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30/1/2 2

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Question numbers 1 to 4 carry 1 mark each.

�% ��/��3, 4, 5, ....., 50 �K��-��&'I-��1M��7��N2�E��

��-.-&'()1M��-��/��ON2��&<()��P��,��-

��.-��/��-��?>�=.��K��&<()

Cards marked with number 3, 4, 5, ....., 50 are placed in a box and mixed

thoroughly. A card is drawn at random from the box. Find the

probability that the selected card bears a perfect square number.



30/1/2 [P.T.O. 3

'% ��:��IAB -��Q�ER��70H�&<�+�CD -��E89�E&<��F<��+

60° ��>1��E&<�+�70H��1S*#D ��&T��E&<()��*AD = 2.54 �E&<I

��E89�E��,�01��P��,��-()U 3 =:BVC�E��-W

��:

In Fig. 1, AB is a 6 m high pole and CD is a ladder inclined at an

angle of 60° to the horizontal and reaches up to a point D of pole. If

AD = 2.54 m, find the length of the ladder. (use 3 =1.73)

Fig. 1

�% ��X�89�E5, 9, 13, ....., 185 ��*�U�+��*��,��WY=��*P��

��,��-()

Find the 9th term from the end (towards the first term) of the

A.P. 5, 9, 13, ....., 185.



30/1/2 4

(% -��1�Z�1S*#P �O ��S$=��=�4��*��6�[��7�-�PA �+�PB 7D�E.��&'()��*

∠ PAB = 50º &<I��∠ AOB P��,��-()

From an external point P, tangents PA and PB are drawn to a circle with

centre O. If ∠ PAB = 50º, then find ∠ AOB.

��01�2�

��&��)�

� ��K��;�:A��J�� \��&<()

Question numbers 5 to 10 carry 2 marks each.

*% �1S*#P ��*][�� 3��"��*][�� *#.#�� &<() ��* �1S*#P, �1S*#��Q(2, –5) �+�

R(–3, 6) ��*?�6+&<I��P ��*][��P��,��-()

The ��coordinate of a point P is twice its "�coordinate. If P is equidistant from

Q (2, –5) and R(–3, 6), find the coordinates of P.

�% ��\��I-��∆ ABC ��.��-��=�4�1��&<��"H#��,H#��AB, BC �+� CA

�� ^��[ �1S*#��D, E �+�F �� 6�[� �� &<() ��*AB, BC �+�CA ��,�01��

�^��[ :\��EI_��E�+�:A��E&<�I��AD, BE �+�CF ��,�01��P��,��-()

��\



30/1/2 [P.T.O. 5

In Fig. 2, a circle is inscribed in a ∆ ABC, such that it touches the sides AB, BC

and CA at points D, E and F respectively. If the lengths of sides AB, BC and CA

are 12 cm, 8 cm and 10 cm respectively, find the lengths of AD, BE and CF.

Fig. 2

+% ��C��IO��S$=��=�4��,AP �+�BP -�E*��6�[��7�-�&'��AP = 5 ��E�+�

∠APB = 60° &<()�E=�AB ��,��1��P��,��-()

��C

In Fig. 3, AP and BP are tangents to a circle with

centre O, such that AP = 5 cm and ∠APB = 60°.

Find the length of chord AB.

Fig. 3



30/1/2 6

,% ��*2

3� = �+� 3� = − -��`��E�E��> 2 7 0�� ++ + = ��?�&'I��+�+��

��P��,��-()

If 2

3� = and 3� = − are roots of the quadratic equation

2 7 0,�� ++ + = find the

values of � and +.

-% =&��#��P��,��- ��"��FI �1S*#�� A(5, 6)− �+� B( 1, 4)− − �� =��

��7�7�/��1��&<()�=H��=��1S*#��*][��HEP��,��-()

Find the ratio in which "�axis divides the line segment joining the points

A(5, 6)− and B( 1, 4).− − Also find the coordinates of the point of division.

�.% ��X�89�E27, 24, 21, .... ��*��-��-��3��.[?S�&��(a

How many terms of the A.P. 27, 24, 21, .... should be taken so that their sum is

zero?

��01��

��&��

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��% ��*-��X�89�E��+�V�*��.LY�+��+�:V�*��.\_Y&<I��

X�89�E��+��*��.P��,��-()

If the sum of first 7 terms of an A.P. is 49 and that of its first 17 terms is 289,

find the sum of first ��terms of the A.P.

�



30/1/2 [P.T.O. 7

�'% L�Eb��-��#��I\:�E��,.&��7��*��&<()��E&T��c�E��#�-��

��C�E��<8/�E-��=�4��=��(ring) 1��&T-��d��e<��-��1��1��

��&<()1��,R��P��,��-()

A well of diameter 4 m is dug 21 m deep. The earth taken out of it has been

spread evenly all around it in the shape of a circular ring of width 3 m to form an

embankment. Find the height of the embankment.

��% �� L ��IABCD -�� =.� &< ��, H#�� :L ��E &<() �J�� H#�� b��

��=�4�1��-.-&'()2��H�.��F�"e��P��,��-()22

7

π =

�E��-

��L

In Fig. 4, ABCD is a square of side 14 cm. Semi�circles are drawn with each side

of square as diameter. Find the area of the shaded region. 22

use 7

π =

Fig. 4



30/1/2 8

�(% �� ; ��I ��=�� - 1�� -�� f�� *[�� .�� &< �� *�� g��!-�� +� -��

��.��1��&<()f��-��Q��EH#��&<�+�3��R��-��.��

&<��b��CB;��E&<()f��#��%�E�F�"e��P��,��-()22

7

π =

�E��-

��;

In Fig. 5, is a decorative block, made up of two solids–a cube and a hemisphere.

The base of the block is a cube of side 6 cm and the hemisphere fixed on the top

has a diameter of 3.5 cm. Find the total surface area of the block. 22

use 7

π =

Fig. 5



30/1/2 [P.T.O. 9

�*% ��Q��I1�E∆ABCI��[Eh�A ��*][��(0, –1) &'�+�H#��AB �+�AC

��i�!�1S*#��D �+�E �� *][��^��[ (1, 0) �+� (0, 1) &'()��*F H#��BC ��

�i�!�1S*#&<��"H#��DEF �+�ABC ��F�"e��P��,��-()

��Q

In Fig. 6, ABC is a triangle coordinates of whose vertex A are (0, –1). D and E

respectively are the mid�points of the sides AB and AC and their coordinates are

(1, 0) and (0, 1) respectively. If F is the mid�point of BC, find the areas of

∆ ABC and ∆ DEF.

Fig. 6



30/1/2 10

��% ��V��I*��PAQ �+�PBQ *[��.��&'()��PAQ, O ��S$=��=�4��H�.&<I

��, �"j�� OP &< �+� �� PBQ, PQ �� b�� 1�� .�� =�4� &<

��S$M &<()��*OP = 10 ��E�+�PQ = 10 ��E��*[��-��7��H�.��

F�"e�� 225 36

π −

�� E &<()

��V

In Fig. 7 are shown two arcs PAQ and PBQ. Arc PAQ is a part of circle with

centre O and radius OP while arc PBQ is a semi�circle drawn on PQ as diameter

with centre M. If OP = PQ = 10 cm show that area of shaded region is

225 3 cm6

π −

.

Fig. 7



30/1/2 [P.T.O. 11

�+% -��E�� [7��-��;A�ER��H=�� [7��+��*��=��>�^��[ 45°

�+�60° &' ) �E�� , R�� P�� ,��- �+� H=� = �E�� 1E�F<�� *?�E HE P��

��,��-()U 3 =:BVC�E��-W

The angles of depression of the top and bottom of a 50 m high building from the

top of a tower are 45° and 60° respectively. Find the height of the tower and the

horizontal distance between the tower and the building. (use 3 =1.73)

�,% � ��-&��,��-(

1 2 2 3

4 ; 1, 2,21 2 2

+ − ++ = − ≠ −

− + −

� � ��

� � �

Solve for � :

1 2 2 3

4 ; 1, 2,21 2 2

+ − ++ = − ≠ −

− + −

� � ��

� � �

�-% *��=�Hk��-��+e��.��()��0��l��,��P��,��-(

� (i) *��C�18/�E�K��l��(

� (ii) *��=��E�K��.Q�+=�V&��(

Two different dice are thrown together. Find the probability of :

(i) getting a number greater than 3 on each die

(ii) getting a total of 6 or 7 of the numbers on two dice



30/1/2 12

'.% C��E�"j��=��-��01=�4�E�[��#��=�^��%��F�"e��LVB:��E2&<()3��

P��,��-()Uπ = 3.14 �E��-W

A right circular cone of radius 3 cm, has a curved surface area of 47.1 cm2. Find

the volume of the cone. (use π = 3.14)

�

��01�(�

��&��/�

� ��K��\:�C:��J�� L��&<()


'�% =��#��89��-��"EE89�E��e��&��.��()��3��"E��

&=��m��E�5M��-��.��>=��#��

��&��.��():;AA��E*?�.��b��&T��-I��"E�.�E38/��8/�

��I��,.��\;A��En��189��*E()=��#��,�?�.��P��,��-()

� �� M��?o�*[��.��&<(a

A passenger, while boarding the plane, slipped from the stairs and got hurt. The

pilot took the passenger in the emergency clinic at the airport for treatment. Due

to this, the plane got delayed by half an hour. To reach the destination 1500 km

away in time, so that the passengers could catch the connecting flight, the speed

of the plane was increased by 250 km/hour than the usual speed. Find the usual

speed of the plane.

What value is depicted in this question?



30/1/2 [P.T.O. 13

''% ��_��I;��E�"j��=��O ��S$��=�4�&<()T -��-��1S*#&<��OT = 13 ��E&<

�+�OT =�4�� 1S*#E ��E&<()��*AB �1S*#E ��=�4��,6�[��7�&<��AB ��,

��1��P��,��-I�1��TP �+�TQ =�4��,*��6�[��7�-�&'()

��_

In Fig. 8, O is the centre of a circle of radius 5 cm. T is a point such that

OT = 13 cm and OT intersects circle at E. If AB is a tangent to the circle at E,

find the length of AB, where TP and TQ are two tangents to the circle.

Fig. 8

'�% �p��,��-��=�4��1�Z!�1S*#�=�4��7D�E.��6�[��7�-��1��&��E&'()

Prove that the lengths of tangents drawn from an external point to a circle are

equal.



30/1/2 14

'(% �p ��,��- �� -�� "H#�I �� [Eh� ( , 2)� � − , ( 2, 2)� �+ + �+� ( 3, )� �+ &<I ��

F�"e��6=��"&<()

Prove that the area of a triangle with vertices ( , 2)� � − , ( 2, 2)� �+ + and ( 3, )� �+

is independent of �.

'*% ��.��7��-��E��-��=�4��1��/��I��_��H�.�� =H��&<I��#��

��&<��K��:I\ICIBBBI_U��YW��E-��K��.�q��&<()��*

�E��,HE�K��q��.��&<��P��,��-��E�(i) ��E

�=h��K��q��.�(ii) C�18/�E�K��q��.�(iii) Y�2��E�K��q��.�()

��Y

A game of chance consists of spinning an arrow on a circular board, divided into

8 equal parts, which comes to rest pointing at one of the numbers 1, 2, 3, ..., 8

(Fig. 9), which are equally likely outcomes. What is the probability that the

arrow will point at (i) an odd number (ii) a number greater than 3 (iii) a number

less than 9.

Fig. 9



30/1/2 [P.T.O. 15

'�% -��E�E 1<o�� -�� E (pulley), ��, �"j�� ;��E &<I �� .*� �� .�� &<()

U��:AW()1<o��-��1S*#C ��E�E1<o��S$O �E��7D��&<

��=&�1S*#P ��&T��&<�&��OP =10 ��E&<()1<o��3H�.��,��1��P��,��-

��1HE��E��+�.�&<()��7��H�.��F�"e��HEP��,��-()

� (π = 3.14 �+� 3 =1.73 �E��-)�

��:A

An elastic belt is placed around the rim of a pulley of radius 5 cm. (Fig. 10)

From one point C on the belt, the elastic belt is pulled directly away from the

centre O of the pulley until it is at P, 10 cm from the point O. Find the length of

the belt that is still in contact with the pulley. Also find the shaded area.

(use π = 3.14 and 3 =1.73)

Fig. 10

'+% R��7#�E-��1�o��E[��#� �� 2�k��,&< ��,��:\CA_B_��E3��,

&<()3��R��E�+��=�4�E��,�"j��-��^��[r\A��E�+�:\��E&<�()1�o��E��,

R�� P�� ,��- �+� 1�o��E �� 1�� .E ��# ��, ��*� �� F�"e�� P�� ,��-()

Uπ = 3.14 �E��-W



30/1/2 16

A bucket open at the top is in the form of a frustum of a cone with a capacity of

12308.8 cm3. The radii of the top and bottom circular ends are 20 cm and 12 cm

respectively. Find the height of the bucket and the area of metal sheet used in

making the bucket. (use π = 3.14)

',% -��E��<�-��7��L�E��<�Y�E��,*?�E��56+�*�� 1S*#��

�E��[7��3k��>�^��[ 60°�+�30°&'()�E��,R��P��,��-()

The angles of elevation of the top of a tower from two points at a distance of 4 m

and 9 m from the base of the tower and in the same straight line with it are 60°

and 30° respectively. Find the height of the tower.

'-% -��"H#�ABC ��,��,��-��BC = 6 ��EIAB = 5 ��E�+�(∠ABC = 60° )

�e��-��"H#��,��,��-��,H#��-�∆ ABC ��,�.�H#��3

4 H�.&<(�)

Construct a triangle ABC in which BC = 6 cm, AB = 5 cm and ∠ABC = 60°.

Then construct another triangle whose sides are 3

4 times the corresponding sides

of ∆ ABC.

�.% -��> �"H#�� QA��E &< �+�3�� >� ��,��1�� \;��E &<() �"H#��

F�"e��P��,��-()

The perimeter of a right triangle is 60 cm. Its hypotenuse is 25 cm. Find the area

of the triangle.

��% -��E��;A�En��,-��.��*�<8/��&<()\��1�*-��#��

=��3��8/��-*�<8/��&<()��*=&�&��QA�E*�<8/��&<�+��.��

��E��;�En��189��*��&<I��1�*�#��=��8/��.�(a

A thief, after committing a theft, runs at a uniform speed of 50 m/ minute. After

2 minutes, a policeman runs to catch him. He goes 60 m in first minute and

increases his speed by 5 m/minute every succeeding minute. After how many

minutes, the policeman will catch the thief?



Strictly Confidential — (For Internal and Restricted Use Only)

Secondary School Certificate Examination

March 2016

Marking Scheme — Mathematics 30/1/1, 30/1/2, 30/1/3

General Instructions:

1. The Marking Scheme provides general guidelines to reduce subjectivity in the marking. The answers

given in the Marking Scheme are suggested answers. The content is thus indicative. If a student has

given any other answer which is different from the one given in the Marking Scheme, but conveys

the meaning, such answers should be given full weightage

2. Evaluation is to be done as per instructions provided in the marking scheme. It should not be done

according to one’s own interpretation or any other consideration — Marking Scheme should be

strictly adhered to and religiously followed.

3. Alternative methods are accepted. Proportional marks are to be awarded.

4. In question (s) on differential equations, constant of integration has to be written.

5. If a candidate has attempted an extra question, marks obtained in the question attempted first should

be retained and the other answer should be scored out.

6. A full scale of marks - 0 to 100 has to be used. Please do not hesitate to award full marks if the

answer deserves it.

7. Separate Marking Scheme for all the three sets has been given.

8. As per orders of the Hon’ble Supreme Court. The candidates would now be permitted to obtain

photocopy of the Answer book on request on payment of the prescribed fee. All examiners/Head

Examiners are once again reminded that they must ensure that evaluation is carried out strictly as

per value points for each answer as given in the Marking Scheme.



QUESTION PAPER CODE 30/1/1

EXPECTED ANSWER/VALUE POINTS

SECTION A

1. ∠APB = 80º1

2

∴ ∠AOB = 100°1

2

2. DB = 3.46 m1

2

∴ DC = 4 m1

2

3. l = 185, d = –41

2

l9 = 153

1

2

4. Possible outcomes are 4, 9, 16, 25, 36, 49, i.e. 6.1

2

∴ P(perfect square number) = 6 1

or48 8

1

2

SECTION B

5.–7 2

– 3a 3

=

⇒ a = 3 1

andb 2

(–3)a 3

= ×

⇒ b = – 6 1

6. Let the point on y-axis be (0, y) and AP: PB = K : 11

2

Therefore 5 – k

0k 1

=+

gives k = 5

Hence required ratio is 5 : 1.1

2

30/1/1 (1)

30/1/1



–4(5) – 6 –13y

6 3 = =

1

2

Hence point on y-axis is –13

0, .3

1

2

7. Let AD = AF = x

∴ DB = BE = 12 – x

and CF = CE = 10 – x

BC = BE + EC ⇒ 8 = 12 – x + 10 – x

⇒ x = 7 1

∴ AD = 7 cm, BE = 5 cm, CF = 3 cm 1

8. Let the point P be (2y, y)1

2

PQ = PR ⇒ 2 2 2 2(2y – 2) (y 5) (2y 3) (y – 6)+ + = + +

1

2

Solving to get y = 81

2

Hence coordinates of point P are (16, 8).1

2

9. Here a = 18, d = –2, Sn = 01

2

Thereforen

[36 (n –1) (–2)] 02

+ = 1

⇒ n = 191

2

10. PA = PB1

2

⇒ ∠PAB = ∠PBA = 60°1

2

∴ ∆PAB is an equilateral triangle.1

2

Hence AB = PA = 5 cm.1

2

(2) 30/1/1

30/1/1



SECTION C

11. Area of square = 196 cm21

2

Area of semicircles AOB + DOC = 22249 154 cm

7× =

1

2

Hence area of two shaded parts (X + Y) = 196 – 154 = 42 cm2 1

Therefore area of four shaded parts = 84 cm2. 1

12. Surface area of block = 22 3.5 3.5 22 3.5 3.5

216 – 27 2 2 7 2 2

× × + × × ×1 1

12 2

+ +

= 225.42 cm2. 1

13. Using Mid Point formula

coordinates of point B are (2, 1)1

2

and coordinates of point C are (0, 3).1

2

Area ∆ABC = 1

[0 2(3 1) 0] 4 sq u.2

+ + + = 1

Coordinates of point F are (1, 2)

Area of ∆DEF = 1

|1(1– 2) 0 1(0 –1) | 1sq u.2

+ + = 1

14. ∠POQ = 60º1

2

Area of segment PAQM = 2100 100 3– cm .

6 4

π

1

Area of semicircle = 225cm

2

π 1

2

Area of shaded region = 25 50

– – 25 3 .2 3

π π

= 225 3 – cm .6

π

1

X O

B

C

A

D

Y

30/1/1 (3)

30/1/1



15. S7 = 49 ⇒ 2a + 6d = 14

1

2

S17

= 289 ⇒ 2a + 16d = 341

2

Solving equations to get a = 1 and d = 2 1

Hence Sn = 2n

[2 (n –1)2] n .2

+ = 1

16. 2x(2x + 3) + (x – 3) + (3x + 9) = 0 1

⇒ 2x2 + 5x + 3 = 0 1

⇒ (x + 1) (2x + 3) = 01

2

⇒ x = –1, x = 3

–2

1

2

17. Volume of earth dug out = π × 2 × 2 × 21 = 264 m3 1

Volume of embankment = π (25 – 4) × h = 66 h m3 1

∴ 66h = 2641

2

⇒ h = 4 m1

2

18. Here r + h = 37 and 2πr(r + h) = 16281 1

2 2+

⇒1628

2 r37

π =

⇒ r = 7 cm1

2

and h = 30 cm.1

2

Hence volume of cylinder = 322

7 7 30 4620 cm7

× × × = 1

(4) 30/1/1

30/1/1



19. Correct Figure1

2

h – 50tan 45º x h – 50

x = ⇒ =

1

2

h htan 60 x

x 3° = ⇒ =

1

2

Hence h

h – 503

=1

2

⇒ h 75 25 3 118.25 m.= + = 1

20. (i) Favourable outcomes are (2, 2) (2, 3) (2, 5) (3, 2) (3, 3) (3, 5) (5, 2) (5, 3) (5, 5)

i.e. 9 outcomes. 1

P(a prime number on each die) = 9 1

or36 4

1

2

(ii) Favourable outcomes are (3, 6) (4, 5) (5, 4) (6, 3) (5, 6) (6, 5)

i.e. 6 outcomes 1

P(a total of 9 or 11) = 6 1

or36 6

1

2

SECTION D

21. Let the usual speed of plane be x km/h.

∴1500 1500 1

–x x 250 2

=+

2

⇒ x2 + 250x – 750000 = 0

(x + 1000) (x – 750) = 0 ⇒ x = 750

Speed of plane = 750 km/h. 1

For writing value 1

22. For correct Given, To prove, construction and figure1

4 22

× =

Correct proof 2

23. Construction of tangent 3

Length of tangent 1

h

x

60°

45°

45°60°

30/1/1 (5)

30/1/1



24. PT 169 – 25 12cm = = and TE = 8 cm1 1

2 2+

Let PA = AE = x

T AT AT AT A

2 = TE2 + EA2 1

⇒ (12 – x)2 = 64 + x2

⇒ x = 3.3 cm. 1

Thus AB = 6.6 cm. 1

25. a(x – b) (x – c) + b(x – a) (x – c) = 2c (x – a) (x – b)1

12

x2(a + b – 2c) + x (–ab – ac – ab – bc + 2ac + 2bc) = 0

x2(a + b – 2c) + x(–2ab + ac + bc) = 01

12

x = ac bc – 2ab

a b – 2c

+

+1

26. Correct Figure 1

80tan 45

y° = ⇒ y = 80

1

2

80tan30

x y° =

+ ⇒ x + y = 80 3

1

2

∴ x = 80( 3 –1) 58.4 m.= 1

Hence speed of bird = 58.4

29.2 m/s.2

= 1

27. Let total time be n minutes

Total distance convered by thief = (100 n) metres1

2

Total distance covered by policeman = 100 + 110 + 120 + ... + (n – 1) terms1

2

∴ 100n = n –1

[200 (n 2)10]2

+ − 1

n2 – 3n – 18 = 01

2

30°45°

B (Bird)D

x

80 m 80 m

CAx y

(6) 30/1/1

30/1/1



(n – 6)(n + 3) = 01

2

⇒ n = 61

2

Policeman took 5 minutes to catch the thief.1

2

28. Area of the triangle =1

t(t 2 – t) (t 2)(t – t 2) (t 3)(t – 2 – t – 2)2

+ + + + + + 2

= 1

[2t 2t 4 – 4t –12]2

+ + 1

= 4 sq. units

which is independent of t. 1

29. (i) Favourable outcomes are 1, 3, 5, 7 i.e. 4 outcomes. 1

∴ P(an odd number) = 4 1

or8 2

1

2

(ii) Favourable outcomes are 4, 5, 6, 7, 8 i.e. 5 outcomes 1

P(a number greater than 3) = 5

8

1

2

(iii) Favarouble outcomes are 1, 2, 3...8

8P(a number less than 9) = 1

8

=

1

30.1

cos 602

θ = ⇒ θ = °1

2

Reflex ∠AOB = 240º1

2

∴ �2 3.14 5 240

ADB 20.93 cm360

× × ×= = 1

Hence length of elastic in contact = 20.93 cm

Now, AP = 5 3 cm

Area 2( OAP OBP) 25 3 43.25 cm∆ + ∆ = =1

2

Area of sector OACB = 225 3.14 120

26.16 cm360

× ×=

1

2

Shaded Area = 43.25 – 26.16 = 17.09 cm2 1

P

A

B

D Oθ C

10 cm

5

30/1/1 (7)

30/1/1



31. Here R = 20, r = 12, V = 12308.8

Therefore 12308.8 = 1

3.14(400 240 144)h3

× + + 1

⇒ h = 15 cm1

2

2 2(20 –12) 15 17 cm= + =l

1

2

Total area of metal sheet used = CSA + base area

= π[(20 + 12) × 17 + 12 × 12] 1

= 2160.32 cm2 1

(8) 30/1/1

30/1/1





SECTION A


2


or48 8

1

2

2. DB = 3.46 m1

2

∴ DC = 4 m1

2

3. l = 185, d = –41

2

l9 = 153

1

2

4. ∠APB = 80º1

2

∴ ∠AOB = 100°1

2

SECTION B


2

PQ = PR ⇒ 2 2 2 2(2y – 2) (y 5) (2y 3) (y – 6)+ + = + +

1

2


2


2

6. Let AD = AF = x

∴ DB = BE = 12 – x


BC = BE + EC ⇒ 8 = 12 – x + 10 – x

⇒ x = 7 1

∴ AD = 7 cm, BE = 5 cm, CF = 3 cm 1

30/1/2 (9)

30/1/2



7. PA = PB1

2

⇒ ∠PAB = ∠PBA = 60°1

2


2


2

8.–7 2

– 3a 3

=

⇒ a = 3 1

andb 2

(–3)a 3

= ×

⇒ b = – 6 1


2

Therefore 5 – k

0k 1

=+

gives k = 5


2

–4(5) – 6 –13y

6 3 = =

1

2


0, .3

1

2

10. Here a = 27, d = – 3, Sn = 01

2

∴ 54 + (n – 1) (–3) = 0 1

⇒ n = 191

2

(10) 30/1/2

30/1/2



SECTION C

11. S7 = 49 ⇒ 2a + 6d = 14

1

2

S17

= 289 ⇒ 2a + 16d = 341

2


Hence Sn = 2n

[2 (n –1)2] n .2

+ = 1



∴ 66h = 2641

2

⇒ h = 4 m1

2


2


7× =

1

2




216 – 27 2 2 7 2 2

× × + × × ×1 1

12 2

+ +

= 225.42 cm2. 1



2


2

Area ∆ABC = 1

[0 2(3 1) 0] 4 sq u.2

+ + + = 1


Area of ∆DEF = 1

|1(1– 2) 0 1(0 –1) | 1sq u.2

+ + = 1

30/1/2 (11)

30/1/2

X O

B

C

A

D

Y



16. ∠POQ = 60º1

2


6 4

π

1


2

π 1

2


– – 25 3 .2 3

π π

= 225 3 – cm .6

π

1

17. Correct Figure1

2

h – 50tan 45º x h – 50

x = ⇒ =

1

2

h htan 60 x

x 3° = ⇒ =

1

2

Hence h

h – 503

=1

2

⇒ h 75 25 3 118.25 m.= + = 1

18.

2 2

2

x 3x 2 x – 3x 2 4x – 8 – 2x – 3

x – 2x x – 2

+ + + +=

+1

(2x2 + 4)(x – 2) = (2x – 11) (x2 + x – 2 )

⇒ 5x2 + 19x – 30 = 0 1

⇒ (5x – 6) (x + 5) = 01

2

⇒ x = –5, 6/51

2

19. (i) Favourable outcomes are (4, 5)(4, 4)(4, 6)(5, 4)(5, 5)(5, 5)(5, 6)(6, 4)(6, 5)(6, 6)

i.e., 9 outcomes 1

P(a number > 3 on each die) = 9 1

or36 4

1

2

h

x

60°

45°

45°60°

(12) 30/1/2

30/1/2



(ii) Favourable outcomes are (1, 5)(2, 4)(3, 3)(4, 2)(5, 1)(1, 6)(2, 5)(3, 4)(4, 3)(5, 2)(6, 1)

i.e. 11 outcomes 1

P(a total of 6 to 7) = 11

36

1

2

20. Here r = 3, πrl = 47.1

∴47.1

5 cm3 3.14

= =×

l 1

2 2h 5 – 3 4 cm= =

1

2

Volume of cone = 1

3.14 3 3 43

× × × ×1

2

= 37.68 cm3 1

SECTION D


∴1500 1500 1

–x x 250 2

=+

2

⇒ x2 + 250x – 750000 = 0

(x + 1000) (x – 750) = 0 ⇒ x = 750


For writing value 1

22. PT 169 – 25 12cm = = and TE = 8 cm1 1

2 2+

Let PA = AE = x

TA2 = TE2 + EA2 1

⇒ (12 – x)2 = 64 + x2

⇒ x = 3.3 cm. 1

Thus AB = 6.6 cm. 1

23. For correct Given, To prove, construction and figure1

4 22

× =

Correct proof 2

30/1/2 (13)

30/1/2




t(t 2 – t) (t 2)(t – t 2) (t 3)(t – 2 – t – 2)2

+ + + + + + 2

= 1

[2t 2t 4 – 4t –12]2

+ + 1

= 4 sq. units




or8 2

1

2



8

1

2



8

=

1

26.1

cos 602

θ = ⇒ θ = °1

2


2

∴ �2 3.14 5 240

ADB 20.93 cm360

× × ×= = 1


Now, AP = 5 3 cm

Area 2( OAP OBP) 25 3 43.25 cm∆ + ∆ = =1

2


26.16 cm360

× ×=

1

2

Shaded Area = 43.25 – 26.16 = 17.09 cm2 1

27. Here R = 20, r = 12, V = 12308.8


3.14(400 240 144)h3

× + + 1

⇒ h = 15 cm1

2

P

A

B

D Oθ C

10 cm

5

(14) 30/1/2

30/1/2



2 2(20 –12) 15 17 cm= + =l

1

2


= π[(20 + 12) × 17 + 12 × 12] 1

= 2160.32 cm2 1

28. This question contains surplus data which does not lead to a unique solution.

Hence 4 marks should be given to every student. 4

29. Correct contruction

30. Here a + b + c = 60, c = 25

∴ a + b = 35 1

Using Pythagorus theorem

a2 + b2 = 625

Using identity (a + b)2 = a2 + b2 + 2ab 1

(35)2 = 625 + 2ab

⇒ ab = 300 1

Area of ∆ABC = 150 cm2 1

31. Let total time be n minutes

Total distance covered by thief = (50 n) metres1

2

Total distance covered by policeman = 60 + 65 + 70 + ... + (n – 2) terms1

2

∴n – 2

50n [120 (n – 3)5]2

= + 1

⇒ n2 – n – 42 = 01

2

(n – 7)(n + 6) = 01

2

∴ n = 71

2

Policeman took 5 minutes to catch the thief.1

2

B

C A

c = 25 cma

b

30/1/2 (15)

30/1/2





SECTION A

1. l = 185, d = –41

2

l9 = 153

1

2


2


or48 8

1

2

3. ∠APB = 80º1

2

∴ ∠AOB = 100°1

2

4. DB = 3.46 m1

2

∴ DC = 4 m1

2

SECTION B


2

Therefore 5 – k

0k 1

=+

gives k = 5


2

–4(5) – 6 –13y

6 3 = =

1

2


0, .3

1

2

(16) 30/1/3

30/1/3



6.–7 2

– 3a 3

=

⇒ a = 3 1

andb 2

(–3)a 3

= ×

⇒ b = – 6 1


2

PQ = PR ⇒ 2 2 2 2(2y – 2) (y 5) (2y 3) (y – 6)+ + = + +

1

2


2


2

8. Let AD = AF = x

∴ DB = BE = 12 – x


BC = BE + EC ⇒ 8 = 12 – x + 10 – x

⇒ x = 7 1

∴ AD = 7 cm, BE = 5 cm, CF = 3 cm 1

9. PA = PB1

2

⇒ ∠PAB = ∠PBA = 60°1

2


2


2

10. Here a = 65, d = –5, Sn = 01

2

130 + (n –1) (– 5) = 0 1

⇒ n = 271

2

30/1/3 (17)

30/1/3



SECTION C



∴ 66h = 2641

2

⇒ h = 4 m1

2

12. S7 = 49 ⇒ 2a + 6d = 14

1

2

S17

= 289 ⇒ 2a + 16d = 341

2


Hence Sn = 2n

[2 (n –1)2] n .2

+ = 1

13. Correct Figure1

2

h – 50tan 45º x h – 50

x = ⇒ =

1

2

h htan 60 x

x 3° = ⇒ =

1

2

Hence h

h – 503

=1

2

⇒ h 75 25 3 118.25 m.= + = 1


2


7× =

1

2




216 – 27 2 2 7 2 2

× × + × × ×1 1

12 2

+ +

= 225.42 cm2. 1

h

x

60°

45°

45°60°

(18) 30/1/3

30/1/3

X O

B

C

A

D

Y





2


2

Area ∆ABC = 1

[0 2(3 1) 0] 4 sq u.2

+ + + = 1


Area of ∆DEF = 1

|1(1– 2) 0 1(0 –1) | 1sq u.2

+ + = 1

17. ∠POQ = 60º1

2


6 4

π

1


2

π 1

2


– – 25 3 .2 3

π π

= 225 3 – cm .6

π

1

18. (i) Number of good shirts = 88 1

P(Ramesh buys the shirt) = 88 22

or100 25

1

2

(ii) Number of shirts without Major defect = 96 1

P(Kewal buys a shirt) = 96 24

or100 25

1

2

19.2 a a b

x x x 1 0a b a

++ + + =

+

a a b ax x x 0

a b a a b

+ + + + =

+ + 1

a a bx x 0

a b a

+ + + =

+ 1

⇒ –a –(a b)

x ,a b a

+=

+1

30/1/3 (19)

30/1/3



20. h = 15.5 – 3.5 = 12 cm1

2

144 12.25 12.5 cm= + =l

1

2

TSA = = = = πrl + 2πr2

= 22 22

3.5 12.5 2 3.5 3.57 7

× × + × × × 1

= 137.5 + 77

= 214.5 cm2 1

SECTION D


∴1500 1500 1

–x x 250 2

=+

2

⇒ x2 + 250x – 750000 = 0

(x + 1000) (x – 750) = 0 ⇒ x = 750


For writing value 1

22. PT 169 – 25 12cm = = and TE = 8 cm1 1

2 2+

Let PA = AE = x

TA2 = TE2 + EA2 1

⇒ (12 – x)2 = 64 + x2

⇒ x = 3.3 cm. 1

Thus AB = 6.6 cm. 1

23. Correct Figure 1

80tan 45

y° = ⇒ y = 80

1

2

80tan30

x y° =

+ ⇒ x + y = 80 3

1

2

∴ x = 80( 3 –1) 58.4 m.= 1

Hence speed of bird = 58.4

29.2 m/s.2

= 1

15.5 cm

h l

3.5

30°45°

B (Bird)D

x

80 m 80 m

CAx y

(20) 30/1/3

30/1/3





or8 2

1

2



8

1

2



8

=

1


t(t 2 – t) (t 2)(t – t 2) (t 3)(t – 2 – t – 2)2

+ + + + + + 2

= 1

[2t 2t 4 – 4t –12]2

+ + 1

= 4 sq. units


26. Here R = 20, r = 12, V = 12308.8


3.14(400 240 144)h3

× + + 1

⇒ h = 15 cm1

2

2 2(20 –12) 15 17 cm= + =l

1

2


= π[(20 + 12) × 17 + 12 × 12] 1

= 2160.32 cm2 1

27.1

cos 602

θ = ⇒ θ = °1

2


2

∴ �2 3.14 5 240

ADB 20.93 cm360

× × ×= = 1


Now, AP = 5 3 cm

P

A

B

D Oθ C

10 cm

5

30/1/3 (21)

30/1/3



Area 2( OAP OBP) 25 3 43.25 cm∆ + ∆ = =

1

2


26.16 cm360

× ×=

1

2

Shaded Area = 43.25 – 26.16 = 17.09 cm2 1

28. Let the three numbers in A.P. be a – d, a, a + d.

3a = 12 ⇒ a = 4. 1

Also (4 – d)3 + 43 + (4 + d)3 = 288 1

⇒ 64 – 48d + 12d2 – d3 + 64 + 64 + 48d + 12d2 + d3 = 288

⇒ 24d2 + 192 = 288

⇒ d2 = 4

d = ± 2 1

The numbers are 2, 4, 6, or 6, 4, 2. 1

29. For correct Given, To prove, construction, figure1

4 22

× =

Correct Proof 2

30. Let the speed while going be x km/h

Therefore 150 150 5

–x x 10 2

=+

2

⇒ x2 + 10x – 600 = 0

⇒ (x + 30)(x – 20) = 0

⇒ x = 20 1

∴ Speed while going = 20 km/h1

2

and speed while returning = 30 km/hr 1

2

31. Correct Construction 4

(22) 30/1/3

30/1/3



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