Karnataka SSLC Question Paper 2021 Maths for Tamil Medium

Question Booklet Serial No. : 11-

Ce°⁄k’ ©¨dL‹m
®æãàí¨Ü~ ÓÜíTæ :
Register Number :
±Ü£ÅPæ - 01 / Paper – 01
GÓ….GÓ….GÇ….Ô. ÊÜááS ±ÜÄàûæ – 2021
SSLC MAIN EXAMINATION – 2021
…Œ⁄æ⁄fl : V⁄{}⁄ + …e¤k´⁄ + —⁄»⁄·¤d …e¤k´⁄
Subjects : MATHEMATICS + SCIENCE + SOCIAL SCIENCE
(ñÜËáÙÜá ÊÜáñÜᤠCíXÉÐ… ÊÜÞ«ÜÂÊÜá / Tamil and English Medium)
(CCE-RF / CCE-RR / CCE-PF / CCE-PR / NSR / NSPR )
—⁄»⁄flæ⁄fl : ∑Ê◊⁄VÊX 10-30 ¬M•⁄ »⁄fl®¤¿‘⁄-1-30 ¡⁄»⁄¡ÊVÊ ] [ Time : 10-30 A.M. to 1-30 P.M.

TEAR HERE TO OPEN THE QUESTION BOOKLET
Jlflo Æ⁄√ÀÊ-V⁄◊⁄ —⁄MSÊ¿ : 40 + 40 + 40 = 120 ] [ Total No. of Questions : 40 + 40 + 40 = 120

Ceœ Jh•Ÿs[ Tœß˚Vl ©¨j’j ß\dL‹m
V⁄¬Œ⁄r @MO⁄V⁄◊⁄fl : 40 + 40 + 40 = 120 ] [ Max. Marks : 40 + 40 + 40 = 120
81 T/E — Pou®/Mathematics
©õnÁºPÐUPõÚ SÔ¨¦PÒ / Instructions to the Students :
1. ©õnÁºPÒ u[PÒ ÷uºÄ ~øÇÄa ^miÀ EÒÍÁõÖ ÷uºÄ
£vöÁsPøÍ Question Booklet ©ØÖ® OMR uõÎÀ AuØöPÚ
JxUP¨£mh £vöÚõ¸ Pmh[PÎÀ GÊu ÷Ásk®.
Write your eleven digit Register Number on the Question Booklet as
allotted in the admission ticket in the space provided at the top right
corner of this front page.
2. C¢u ÂÚõzuõÒ ¤ß¦Ó©õP ‰i øÁUP¨£mkÒÍx (Sealed by
reverse jacket). }[PÒ ÷uºÄ Bµ®¤US®ö£õÊx Á»UøP¨ £UP
Kµzøu ¤›zx vÓUP ÷Ásk®. ÂÚõUPøÍ EÒÍhUQ¯
AøÚzx¨ £UP[PЮ \›¯õPÄ® ¤›UPõ¨£hõ©¾® EÒÍuõ GÚ
\›£õºUPÄ®.
This Question Booklet has been sealed by reverse jacket. You have to cut
on the right side to open the Question Booklet at the time of
commencement of the examination. Check whether all the pages of the
Question Booklet are intact.
3. OMR uõÒPÒ uÛzuÛ¯õP ÁÇ[P¨£k®.
OMR Sheet will be provided subject-wise separately.
4. C¢u ÂÚõzuõÒ öuõS¨¦ ‰ßÖ £SvPÍõP ¤›zxÒÍx & Pou®,
AÔ¯À ©ØÖ® \‰P AÔ¯À.
Tear here

This set of Question Booklets consists of three core subjects and each
subject has separate Question Booklet.
1115 ( ) 1 of 32

5. JÆöÁõ¸ ¤›Â¾® 40 ÂÚõUPÒ öPõkUP¨ £mkÒÍx. ö©õzu©õP
120 ÂÚõUPÒ EÒÍÚ.
(i) Pou® & ÂÚõ GsPÒ 1 to 40
(ii) AÔ¯À & ÂÚõ GsPÒ 41 to 80
(iii) \‰P AÔ¯À & ÂÚõ GsPÒ 81 to 120
40 questions are provided against each subject. This set of Question
Booklets contains 120 questions in all.
(i) Mathematics – Question Numbers 1 to 40
(ii) Science – Question Numbers 41 to 80
(iii) Social Science – Question Numbers 81 to 120
6. JÆöÁõ¸ ÂÚõÂØS® J¸ ©v¨ö£s ÁÇ[P¨£k®. AøÚzx
ÂÚõÂØS® Pmhõ¯©õP Âøh¯ÎUP ÷Ásk®. JÆöÁõ¸ \›¯õÚ
ÂÚõÂØS® J¸ ©v¨ö£s ÁÇ[P¨£k®. uÁÓõÚ ÂøhPÐUS
(negative marks) Gvº©øÓ ©v¨ö£sPÒ CÀø».
Each question carries one mark. Answering all the questions is
compulsory and each correct answer will be awarded one mark. There
will be no negative marking for wrong answers.
7. ÷uºÂß ÷£õx PÁÛUP ÷Ási¯øÁ
a) ÂÚõzuõøÍ PÁÚ©õP¨ £izx¨ £õºUPÄ®.
b) Ámhzøu •Êø©¯õP {µ¨£, P¸¨¦ AÀ»x }» Ásn ø©
Eøh¯ GÊx÷Põø»¨ £¯ß£kzu»õ®. OMR uõÎÀ Ámhzøu
{µ¨¦® ÷£õx, PÁÚ©õP {µ¨£ ÷Ásk®. {µ¨¤¯ ¤ßÚº
Aøu ©õØÓ÷Áõ, \› ö\´Á÷uõ, ÷ÁÖ QÖUPÀP÷Íõ Ch®
ö£ÖuÀ Thõx.
GkzxUPõmk : 20 B® ÂÚõÂØPõÚ \›¯õÚ Âøh Auß
Á›ø\°À EÒÍ A, B, C, D CÀ C GߣuõP C¸¨¤ß,
öPõkUP¨£mkÒÍ |õßS Ámh[PÎÀ C GßÓ GÊzx EÒÍ
Ámhzøu }»® AÀ»x P¸¨¦ {Ó ø© öPõsh GÊx÷Põ»õÀ
{ǼhÄ® ÷Ásk®.
ÂÚõ Gs : 20) A B C D (Cx GkzxUPõmk ©mk÷©)

c) OMR uõÒPøÍ ©hUP÷Áõ, ÷\u® ÂøÍÂUP÷Áõ Thõx.
AÆÁõÖ ÷\u® ÂøÍÂzx ©v¨ö£sPÒ SøÓ¢uõÀ AuØS
©õnÁºPÒ uõß ö£õÖ¨÷£ØP ÷Ásk®.
1115 ( ) 2 of 32

During the examination,

a) Read the questions carefully.

b) Completely darken / shade the relevant circle against Question

Number in the OMR Sheet using blue / black ball point pen. Do not

try to alter the entry and not to do any stray marks on OMR Sheet.

Example : In the question booklet, if C is the correct answer for

Question No. 20, then in the OMR Sheet shade the option C using

blue / black ball point pen as follows.

Question No. 20) A B C D ( This is an example only )

c) Do not fold, tear, wrinkle or staple on the OMR Sheet.

8. J¸ ÂÚõÂØS, JßÔØUS® ÷©Ø¨£mh (circle) Ámh[PøÍ
{µ¨¤ÚõÀ, A¢u Âøh uÁÓõÚuõP P¸v AuØS ©v¨ö£s
öPõkUP¨£hõx/ÁÇ[P¨£hõx.

If more than one circle is shaded for a given question, such answer is

treated as wrong and no marks will be given.

9. ©õnÁºPЮ, Aøu PsPõo¨£õ͸® OMR uõÎÀ AuØöPÚ
JxUP¨£mh Ch[PÎÀ øPö¯õ¨£® Ch÷Ásk®.

Student and Room Invigilator should sign in the OMR Sheet in the space

provided.

1115 ( ) 3 of 32

10. ©õnÁºPÒ ÷uºÄ •izuÄhß öÁÎ÷¯ÖÁuØS •ß£õP OMR
ÂøhzuõÒPøÍ (subject-wise) £SvÁõ›¯õP \›£õºzx
÷uºÁõÍ›h® J¨£øhUP ÷Ásk®.
Candidate should return the subject-wise answered OMR Sheet to the
Room Invigilator before leaving the examination hall.

11. ÷uºÂß ÷£õx ©õnÁºPÒ u[PÐøh¯ (Rough work) ÷ÁÖ
SÔ¨¦PøÍ ÂÚõzuõÎß Pøh] £UPzvÀ AuØöPÚ JxUP¨£mh
ChzvÀ ©mk÷© GÊv¨ £õºUP ÷Ásk®.
Rough work can be done in the space provided at the end of the Question
Booklet.

12. ©õnÁºPÒ PõÀS÷»mhº, øP÷£], ì©õºm PiPõµ® ©ØÖ® CßÚ
¤Ó ªßÚÝ E£Pµn[PøÍ ÷uºÂß ö£õÊx EÒ÷Í Gkzx ö\À»
AÝ©v°Àø».
Calculators, Mobiles, Smart Watches and any other electronic equipment
are not allowed inside the examination hall.

1115 ( ) 4 of 32

£õh® : Pou®
Subject : MATHEMATICS
RÌU Põq® ÂÚõUPÒ/•ØÖ¨ ö£Óõu öuõhºPÐUPõÚ |õßS
ÂøhPÐÒ \›¯õÚ ÂøhUPõÚ SÔ±møhz ÷uº¢öukzx
öPõkUP¨£mkÒÍ OMR uõÎÀ P¸¨¦ AÀ»x }» Ásn ø©
öPõsh GÊx ÷Põ»õÀ SÔ¨¤mh Ámhzøu {µ¨¦P : 40 × 1 = 40

Four choices are given for each of the following questions / incomplete
statements. Choose the correct answer among them and shade the correct
option in the OMR Answer Sheet given to you with a black / blue ball point
pen. 40 × 1 = 40
1. J¸ Tmkzöuõhº Á›ø\°ß n Áx EÖ¨¦ an = 4n + 5 GÚ öPõsk
Azöuõh›ß 5Áx EÖ¨¦
(A) 20 (B) 14
(C) 25 (D) 24
The nth term of an Arithmetic Progression is an = 4n + 5. Then its 5th

term is
(A) 20 (B) 14
(C) 25 (D) 24
2. 5 x 2 = 2 ( 2x + 3 ) GßÓ C¸£ia \©ß£õmøh ö£õx ÁiÁzvÀ GÊx®
÷£õx AÀ»x ÂÁ›US® ÷£õx QøhUS® uÛ EÖ¨¦ AÀ»x
{ø»¯õÚ (constant term) EÖ¨¦
(A) 5 (B) 6
(C) 4 (D) –6
When the quadratic equation 5 x 2 = 2 ( 2x + 3 ) is expressed in the
standard form, the constant term obtained is
(A) 5 (B) 6
(C) 4 (D) –6

1115 ( ) 5 of 32

81-T/E
T/E ( RF/RR/PF/PR/NSR/NSPR
RF/RR/PF/PR/ )

3. x – 2y = 0 ©ØÖ® 3x
3 + 4y – 20 = 0 GߣøÁPÍõP C¸¨£x
(A) öÁmkU ÷PõkPÒ
(B) Jß÷Óõh JßÖ ö£õ¸¢x® ÷PõkPÒ
(C) Cøn¯õÚ ÷PõkPÒ
(D) JßÖUöPõßÖ ö\[SzuõÚ
ö ÷PõkPÒ
x – 2y = 0 and 3x + 4y
4 – 20 = 0 are
(A) Intersecting lines (B) Coincident lines
(C) Parallel lines (D) Perpendicular lines
4. Áøµ£hzvÀ Põmh¨£mhx ÷£õ» C¸UPUTi¯ ÷PõkPøÍ öPõsk
EÒÍ \©ß£õkPÒ
©ß£õkPÒ

(A) x + y = 1 ©ØÖ® 2x – y = 1

(B) 2x + y = 2 ©ØÖ® x + y = 2

(C) 2x – y = 2 ©ØÖ® 4x – y = 4

(D) y – x = 0 ©ØÖ® x – y = 1

1115 ( ) 6 of 32

81-T/E
T/E ( RF/RR/PF/PR/NSR/NSPR
RF/RR/PF/PR/ )

The pair of equations of lines as shown in the graph are

(A) x + y = 1 and 2x
2 –y=1

(B) 2x + y = 2 and x + y = 2

(C) 2x – y = 2 and 4x
4 –y=4

(D) y – x = 0 and x – y = 1

5. a1x + b1y + c1 = 0 ©ØÖ® a 2 x + b 2y + c 2 = 0 GßÓ Cµsk ©õÔPøÍU

öPõskÒÍ J¸ ÷\õi
÷ J¸£ia \©ß£õkPÒ
©ß£õkPÒ Cøn¯õÚ ÷PõkPÍõP
C¸UQÓx GÛÀ AuØPõÚ \›¯õÚ
›¯õÚ J¨¤mk ÂQu öuõhº¨¦
a1 b1 c1 a1 b1 c1
(A) = ≠ (B) = =
a2 b2 c2 a2 b2 c2

a1 b1 a1 b1
(C) ≠ (D) =
a2 b2 b2 a2

1115 ( ) 7 of 32

If the pair of linear equations in two variables a1x + b1y + c1 = 0 and

a 2 x + b2y + c 2 = 0 are parallel lines then the correct relation of their

coefficients is
a1 b1 c1 a1 b1 c1
(A) = ≠ (B) = =
a2 b2 c2 a2 b2 c2

a1 b1 a1 b1
(C) ≠ (D) =
a2 b2 b2 a2

6. 2x + 3y + 7 = 0 ©ØÖ® ax + by + 14 = 0 GßÓ ÷PõkPÎß ÷\õi¯õÚx
Jß÷Óõh JßÖ ö£õ¸¢xQÓx GÛÀ ‘a’ ©ØÖ® ‘b’ US \©©õÚ
©v¨¦PÒ •øÓ÷¯
(A) 2 ©ØÖ® 3 (B) 3 ©ØÖ® 2

(C) 4 ©ØÖ® 6 (D) 1 ©ØÖ® 2

If the pair of lines 2x + 3y + 7 = 0 and ax + by + 14 = 0 are coincident

lines then the values of ‘a’ and ‘b’ are respectively equal to

(A) 2 and 3 (B) 3 and 2

(C) 4 and 6 (D) 1 and 2

7. ¤ß Á¸£øÁPÎÀ Gx Tmkzöuõhº Á›ø\°À EÒÍx ?

(A) 1, – 1, – 2, ..........

(B) 1, 5, 9, ..............

(C) 2, – 2, 2, – 2, ............

(D) 1, 2, 4, 8, ..........

1115 ( ) 8 of 32

Which of the following is an Arithmetic Progression ?

(A) 1, – 1, – 2, ..........

(B) 1, 5, 9, ..............

(C) 2, – 2, 2, – 2, ............

(D) 1, 2, 4, 8, ..........

8. – 3, – 1, 1, 3, ....... GßÓ Tmköuõhº Á›ø\°À 11Áx EÖ¨¦

(A) 23 (B) – 23

(C) – 17 (D) 17

The 11th term of the Arithmetic Progression – 3, – 1, 1, 3, ....... is

(A) 23 (B) – 23

(C) – 17 (D) 17

9. J¸ Tmkzöuõhº Á›ø\°À •uÀ 10 EÖ¨¦PÎß TkuÀ 155 BS®
©ØÖ® A÷u Tmkzöuõh›À •uÀ 9 EÖ¨¦PÎß TkuÀ 126 GÛÀ
A¢u öuõh›À 10Áx EÖ¨¦

(A) 27 (B) 126

(C) 29 (D) 25

The sum of the first 10 terms of an Arithmetic Progression is 155 and the
sum of the first 9 terms of the same progression is 126 then the 10th term
of the progression is

(A) 27 (B) 126

(C) 29 (D) 25
1115 ( ) 9 of 32

10. 2 x 2 + ax + 6 = 0 GßÓ \©ß£õmiß J¸ ‰»® BÚx 2 GÛÀ ‘a’ Cß
©v¨¦
7
(A) 7 (B)
2
7
(C) –7 (D) −
2

If one root of the equation 2 x 2 + ax + 6 = 0 is 2, then the value of ‘a’ is
7
(A) 7 (B)
2
7
(C) –7 (D) −
2

11. px 2 + qx + r = 0 GßÓ C¸£ia \©ß£õmiß ¤›¨¦ Gs (discriminant)

BP C¸¨£x
(A) q 2 − 4 pr (B) q 2 + 4 pr
2
(C) p − 4 pr (D) p 2 + 4qr

The discriminant of the Quadratic equation px 2 + qx + r = 0 is

(A) q 2 − 4 pr (B) q 2 + 4 pr
2
(C) p − 4 pr (D) p 2 + 4qr

12. 4, x, 10 Gß£Ú J¸ Tmkzöuõhº Á›ø\°À EÒÍöuÛÀ x Cß
©v¨¦
(A) 14 (B) –6

(C) –7 (D) 7

If 4, x, 10 are in Arithmetic Progression the value of x is

(A) 14 (B) –6

(C) –7 (D) 7

1115 ( ) 10 of 32

13. ax 2 + bx + c = 0 GßÓ C¸£ia \©ß£õmiß ‰»[PÒ

−b ± b 2 − 4ac
(A) x =
2a

−b ± b 2 + 4ac
(B) x =
2a

−b − b 2 − 4c
(C) x =
2a

−b + b 2 − 4ac
(D) x =
2a

The roots of the quadratic equation ax 2 + bx + c = 0 are

−b ± b 2 − 4ac
(A) x =
2a

−b ± b 2 + 4ac
(B) x =
2a

−b − b 2 − 4c
(C) x =
2a

−b + b 2 − 4ac
(D) x =
2a

14. (x – 3) (x + 2) = 0 GßÓ \©ß£õmiß ‰»[PÒ

(A) – 3, 2 (B) 3, – 2

(C) – 3, – 2 (D) 3, 2

The roots of the equation ( x – 3 ) ( x + 2 ) = 0 are

(A) – 3, 2 (B) 3, – 2

(C) – 3, – 2 (D) 3, 2

1115 ( ) 11 of 32

15. öuõhºa]¯õP EÒÍ Cµsk •ÊUPÎß (integer) TkuÀ 27 GÛÀ
A¢u •ÊUPÒ
(A) 7 ©ØÖ® 20 (B) 13 ©ØÖ® 14

(C) 1 ©ØÖ® 26 (D) – 13 ©ØÖ® – 14

If the sum of two consecutive integers is 27, then the integers are

(A) 7 and 20 (B) 13 and 14

(C) 1 and 26 (D) – 13 and – 14

16. £hzvÀ sin θ Âß ©v¨¦

1 3
(A) (B)
2 2
2
(C) 3 (D)
3

In the figure, the value of sin θ is

1 3
(A) (B)
2 2
2
(C) 3 (D)
3

1115 ( ) 12 of 32

17. ( sin 30° + cos 60° – tan 45° ) Cß ©v¨¦

(A) 1 (B) –1

(C) 2 (D) 0

The value of ( sin 30° + cos 60° – tan 45° ) is

(A) 1 (B) –1

(C) 2 (D) 0

18. 3 + sec 2 θ US \©®

(A) 4 + tan 2 θ (B) 4 + cot 2 θ

(C) 2 + cot 2 θ (D) 3 + cot 2 θ

3 + sec 2 θ is equal to

(A) 4 + tan 2 θ (B) 4 + cot 2 θ

(C) 2 + cot 2 θ (D) 3 + cot 2 θ

19. J¸ ÷Põ¦µzvß Ai¨£Sv°¼¸¢x 30 «. yµzvÀ uøµ©mhzvÀ
EÒÍ J¸ ¦Òΰ¼¸¢x, ÷Põ¦µzvß Ea]°ß HØÓU÷Põn® 30°
BP EÒÍx. A¨£i¯õÚõÀ ÷Põ¦µzvß E¯µ®
(A) 10 «. (B) 30 «.

(C) 10 3 «. (D) 30 3 «.

The angle of elevation of the top of a tower from a point on the ground,
which is 30 metres away from the foot of the tower, is 30°. Then the height
of the tower is

(A) 10 m (B) 30 m

(C) 10 3 m (D) 30 3 m

1115 ( ) 13 of 32

20. ( sin θ × cosec θ ) Cß ©v¨¦

(A) 2 (B) 1

1 3
(C) – (D)
2 2

The value of ( sin θ × cosec θ ) is

(A) 2 (B) 1

1 3
(C) – (D)
2 2

21. A ( x1 , y1 ) ©ØÖ® B ( x 2 , y2 ) GßÓ ¦ÒÎPøÍ CønUS®

÷Põmkzxsiß ø©¯¨¦ÒÎø¯ Psk¤iUS® `zvµ®

 x 2 + x1 y 2 + y1   x 2 − x1 y 2 − y1 
(A)  ,  (B)  , 
 2 2   2 2 
   

 x 2 + y2 x1 + y1   x 2 + x1 y 2 + y1 
(C)  ,  (D)  , 
 3 3   3 3 
   

The formula to find the mid-point of the line segment joining the points
A ( x1 , y1 ) and B ( x 2 , y 2 ) is

 x 2 + x1 y 2 + y1   x 2 − x1 y 2 − y1 
(A)  ,  (B)  , 
 2 2   2 2 
   

 x 2 + y2 x1 + y1   x 2 + x1 y 2 + y1 
(C)  ,  (D)  , 
 3 3   3 3 
   

1115 ( ) 14 of 32

22. ( x1 , y1 ) ©ØÖ® - ( x 2 , y 2 ) GßÓ ¦ÒÎPÐUS Cøh÷¯²ÒÍ yµ®

(A) ( x1 − x 2 )2 + ( y1 − y 2 )2 (B) ( x 2 − x1 )2 − ( y 2 − y1 )2

(C) ( x1 + x 2 )2 − ( y1 + y 2 )2 (D) ( x 2 + x1 )2 + ( y 2 + y1 )2

The distance between the points ( x1 , y1 ) and ( x 2 , y 2 ) is

(A) ( x1 − x 2 )2 + ( y1 − y 2 )2 (B) ( x 2 − x1 )2 − ( y 2 − y1 )2

(C) ( x1 + x 2 )2 − ( y1 + y 2 )2 (D) ( x 2 + x1 )2 + ( y 2 + y1 )2

23. Põn¨£mh uµÄ ©v¨¦PÎÀ v¸®£v¸®£ ªP AvP •øÓ°À
Á¸QßÓ ©v¨¦US
(A) \µõ\› (Mean) (B) •Pk (Mode)
(C) Cøh{ø» (Median) (D) Ãa_ (Range)
The value among the observations of most repeated scores of the data is

(A) the mean (B) the mode

(C) the median (D) the range

24. ¤ßÁ¸® ©v¨¦PÐUS \µõ\›

©v¨¦PÒ 1 3 5 7

(A) 16 (B) 5
(C) 1.6 (D) 4

The Mean of the following scores is

Marks 1 3 5 7

(A) 16 (B) 5
(C) 1.6 (D) 4

1115 ( ) 15 of 32

25. öPõkUP¨£mhøÁPÎÀ \µõ\›, •Pk ©ØÖ® Cøh{ø»¯ÍÄ
CøÁUPÐUPõÚ öuõhº¦

(A) 3 Cøh{ø»¯ÍÄ = 2 \µõ\› + •Pk

(B) 3 \µõ\› = 2 Cøh{ø»¯ÍÄ + •Pk

(C) \µõ\› = 3 Cøh{ø»¯ÍÄ + •Pk

(D) •Pk = 3 \µõ\› + 2 Cøh{ø»¯ÍÄ

The relation among the Mean, Mode and Median is

(A) 3 Median = 2 Mean + Mode (B) 3 Mean = 2 Median + Mode

(C) Mean = 3 Median + Mode (D) Mode = 3 Mean + 2 Median

26. Kº E¸øÍø¯ ö©ÊPõÀ (wax) ö\´¯¨£mkÒÍx. Aøu E¸UQ
•ÊÁx©õP J¸ ÷PõÍ©õP ©Ö Aa_ ö\´uõÀ Auß PÚ¯ÍÄ

(A) E¸øÍ°ß PÚ¯ÍÄ ÷£õ» 2 ©h[SPÒ

(B) E¸øÍ°ß PÚ¯ÍÂÀ £õv

(C) E¸øÍ°ß PÚ¯ÍÄ ÷£õ» 3 ©h[SPÒ
(D) E¸øÍ°ß PÚ¯ÍÄUS \©®

A cylinder made of wax is melted and recast completly into a sphere. Then
the volume of the sphere is

(A) two times the volume of the cylinder

(B) half the volume of the cylinder

(C) 3 times the volume of the cylinder

(D) equal to the volume of the cylinder

1115 ( ) 16 of 32

27. ¤›Ä CøhöÁ롧 |k ¦ÒÎ AÀ»x ø©¯¨¦ÒÎ Psk¤iUS®
`zvµ®
¤ ›Âß ÷© À GÀø» − ¤ ›Âß RÌ GÀø»
(A)
2
¤ ›Âß ÷© À GÀø» × ¤ ›Âß RÌ GÀø»
(B)
3
¤ ›Âß ÷© À GÀø» + ¤ ›Âß RÌ GÀø»
(C)
2
¤ ›Âß ÷© À GÀø» + ¤ ›Âß RÌ GÀø»
(D)
3
The formula to find the mid-point of the class interval is
Upper limit − lower limit
(A)
2
Upper limit × lower limit
(B)
3
Upper limit + lower limit
(C)
2
Upper limit + lower limit
(D)
3
28. ∆ ABC CÀ XY || BC GÛÀ

AX AC
(A) =
AB AY
AX AY
(B) =
BX CY
AX XY
(C) =
BX AY
AB AC
(D) =
BX AY

1115 ( ) 17 of 32

In the ∆ ABC, XY || BC then

AX AC AX AY
(A) = (B) =
AB AY BX CY
AX XY AB AC
(C) = (D) =
BX AY BX AY
29. öPõkzxÒÍ Cµsk •U÷Põn[PøÍ PÁÛUPÄ® ©ØÖ® ¤ßÁ¸®
AÍÄPÎÀ DF Cß }Ízøu PshÔP :

(A) 6 2 ö\.«. (B) 3 2 ö\.«.
(C) 4.2 ö\.«. (D) 8.4 ö\.«.
Observe the given two triangles and then identify the length of DF in the
following :

(A) 6 2 cm (B) 3 2 cm
(C) 4.2 cm (D) 8.4 cm

1115 ( ) 18 of 32

30. ∆ ABC ~ ∆ PQR BP EÒÍx. ∆ ABC Cß £µ¨£ÍÄ = 64 ö\.«. 2 ©ØÖ®

∆ PQR Cß £µ¨£ÍÄ = 100 ö\.«. 2 , AB = 8 ö\.«. GÛÀ PQ Cß }Í®

(A) 12 ö\.«. (B) 15 ö\.«.

(C) 10 ö\.«. (D) 8 ö\.«.

∆ ABC ~ ∆ PQR. Area of ∆ ABC = 64 cm 2 and the area of ∆ PQR = 100 cm 2 .

If AB = 8 cm then the length of PQ is

(A) 12 cm (B) 15 cm

(C) 10 cm (D) 8 cm

31. ∆ ABC CÀ B = 90° ©ØÖ® BD ⊥ AC. AB = 6 ö\.«., BC = 8 ö\.«.

GÛÀ CD Cß }Í®

(A) 10 ö\.«. (B) 6.4 ö\.«.

(C) 4.8 ö\.«. (D) 3.6 ö\.«.

1115 ( ) 19 of 32

In the ∆ ABC, B = 90° and BD ⊥ AC. If AB = 6 cm, BC = 8 cm then

the length of CD is

(A) 10 cm (B) 6.4 cm

(C) 4.8 cm (D) 3.6 cm

32. öPõkzxÒÍ £mzvÀ "O' øÁ ø©¯©õP EÒÍ J¸ ÁmhzvØS AT J¸
öuõk÷Põk. OT = 4 ö\.«. ¦ÒÎ "A'°À AT öuõk÷Põk
Áøµ¯¨£mkÒÍx. OTA = 30° GÛÀ AT °ß }Í®

(A) 4 ö\.«. (B) 2 ö\.«.

(C) 2 3 ö\.«. (D) 4 3 ö\.«.

1115 ( ) 20 of 32

In the given figure AT is a tangent drawn at the point A to the circle with

centre O such that OT = 4 cm. If ∠ OTA = 30° then AT is

(A) 4 cm (B) 2 cm

(C) 2 3 cm (D) 4 3 cm

33. öPõkzxÒÍ £hzvÀ "O' øÁ ø©¯©õP EÒÍ J¸ ÁmhzvØS PA,
PBC ©ØÖ® CD Gß£Ú öuõk÷PõkPÒ. PC = 8 ö\.«. ©ØÖ®
AP = 5 ö\.«. GÛÀ CD öuõk÷Põmiß }Í®

(A) 5 ö\.«. (B) 3 ö\.«.

(C) 8 ö\.«. (D) 13 ö\.«.

1115 ( ) 21 of 32

In the given figure PA, PBC and CD are the tangents to a circle with
centre O. If PC = 8 cm and AP = 5 cm, the length of the tangent CD is

(A) 5 cm (B) 3 cm

(C) 8 cm (D) 13 cm
34. ¤ß Á¸£øÁPÎÀ uÁÓõÚ TØÖ AÀ»x ÁõUQ¯® (wrong statement)
(A) J¸ öuõk÷Põk J¸ ÁmhzvØS \›¯õP J¸ ¦ÒΰÀ
öuõmkUöPõÒQÓx
(B) J¸ ÷|ºU÷Põk J¸ ÁmhzvØS Áøµ²® ÷£õx A¢u ÷Põk J¸
¦ÒÎ ÁȯõP £õ´¢x ö\À¾®
(C) ÁmhzvØS AuÝøh öuõk÷PõmhõÀ öuõk® ö£õxÁõÚ
¦ÒÎø¯ öuõk® ¦ÒÎ GßÓøÇUQ÷Óõ®
(D) H÷uÝ® J¸ ¦ÒΰÀ ÁmhzvØS Áøµ¯¨£k® öuõk÷Põk
BÚx BµzvØS ö\[SzuõP C¸QÓx
The wrong statement in the following is

(A) a tangent to a circle touches the circle exactly at one point

(B) when a straight line is drawn to a circle it always passes through a
point on the circle

(C) the point common to the circle and its tangent is called the point of
contact

(D) the tangent drawn at any point to a circle is perpendicular to the
radius drawn at the point of contact

1115 ( ) 22 of 32

35. öPõkzxÒÍ £hzvÀ J¸ ÁmhzvØS öÁΰÀ EÒÍ ‘T ’ GßÓ J¸
¦Òΰ¼¸¢x A¢u ÁmhzvØS Cµsk öuõk÷PõkPÒ Áøµ²®
÷£õx ¤ß£ØÓUTi¯ Akzu £i{ø» (next step) Gx?

(A)

(B)

(C)

(D)

1115 ( ) 23 of 32

Which is the next step of construction while constructing a pair of

tangents to a circle from an external point ‘T ’, given in the figure ?

(A)

(B)

(C)

(D)

1115 ( ) 24 of 32

36. J¸ ÷PõÍzvß ¦Ó¨£µ¨£ÍÄ 616 \xµ ö\.«. A÷u ÷PõÍzvß
Bµzvß AÍÄ
(A) 49 ö\.«. (B) 14 ö\.«.

(C) 21 ö\.«. (D) 7 ö\.«.

The surface area of a sphere is 616 sq.cm. Then the radius of the same
sphere is

(A) 49 cm (B) 14 cm

(C) 21 cm (D) 7 cm
37. £hzvÀ Põmi²ÒÍx ÷£õ» C¸US® J¸ T®¤ß PÚ¯ÍÄ

(A) πr 2 h (B) πr ( r + l )
1
(C) πr 2 h (D) πr l
3
The volume of a cone as shown in the figure is

(A) πr 2 h (B) πr ( r + l )
1
(C) πr 2 h (D) πr l
3
1115 ( ) 25 of 32

38. J¸ •øÚ (one end) AÀ»x J¸ £UP® vÓUP¨£mkÒÍ Ai¨£UP®
÷|ºÁmh Kº E¸øÍ ÁiÁ P»Ûß (£õzvµ®) Bµ® r ö\.«. ©ØÖ®
Auß E¯µ® h ö\.«. GÛÀ A¢u P»Ûß ö©õzu ¦Ó£µ¨£ÍÄ
Psk¤iUS® `zvµ®

(A) ( π r 2 + 2πrh ) ö\.«. 2 (B) 2πrh ö\.«. 2

1
(C) π r 2h ö\.«. 3 (D) ( π r 2 + h ) ö\.«. 2
3

The formula to find the total surface area of a right circular based

cylindrical vessel of base radius r cm and height h cm opened at one end is

(A) ( π r 2 + 2πrh ) cm 2 (B) 2πrh cm 2

1
(C) π r 2h cm 3 (D) ( π r 2 + h ) cm 2
3

39. £hzvÀ Põmh¨£mhx ÷£õ¾ÒÍ J¸ T®¤ß CøhUPshzvß
ÁøÍÄ £µ¨£ÍÄ Psk¤iUS® `zvµ®

1 1
(A) π l ( r1 + r 2 ) (B) π h ( r12 + r2 2 + r1 r 2 )
3 3

(C) πl ( r1 + r2 ) (D) π l ( r1 − r2 )

1115 ( ) 26 of 32

To find the curved surface area of a frustum of a cone as shown in the

figure the formula used is

1 1
(A) π l ( r1 + r 2 ) (B) π h ( r12 + r2 2 + r1 r 2 )
3 3

(C) πl ( r1 + r2 ) (D) π l ( r1 − r2 )

40. J¸ vs© Aøµ ÷PõÍzvß ö©õzu ¦Ó¨£µ¨£ÍÄ 462 ö\.«.2.
AuÝøh ÁøÍÄ £µ¨£ÍÄ 308 ö\.«.2 GÛÀ A¢u Aøµ÷PõÍzvß
Ai¨£UP¨£µ¨£ÍÄ BP EÒÍøÁ
(A) 308 ö\.«. 2 (B) 231 ö\.«. 2

(C) 154 ö\.«. 2 (D) 1078 ö\.«. 2

The total surface area of solid hemisphere is 462 cm 2 . If the curved

surface area of it is 308 cm 2 , then the area of the base of the hemisphere

is

(A) 308 cm 2 (B) 231 cm 2

(C) 154 cm 2 (D) 1078 cm 2

1115 ( ) 27 of 32

(©ØÓ ÷Áø»PøÍ ö\´x¨£õº£uØPõP JxUP¨£mhªh®)
( SPACE FOR ROUGH WORK )

1115 ( ) 28 of 32

(©ØÓ ÷Áø»PøÍ ö\´x¨£õº£uØPõP JxUP¨£mhªh®)
( SPACE FOR ROUGH WORK )

1115 ( ) 29 of 32

(©ØÓ ÷Áø»PøÍ ö\´x¨£õº£uØPõP JxUP¨£mhªh®)
( SPACE FOR ROUGH WORK )

1115 ( ) 30 of 32

(©ØÓ ÷Áø»PøÍ ö\´x¨£õº£uØPõP JxUP¨£mhªh®)
( SPACE FOR ROUGH WORK )

1115 ( ) 31 of 32

1115 ( ) 32 of 32