SOLUTIONS

SECTION – A

1. (C)

Roots=-2,-3

We know that, sum of roots=-b/a

=>(-2)+(-3)=-k

=>k=5 (1)

2. (D)

The remainder theorem tells us that if a polynomial p(x) is divided by (x-a) then the

remainder is equal to p(a).

So, remainder when p(x) is divided by (x-2) is p(2) and when q(x) is divided by x-1 is

q(1).

p(2) = 32-24+8+4-1 = 19

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

So, 2 x [sum of the remainders] = 2 x [19+1] = 2 x [20] = 40

(1)

3. (B)

Two triangles will be congruent by SAS axiom if 2 sides and the included angle of

one triangle are equal to the two sides and the included angle of the other triangle.

So,AC=DE. (1)

4. (C)

According to 5th postulate of Euclid “if the sum of the two angles marked is less than

180 degrees then lines will eventually intersect on that side on which the sum of angles is less than two right angles.

Here the sum of the two interior angles is 120o which is less than two right angles.

Hence, the two lines will meet on the side where the sum if equal to 120o. (1)

5. (C)

The area of a triangle by heron’s formula = s(s a)(s b)(s c)− − − (1/2)

And area of a triangle = ½ × Base × Altitude.

1s(s a)(s b)(s c) Base Altitude

2

a b c 6 8 10s 12

2 2

112(12 6)(12 8)(12 10) 8 Altitude

2

− − − = × ×

+ + + += = =

− − − = × ×

Therefore, Altitude = 6 cm (1/2)

6. (C)

Apply the formula of (a + b)2

2( 5 6) 5 6 2 30

11 2 30

On compairing a b 30 & 11 2 30

+ = + +

= +

+ +

We get,

A = 11 & b = 2 (1)

7. (D)

s = perimeter/2 =150

Perimeter= 300 = 15x

x = 20

Then, a = 2x20=40

b = 6x20=120

c =7x20=140

2

Area of triangle s(s a)(s b)(s c)

150(150 40)(150 120)(150 140)

150(110)(30)(10) 300 55 cm

= − − −

= − − −

= =

(1)

8. (C)

2

2

x +9x+14

=x +7x+2x+14

=x(x+7)+2(x+7)

=(x+2)(x+7)

So the dimensions are (x+2) and (x+7).

So, by substituting x=2,

We get the dimensions to be 9 and 4.

(1)

SECTION – B

9. ( )23 343

= (343)-2/3 (½)

= [(7)3}-2/3 (½)

= 3 2 /37× −

(½)

= 27− = 1

49 (½)

10. ∠ SPR + ∠ QPR = 180o (Linear Par) ⇒ ∠ 130o + ∠ QPR = 180o

= ∠ QPR = 450o (1)

In ∆ PQR ∠ TQP = ∠ QPR + ∠ PRQ [Exterior angle property] 110 = 45 + ∠ PRQ ⇒ ∠ PRQ = 110 – 45 = 65o (1)

11. 7 2 x2 – 10x - 4 2

⇒ 7 2 x2 – 14x + 4x - 4 2 (1)

= 7 2 (x - 2 ) + 4 (x - 2 ) (1/2)

= (7 2x 4)(x 2)+ − (½)

12. (a + b +c)2 = a2 + b2 + c2 +2 (ab + bc + ca) (½)

(7)2 = a2 + b2 + c2 + 2(20) (½)

49 = a2 + b2 + c2 + 40 (½)

∴ a2 + b2 + c2 = 49 – 40 = 9 Ans (½)

13. Since AB || CD and EF is the transversal

∴ ∠ BPQ + ∠ PQD = 180o [Consecutive interior angles are supplementary] (1)

= 5x – 20 + 2x – 10 = 180o (1/2)

= 7x – 30 = 180o

= x = 30o ans (½)

OR

In ∆ AOB ,∠ B < ∠ A (given)

OA < OB (sides opposite to larger angles in longer)…..(i) (½)

In ∆DOC

∠ C < ∠ D

⇒ OD < OC …(ii) (½)

Adding (i) and (ii)

OA + OD < OB + OC (½)

⇒ AD < BC (½)

14. (i) Coordinates of A are (-7,3) (½)

(ii) Abscissa of point D is 4. (½)

(iii) Point is B. (½)

(iv) Coordinates of C are (-3,-2) (½)

Section – C

15.

.

( )

( )

x 3 8

1 1

x 3 8

= +

=+

( )

( )

( )

( )

( )

( )( )

( )

( ) ( )

22

2

2

2 2

2

2

3 81 1

x 3 8 3 8

3 8 3 83 8 (1Mark)

9 83 8

1x 3 8 3 8 6

x

1x 6 36 (1Mark)

x

1 1 1 1x x 2x. x 2

x x xx

36 2 34 (1Mark)

−= ×

+ −

− −= = = −

− −

+ = + + − =

+ = =

+ = + − = + −

= − =

OR

Let x = 5.347 = 5.34747 …(1) (½)

Multiplying (i) by 10,we get

10x = 53.4747 …(ii) (½)

Multiplying (ii) by 100, we get

1000x=5347.47…..(iii) (1/2)

Subtracting (ii) from (iii),we get

1000x – 10x = 5347.47 … - 53.47… (½)

990x = 5294

x = 5294 2647

990 495= (½)

∴ 5.347 = 2647

ans495

(½)

16. Let a = x – 3y, b = 3y – 7z, c = 7z – x (½)

a + b + c = x – 3y + 3y – 7z + 7z – x = 0 (½)

Since a + b + c = 0. Therefore,a3 + b3 + c3 = 3abc (1)

Hence (x – 3y)3 + (3y – 7z)3 + (7z – x)3 = 3(x – y) (3y – 7z) (7z – x) ans (1)

17. 3 3 3

2 2a 8b 27c 18 2 abc+ − +

= ( ) ( ) ( ) ( ) ( ) ( )3 3 3

2a 2b 3 3 2a 2b 3c+ + − − − (1)

= ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( )2 2 2

2a 2b 3c 2a 2b 3c 2a 2b 2b 3c ( 3c) 2a + − + + − − − − − −

(1)

∵ a3 + b3 + c3 – 3abc = (a + b + c) (a2 + b2 + c2 – ab – bc – ca) (½)

= ( ) ( )2 2 22a 2b 3c 2a 4b 9c 2 ab 6bc 3 2 ca+ − + + − + + Ans (½)

18. Proof : AB = AC (Given)

AD = AB (Given)

∴ AD = AC [Line segments equal to same line segments are equal] (½)

In ∆ ABC AB = AC ⇒ ∠ B = ∠ ACB –---(i) [angle opposite equal sides are equal ] (½)

In ∆ ACD AC = AD ⇒ ∠ D = ∠ ACD – (ii) [angle opposite equal sides are equal] (½)

Adding (i) & (ii) we have

∠ B + ∠ D = ∠ ACB + ∠ ACD = ∠ C …(iii) (½)

In ∆ BCD ∠ B + ∠ C + ∠ D = 180o (Angle sum property) ⇒ ∠ C + ∠ C = 180o

⇒ 2∠ C = 180o

⇒ ∠ C = 90o (1)

19. In ∆ ADC, AD = AC (Given) => ∠ ADC = ∠ ACD – (i) [angle opposite equal sides are equal] (½)

∠ ADC > ∠ ABD – (ii)[exterior angle is greater than interior opposite angle] (½) From (i) and (ii)

∠ ACD > ∠ ABD (½)

i.e ∠ ACB > ∠ ABC (½)

i.e In ∆ ABC ∠ C > ∠ B ⇒ AB > AC [ sides opp greater angle is longer] (½)

⇒ AB > AD [∵ AC = AD] (½)

20. Area of triangle

a = 15, b = 14, c = 13

s = a b c 15 14 13

21 cm2 2

+ + + += = (½)

Area of ∆ = ( ) ( ) ( )s s a s b s c− − − (½)

= 21 6 7 8× × ×

= 84 cm2

By the given condition

Area of parallelogram = area of triangle

base × height = 84 (½)

15 × height = 84

Height = 84

5.6 cm15=

∴ Height of parallelogram is 5.6 cm (½)

21. In ∆ XYZ ∠ X + ∠ Y + ∠ Z = 180o 72o + ∠ Y + 46o = 180o (1)

∠ Y = 62o

∠ OYZ = 1

2 ∠ Y [ OY is bisec tor of Y∠∵ ]

∴ ∠ OYZ = 1

2 × 62o = 31o (½)

∠ OZY =1

2 ∠ Z [∵ OZ is bisector of Z∠ ]

= o o1

46 232× = (½)

In ∆OYZ ∠ OYZ + ∠ YOZ + ∠ OZY = 180o [Angle sum property ] 31o + ∠ YOZ + 23o = 180o (1)

∠ YOZ = 126o OR

∠ DAC = ½ ∠ A [∵ AD is bisector of ∠ A]

1OAC A

2⇒ ∠ = ∠ --- (i) [∵ ∠ OAC = ∠ DAC] (½)

∠ ECA = 1

C2∠ [∵ CE is angle bisector of C∠ ]

⇒ ∠ OCA = 1

C2∠ ---(ii) [ OCA ECA∠ = ∠∵ ] (½)

In ∆ ABC ∠ A + ∠ B + ∠ C = 180o (Angle sum property) (½)

∠ A + ∠ C + 90o = 180o ∠ A + ∠ C = 90o (½)

o1 1A C 45

2 2∠ + ∠ = --- (iii)

oOAC OCA 45⇒ ∠ + ∠ = --- (iv)

In ∆OAC ∠ AOC + ∠ OAC + ∠ OCA = 180o (Anre sum property ) ∠ AOC + 45o = 180o ∠ AOC = 135o ans (½)

22. In rt BEC & CFB∆

∠ BEC = ∠ CFB (90o each)

BE = CF ( given)

BC = BC (Common)

( )BEC CFB RHS congruence∴ ∆ ≅ ∆ (½)

∠ BCE = ∠ CBF (CPCT) (½)

=> ∠ BCA = ∠ CBA (same angle) (½)

In ∆ ABC ∠ BCA = ∠ CBA Proved

AB = AC [sides opposite equal angles are equal] (½)

23.

11 7 11 7 11 7

11 7 11 7 11 7

11 7 2 77

11 7

18 2 77

4

9 177 (1Mark)

2 2

+ + += ×

− − +

+ +=

−

+=

= +

Comparing with a - 77 b

a = 9

2 , b =

1

2 (2 Marks)

OR

Consider

2 5a 3 b 5

3 5 3 5+ = +

+ −

( ) ( )

( ) ( )

2 3 5 5 3 5

3 5 3 5

− + +=

+ − [1]

2 3 2 5 5 3 5 5

3 5

− + +=

−

7 3 3 5 7 3

3 52 2 2

+ − −= = +

− [1]

By comparing, a = 7 3, b

2 2

− −= [1]

24.

(1 Mark)

Plot the points (0,2) on y-axis and (2,0) on x-axis such that (0,0) is at origin.

AC = BC = OB = OA = 2 units (1 Mark)

Area of square = side x side

=2 x 2 = 4 square units (1 Mark)

SECTION-D

25.

7 3 5 7 3 5

3 5 3 5

+ −−

+ −

= ( ) ( ) ( ) ( )

( )( )

7 3 5 3 5 7 3 5 3 5

3 5 3 5

+ − − − +

− + (1)

= ( )

( ) ( )22

21 7 5 9 5 15 21 7 5 9 5 15

3 5

− + − − + − −

−

= 6 2 5 6 2 5

4

+ − + (1)

= 5 (1/2)

∴ 7 3 5 7 3 5

a 5b3 5 3 5

+ −− = +

+ −

⇒ 5 a 5b= + (½)

0 5 a 5b⇒ + = + (½)

a 0, b 1⇒ = = ans. (½)

OR

a = 7 - 4 3

1 1 1 7 4 3

a 7 4 3 7 4 3 7 4 3

+= = ×

− − + (½)

=

( ) ( )22

7 4 3

7 4 3

+

− (1)

= 7 + 4 3

Now ( ) ( )2 2

21 1 1a a 2 a

a a a

+ = + + (1)

= a + 1

2a+

= 7 - 4 3 7 4 3 2+ + +

= 16 (1)

∴ 1

a 16 4a

+ = = ans. (½)

26. p(x) =x4 – 2x3 + 3x2 – ax + 3a – 7

By given conditions

p(-1) = 19 (½) ⇒ (-1)4 – 2(-1)3 + 3(-1)2 – a(-1) + 3a – 7 = 19 (½)

⇒ 4a – 1 = 19 (½)

⇒ a = 5 ans

(½)

∴ p(x) = x4 – 2x3 + 3x2 – 5x + 15 – 7

= x4 – 2x3 + 3x2 – 5x + 8 (½)

Sum p(x) is divided by (x + 2)

∴ Remainder = p(-2) (½)

= (-2)4 – 2(-2)3 + 3(-2)2 – 5(-2) + 8 (½)

= 62 ans (½)

OR

Let f(x) = x4 + ax3 – 3x2 + 2x + b

Since (x + 1) is factor of f(x)

∴ f(-1) = 0 (1/2)

⇒ (-1)4 + a(-1)3 – 3(-1)2 + 2(-1) + b = 0

⇒ 1-a + b = 4 …(i) (1)

Since (x-1) is a factor of f(x) ∴ f(1)=0 (1/2)

=> (1)4 + a(1)3 – 3(1)2 + 2(1) + b = 0

=> a+b=0 …(ii) (1)

Adding (i) & (iii) we have

-a + b + a + b = 4

b = 2 (1/2)

substituting the value of b in (i) we have -a + 2 = 4 ⇒ a = -2 (½)

∴ a = -2, b = 2 ans

27. Given: Two lines AB and CD which intersect each other at O. (1/2)

To prove ∠ AOC = ∠ BOD (½)

∠ AOD = ∠ BOC

(1/2)

Proof: Since AB is line and ray OD stands on it

∴ ∠ AOD + ∠ BOD= 180o ----(1)(Linear pair axiom) (½)

Since CD is a line and ray OA stands on it

∴ ∠ AOC + ∠ AOD = 180o ---(2)(linear pair axiom ) (½)

From (i) & (ii)

∠ AOC + ∠ AOD = ∠ AOD + ∠ BOD � ∠ AOC = ∠ BOD (1)

Similarly we can prove that

∠ BOC = ∠ AOD (1/2)

28. Since BO is angle bisector of ∠ CBE

∠ CBO = 1

2 ∠ CBE – (i) (½)

Since CO is angle bisector of ∠ BCD

∴ ∠ BCD = 1

2 ∠ BCD – (ii) (½)

∠ ABC + ∠ CBE = 180o (Linear pair axiom)

∠ CBE = 180o - ∠ ABC – (iii)

∴ ∠ CBO = ( )o o1 1180 ABC 90 ABC

2 2− ∠ = − ∠ - (iv) (using(i) and (iii)) (1)

∠ ACB + ∠ BCD = 180o (linear pair axiom) ∠ BCD = 180

o - ∠ ACB -----(v)

∠ BCO = 1

2 (180o -∠ ACB) = 90o -

1

2 (∠ ACB )--– (vi) (using (ii) and (iv)) (½)

In ∆OBC

∠ BOC + ∠ CBO + ∠ BCO = 180o (angle sum property) ∴ ∠ BOC = 180o - ∠ CBO - ∠ BCO

= 180o – (90o - 1

ABC2∠ ) – (90o -

1ACB

2∠ ) [using (iv)to (vii)

= o o o1 1

180 90 ABC 90 ACB2 2

− + ∠ − + ∠ (1)

= ( )1

ABC ACB2∠ + ∠

= ( )o1180 A

2− ∠

oA ABC ACB 180

∠ + ∠ + ∠ =

∵

=o

90 − A∠ ans (½)

29. (i) (3x – 5y – 4) (9x2 + 25y2 + 15xy + 12x – 20y + 16)

= (3x + (-5y) + (-4)) [(3x)2 + (-5y)2 + (-4)2 – (3x) (-5y) – (-5y) (-4) – (3x) (-4)]

(1/2)

= (3x)3 + (-5y)3 + (-4)3 – 3(3x) (-5y) (-4) (½)

[(a + b + c) (a2 + b2 + c2) – ab – bc – ca) = a3 + b3 + c3 – 3abc] (½)

= 27x3 – 125y3 – 64 + 180 xy ans (1/2)

(ii) a2 + b2 – 2 (ab – ac + bc)

= a2 + b2 – 2ab + 2ac – 2bc (½)

= (a2 + b2 -2ab) + 2c (a- b) (½)

= (a – b)2 + 2c (a – b) (½)

= (a – b) (a – b + 2c) Ans (½)

30. In ∆ ABC AD = AE

∠ ADE = ∠ AED (angle opp equal sides are equal) ∠ ADC = ∠ AED (Same angle)…..(i) (1) ∠ BAD = ∠ CAE (Given) Adding ∠ DAE an both sides we get ∠ BAD + ∠ DAE = ∠ CAE + ∠ DAE ∠ BAE = ∠ CAD (1)

In ∆ BAE + ∆CAD ∠ AEB = ∠ ADC (proved in(i))

AE = AD (Given) (1)

∠ BAE = ∠ CAD (proved) ∴ ∆ BAE CAD (ASA Criterion)≅ ∆ (½)

AB = AC (CPCT ) proved (½)

31.

1 1 3 8 3 8 13 8 Mark

9 8 23 8 3 8 3 8

1 1 8 7 8 7 18 7 Mark

8 7 28 7 8 7 8 7

1 1 7 6 7 6 17 6 Mark

7 6 27 6 7 6 7 6

1 1 6 5 6 5 16 5 Mark

6 5 26 5 6 5 6 5

1 1 5 2 5 2 15 2 Mark

5 4 25 2 5 2 5 2

1

3 8

+ + = × = = +

−− − +

+ + = × = = +

−− − +

+ + = × = = +

−− − +

+ + = × = = +

−− − +

+ + = × = = +

−− − +

−−

( ) ( ) ( ) ( )

1 1 1 1

8 7 7 6 6 5 5 2

3 8 8 7 7 6 6 5 5 2 (1Mark)

15 Mark

2

+ − +− − − −

= + − + + + − + + +

=

32.

( )1 5 3 3 2

Let 2x 5y u, y z v and z x w (1 Mark)3 3 4 4 3

+ = − + = − + =

( )

( )

( )

( ) ( ) ( )

3 33

1 5 3 3 2Then, u v w 2x 5y y z z x 0 (1 Mark)

3 3 4 4 3

1 5 3 3 22x 5y y z z x

27 3 4 4 3

1 5 3 3 23 2x 5y y z z x (1 Mark)3 3 4 4 3

12x 5y 20y 9z 9z 8x (1 Mark)

144

+ + = + − + − − =

+ + − + − +

= − + − + +

= + − +

33.

(1+1/2 Mark)

Measure AC and BC with the help of a scale (or using Pythagoras theorem)

Here OA = OB = 3 units and OC = 4 units

From Pythagoras theorem,

AC = BC = 2 24 3 5+ = units and BC = 6 units. (1 Mark)

Therefore, perimeter of triangle ABC = AB + BC + CA

= 5 + 6 + 5 = 16 units (1+1/2 Mark)

34.(i)∠C= ∠A... angles opposite to equal sides are equal 1Mark

2

But ∠B= ∠ A/2 given 1Mark

2

⇒ ∠A > ∠B

⇒ ∠C > ∠B as ∠A= ∠C 1Mark

2

⇒AB > AC....side opposite to bigger angle is longer. 1Mark

2

(ii) We see that, AB < AC and AB < BC

So AB is the shortest side of the triangle

∠ C < ∠ B...as AB <AC...(i) 1Mark

2

∠ C < ∠ A...as AB < BC...(ii) 1Mark

2

Also, ∠ A < ∠ B...as BC <AC...(iii) 1Mark

2

∠ C < ∠ A< ∠ B 1Mark

2