 2x – 1 = q. ½ [CBSE Marking Scheme. 2012] . 2012] 10. Since. Euclid's axiom : If C be the mid point of a line segment AB.A. 1 AC = AB ½ 2 A D C B 1 AD = AC ½ 2 1 1  AD = AB  ½ 2  2  1 AD = AB. 7. ½ C H 80° F So.

[CBSE Marking Scheme. x = 3 + 2 2 1 1 = x 3+2 2 ½ 1 3−2 2 = × ½ 3+2 2 3−2 2 3−2 2 = = 3−2 2 9−8 3  1 3  x − x  = {3 + 2 2 − (3 − 2 2 )} 1   ( ) 3 = 4 2 ½ = 128 2 ½ OR x = 9 + 4 5 = 5 + 4 + 4 5 ( 5 ) + (2) 2 2 = +2×2× 5 x = ( 5 + 2 ) 2 x = 5 + 2 1 1 = 1 ( × 5 −2 ) x ( 5 + 2) ( 5 − 2) 5 −2 = = 5 −2 1 5−4 x− 1 x = ( 5 +2 − ) ( 5 −2 ) = 5 + 2 – 5 + 2 = 4 1 [CBSE Marking Scheme.48 | OSWAAL CBSE (CCE) Term-1. (1 + 2 ) ( + 2+ 3 + ) ( 3+ 5 ) 1 ( 2 − 1) 1 ( 3− 2 ) 2 ( 5 − 3) = × + × + × 1 ( 2 + 1) ( 2 − 1) ( 3+ 2 ) ( 3 − 2) ( 5 + 3) ( 5 − 3) = ( 2 − 1) + ( 3 − 2 ) + 2 ( 5 − 3 ) 1 2 −1 3−2 5−3 = 2 –1+ 3 – 2 + 5 – 3 ½ = 5 – 1. Mathematics. 2012] 1 1 2 12. 2012] ½ . Class – 9 SECTION ‘C’ 11.

2 2 z – 2x 1 = (– 2 x + y + 2 2 )2 = (– 2 x + y + 2 2 z) (– 2 x + y + 2 2 z) 1 (x2 – 4x)(x2 – 4x – 1) – 20 = (x2 – 4x)2 – (x2 – 4x) – 20 14. A B 45º xº 1 l m 2 30º ½ C D ∠1 = 45° (Alternate angles) ½ ∠2 = 30° (Alternate angles) ½ x = 360° – (∠1 + ∠2) ½ = 360° – (∠45° + ∠30°) ½ = 360° – 75° = 285° ½ OR ∠1 = 35° (Corresponding angles) ½ ∠QHP = 35° (V.y + 2. ½ = (x2 – 4x)2 –5(x2 – 4x) + 4(x2 – 4x) – 20 ½ = (x2 – 4x)(x2 – 4x –5) + 4(x2 – 4x – 5) ½ = (x2 – 4x –5)(x2 – 4x + 4) = (x2 – 5x + x – 5)[(x)2 – 2 × 2 × x + (2)2] ½ = [x(x – 5) + 1(x –5)][x – 2]2 ½ = (x – 5)(x + 1) (x – 2)(x – 2). Construction : 1 Draw l || AB. ½ 15. 2012] OR Identity 2x2 + y2 + 8z2 – 2 2 xy + 4 2 yz – 8xz 1 = (– 2 x)2 + y2 + (2 2 z)2 + 2(– ( ) 2 x).O. x + y + 4 = 0 We know that a3 + b3 + c3 – 3abc = (a + b + c) (a2 + b2 + c2 – ab – bc – ca) x3 + y3 – 12xy + 64 = (x3) + (y3) + (4)3 – 3 × (x) (y) (4) 1 2 2 = (x + y + 4) (x – y + 16 – xy – 4y – 4x) 1 2 2 = 0 × (x + y + 16 – xy – 4y – 4x) 1 = 0 [CBSE Marking Scheme.2 2 z + 2. Solutions | 49 13. Given.A.) ½ x° ½ .y.

∴ OA > OB .c. (By c. 2012] 18. Proof : In ∆PAM and ∆QBM AM = BM (Given) ½ ∠1 = ∠2 (V. Given. ∠B > ∠A and ∠C > ∠D.A.50 | OSWAAL CBSE (CCE) Term-1.) ½ ∠3 = ∠4 = 90° ∴ ∆PAM ≅ ∆QBM (AAS cong.) ∴ BD = AC.(1) 1 OD > OC (Side opposite to greater angle is longer) .. 1 [CBSE Marking Scheme.(2) 1 Adding 1 and (2). we get OA + OD > OB + OC AD > BC. ∠PQH + ∠QHP + ∠HPQ = 180° ½ ∠PQH + 35° + 90° = 180° ½ ∠PQH = 55° ½ [CBSE Marking Scheme.t. Mathematics......) 1 [CBSE Marking Scheme. 2012] 16..c. Class – 9 In ∆PQH. 2012] 17.p.p..) 1½ ∴ PA = BQ (By c. 2012] . BC = AD (Given) ∠CBA = ∠BAD (Given) AB = AB 1 ∴ ∆ABC ≅ ∆BAD (By SAS cong. Proof : In ∆ABC and ∆BAD...t) ½ 3 1 2 4 [CBSE Marking Scheme..O..

½ (iii) A balanced ratio leads to good result. c = 25k ½ Perimeter = 540 cm a + b + c = 540 12k + 17k + 25k = 540 54k = 540 540 k = = 10 ½ 54 Hence a = 12 × 10 = 120 cm. b and c. [CBSE Marking Scheme. ½ SECTION ‘D’ 1 × ( ) 2+ 3 + 4 = 2+ 3+ 4 = 2+ 3+ 4 1 21. b = 17k. Solutions | 51 19. ( 2+ 3 − 4) ( 2+ 3)+ 4 ( 2+ 3 ) 2 −4 (2 + 3 + 2 6 ) − 4 = ( 2+ 3+ 4 ) × (1 − 2 6 ) 1 1+ 2 6 (1 − 2 6 ) . (Angle sum property) ½ = 180° – 162° ½ = 18°. ½ (ii) Heron's formula. c = 25 × 10 = 250 cm 1 Now. 2012] ½ 20. (i) Let the sides of the triangle be a. s = × 540 = 270 cm 2 (s – a) = (270 – 120) cm = 150 cm (s – b) = (270 – 170) cm = 100 cm (s – c) = (270 – 250) cm = 20 cm ½ \ Area of triangle = s( s − a)( s − b)( s − c ) 2 = 270 × 150 × 100 × 20 cm 2 = 100 27 × 15 × 10 × 2 cm = 100 3 × 3 × 3 × 3 × 5 × 2 × 5 × 2 cm2 = 100 × 3 × 3 × 5 × 2 cm2 = 9000 cm2. a : b : c = 12 : 17 : 25 a b c = = = k (say) 12 17 25 ⇒ a = 12k. B D M ax Q y b P 88° 110° A C y = 88° (Corresponding angles) ½ a = 180° – 88° = 92° (linear pair) ½ b = 180° – 110° = 70° (linear pair) ½ x = 180° – (92 + 70)°. b = 17 × 10 = 170 cm.

. Mathematics.52 | OSWAAL CBSE (CCE) Term-1.(i) 2 1 2 5 + 21 = × 1 x 5 − 21 5 + 21 = 2 5 + 21 = 5 + 21 ( ) ... (iv) and (v).. we get 1 5 − 21 5 + 21 x+ = + = 5 .(iii) x 2 2 Squaring both sides of (iii).(iv) 1 x2 x x Raising to power 3 on both sides of (iii). 1 . Class – 9 2 + 3 + 4 − 2 12 − 2 18 − 4 6 = ½ ( ) 2 12 − 2 6 2 + 3 + 4 −4 3 −6 2 −4 6 = ½ 1 − 24 −5 2 − 3 3 – 4 6 + 2 = ½ −23 +5 2 + 3 3 + 4 6 − 2 = ½ 23 OR 5 − 21 x = . we get 1 1  1 x3 + + 3(x)    x +  = (5)3 x 3 x  x 1 ⇒ x3 + + 3(5) = 125 (using eqn..(ii) 25 − 21 2 Adding (i) and (ii). (iii).. we get 1 1 1 x2 + +2×x× = 25 ⇒ x2 + 2 = 23 ..(v) 1 x3 Using eqns.. we get  3 1   2 1   1  x + 3  − 5  x + 2  +  x +  = 110 – 5(23) + 5  x   x   x = 115 – 115 = 0.. iii) x3 1 ⇒ x3 + = 110 ..

(i) p(–1) = 0 1 3 2 ⇒ a(– 1) + (–1) – 2(–1) + b = 0 = – a + 1 + 2 + b = 0 1 b = a – 3 . we get 3n – 3m = – 6 ½ – 3(m – n) = – 6 m – n = 2. ½ 3 2 p(1) = 0 ⇒ a(1) + (1) – 2(1) + b = 0 ⇒ a + 1 – 2 + b = 0 ⇒ a + b = 1 . b = –1 ½ 3 2 2 2x + x – 2x – 1 = (x – 1) (2x + 1) Third factor is 2x + 1. . p(y) = y3 – 2y2 – 29y – 42 p(–2) = (–2)3 – 2(–2)2 –29 (– 2) – 42 = – 8 – 8 + 58 – 42 = – 58 + 58 = 0 ∴ y + 2 is a factor of the given polynomial 1 2 p(y) = (y + 2) (y – 4y – 21) 1 g(y) = y2 – 4y – 21 1½ = (y – 7) ( y + 3) ½ ∴ The factors are (y + 2)... 25. ½ [CBSE Marking Scheme.. (y + 3). Solutions | 53 9 n + 1 × ( 3− n/2 ) − 27 n −2 1 22. 1 = 85 + 2(36) 1 = 85 + 72 = 157 1 (a + b + c) = ± 157 1 3 2 24.. = ( 3m × 2 ) 3 729 2 n +1 n 3 n (3 ) × 3 − (3 ) 1 3m 3 = 6 1 3 ×2 3 32 n + 2 + n − 33 n 1 3m = 6 1 3 ×8 3 33 n [32 − 1] 1 3m = 6 ½ 3 ×8 3 8 33n – 3m × = 3–6 ½ 8 Comparing powers on both sides. p(x) = ax + x – 2x + b.(ii) ½ 2a – 3 = 1 2a = 4 ½ a = 2. 2012] (a + b + c)2 = a2 + b2 + c2 + 2ab + 2bc + 2ca 23. (y – 7).

3) 3 2 (4. Class – 9 26. 1 27.1) 1 3 2 1 x' x 1 2 3 4 5 6 –1 B(1. x = 42° (Alternate angles) 1 ∠1 = 24° + 42°= 66° (Exterior angles) ½ A B x y 1 E z 24º 42º C D y = ∠1 = 66° (∴ CE = AC ) 1 z = 180° – (x + y) = 180° – 132° 1 = 48°.54 | OSWAAL CBSE (CCE) Term-1.3) D(7.–1) C(7. to greatest angle is the longest) 1 Also in DABC. ½ 28. AD = BD ⇒ ∠ABD = ∠DAB = 59° (Angles opp. AB > BD (Side opp. The given figure ABCD is a rectangle. 59° + 59° + ∠ADB = 180° ∠ADB = 180° – 118° = 62° 1 ∠ACD = 62° – 32° = 30° (Exterior angle is equal to the sum of interior opposite angles) In DABD. AB < AC ⇒ BD < AC 1 . 1). Mathematics. y 4 A(1.–1) –2 –3 y' 3 It is clear from the graph that co-ordinate of the point of intersection of the diagonals is (4. to equal sides are equal) 1 In DABD.

p.p.) ½ .p.c.O.p.t. ½ In DEAD and DECD.) Proved. AD = CD (Given) ∠ADE = ∠CDE = 90° ½ and DE = DE (Common) So.t. Solutions | 55 29.) ½ ∠3 + ∠4 = 180° \ ∠4 = 90° \ ∠ADE = ∠4 (V.t. ∠DAB = ∠CBA (Given) AB = AB (Common) 1 AD = BC (Given) ∴ ∆ABD ≅ ∆BAC (By SAS) Proved. (By c. DEAD ≅ DECD (By SAS cong. E D C A 1 2 3 4 B In ∠ABC. 1 (iii) ∠ABD = ∠BAC. (i) In ∆ABD and ∆BAC. 1 [CBSE Marking Scheme. (By c. 1 (ii) ∴ BD = AC (By c.) ∠3 = ∠4 (By c.c.t.) Proved. AB = BC ∠1 = ∠2 AD = CD 1 \ DABD ≅ DCBD (By SAS cong. AB = BC (Given) 1 \ ∠1 = ∠2 (Angles opposite to equal sides) In DABD and DCBD.A) \ ∠ADE = 90° Proved.c.c.) ⇒ AE = CE. 2012] 30.

a = 28 cm.45 cm2 (approx.) 1 Area of 16 tiles = (88.56 | OSWAAL CBSE (CCE) Term-1. Mathematics. Class – 9 31.2 cm2 1 1 Cost of polishing = ` 1411.2 × 16) cm2 = 1411. For one triangular tile.2 × (as 50 paise = ` ½) 2 = ` 705.) 1 qqq . c = 35 cm B 28 cm 9 cm 35 cm A 28 cm a+b+c s = 2 28 + 9 + 35 = = 36 cm 2 Area of one tile = s( s − a)( s − b)( s − c ) 1 36(36 − 28)(36 − 9)(36 − 35) cm2 = = 36 × 8 × 27 × 1 cm2 = 6 × 6 × 2 × 2 × 2 × 3 × 3 × 3 cm2 = 6 × 2 × 3 6 cm2 = 36 6 cm2 = 36 × 2.60 (approx. b = 9 cm.