SOLUTIONS

SECTION – A

1. (D) If x is an irrational number, then x can be any positive real number. (1)

2. (C) (x-1) – (x2 -1) = (x-1) – (x-1)(x+1) = (x-1){1 – (x+1)} =(x-1)(-x)

So, (x-1) is a factor. (1)

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

the remainder is equal to p(a) which is non zero since (x-a) is not a factor of p(x).

(1)

4. (C) 108o

Let ∠POQ = 2x

Then ∠QOR = 3x

Since PR, QS intersect at O, so these angles must form a linear pair.

So, 2x + 3x = 180o

i.e x = 36o

∠QOR=3x =108o

∠POS= ∠QOR (vertically opposite angles)

So, ∠POS = 108o (1)

5. (A) 0.8 cm

The sum of two sides of a triangle must be greater than the third

And 12 + 0.8 = 12.8 < 13 (1)

6. (A) Only when 0.5 is added to the given equation it will become a perfect square

(1)

7. (B) Area (1)

8. (B) Perimeter of rhombus = 4 × Side= 32

Side= 8

The cost of painting the square= Area of square × the rate of painting per cm2.

= 64 x 5 = Rs. 320 (1)

SECTION-B

9. We can write 2 as 4 and 3 as 9 . (1)

5 and 6 both lie between 4 and 9

i.e. the irrational numbers 5 and 6 lie between 2 and 3 (1)

10. Given ∠ A > 60, ∠ B = 40o

∴ ∠ C = 180o – 60o – 40o – 80o

∴ ∠ B < ∠ A < ∠ C (1)

⇒ ∠ B in the smallest

So, the side opposite to ∠ B in smallest

∴ AC is smallest side. (1)

OR

∠ AOC = ∠ COD + ∠ AOD = 150o + 30o = 180o

∴ AOC is a straight line. (1)

∠ BOD = ∠ BOA + ∠ AOD = 150o + 30o = 180o (1)

∴ BD is also a straight line.

11. One of Euclid’s axioms states that.

Things which coincide with one another are equal to one another.

If two circles are equal then the circles coincide (1)

i.e their centres coincide and their boundaries coincide.

Therefore, their radii are equal (1)

12. 303 + 203 - 503

Wow 30 + 20 + (-50) = 0. (1)

x ∴ 303 + 203 + (-50)3 = 3 × 30 × 20 × (-50)

= -9000 (1)

13. Hare the coordinates of the point are given by:

P (3,0) (½)

Q (1,4) (½)

R (-3,2) (½)

S (0,-4) (½)

14. x +1 = 0 ⇒ x = -1 (1)

Let p(x) = x11 +1

⇒p (-1) = (-1)11 +1 = 0

The remainder when x11 +1 is divided by (x+1) is 0. (1)

SECTION-C

15.

( ) ( )

( ) ( )

( ) ( )

( ) ( )

( ) ( )

( ) ( )

3 3

2 5

5 4

4 3

3 31 1

2 5

5 41 1

4 3

3 31 1

2 52 5

5 41 1

4 34 3

3 3

5 4

25 243

16 8

25 243

(1)

16 8

5 3

2 2

5 3(1)

2 2

125 27

32 16

3375(1)

512

×

×

×

=

×

×

=

×

×=

×

×=

×

=

16. (x – 1) is a factor of kx2 – 3x + k.

∴ k(1)2 – 3.1 + k = 0

⇒ k – 3 + k = 0

⇒ 2k – 3 ⇒ k = 3

2 (1)

For this value of k, polynomial is

23 3x 3x

2 2− +

= { }23x 2x 1

2− +

= ( ) ( ) ( )23 3

x 1 x 1 x 12 2

− = − −

So another factor is (x – 1) (1)

17. To simplify this, taking 2 2 2 2x y a and x y b+ = − = , we get

( ) ( ) ( ) ( )2 2

2 2 2 2 2 2 2 2x y 2 x y x y x y+ − + − + − = 2 2a 2ab b− +

= ( )2

a b−

( ) ( )2

2 2 2 2x y x y = + − −

2

2 2 2 2x y x y = + − +

( )2

2 42y 4y= =

OR

Given x + y +z = 9

Let a = 3 - x

b = 3 – y

c = 3 – z

a + b + c = 9 – (x + y + z) = 9 – 9 = 0 (2)

( ) ( ) ( )23 3

x 1 x 1 x 12 2

− = − −

∴ We have

a3 + b3 + c3 = 3abc

⇒ (3 – x)3 + (3 – y)3 + (3 – z)3 = 3 (3 – x) (3 – y) (3 – z)

⇒ (3 – x)3 + (3 – y)3 + (3 – z)3 – 3 (3 – x) (3 – y) (3 – z) = 0 (1)

18.

Let the diagonals of parallelogram ABCD meet at O. Consider ∆CDO and ∆CBO

DO = BO (given)

CO is a common side

∠ COD =∠ COB (each given 90 degrees) (1)

⇒ CDO CBO∆ ≅ ∆ (SAS congruency criterion)

⇒CD=CB...CPCT (1)

So we have a parallelogram ABCD whose adjacent sides are equal.

⇒ ABCD is a rhombus. (Using definition of a rhombus) (1)

19.

∠ AEB = ∠ EAC + ∠ ACE

[The measure of exterior angle in equal is the sum of interior opposite angles].

∠ xo = 50o + 90o = 180o (1)

Now, in ∆ BDC,

yo + 30o + 90 = 180o (1)

⇒ yo = 180o – 120o = 60o (1)

20. Given: AP⊥ l & PR⊥ P To show: AR >AQ

Const: Take a point S on PR such that PS = PQ (½)

Join AS

In ∆APQ &∆ APS ∠ APQ =∠ APS (each 90o)

AP = AP (common)

PQ = PS ( by const)

∆APQ ≅ ∆APS (by SAS cong thm) (1)

∠ AQP =∠ ASP (by cpct) (½)

∠ ASP > ∠ ARS (exterior angle of ∆ASR ) (½)

So, ∠ AQP > ∠ ARS (as ∠ AQP =∠ ASP )

AR > AQ (side opposite to greater angle in ∆AQR is greater) (½)

21. ∠ CEF = 90o

⇒ ∠ Z + 55o = 90o ⇒ ∠ z = 90o – 55o = 35o (1)

Also ∠ y + 55o = 180o

[Sum of co-interior angles on same side of parallel line is 180o]

⇒ ∠ y = 180o – 55o

= 125o (1)

Again, ∠ x = ∠ y [Corresponding angles]

⇒ ∠ x = 125o (1)

22. a + b 3 = 3 1 3 1 3 1

3 1 3 1 3 1

− − − = × + + − (½)

= ( )

( )( ) ( )

2

2

3 1 3 1 2 3 4 2 3

23 1 3 1 3

− + − −= =

− +

= 2 - 3 (1½)

Comparing both sides, we get

a = 2, b = -1 (1)

OR

( )( )( ) ( )

3 5 3 2 5 33 2

5 3 5 3 5 3 5 3

+ + −+ =

− + + −

( )22

15 3 3 10 2 3

5 3

+ + −=

−

(1)

= 25 3 25 3

25 3 22

+ +=

− (2)

23. Let the base of the isosceles triangle be b = 8 and the equal side be a.

Perimeter = 18 cm therefore 18 = 8 + 2a

Hence a = 5 cm (1)

Now a b c 8 5 5

s 92 2

+ + + += = =

s a 9 5 4− = − = , s b 9 8 1− = − = and s c s a 4− = − =

Area of the triangle = ( ) ( ) ( )s s a s b s c 9.4.1.4 144 12− − − = = =

(1)

Area of the rectangular board = Length x Breadth = 16.9 =144 sq cm

So the number of triangular pieces is = 144

1212

= (1)

Hence 12 pieces are required for the puzzle.

24.

PQ is a straight line

∠POB +∠BOQ = 180o (Linear pair)

∠POA+∠AOQ = 180o (Linear pair)

∴ ∠POB +∠BOQ = ∠POA+∠AOQ... (i) (1)

But, OP bisects ∠AOB

∴ ∠POA = ∠POB (1)

Substituting in (i), we have

∠BOQ = ∠AOQ (1)

OR

CD EF and BE is transversal line�

oy 180 55 125 Sum of Angle of the same side of transveral is180

∠ = − =

[1]

y x

AB CD,BE is transvera line

∠ = ∠

� [1]

o o oz 90 55 35∠ = − = [1]

SECTION-D

25. a12 x4 – a4 x12

= a4 x4 (a8 – x8) (1)

= a4x4 [(a4)2 – (x4)2]

= a4x4 (a4 + x4) (a4 – x4) (1)

= a4x4 (a4 + x4) [(a2)2 – (x2)2]

= a4x4 [a4 + x4] [a2 + x2] [a2 – x2] (1)

= a4x4 (a4 + x4)(a2 + x2) (a + x) (a – x) (1)

26.

In ∆ ABC and ∆ LAC

∠ BAC = ∠ ALC [90o each] (1)

∠ ACB = ∠ LCA [Common angle] (1)

∴ ∠ ABC = ∠ LAC [Third angle of the triangles ABC & LAC] (1)

∴ ∠ CAL = ∠ ABC (1)

27. (i) (20)2 + (-8)3 + (-12)3

Let a = 20, b = -8, c = -12

a + b + c = 20 + (-8) + (-12) = 20 – 20 = 0 (1)

∴ a3 + b3 + c3 = 3abc

⇒ (20)3 + (-8)3 + (-12)3 = 3 × 20 × (-8) × (-12) = 5760 (1)

(ii) 105 × 95

We can write 105 × 95 = (100 + 5) (100 – 5) (1)

Using identity, (a + b) (a – b) = a2 – b2

105 × 95 = (100+ 5 ) (100 – 5)

= 1002 - 52

= 10000 – 25

= 9975 (1)

28.

( )

( )

( )

( )

( )

( ) ( ) ( ) ( ) ( )

( )

1 1 3 8 3 8 13 829 83 8 3 8 3 8

1 1 8 7 8 7 18 728 78 7 8 7 8 7

1 1 7 6 7 6 17 627 67 6 7 6 7 6

1 1 6 5 6 5 16 526 56 5 6 5 6 5

1 1 5 2 5 2 15 225 45 2 5 2 5 2

1 1 1 1 1

3 8 8 7 7 6 6 5 5 2

3 8 8 7 7 6 6 5 5 2 1

152

+ += × = = +

−− − +

+ += × = = +

−− − +

+ += × = = +

−− − +

+ += × = = +

−− − +

+ += × = = +

−− − +

− + − +− − − − −

= + − + + + − + + +

=

OR

4 5 4 5

4 5 4 5

+ −+

− +

Rationalizing the denominator,

4 5 4 5 4 5 4 5

4 5 4 5 1 5 4 5

× + − − × + − + + − (2)

= ( ) ( )

( )( ) ( )

2 24 5 4 5

4 5 4 5 4 5 4 5

+ − +

+ + + −

= ( )

( )

( )

( )

2 22 2

2 22 2

4 5 2.4. 5 4 5 2.4. 5

4 5 4 5

+ + + −+

− −

(1)

= 16 5 8 5 16 5 8 5

16 5 16 5

+ + + −+

− −

= 21 8 5 21 8 5

11 11

+ −+

= 21 8 5 21 8 5 42

11 11

+ + −= (1)

29. Let O be the position of car when it starts.

Let A, B, C and D is the positions of car after each consecutive hour.

Let us take the scale to be 1 unit = 4 km.

To trace the path of the car we first need to tabulate the distance moved in each

direction. The (x, y) coordinate after each consecutive hour can be computed as

follows:

Point At origin O A B C D

x (in km) 0 0 16 16 16 – 20 = -4

y (in km) 0 20 20 20 – 24 = -4 -4

(2)

Now plotting the various points on the Cartesian plane gives:

(1)

Coordinates of the point D which is the final position of the car (-4,-4)

So, Distance of car from x axis = 4 km (in negative x direction)

And distance of car from y axis = 4 km (in negative y direction) (1)

30. Let f(x) = x3 – ax2 – 13x + b

f(1) = 13 – a.12 – 13.1 + b = b – a – 12

Since x – 1 is a factor of f(x)

So b – a – 12 = 0 ---- (i) (1)

f(-3) = (-3)3 – a(-3)2 – 13(-3) + b = b – 9a + 12

Since x + 3 is a factor of f(x)

So b - 9a + 12 = 0 ---- (ii) (1)

Subtracting (ii) from (i), we get

8a – 24 = 0 ⇒ a = 24 / 8 = 3 (1)

(i) gives

b = a + 12 = 3 + 12 = 15 (1)

OR

Let f(x) = 2x4 – 5x3 + 2x2 – x + 2

and g(x) = x2 -3x + 2 = x2 -2x -x+ 2

= x(x-2) – 1 (x-2) = (x – 1) (x – 2) (1)

We show that (x-1) and (x-2) are factors of f(x)

i.e. f(1) = 0 and f(2) = 0

f(1) = 2(1)4 – 5(1)3 + 2(1)2-1 + 2 = 2 – 5 + 2 – 1 + 2 = 0

So (x – 1) is a factor is f(x) (1)

f(2) = 2(2)4 – 5(2)3 + 2(2)2-2 + 2 = 32 – 40 +8 – 2 + 2 = 0

So (x – 2) is a factor of f(x) (1)

Thus, f(x) is div by x – 1 and x – 2

i.e. f(x) is divisible by (x – 1) (x – 2) = g(x) (1)

31. ∠ 1 = ∠ 2 [as OB bisects ∠PBC]

and ∠ 3 = ∠ 4 [as OC bisects ∠QCB] (½)

∠ PBC = ∠ A + ∠ ACB [the measure of exterior angle is equal to sum of

opposite interior angles] (½)

⇒ 2∠ 2 = ∠ A + ∠ ACB

⇒ ∠ 2 = 1 1A ACB

2 2∠ + ∠ (½)

∠ BCQ = ∠ A + ∠ ABC [the measure of exterior angle is equal to sum of

opposite interior angles] (½)

⇒ 2∠ 3 = 1 1

A ABC2 2∠ + ∠ (½)

⇒ ∠ 3 = 1 1

A ABC2 2∠ + ∠ (½)

∠ 2 + ∠ 3 + ∠ BOC = 180o

And ∠ 2 + ∠ 3 = 1 1A ACB

2 2∠ + ∠

1 1A ABC

2 2+ ∠ + ∠

o o1 1 1 1 1A ABC ACB A x180 A 90 A

2 2 2 2 2= ∠ + ∠ + ∠ + ∠ = + ∠ = + ∠

( )o o o

o

1BOC 180 2 3 180 90 A

2

190 A

2

∴ ∠ = − ∠ + ∠ = − − ∠

= − ∠ (1)

32.

�PQRS is a square

∴ PQ = QR = RS = SP … (i)

Also ∠ RSP = ∠ SRQ = ∠ ROP = ∠ SPQ = 90o ...(ii)

Also ∆ TSR is equilateral

TS = TR = SR … (iii) (½)

Also ∠ STR = ∠ TSR = ∠ TRS = 60o

From (i) and (iii)

SP = ST

Also ∠ TSP = ∠ RSP + ∠ TSR = 90o + 60o = 150o (½)

Similarly ∠ TRQ = 150o (½)

In ∆ TSP and ∆ TRQ,

PS = QR [by (i)] (½)

∠ TSP=∠ TRQ [both 150o] (½)

TS = TR [by (ii)] (½)

∴ ∆ TSP ≅ ∆ TRQ [by SAS criterion]

∴ (i) ∠ PST = ∠ QRT [both 150o]

(ii) PT = QT [CPCTC] (½)

(iii) ∠ QTR = ( ) ( )0 o o o1 1180 TRQ 180 150 15

2 2−∠ = − = (½)

33. Given

x = 3 2

3 2

+

−

x2 =

23 2

3 2

+ −

= ( )

( )

( ) ( )

( ) ( )

2 2 2

2 2 2

3 2 3 2 2. 3. 2

3 2 3 2 2. 3 . 2

+ + +=

− + −

= 3 2 2 6 5 2 6

3 2 2 6 5 2 6

+ + +=

+ − − (½)

Rationalizing the denominator,

x2 = ( )

( ) ( )( )

( )

2 22

22

5 2 6 5 2 6 2.5.2 65 2 6 5 2 6

5 2 6 5 2 6 5 2 6 5 2 6 5 2 6

+ + + + + = = − + − + −

= 25 24 20 6

49 20 625 24

+ += +

−. (1)

y = 3 2

3 2

−

+

∴y2 = ( )

( )

( ) ( )

( ) ( )

2 2 22

2 2 2

3 2 3 2 2. 3 . 23 2

3 2 3 2 3 2 2. 3. 2

− + − − = = + + + +

3 2 2 6 5 2 6

3 2 2 6 5 2 6

+ − −= =

+ + + (½)

Rationalizing the denominator,

y2 = ( )

( ) ( )

25 2 65 2 6 5 2 6

5 2 6 5 2 6 5 2 6 5 2 6

− − − = + − + −

= ( )

( )

22

22

5 2 6 2.5.2 6

5 2 6

+ −

−

= 25 25 20 6

49 20 625 24

+ −= −

− (1)

∴ x2 + y2 = (49 + 20 6 ) + (49 - 20 6 ) = 98 (1)

34. Given: ABC is an equilateral triangle. AD, BE and CF are the medians on the sides

BC, CA and AB respectively.

(½)

To prove: AD = BE = CF. (½)

Proof: ∆ ABC is equilateral

∴ AB = BC = CA

∴ 1 1AB AC BF CE

2 2= ⇒ = …(i) (1)

In ∆ EBC and ∆ FCB,

CE = BF (by (i)

∠ A = ∠ B = ∠ C = 60o

⇒ ∠ ECB = ∠ FBC (both 60o)

BC = BC [Common side]

∴ EBC FCB∆ ≅ ∆ [by SAS] (1)

∴ BE = CF [CPCT] (½)

Similarly it can be shown CF = AD

and BE = AD

Hence AD = BE = CF (½)