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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)

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

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. 2012] OR 1 x+ =5 x On squaring both sides. 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. 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] SECTION ‘D’ 2− 5 2+ 5 21.

216x3 – 125y3 = (6x)3 – (5y)3 = (6x – 5y) [(6x)2 + (5y)2 + 6x × 5y] = (6x – 5y) (36x2 + 25y2 + 30xy). ½ 3 3 3 3 3 25. 1½ 27. 2 [C. 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. ½ 26. 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.B. (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). Class – 9 22.8 | OSWAAL CBSE (CCE) Term-1. RHS = (x – y) (x2 + y2 + xy) = x3 + xy2 + x2y – x2y – y3 – xy2 = x3 – y3 = LHS 2 Now. 2012] 23.E.S. (i) 2½ (ii) Graph is a straight line. ∠PYZ + ∠ZYX = 180° ⇒ 2x + 64° = 180° ⇒ x = 58° 1 1 ∠XYQ = 64° + 58° = 122° 1 Reflex ∠QYP = 360° – x = 360° – 58° = 302° 1 . Marking Scheme. Mathematics.

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

Class – 9 30.5 − 1) cm2 .5 cm2 2 = 4.5 − 5)(5. c = 1 cm a+b+ c 5+5+1 s = = cm 2 2 = 5. a = 5 cm. ∠M = ∠1 + 70° 1 90° = ∠1 + 70° ∠1 = 20° ∠1 + ∠2 = ∠3 (AN is angle bisector ) 1 20° + ∠2 = ∠3 In ∆AMC.5 cm.5 − 5)(5. 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 × 1.5 cm2 For (Shade 1). In ∆ABM. Mathematics. Ext. ∠MAN = 25° 1½ 31.5(5.10 | OSWAAL CBSE (CCE) Term-1.5 cm2 1 For (Shade II). Area = s( s − a)( s − b)( s − c ) = 5. 6.5 × 1 = 6. ∠AMC + ∠ACM + ∠2 + ∠3 = 180° 90° + 20° + ∠2 + 20° + ∠2 = 180° ½ 2∠2 = 50° ∠2 = 25°. b = 5 cm.

5 + 4.732 = 1.5 × 0. ∆ADM.5 + 6.3 cm2 Area of total paper used = (2.866 cm2 = 1.5 × 0.5 cm 2 = 11 × 0.32 cm2 = 2.5 cm For area (Shade III).5 × 9 cm = 0.3 cm2 1 qqq . ∆ADM In 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 1.75 × 3.5 × 0.5 × 0.49 cm2 1 2 = 2.5 ) cm2 = 19.5 × 3 11 cm2 = 0.5 + 1.5 × 0.5 × 0. 2 1 3 h2 = 12 –   = 2 4 3 ∴ h = cm.5 × 4. consider rt.5 × 0.5 × cm2 1 2 1.3 + 4. Solutions | 11 2 = 5.

M. 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. L. < < < < ½ 35 35 35 35 35 . (D) (2x2 – 5) (4 + 3x2) = 8x2 + 6x4 – 20 – 15x2 = 6x4 –7x2 – 20 Hence. of 5 and 7 is 35. (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) 2+ 5 = +2 = 2 + 5 + 2 10 = 7 + 2 10 = Irrational number. 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 3. coefficient of x2 is –7. 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.C. 1 SECTION ‘B’ 5.

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

Mathematics. 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..14 | OSWAAL CBSE (CCE) Term-1. 2012] OR We know that ∠A + ∠B + ∠C = 180° . 40° + ∠B + 115° = 180° ∠B = 25° ½ [CBSE Marking Scheme. 2012] 10.. Class – 9 [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 . Given..(i) 65° + ∠C = 180° 1 ∠C = 115° Again by (i). ∠A + 140° = 180° ∠A = 40° ½ Again by (i).

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. 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] 13. 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. 2012] 12. 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. Given.

Mathematics.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. ∠APD + ∠DAP + ∠ADP = 180° ∠A ∠D ∠APD + + = 180° 1 2 2 . ½ 15.16 | OSWAAL CBSE (CCE) Term-1. Class – 9 10  100 + 405  = 9  81  5050 = = 6.927 1 729 [CBSE Marking Scheme.O. 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. ∠1 = ∠3 = 2x + 90° (V. 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.

.(i) 1 Given that. we get ∆ABC ≅ ∆AED (By AAS rule) 1 17. 2011. 2∠APD = ∠B + ∠C.(i) ½ By ASP.. 2010. A 1 2 B D C E Let ∠DAC be ∠3 ∠1 = ∠2 (Given) ∠1 + ∠3 = ∠2 + ∠3 ∠BAC = ∠EAD .(ii) ½ On comparing (i) and (ii). BD = CE BD + DC = CE + DC BC = DE . ½ [CBSE Marking Scheme. Proved.(iii) From (i)... ∠1 = ∠2) ½ ∴ ∠ADC > ∠ADB (Exterior angle property of triangle) 1½ .. ∠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) .. ∠B + ∠C = 360° – (∠A + ∠D) ... (ii) and (iii).(ii) 1 ∠B = ∠E (Given) . 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.. 2012] 16.. Solutions | 17 1 Given.

Produce CB to a point D such that BC = BD and join AD..(iii) 1 and ∠ACD = ∠ADB = 2x [From (ii). OP ⊥ AB ⇒ ∠POA = 90° ½ Let x = a..p.) .e.(i) and AC = AD .18 | OSWAAL CBSE (CCE) Term-1. Class – 9 18. ∠CAB = ∠DAB (c... Mathematics. y = 3a. 1 19. ∆ACD is an equilateral triangle.. A x D B C In ∆ABC and ∆ABD.e. i.(iv) i..(ii) Thus. 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 .t. AC = 2BC (Since BC = BD). AC = AD] . ∆ABC ≅ ∆ABD (SAS rule) So..c. [From (iii) and (iv)] or AC = CD.. ∠CAD = ∠CAB + ∠BAD = x + x = 2x [From (i)] ..

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. 2012] SECTION ‘D’ 21.142857. ½ 7 ) 1. Solutions | 19 20.S. Marking Scheme. a = 120 m. 0. We have..E.00000000 7 30 ..B. 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. b = 170 m. ½ [C. b = –5 1 22.

We can take the point C(2.142857 7 7 = 0. x + y = p. Mathematics. –1) and B(4. 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).571428 ½ 5 1 = 5 × = 5 × 0. (i) Plot the points A(1. . Let.285714 ½ 3 1 = 3 × = 3 × 0. (z + x) – (z + x). y + z = q.714285 ½ 3 2 3 2 23. 1) between the points A and B. we get the line segment AB. 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. Joining these points. (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. without actually doing the long division 7 7 7 7 7 as follows : 2 1 = 2 × = 2 × 0. Class – 9 28 20 14 60 56 1½ 40 35 50 49 1 2 3 4 5 6 Yes.428571 ½ 4 1 = 4 × = 4 × 0. . 5). . . 1 .142857 7 7 = 0.20 | OSWAAL CBSE (CCE) Term-1. 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 26.142857 7 7 = 0. we can predict the decimal expansions of .

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

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

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

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. 1 qqq . Mathematics.24 | OSWAAL CBSE (CCE) Term-1.

– 0. (a2 + a + 1) 1 3..4152 1 ⇒ 625 [CBSE Marking Scheme.. 2012] ...777. 10x = 7.777.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.. (C) Let x = 0... 1 SECTION ‘B’ 2157 5. (A) x = 360 – (120 + 140) = 360 – 260 = 100 4.8 MATHEMATICS Oswaal CBSE (CCE) Class -9. SA-1 Examination Sample Question Paper Self Assessment__________________________________ Time : 3 Hours Maximum Marks : 90 SECTION ‘A’ 1.777. (D) a3 – 1 = (a – 1). = 625 2157 × 16 = 625 × 16 34512 = 10000 2157 = 3. 10x – x = 7.....777. 9x = 7 7 x = 1 9 2.

2012] 7. Given. 1 13 . a = 5 cm. 2012] 10. 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°. 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. Class – 9 6. 249 × 251 = (250 – 1) × (250 + 1) 1 2 2 = (250) – (1) = 62500 – 1 = 62499 1 [CBSE Marking Scheme. ½ 3 1 ∠2 = ∠1 + of right angle 3 [CBSE Marking Scheme. (7p)3 – (11b)3 = (7p – 11b)[(7p)2 + (7p)(11b) + (11b)2] 1½ 2 2 = (7p – 11b)(49p + 77pb + 121b ) ½ 8. AE = DF (Given) ∴ AB = CD (Things which are double of the same thing are equal to one another) 1 9.26 | OSWAAL CBSE (CCE) Term-1. AB = 2AE (E is the mid point of AB) ½ CD = 2DF (F is the mid point of CD) ½ Also. b = 12 cm. Area = 30 1  Area = × 13 × h = 30 2 2 × 30 ∴ h = 13 60 = cm. Mathematics.

..000 ½ 325 65 x = = .2828..  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.. Solutions | 27 SECTION ‘C’ — 11.2828..  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. 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 .282828. ½ 1000x – 10x = 328. ½ 990x = 325.3 28 = 0.. ½ 1000x = 328. 2012] 1 1 −   1 1   4 2    1 2  1    4    1  – 3  1  – 3       1 3× − 3   1 3× − 3    12. Given... Let x = 0.282828.. – 3..... ½ 990 198 [CBSE Marking Scheme. x + =3 x 3  1 3 1  1  x +  = x + 3 + 3 x +  ½  x x  x 3  1 3 But.3282828 ½ 10x = 3..

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

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

 a  x 2  2  2    x  x  1 1 1 = ( x ) . Mathematics. 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 . 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. OR 1 1 1  xa  a+b  xb  b+c  xc  c+a 2 2 2  b  .  c  .30 | OSWAAL CBSE (CCE) Term-1.x 1 = xa–b. xb–c.x .( 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 . 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. xc–a ½ = xa–b+b–c+c–a 1 0 = x = 1 ½ 1 (2 + 3) 22.( a 2 – b2 a + b x ) .

. 0). (i) Draw X′OX and Y′OY as the co-ordinate axes and mark their point of intersection O as the origin (0. Hence. 1 [CBSE Marking Scheme. 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 26. 2 Hence. 2012] 23. then p(–1) = (–1)3 + (–1)2 – 4(– 1) – 4 = – 1 + 1 + 4 – 4 = 0. x3 + x2 – 4x – 4 = (x + 1) (x2 – 4) 1 = (x + 1) (x + 2) (x – 2). (x + 1) is a factor of this polynomial. 1 2 24. (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. x3 + x2 – 4x – 4 Now put x = – 1. 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 . 1 25.

we take 2 units on OX′ and then 8 units parallel to OY to obtain the point A(– 2.25).7) 6 5 4 3 D(1.32 | OSWAAL CBSE (CCE) Term-1. – 1). to equal sides of a triangle are equal) Also. we take 1 unit on OX and then 3 units parallel to OY to obtain the point D(1. ½ (iii) Co-ordination among people is good for progress. ½ 27. ½ In order to plot (0. we take 3 units on OX and then 1 unit below x-axis parallel to OY′ to obtain the point E(3. –1.8) 8 B 7 (–1. –1) ½ (ii) Co-ordinate geometry. 8). ∠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 . 3) ½ In order to plot (3.3) 2 1 1 2 3 4 5 6 7 8 x' x –5 –4 –3 –2 –1–1O E(3. Mathematics. ∠A = 90° AB = AC ∠B = ∠C ( Q Angles opp. 8).–1) C –2 (0. 7).–1.25 units below x-axis on the y-axis to obtain C(0.25) ½ In order to plot (1. we plot the point B(–1. –1. Class – 9 y A(–2. A 90° B C 1 We have.25) –3 –4 –5 y' ½ In order to plot the points (– 2. we take 1. ½ Similarly. 3).

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

Mathematics. 000 × 1320 ×  m  12  = ` 16. Class – 9 31. 1 qqq .34 | OSWAAL CBSE (CCE) Term-1.50. b = 22 cm.000. 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.  3  2 Total rent for 3 months = `  5. Let a = 122 cm.

the remainder is f(– 1) = (–1)11 + 101 1 = –1 + 101 = 100.707 + 3. 4. SA-1 Examination Sample Question Paper Self Assessment___________________________________ Time : 3 Hours Maximum Marks : 90 SECTION ‘A’ 1.707 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. (C) (x. 1 2 2 3.9 MATHEMATICS Oswaal CBSE (CCE) Class -9.414 == = 0. (D) Put x + 1 = 0 ⇒ x = –1 in given expression. 1 SECTION ‘B’ 1 1 2 5. x– = 3 x Squaring both sides. 0) 1 π π 2. 2012] 1 6. (D) Constant polynomal is 7.S O L U T I O N S SAMPLE QUESTION PAPER . (D) Coefficient of x in the expression x2 + x – 7 is . = × (Rationalising) ½ 2 2 2 2 1.848 1 2 [CBSE Marking Scheme.141 = 3. ½ x .

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

2012] . = 1 3x 2 2 x 2 4 2x–2x.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.   . we get 4q2x2 + p2 – 4pqx = p2 – 4q2 ½ 2 2 4q(qx – px) = –4q ½ 2 qx – px + q = 0.  = 1 3 2 16 2 x 32 x 34 . 2012] SECTION ‘C’ x 2x 2 3 81 11. ½ [CBSE Marking Scheme. ½ [CBSE Marking Scheme. Solutions | 37 3 2 10. 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.

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. 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).38 | OSWAAL CBSE (CCE) Term-1.(ii) ½ Since remainders are same.. Mathematics. 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 . we have. (i) and (ii).. then from eqs..(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 . 2012] 13. Class – 9 12.. . Let p(x) = 3x3 + x2 – 20x + 12 Given (3x – 2) is a factor of p(x).

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). Given AB || CD. AB = AC ⇒ = ½ 2 2 ⇒ AE = AF. CD || EF. Solutions | 39 15. 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°. A E F B C . 2012] AB AC 16.

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

3 sq. =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 19.732 1 = 43.cm ½ 2 ∴ Area of unshaded portion = 43.3 – 24 = 19. 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°. 2012] 20.3 sq. 1 [CBSE Marking Scheme. cm ½ SECTION ‘D’ 1 21. y – x = 50° (½+ ½) ∴ (x + y) : (y – x) = 21 : 5. cm In ∆BDC. 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.

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 5.2 + 2. quotient = 2x3 + x2 + 3x – 5 and remainder = 47. 2012] (1 + x + y–1)–1 + (1 + y + z–1)–1 + (1 + z + x–1)–1 22.2) = = 5.4 1 5 [CBSE Marking Scheme. 1 4 3 2 Verificaiton : Let p(x) = 6x + 11x + 13x – 3x + 27 Zero of 3x + 4 is – 4/3 . Mathematics. 2012] 23. 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 . 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.42 | OSWAAL CBSE (CCE) Term-1.(3. 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.

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. Solutions | 43 By remainder theorem. (i) Plot the point M(5. 2012] 25. 2012] 26. 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. ½ 2 2 2 2 24. 1 [CBSE Marking Scheme. 2 . – 3) on the graph paper. 1 19 [CBSE Marking Scheme. (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). 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. q – p = 10 \ – 15 + 9a – 8 + 10a = 10 19a – 23 = 10 19a = 33 33 a = .

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

For one triangular piece. AQ = Hyp. we get DP + PQ = PQ + CQ 1 ⇒ DQ = CP. (c. 31.  Hyp. BT (Given) Side DQ = Side CP (Proved above) 1 ∴ ∆ADQ ≅ ∆BPC (RHS) ∴ ∠DAQ = ∠CBP. 1 qqq .) 1 Proved.t.c. a = 20 cm. Solutions | 45 30. b = 50 cm.p. Now. 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. in right triangles ADQ and BPC. 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.

.. 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)...... (C) Put x + 1 = 0 ⇒ x = – 1 in given expression.237237237237.) – (0........ 1000x = 237. SA-1 Examination Sample Question Paper Self Assessment__________________________________ Time : 3 Hours Maximum Marks : 90 SECTION ‘A’ 1 1 1....) ½ 999x = 237 ½ 237 79 x = = ½ 999 333 6. S O L U T I O N S SAMPLE QUESTION PAPER-10 MATHEMATICS Oswaal CBSE (CCE) Class -9.237237..237237237. The remainder is : p(–1) = x31 + 31 = (– 1)31 + 31 = – 1+ 31 = 30 1  1  3 3 1 3 4. (B) The degree of the polynomial 2x3 + x2 – 2x + 34 is 3...237 = 0. rationalising factor is 2. ½ ... 1 3..237237. Let x = 0...... (B) = 50 5×5×2 1 2 2 = × = 1 5 2 2 10 So.. ½ 1000x – x = (237.... 2.... (B)  x + 2   x + 2  = x2 + x+ x+     2 2 4 3 = x2 + 2x + 1 4 SECTION ‘B’ 5.

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

[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. 2012] ½ . (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. Class – 9 SECTION ‘C’ 11.48 | OSWAAL CBSE (CCE) Term-1. Mathematics. 2012] 1 1 2 12.

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.O. Construction : 1 Draw l || AB.y.) ½ x° ½ . 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). Solutions | 49 13.y + 2. Given. ½ 15. 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. 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).2 2 z + 2.A. 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.

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

2012] ½ 20. [CBSE Marking Scheme. 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 = 17 × 10 = 170 cm. ½ (ii) Heron's formula. ( 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 ) . 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)°. a : b : c = 12 : 17 : 25 a b c = = = k (say) 12 17 25 ⇒ a = 12k. Solutions | 51 19. c = 25 × 10 = 250 cm 1 Now. b = 17k. ½ (iii) A balanced ratio leads to good result. 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. (Angle sum property) ½ = 180° – 162° ½ = 18°. b and c. ½ SECTION ‘D’ 1 × ( ) 2+ 3 + 4 = 2+ 3+ 4 = 2+ 3+ 4 1 21. (i) Let the sides of the triangle be a.

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

= ( 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(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). Solutions | 53 9 n + 1 × ( 3− n/2 ) − 27 n −2 1 22. p(x) = ax + x – 2x + b..(i) p(–1) = 0 1 3 2 ⇒ a(– 1) + (–1) – 2(–1) + b = 0 = – a + 1 + 2 + b = 0 1 b = a – 3 . (y – 7). . ½ [CBSE Marking Scheme. 2012] (a + b + c)2 = a2 + b2 + c2 + 2ab + 2bc + 2ca 23. (y + 3). ½ 3 2 p(1) = 0 ⇒ a(1) + (1) – 2(1) + b = 0 ⇒ a + 1 – 2 + b = 0 ⇒ a + b = 1 .(ii) ½ 2a – 3 = 1 2a = 4 ½ a = 2. 25.. 1 = 85 + 2(36) 1 = 85 + 72 = 157 1 (a + b + c) = ± 157 1 3 2 24. b = –1 ½ 3 2 2 2x + x – 2x – 1 = (x – 1) (2x + 1) Third factor is 2x + 1.. we get 3n – 3m = – 6 ½ – 3(m – n) = – 6 m – n = 2.

3) 3 2 (4.54 | OSWAAL CBSE (CCE) Term-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.–1) C(7. ½ 28. 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°. The given figure ABCD is a rectangle. AB > BD (Side opp. AB < AC ⇒ BD < AC 1 .3) D(7. 1).1) 1 3 2 1 x' x 1 2 3 4 5 6 –1 B(1. 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. AD = BD ⇒ ∠ABD = ∠DAB = 59° (Angles opp. Mathematics. y 4 A(1. Class – 9 26. 1 27. to equal sides are equal) 1 In DABD. to greatest angle is the longest) 1 Also in DABC.

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

2 × 16) cm2 = 1411.2 cm2 1 1 Cost of polishing = ` 1411.60 (approx. Mathematics. 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 × 2.45 cm2 (approx.56 | OSWAAL CBSE (CCE) Term-1.) 1 qqq . Class – 9 31.2 × (as 50 paise = ` ½) 2 = ` 705. a = 28 cm. For one triangular tile. b = 9 cm.) 1 Area of 16 tiles = (88.