SOLUTIONS

SECTION – A

1. (D) ( )1 1 1 3 3 3(25) x (5) = 125 =5

(1)

2: (B)

2

2 2

3The area of equilateral traingle: a ,where a is side.

4

3(14) 49 3 cm

4⇒ =

(1)

3(A) Since AOB is a straight line,

AOB 180

x 10 x x 20 180

3x 150

x 50

∠ = °

⇒ + ° + + + ° = °

⇒ = °

⇒ = °

(1)

4 (B) (1)

∠OBC= ∠OCB = 25..(Since OB,OC are bisectors.)

So, by angle sum property in triangle OBC,

∠O + 250+250 = 1800

∠O =130o

(1)

5(A)

Degree of a polynomial is the degree of the leading term.

Here leading term is x4 and its degree is 4.

6(D) (1)

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

7(A)

The coefficient of x2 in (7x-5) (5x2 - 3x +1) = -25 – 21 = - 46

(1)

8: (D)

Semi perimeter,a b c 122 120 22

s , s , s 132 m2 2

+ + + += = =

(1)

Now, area of the triangle = s(s a)(s b)(s c)− − −

= 132(132 122)(132 120)(132 22)− − −

= 132 10 12 110× × × = 1320 m2

Rent paid by the company = Rs. 5000 × 1320 × 3

12 = Rs.1650000

SECTION – B

9. ∠ QPR = ∠ PRT ⇒ ∠ PRT = 64

∠ SRT + ∠ PRT + ∠ QRP = 180 (1)

67o + 64o + ∠ QPR = 180o ∠ QRP = 180o – 131 (½)

∠ QRP = 49o (½)

10. 70o + 30o + ∠ AEB = 180o

∠ AEB = 80o (1)

∠ AEB = ∠ CED = 80o (½)

∠ CED + ∠ EDC + 90 = 180o 80o + 90o + ∠ EDC = 180 ∠ EDC = 10o (½)

11. x = 2 - 3

( ) ( )22

2 3 2 31 1

x 4 32 3 2 3

+ += = =

−− −

= 2 + 3 (1)

x - ( ) ( )12 3 2 3 2 3

x= − − + = − (½)

( )3

31x 2 3 24 3

x

− = − = − (½)

OR

1 1

5 2 3 5 2 3+

− +

= ( ) ( )5 2 3 5 2 3

5 2 3 5 2 3

+ + −

− + (1)

=

( ) ( )22

10

5 2 3−

= 10 10

25 12 13=

− (1)

12. 12x2 – 7x + 1

12x2 – 4x – 3x + 1 (splitting middle term)

(1)

4x (3x – 1) – 1(3x – 1) (Taking common factors)

(½)

(3x – 1) (4x – 1) (½)

13. AC = 2a

AO = a

OC2 = AC2 – AO2 (By Pythagoras thm) (½)

= 4a2 –a2

OC = a 3 (1)

So,coordinates of C are (0, a 3 )

Similarly,coordinates of D are (0,- a 3 )

14. 2

2 2

2 2 2

2 2 2 2

2 2 2

2 2 2

2 2 2

2

2 2

b

a b a

b a b a

a b a a b a

b a b a(1)

a b a

b a b a

b

a b a (1)

+ +

+ −= ×

+ + + −

+ − =

+ −

+ − =

= + −

SECTION – C

15. 2 1 2 1 2 1

2 1 2 1 2 1

− − −= ×

+ + −

( ) ( )

( ) ( )

( )( )

( )

2

2

2 1 2 1 (1)

2 1 2 1

2 12 1 (1)

2 1

2 1 1.4142 1 0.4142 (1)

− −=

− +

−= = −

−

= − = − =

16. Let x = -32

y = 18

z = 14

x + y + z = -32 + 18 + 14 = 0

1

12

⇒ x3 + y3 + z2 = 3xyz

(½) ⇒ (-32)3 + (18)3 + (14)3 = 3(-32) (18) (14)

(1)

= -24192

OR

(a + b + c)2 + (a – b – c)2 + (a + b – c)2

= 2 2 2a b c 2ab+ + + 2bc+ 2ca+

2 2 2a b c 2ab+ + + −

- 2bc2 2 2

2ac a b c 2ab 2bc 2ac+ + + + + − − (2)

= 3a2 + 3b2 + 3c2 + 2ac + 2ab – 2bc (1)

17. Let x2 – 1 = A

Then 4A2 + 13A – 12 = 0

4A2 + 16A – 3A – 12 = 0 (1)

4A (A + 4) – 3(A + 3)= 0

(4A – 3) (A + 4) = 0 (1)

(4x2 + 1) (x2 – 5) (1)

18. Third side = 42 – (18 + 10)

= 14 cm (1)

2S= 42 ⇒ S = 21 cm

(½)

( ) ( ) ( )s s a s b s c∆ = − − −

= ( ) ( ) ( )21 7 3 11

= 21 11 cm2 (1)

19.

∠ BPR=∠DQR (Each given 90 degrees)

∠ BRP=∠DRQ (vertically opposite angles) (1)

BP = DQ (given)

⇒ BPR DQR∆ ≅ ∆ (By ASA congruency criterion) (1)

⇒BR = DR (By CPCT)

⇒AC bisects BD at R (1)

20. In ∆ ABL, ∠ BAL + ∠ ALB + ∠ B = 180o ∠ BAL + 90 + ∠ B = 180o

∠ BAL = 90o - ∠ B

(1)

In ∆ ABC ∠ A + ∠ B + ∠ C = 180o 90 + ∠ B + ∠ C = 180o ∠ B + ∠ C = 90o ∠ C = 90o - ∠ B ∠ ACB = 90 - ∠ B ⇒ ∠ BAL = ∠ ACB (2)

OR

Here, AB = AC

Therefore, ∠ ABE = ∠ ACE (angles apposite the equal sides are equal) (1)

Also, BE = CD

( )ABE ACD SAS⇒ ∆ ≅ ∆ (1)

( )AD AE CPCT⇒ = (1)

21.

(1)

∠ A + ∠ B + ∠ C = 180o

35o + 65o + ∠ A = 180o ∠ A = 80o

∠ BAD + 35o + ∠ BDA = 180o

∠ BAD = 1

BAC2∠

= 40o

40o + 35o +∠ BDA = 180

∠ BAD = 1

BAC2∠

= 40o

40o + 35o + ∠ BDA = 180o 1

12

∠ BDA = 105o ∠ ADC = 180o – 105o = 75o (1/2)

22. Given: ∠ ABC = 80o, ∠ CFD = 45o, ∠ CDF = 35o To prove :- e 11m

Proof :- ∠ FCD + ∠ CDF + ∠ DFD = 180o ∠ FCD + 45o + 35o = 180o ∠ FCD = 100o (1)

∠ EDC + ∠ FCD = 180o 100o + ∠ FCD = 180o ∠ FCD = 80o (1)

∠ FCD = ∠ ABC = 0

l m⇒ � (1)

OR

∠ BCD = ∠ BCE + ∠ DCE = 30o + 35o = 65o ∠ ABC = 65o ∠ ABC = ∠ BCD, AB || CD (1)

∠ ECD + ∠ CEF = 35 + 145 = 180o (1)

CD || EF (1)

AB || CD, CD || EF ⇒ AB || EF (1)

23.

( ) ( )

( ) ( )33 3

3

1x

3 2 2

1 3 2 2

3 2 2 3 2 2

3 2 2 13 2 2

9 8 2

1and y=

3 2 2

1 3 2 2

3 2 2 3 2 2

3 2 2 13 2 2

9 8 2

1x y 3 2 2 3 2 2 6

2

1 1 1xy 1

9 8 23 2 2 3 2 2

x y x y 3xy x y

6 3 1 6 216 18 198 (1)

=−

+= ×

− +

+ = = +

−

+

−= ×

+ −

− = = −

−

⇒ + = + + − =

= = =

− − +

+ = + − +

= − × × = − =

24. Consider a pentagon ABCDE.

Construction: Join A to C and to D dividing the pentagon into three triangles.

In triangle ABC,

∠BAC+∠B+∠BCA =180° (angle sum property)....(i) (1)

In triangle ACD,

∠DAC+∠ACD+∠CDA =180° (angle sum property)...(ii) (1 )

In triangle ADE,

∠EAD+ ∠ADE+ ∠E = 180° (angle sum property)...(iii)

Adding the three equations,

∠BAC+∠B+∠BCA + ∠DAC+∠ACD+∠CDA + ∠EAD+ ∠ADE+ ∠E = 540o

(∠BAC + ∠DAC +∠EAD)+ ∠B +(∠BCA + ∠ACD)+ (∠CDA +∠ADE) +∠E = 540o

∠A+∠B+∠C+∠D+∠E = 540° (1)

Hence proved.

SECTION - D

25.x2 – 3x+ 2

= x2 – 2x – x + 1

= x (x – 2) -1 (x – 2)

= (x – 2) (x – 1) (1)

If (x – 2) is a factor of x4 – ax2 + b then

(2)4 – a(2)2 + b = 0

-4a + b = 16 …(i) (1)

If (x – 1) is a factor x4 – ax2 + b then

(1)4 – a(1)2 + b = 0

-a + b = 1 …(ii)

Solving (i) and (ii)

b = 4, a = 5 (1)

OR

ax2 + (4a2 – 2b) x – 12ab

ax2 + 4a2x – 3bx – 12ab (1)

ax (x + 4a) -3b (x + 4a) (2)

(x + 4a) (ax – 3b) (1)

26.

(1 Mark)

If the abscissa of Anil’s position is at 0 then the coordinates of his position are of the

form (0,y). (1 Mark)

Since Ram is at a distance of 8 units from Shyam then Anil is also at a distance of 8

units in opposite direction. (1 Mark)

Therefore, Anil’s position is (0,-8) (1 Mark)

27. ( )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

+ + = + − + − − =

+ + − + − +

= − + − + +

= + − +

28 We know ∠A+∠B+ ∠C = 180° (Angle sum property)

∠A - ∠B = 15° (Given)

2∠A + ∠C = 195….(i)

∠A - ∠B = 15° , ∠B - ∠C = 30°⇒∠A - ∠C = 45°….(ii) (2 Marks)

Adding (i) and (ii) , we get 3∠A = 240⇒ ∠A = 80o

∠B = 65o , ∠C = 35o (2 Marks)

SECTION – D

29.

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

+ − +− − − −

= + − + + + − + + +

=

30. x3 + y3 + z3 – 3xyz = (x + y + z) (x2 + y2 + z2 – xy – yz – zx) (1)

= ( ) ( )2 2 21x y x 2x 2y 2z 2xy 2yz 2zx

2+ + + + − − −

=

( ) ( ) ( ) ( )2 2 2 2 2 21x y x x y 2xy y z 2yz z x 2zx

2

+ + + − + + + + −

(1)

= ( ) ( ) ( ) ( )2 2 21

x y z x y y z z x2

+ + − + − + −

(1)

31.

Draw AB as 9.3 cm

Marke a point c on AB prolened such that BC = 1

Find mid point of AC. Let it is0. (1)

Taking O as centre and OC = OA as radius draw

Semi circle. Draw line passing through B and perpendicular to OB. If cuts semi circle at D. (2)

Taking B as centre and BD as radius draw an arc cutting OC produced at E. Point

E represents 9.3 (1)

32. ∵ ∠ AEO = ∠ CDO

∴ ∠ CEB = ∠ ADB (linear pair)

ABD CBE∆ ≅ ∆ 1

22

33.

From ∆ ABC ∠ 1 = ∠ 2 …(i)

∠ 3 = ∠ 4 …(ii)

From ∆ BCD by Angle Sum Property,

∠ 1+∠ 2+∠ 3 + ∠ 4=180 o

∠ BCD = 90o …(2)

34. Make a quadrilateral PQRS as shown in figure.

Given: PQRS is a quadrilateral.

Diagonals intersect at O.

To Prove: PQ+QR+PS+SR > PR+QS

Proof: In ∆PQR

PQ+QR>PR (Sum of two sides of triangle is greater than the third side) (1)

Similarly, In ∆PSR, PS+SR>PR

In ∆PQS, PS+PQ>QS and in ∆QRS we have QR+SR>QS

Now we have Since sum of two sides is always greater than the third side:

PQ+QR>PR (1)

PS+SR>PR

PS+PQ>QS

QR+SR>QS (1)

After adding above inequalities we get

2(PQ+QR+PS+SR) > 2(PR+QS)

i.e PQ+QR+PS+SR>PR+QS. (1)

OR

Given :- S is any point in the interior of ∆ PQR To Prove:- SQ + SR < PQ + PR

Construction : - Produce of s to meet PR at T (½)

Proof : - In ∆ PQT, PQ + PT > QT ⇒ PQ + PT > QS + ST …(i) (½)

In ∆ RST, ST + TR > ST …(ii) (½)

Adding (i) and (ii)

PQ + PT + ST + TR > SQ + ST + SR

PQ + (PT + TR) >. SQ + SR

SQ + SR < PQ + PR 1

12