SUB: MATH TOPIC: Time Allowed: 3 hours [Maximum Marks: 80] Question numbers 1 to 10 carry one mark each. 1. Find the value of k so that the following system of equations has on

solution: 3x − y − 5 = 0; 6x − 2y − k = 0 Ans. 3x − y = 5 (1) 6x − 2y = k (2) Multiplying (1) by (2) we get 6x − 2y = 10 (3) Now equation will have no solution if k ≠ 10. 2. The nth term of an A.P. is 6n + 2. Find its common difference Ans. Let tn = 6n + 2 then tn+1 = 6(x + 1) + 2 = 6n + 8 d = (6n + 8) − (6n + 2) = 6 3. In Fig 1 AD = 4 cm, BD = 3 cm and CB = 12 cm find cotθ.

C

A

B D

Fig. 1

θ

Ans. AB2 = BD2 + AD2 = 32 + 42 = 9 + 16 = 25 ⇒ AB = 5

cotθ = 512

ABCB

= . 4. Write the zeroes of the polynomial x2 − x − 6. Ans. x2 − x − 6 = 0 ⇒ x2 − 3x + 2x − 6 = 0 x(x − 3) + 2(x − 3) = 0 (x + 2) (x − 3) = 0 ⇒ x + 2 = 0, x − 3 = 0 ⇒ x = −2, x = 3.

5. If qp

is rational number (q ≠ 0), what is condition on q so that the

decimal representation of qp

is terminating?

Ans. If qp

is a rational number which is terminating then q must be of the form 2n5m where n and m are non negative integers.

6. From a well shuffled pack of cards, a card is drawn at random. Find

the probability of getting a black queen. Ans. Number of black queen = 2 Total number of cards = 52

∴ Probability of getting black queen = 261

522

= . 7. Which measure of central tendency is given by the x−coordinate of

the point of intersection of the “more than ogive” and less than ogive”?

Ans. Measure of central tendency given is median.

8. In Fig. 4, O is the centre of a circle. The area of sector OAPB is 185

of the area of the circle. Find x.

x

O

A B P

Ans. Area of sector OAPB = π2x

× total area of circle where x is in radian Let A be total area of the circle then

Area of sector OAPB = A×185

AAx×=×

π 185

2

185

2x

=π ⇒ 18

10π=x

9. In Fig. 3, PQ = 24 cm, QR = 26 cm, ∠PAR = 90o, PA = 6 cm and AR

= 8 cm. Find ∠QPR.

A 90o

Fig. 3

P

R

Q

Ans. PR2 = AP2 + AR2 = 62 + 82 = 36 + 64 = 100 PR = 10 Since PQ2 + PR2 = QR2 242 + 102 = (26)2 576 + 100 = 676 676 = 676 ⇒ ∆PQR is right angled triangle ⇒ ∠QPR = 90o. 10. In Fig. 2, P and Q are points on the sides AB and AC respectively of

∆ABC such that AP = 3.5 cm, PB = 7 cm, AQ = 3 cm and QC = 6 cm. If PQ = 4.5 cm find BC,

C

Q

A

P

B Fig. 2

Ans. Since 21

PBAP

QCAQ

== ⇒ CB

PQABAP

ACAQ

==+

==31

211

’ Q (PQ is parallel to BC) ∴ Also BC = 3.PQ = 3 × 4.5 = 13.5 cm

SECTION−B

Question numbers 11 to 15 carry 2 marks each. 11. For what value of p, are points (2, 1), (p, −1) and (−1, 3) collinear? Ans. Since the given points are collinear, the area of the triangle formed by them

must be 0.

Hence ∆ ≡ ( ) ( ) ( )[ ] 0yyxyyxyyx21

213132321 =−+−+−

⇒ ( ) ( ) ( )( )[ ] 0212p4221

=−++−

⇒ [ ] 02p2821

=−+− ⇒ p = 5. 12. Without using trigonometrical tables, evaluate the following :

[ ]ooooo

o

80tan50tan40tan30tan10tan372cos18sin

+ο .

Ans. 372sin18sin

o

o

+ [tan10o tan30otan40otan50otan80o]

= ( ) 31890cos

18sinoo

o

+− [tan(90 − 80) tan30otan(90 − 50) tan50o tan80o]

[ ]ooooo

o

o

80tan50tan50cot30tan80cot318sin18sin

+=

= 211

31.31 =+=+

13. Find the zeroes of the quadratic polynomial 6x2 − 3 − 7x and verify the

relationship between the zeroes and the coefficients of the polynomial.

Ans. 6x2 − 3 − 7x = 6x2 − 9x + 2x − 3 = 3x (2x − 3) + 1 (2x − 3) = (3x + 1) (2x − 3)

Hence zeroes of the quadratic polynomials are 31

− and 23

Sum of the zeroes of quadratic polynomial = 23

31

+−

= 67

(1)

Product of zeroes of quadratic polynomial = ⎟⎠⎞

⎜⎝⎛×⎟

⎠⎞

⎜⎝⎛−

23

31

= − 21

(2)

Sum of zeroes = − 67

67

xoftcoefficienxoftcoefficien2 =⎟

⎠⎞

⎜⎝⎛−−=⎟⎟

⎠

⎞⎜⎜⎝

⎛

and product of zeroes = ⎟⎟⎠

⎞⎜⎜⎝

⎛2

0

xoftcoefficienxoftcoefficien

= 21

63

−=− . 14. Adie is thrown once. Find the probability of getting (i) an even prime number (ii) a multiple of 3. Ans. (i) even prime is 2

Probability = 61

(ii) Multiple of 3 = 3, 6

Probability = 31

62

=

15. ABC is an isosceles triangle, in which AB = AC, circumscribed about a circle. Show that BC is bisected at the point of contact

OR In Fig. 5, a circle is inscribed in a quadrilateral ABCD in which ∠B =

90o. If AD = 23 cm, AB = 29 cm and DS = 5 cm, find the radius (r) of the circle.

O

r

r

B

Q

P S

C

D

R

A

Fig. 5 Ans. Since AB = AC, ∴ ∠ABD = ∠ACD = 2� (let) ∠ODB = ∠ODC = 90o, as the line BC is tangent to the circle.

θ θ θ

θ

A

B C D

O

∠OBD = ∠OCD = θ, as line BO and CO are the angle bisectors Now consider ∆BOD and ∆COD ∠ODB = ∠OCD = θ and ∠ODB = ∠OCD = 90o and side OD is common Hence ∆OBD ≈ ∆OCD ∴ BD = DC or D is the mid point of BC.

OR

O

r

r

B

Q

P S

C

D

R

A

Since DR = DS ∴ DR = 5 cm ∴ AR = AD − DR = 23 − 5 = 18 cm AR = AQ ∴ BQ = AB − AQ = 29 − 18 = 11 cm Now OQ = BQ = BP = OP (QOPBQ is square) ∴ radius = 11 cm.

SECTION−C

Question numbers 16 to 25 carry 3 marks each. 16. Prove that :

1cos1cos

coscotcoscot

+−

=+−

ecAecA

AAAA

OR Prove that

2)sectan1)(coscot1( =++−+ AAecAA

Ans. L.H.S. = Acos

AsinAcos

AcosAsinAcos

AcosAcotAcosAcot

+

−=

+−

= ( )( )Asin1Acos

Asin1Acos+−

= 1

Asin1

1Asin

1

+

−

dividing Nr and Dr by sinA

= =+−

1ecAcos1ecAcos

R.H.S

OR L.H.S = (1 + cotA − cosecA) (1 + tanA + secA)

= ⎟⎠⎞

⎜⎝⎛ ++⎟

⎠⎞

⎜⎝⎛ −+

Acos1

AcosAsin1

Asin1

AsinAcos1

= ( )( )

AcosAsin1AsinAcos1AcosAsin ++−+

= ( )

AcosAsin1AcosAsin 2 −+

= AcosAsin1AcosAsin2AcosAsin 22 −++

= AcosAsinAcosAsin2

= 2 = R.H.S. 17. Find the 10th term from the end of the A.P. 8, 10, 12, …., 126. Ans.. Given A.P. can be written as 126, 124, 122, …., 12, 10, 8 a = 126, d = 124 − 126 = −2 10th term = a + (n − 1)d = 126 + (10 − 1) (−2) = 126 − 18 = 108 18. Represent the following system of linear equations graphically. From

the graph, find the points where the lines intersect y-axis. 3x + y-5 = 0; 2x – y –5 =0 Ans. Given line 3x + y − 5 = 0 y = 5 − 3x (1)

y = 5 − 3x 5 2 −1 −4 −7 x 0 1 2 3 4

Second line 2x − y − 5 = 0 y = 2 x − 5 (2)

y = 2x − 5 −5 −3 −1 1 −3 x 0 1 2 3 4

(1) and (2) can be represents

1 2 3 4 5 6 –1–2–3 –4 –5–6

1

2

3

4

6

5

–1

–2

–3

–4

3x + y − 5 = 0

(5/3, 0) X

Y

Y′

X′

–5

line 2x − y − 5 = 0

(5/2, 0)

(0, 5)

(0, −5)

Here line 2x− y − 5 = 0 cuts y axis at (0, −5) line 3x + y − 5 = 0 cut y axis at (0, 5) 19. Find the roots of the following equation:

7,4;

3011

71

41

−≠=−

−+

xxx

Ans. 7,4x;3011

7x1

4x1

−≠=−

−+

( ) ( )( )( ) 30

117x4x4x7x

=−++−−

⇒ ( )( ) 3011

7x4x11

=−+

−

⇒ (x + 4) (x − 7) = −30 ⇒ x2 − 3x + 2 = 0 ⇒ (x − 1) (x − 2) = 0 x = 1 and x = 2.

Both x = 1 and x = 2 are satisfying the given equation hence the solution of the equation.

20. Show that 2 − 3 is an irrational number.

Ans. Let 2 − 3 is rational then 2 − 3 = qp

where p, q are integers and q ≠ 0 and also p and q have no common factor.

⇒ qp23 −=

qpq23 −

=

⇒ 3 is rational This is a contradiction Hence 2 − 3 is irrational.

21. In figure 6, find the perimeter of shaded region where ADC, AEB and

BFC are semi-circles on diameters AC, AB and BC respectively.

2.8cm A B C

D

E

F

1.4cm

Figure 6

OR Find the area of the shaded region in figure 7, where ABCD is a square side 14

A B

C D

Figure 7

Ans. Perimeter of shaded region in AEBFCDA = ar (AEB) + ar (BFC ) ar (CDA)

2)(

2)(

2)( ACBCAB π

+π

+π

Where AB, BC and AC are diameter of the different circles and we know

semi-perimeter of a circle is 2diameter×π

∴ perimeter required = 2)4.18.2(

24.1

28.2 +π

+×π

+×π

= 2π

. (2.8 + 1.4 + 2.8 + 1.4)

= 2π

× (8.4) = π × 4.2

= 722

×4.2 = 22 ×0.6 =13.2 ans.

OR

AD = 21 × diameter of the given circles = 14

∴ diameter of any given circle = 214

= 7 = D (let) Now, area of shaded region = Area of square − 4 × Area of any circle drawn inside

= (14)2 − 4 × 4D2π

= 142 − π × 72 = (2 × 7)2 − π × 72 = 72 (22 − π)

= 49 × ⎟⎠⎞

⎜⎝⎛ −

7224

= ⎟⎠⎞

⎜⎝⎛ −

×7

222849

= 7 × 6 = 42 sq. units. 22. If the distances of P(x, y) from the points A(3, 6) and B(-3, 4) are

equal. Prove that 3x + y = 5

Ans. Applying distance formula,

22 )6()3( −+−= yxPA 22 )4()3( −++= yxPB PA = PB

∴ 2222 )4()3()6()3( −++=−+− yxyx On squaring both sides, (x – 3)2 + (y – 6)2 = (x + 3)2 + (y – 4)2 ⇒ x2 – 6x + 9 + y2 – 12y + 36 = x2 + 6x + 9 + y2 –8y + 16 12x + 4y – 20 =0 ⇒ 3x + y = 5 proved 23. If the diagonals of a quadrilateral divide each other proportionally,

prove that it is a trapezium.

OR (Two ∆’s ABC and DBC are on the same base BC and on the same

side of BC in which ∠A= ∠D=900. If CA and BD meet each other at E, show that AE. EC = BE. ED.

Ans.

A B

F G

C D

E

Let ABCD be any quadrilateral . Now, AC and BD are its diagonal, which are cutting each other at some point E.

Now EDBE

ECAE

= (given ) (1) Construction: FG is drawn parallel to DC passing through E.

In triangle BDC, GCBG

EDBE

= { Using Thale’s theorem) (2) As FG | | DC ⇒ EG | | DC Now, in triangle CBA,

ECAE

may be equal to GCBG

using equation (i) and (ii)

so, GCBG

ECAE

= ⇒ EG | | AB but i.e. FG | | AB

But FG is drawn parallel to DC So, AB | | DC (as, if two lines are parallel to any third given line, then they are parallel to each other) ⇒ ABCD is a trapezium.

OR In this figure ABCD are concyclic ∠BAC = ∠BDC = 90o ∠AEB = ∠DEC

A D

B C E

90o90o

∴ ∆AEB and ∆DEC are similar

ECBE

EDAE

= ∴ AE . EC = BE . ED.

24. Construct a ∆ ABC in which AB = 6.5 cm., ∠ B = 600 and BC = 5.5 cm. Also construct a triangle AB′C′ similar to ∆ ABC, whose each side

is 23

times the corresponding side of the ∆ABC.

60o

B A

C

5.5

6.5

60o

BA AD

C

C′

E

B′

Ans. Extend AB to D so that AB = BD and extend AC to E so that AC = CE Now bisect BD at B′ and CE at C′ join B′C′ Now required triangle is AB′C′. 25. Determine the ratio in which the line 3x + 4y – 9=0 divides the line

segment joining the points (1, 3) and (2, 7) Ans.

Line; 3x + 4y-9=0

A(1, 3) B(2, 7)M

λ 1

Let the line 3x + 4y−9=0 divides the line segments joining A(1, 3) & B(2, 7)

in ratio λ : 1 at point M

λ 1

A(1,3) B(2,7)M

∴ Co−ordinate of M≡ ⎟⎠⎞

⎜⎝⎛

+λ+λ

+λ+λ

137,

112

Since M lies on the line 3x + 4y –9=0 So co−ordinate of M satisfies 3x + 4y –9=0

09

1374

1123 =−⎟

⎠⎞

⎜⎝⎛

+λ+λ

+⎟⎠⎞

⎜⎝⎛

+λ+λ

⇒ 6λ + 3 + 28λ + 12 − 9λ −9=0 ⇒ 25λ + 6=0

⇒ λ = 256−

Line 3x + 4y – 9=0 Divides line joining the points A & B in 6 : 25 externally.

SECTION−D

Question numbers 26 to 30 carry 6 marks each.

26. A statue 1.46 m tall, stands on the top of a pedestal. From a point on the ground, the angle of elevation of the top of the statue is 600 and from the same point the angle of elevation of the top of the pedestal is

450. Find the height of the pedestal (use 73.13 = ) Ans. Let PQ = h, then PS = h RQ =1.46

In ∆PRS, tan60o = PSPR

h46.1h3 +

= 46.1hh3 +=

( )1346.1h−

= = 173.1

46.1−

= 73.046.1

h = 73146

h = 2. 27. In a class test, the sum of the marks obtained by P in Mathematics

and Science is 28. Had he got 3 more marks in Mathematics and 4 marks less in Science, the product of marks obtained in the two subjects would have been 180. Find the marks obtained in the two subjects separately

OR

The sum of the areas of two squares is 640 m2. If the difference in their perimeters be 64m, find the sides of the two squares.

Ans. Let the marks obtained in mathematics be M and marks obtained in Science be S. Now, M + S = 28 (given) M = 28 − S (i) Now, it is also given that (M+3) × (S-4)=180 (ii) Now, using equation (i) in equation (ii), we have ((28-S)+3) × (S-4)=180 (31-S) ×(S-4)=180 31S – 31 ×4 – S2+4S = 180

⇒ � S2 – 35S +304=0

S2 – 16S –19S +304=0 S(S – 16) – 19(S – 16) = 0 ⇒ S = 16 or 19 So, (i) , M =12 or 9 ∴ Marks obtained may be (12, 16) or (9, 19) in mathematics and science respectively.

OR

Let a be m length of the side of m first square and b be the length of m side of the second square.

Now a2 + b2 = 640 (i) (given) Let a be m length of the side of m bigger square.

4a – 4b = 64 a – b = 160 ⇒ b = a – 16 (ii) putting m V value of b in terms of a from equation (ii), in equation (i), we get a2 + (a – 16)2 = 640 a2 + a2 – 32a + 256 = 640 2a2 –32a –384 = 0 a2 – 16a – 192 = 0 a2 – 24a + 8a –192 =0 ⇒ a(a – 24) + 8(a – 24) = 0 (a – 24) ( a + 8)=0 a = 24 or a = - 8, (not allowed) ∴ a =24 and then b = 24-16 =8 so given length of the sides of the two squares are 24 and 8 28. 100 surnames were randomly picked up from a local telephone

directory and the distribution of number of letters of the English alphabet in the surnames was obtained as follows: No. of letters 1−4 4−7 7−10 10−13 13−16 16−19 Number of surnames

6 30 40 16 4 4

Determine the median and mean number of letters is the surnames. Also find the modal size of surname.

Ans.

No. of letter Number of surnames (fi)

xi fixi

1-4 4-7 7-10

6 30 40

2.5 5.5 8.5

15 165 340

10-13 13-16 16−19

16 4 4

11.5 14.5 17.5

184 58 70

Total Σfi = 100 Σfixi = 832

Mean ( ) 32.8

100832

fxfxi

ii ==∑

∑=

No. of letters frequency Cumulative frequency 1-4 4-7 7-10 10-13 13-16 16−19

6 30 40 16 4 4

6 36 76 92 96 100

n = 100 ⇒ 502

1002n

== ∴ l = 7 cf = 36 f = 40 h = 3

Median =

hf

cf2n

l ×⎟⎟⎟⎟

⎠

⎞

⎜⎜⎜⎜

⎝

⎛ −+

=

340

362

100

7 ×⎟⎟⎟⎟

⎠

⎞

⎜⎜⎜⎜

⎝

⎛ −+

= 8.05.

No. of letters frequency 1-4 4-7 7-10 10-13 13-16 16−19

6 30 40 16 4 4

Model class = 7 − 10 l = 7 h = 3 f1 = 40 f0 = 30 f2 = 16

Mode = h

fff2ff

l201

01 ×⎟⎟⎠

⎞⎜⎜⎝

⎛−−

−+

= 3

163040230407 ×⎟

⎠⎞

⎜⎝⎛

−−×−

+

= 7.88.

29. Prove that the ratio of the areas of two similar triangles is equal to the ratio of squares of their corresponding sides.

Using the above result, prove the following:

In a ∆ ABC, XY is parallel to BC and it divides ∆ABC into two parts

of equal area. Prove that 212 −

=ABBX

Ans. Let ∆ ABC similar to ∆DEF ∆ABC ~ DEF ∠ 1 = ∠3 ∠ AGB = ∠DHE = 900 each ∴ By AA property ∆ ABG ~ ∆DEH

A

B C G1 2

∴ DEAB

DHAG

= (1)

D

E F H3 4

Now DFAB

EFBC

DHEF

AGBC

DEFArABCAr

×=×

×=

∆(∆

2121

))(

from (1)

Now Q )()(

DEFArABCAr

∆∆

= 2

2

EFBC

EFBC

EFBC

=×

2

2

2

2

DFAC

DEAB

== Q ∆ABC ~ ∆ DEF

Second part

21

=ABCArAXYAr

(given )

∴ 21

2 =ABAX

(Proved above)

A

X Y

B C

∴ 212

2111

21

21 −

=−

=−=−⇒−=−

⇒=AB

AXABABAX

ABAX

ABAX

212

ABBX −

=⇒

30. A bucket made up of a metal sheet is in the form of a frustum of a

cone of height 16 cm with diameters of its lower and upper ends as 16 cm and 40 cm respectively. Find the volume of the bucket. Also find the cost of the bucket if the cost of metal sheet used is Rs 20 per 100 cm2. (use π = 3.14)

OR A farmer connects a pipe of internal diameter 20 cm from a canal into

a cylindrical tank in his field which is 10 m in diameter and 2 m deep. If water flows through the pipe at the rate of 6 km/h., in how much time will the tank be filled?

30. Let OM = h ∆OND ~ ∆OMB

∴ hh16

820

OMLN

BMDN +

=⇒= ⇒ 20h = 128 + 8h ⇒ h = 10.67 cm

C D 40 cm

16 cm

h

l1

l2

M r2

16 cm

N

r1

O

A B

Volume of cone = hr31 2π

Volume of bucket = [ ]67.10867.262031 22 ×−×π

= 10451.09 cm3 Curved surface of bucket = πrl Area of base = πr2 ∴ total surface of bucket = πrl + πr2 ∴ Curved surface of bucket = π × 20 × (l1) − π × 8 × (l2)

where a34.3367.2620l 221 =+=

and l2 = a34.1367.108 22 =+ = 20 × 3.14 × 33.34 − 8 × 3.14 × 13.34 = 1758.65 Base area of bucket = π × 82 = 200.96 Total surface of bucket = 1758.65 + 200.96 = 1959.61 cm2

Cost of metal = 92.39161.195910020

=×

OR

Volume of tank = πr2h = π × 52 × 2 = 50π m3 The volume of tater collected in the tank in 1 hr. = 6 × 1000 × π (0.1)2 = 60 π m3

∴ Total time taken = .hr65

6050

=ππ

= 50 min.