Karnataka SSLC Question Paper 2021 Maths for Telugu Medium

CÈÉí¨Ü PÜñܤÄÔ
Question Booklet Serial No. : 11-

®æãàí¨Ü~ ÓÜíTæ :
Register Number :

±Ü£ÅPæ - 01 / Paper – 01
GÓ….GÓ….GÇ….Ô. ÊÜááS ±ÜÄàûæ – 2021
SSLC MAIN EXAMINATION – 2021
+ +
Subjects : MATHEMATICS + SCIENCE + SOCIAL SCIENCE
ñæÆáWÜá ÊÜáñÜᤠCíXÉÐ… ÊÜÞ«ÜÂÊÜá / Telugu and English Medium

TEAR HERE TO OPEN THE QUESTION BOOKLET
CCE-RF / CCE-RR / CCE-PF / CCE-PR / NSR / NSPR
: 10-30 1-30 ] [ Time : 10-30 A.M. to 1-30 P.M.
: 40 + 40 + 40 = 120 ] [ Total No. of Questions : 40 + 40 + 40 = 120
: 40 + 40 + 40 = 120 ] [ Max. Marks : 40 + 40 + 40 = 120
81-L/E — /Mathematics
ѧéÅÆæÿ$¦ËMæü$ çÜ*^èþ¯@þË$ / Instructions to the Students :
1. Ò$Mæü$ CÐ@þÓºyìþ¯@þ àÌŒýsìýMðür$tÌZ° 11 A…MðüË ÇhçÜtÆæÿ$ ¯@þ…ºÆæÿ$¯@þ$ {ç³Ô¶ý²ç³#çÜ¢Mæü…ÌZ çܵçÙt…V>
Æ>Ķý$…yìþ. D õ³hÌZ C_a¯@þ Mæü$yìþOÐðþç³# Ð@þÊËV> ÇhçÜtÆŠÿ ¯@þ…ºÆŠÿ¯@þ$ {ÐéĶý$Ð@þÌñý¯@þ$.
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. {ç³Ô¶ý²ç³#çÜ¢Mæü…
{ç³Ô¶ý²ç³#çÜ¢Mæü… {ç³Ô¶ý²ç³#çÜ¢Mæü…

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. ÑçÙĶý*¯@þ$ÝëÆæÿ…V> OMR ç³{™éË$ {ç³™óþÅMæü…V> CÐ@þÓºyæþ$¯@þ$.
OMR Sheet will be provided subject-wise separately.
4. D {ç³Ô¶ý²ç³#çÜ¢Mæü… Ð@þÊyæþ$ MøÆŠÿ ÑçÙĶý*˯@þ$ MæüÍWĶý¬…yìþ, {糆 ÑçÙĶý*°Mìü {ç³™óþÅMæü…V>
{ç³Ô¶ý²ç³#çÜ¢Mæü… MæüÍWÐ@þ#…¨.
This set of Question Booklets consists of three core subjects and each
Tear here

subject has separate Question Booklet.

1113 () 1 of 32

5. {糆 ÑçÙĶý*°Mìü 40 {ç³Ô¶ý²Ë^ö糚¯@þ D {ç³Ô¶ý²ç³#çÜ¢Mæü… Ððþ¬™èþ¢… 120 {ç³Ô¶ý²Ë$ MæüÍWÐ@þ#…¨.
(i) Væü×ìý™èþ… & {ç³Ô¶ý²çÜ…QÅ 1 ¯@þ$…yìþ 40 Ð@þÆæÿMæü$
(ii) Ñgêq¯@þÐ@þ¬ & {ç³Ô¶ý²çÜ…QÅ 41 ¯@þ$…yìþ 80 Ð@þÆæÿMæü$
(iii) çÜÐ@þ*f Ñgêq¯@þ… & {ç³Ô¶ý²çÜ…QÅ 81 ¯@þ$…yìþ 120 Ð@þÆæÿMæü$
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. A°² {ç³Ô¶ý²ËMæü$ ™èþç³µ° çÜÇV> çÜÐ@þ*«§é¯éË$ Æ>Ķý$…yìþ. {糆 {ç³Ô¶ý²Mæü$ JMæü Ð@þ*Ææÿ$P CÐ@þÓºyæþ$¯@þ$.
¯ðþVðüsìýÐŒþ Ð@þ*Ææÿ$PË$ CÐ@þÓºyæþÐ@þ#. {糆 çÜÇOÄñý$¯@þ çÜÐ@þ*«§é¯é°Mìü JMæü Ð@þ*Ææÿ$P CÐ@þÓºyæþ$¯@þ$.
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. ç³È„æü fÆæÿ$Væü$ çÜÐ@þ$Ķý$…ÌZ,
G {ç³Ô¶ý²¯@þ$ gê¶Væü™èþ¢V> ^èþ§æþÐéÍ.
¼ OMR ç³{™èþ…ÌZ {糆 {ç³Ô¶ý²Mæü$ çÜÇOÄñý$¯@þ çÜÐ@þ*«§é¯é°² Væü$Ç¢…^èþ$rMæü$ ¯éË$Væü$ Ð@þ–™é¢Ë$
CÐ@þÓºyìþ¯@þÑ. ѧéÅÆæÿ$¦Ë$ çÜÇOÄñý$¯@þ çÜÐ@þ*«§é¯@þÐ@þ¬¯@þ$ ±Í / ¯@þË$ç³# »êÌŒý ´ëÆÿ¬…sŒý
ò³¯@þ$²™ø HO§ðþ¯@þ JMæü Ð@þ–™é¢°² Ð@þ*{™èþÐóþ$ õÙyŠþ ^ðþĶý*ÅÍ. õÙyìþ…VŠü çÜ…ç³NÆæÿ~…V> E…yéÍ. JMæü
ÝëÇ õÙyŠþ ^óþíܯ@þ ™èþÆæÿ$Ðé™èþ §é°° Ð@þ*ÆæÿaMæü*yæþ§æþ$. ¨§æþªMæü*yæþ§æþ$. OMR ç³{™èþ… Ò$§æþ
GÌê…sìý X™èþË$ E…yæþÆ>§æþ$.
E§éçßýÆæÿ×ýMæü$ {ç³Ô¶ý²ç³#çÜ¢Mæü…ÌZ 20 Ð@þ {ç³Ô¶ý²Mæü$ C çÜÇOÄñý$¯@þ çÜÐ@þ*«§é¯@þ… AÆÿ¬¯@þ³ç #yæþ$
OMR fÐéº$ ç³{™èþ…ÌZ ¶Mìü…§æþ ^èþ*糺yìþ¯@þ Ñ«§æþ…V> ±Í / ¯@þË$ç³# »êÌŒý´ëÆÿ¬…sŒý
ò³¯Œþ™ø õÙyŠþ ^ðþĶý$Å…yìþ.
{ç³Ô¶ý²çÜ…QÅ 20) A B C D (C¨ E§éçßýÆæÿ×ý Ð@þ*{™èþÐóþ$)

Ô OMR ç³{™èþ…¯@þ$ Ð@þ$yæþ™èþ ò³rtÐ@þ§æþ$ª. JMæüÐóþâ¶ý Ð@þ$yæþ™èþ ò³sìýt¯é, ´ëyæþ$^óþíܯé A¨ ç³NÇ¢V>
ѧéÅÆæÿ$¦Ë »ê«§æþÅ™èþ.
1113 () 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. JMæü {ç³Ô¶ý²Mæü$ JMæüsìý Mæü…sôý GMæü$PÐ@þ Ð@þ–™é¢Ë¯@þ$ õÙyŠþ ^óþÜí ¯@þrÏÆÿ¬™óþ §é°° ™èþ糚V> ¿êÑ…_,
Ð@þ*Ææÿ$PË$ CÐ@þÓºyæþÐ@þ#.

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

treated as wrong and no marks will be given.

9. ѧéÅǦ Ð@þ$ÇĶý¬ Væü¨Oò³ A«¨M>Ç C§æþªÆæÿ* OMR ç³{†MæüÌZ CÐ@þÓºyìþ¯@þ çܦ˅ÌZ çÜ…™èþMæü…
^ðþĶý*ÅÍ.

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

provided.

1113 () 3 of 32

10. ѧéÅÆæÿ$¦Ë$ OMR fÐéº$ ç³{™èþ… Væü¨ Oò³ A«¨M>ÇMìü C_a¯@þ ™èþÆæÿ$Ðé™èþ Ð@þ*{™èþÐóþ$ Væü¨¯@þ$…yìþ
ºÄ¶ý$rMæü$ Ððþâ¶ýåÐ@þÌñý¯@þ$.
Candidate should return the subject-wise answered OMR Sheet to the

Room Invigilator before leaving the examination hall.

11. {ç³Ô¶ý²ç³#çÜ¢Mæü…ÌZ _Ð@þÇ õ³hÌZ _™èþ$¢ç³° ^ðþĶý$ÅÐ@þÌñý¯@þ$.

Rough work can be done in the space provided at the end of the Question

Booklet.

12. M>Ë$PÌôýrÆæÿ$Ï, Ððþ¬O»ñýÌŒý ¸ù¯@þ$Ï, ÝëÃÆŠÿt Ðé^Œþ Ð@þ$ÇĶý¬ C™èþÆæÿ GË[M>t°MŠü ç³ÇMæüÆ>Ë$ ç³È„> àË$Mæü$
A¯@þ$Ð@þ$†…^èþºyæþÐ@þ#.

Calculators, Mobiles, Smart Watches and any other electronic equipment
are not allowed inside the examination hall.

1113 () 4 of 32

Subject : MATHEMATICS
“Mìü…§æþ CÐ@þÓºyìþ¯@þ {糆 {ç³Ô¶ý²Mæü$ ÌôýMæü AçÜ…ç³NÆæÿ~ ÐéMæüÅÐ@þ¬¯@þMæü$ ¯éË$Væü$ {ç³™éÅÐ@þ*²Ä¶ý*Ë$
CÐ@þÓºyézÆÿ¬. ÐésìýÌZ çÜÇOÄñý$¯@þ çÜÐ@þ*«§é¯@þÐ@þ¬¯@þ$ G¯@þ$²Mö° Ò$Mæü$ CÐ@þÓºyìþ¯@þ OMR E™èþ¢Ææÿ
ç³{†MæüÌZ ±Í / ¯@þË$ç³# »êÌŒý ´ëÄñý$…sŒý ò³¯Œþ™ø çÜÇOÄñý$¯@þ çÜÐ@þ*«§é¯@þÐ@þ¬¯@þMæü$ béĶý$ ^óþĶý$…yìþ
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. JMæü A…Mæü¶ÔóýÉìþ Äñý¬MæüP nÐ@þ ³ç §æþ… an = 4n + 5 AÆÿ¬¯@þ §é° Äñý¬MæüP 5Ð@þ 糧æþ…
(A) 20 (B) 14

(C) 25 (D) 24
The nth term of an Arithmetic Progression is a n = 4n + 5. Then its 5th

term is

(A) 20 (B) 14

(C) 25 (D) 24
2. 5 x 2 = 2 ( 2x + 3 ) Ð@þÆæÿYçÜÒ$MæüÆæÿ×ýÐ@þ¬¯@þ$ {´ëÐ@þ*×ìýMæü Ææÿ*ç³…ÌZ Æ>íܯ@þç³#yæþ$ §é° Äñý¬MæüP
íܦÆæÿ³ç §æþ…
(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

1113 () 5 of 32

3. x – 2y = 0 Ð@þ$ÇĶý¬ 3x + 4y – 20 = 0 çÜÆæÿâ¶ýçÜÒ$MæüÆæÿ×ê˯@þ$ Üç *_…^èþ$ çÜÆæÿâ¶ý ÆóÿQË$
(A) ç³ÆæÿçܵÆæÿ… Q…yìþ…^èþ$Mæü$…sêÆÿ¬ (B) HMîü¿¶ýÑÝë¢Æÿ¬
(C) çÜÐ@þ*…™èþÆæÿ…V> E…sêÆÿ¬ (D) ç³ÆæÿçܵÆæÿ Ë…º…V> E…sêÆÿ¬
x – 2y = 0 and 3x + 4y – 20 = 0 are

(A) Intersecting lines (B) Coincident lines

(C) Parallel lines (D) Perpendicular lines
4. CÐ@þÓºyìþ¯@þ ¶V>çœ#ÌZ° çÜÆæÿâ¶ý çÜÒ$MæüÆæÿ×êË f™èþ

(A) x+y=1 Ð@þ$ÇĶý¬ 2x – y = 1
(B) 2x + y = 2 Ð@þ$ÇĶý¬ x + y = 2
(C) 2x – y = 2 Ð@þ$ÇĶý¬ 4x – y = 4
(D) y–x=0 Ð@þ$ÇĶý¬ x – y = 1

1113 () 6 of 32

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

(A) x + y = 1 and 2x – y = 1

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

(C) 2x – y = 2 and 4x – y = 4

(D) y – x = 0 and x–y=1

5. a1x  b1y  c1  0 Ð@þ$ÇĶý¬ a 2x  b2y  c 2  0 çÜÆæÿâ¶ýçÜÒ$MæüÆæÿ×êË f™èþ¯@þ$
çÜ*_…^èþ$ çÜÆæÿâ¶ýÆóÿQË$ çÜÐ@þ*…™èþÆæÿ…V> E…sôý Ðésìý çÜçßýVæü$×ýM>Ë Ð@þ$«§æþÅçÜ…º…«§æþ…
a1 b1 c1 a1 b1 c1
(A)   (B)  
a2 b2 c2 a2 b2 c2

a1 b1 a1 b
(C)  (D)  1
a2 b2 b2 a2

1113 () 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 b
(C)  (D)  1
a2 b2 b2 a2

6. 2x + 3y + 7 = 0 Ð@þ$ÇĶý¬ ax + by + 14 = 0 çÜÆæÿâ¶ýçÜÒ$MæüÆæÿ×ê˯@þ$ çÜ*_…^èþ$
çÜÆæÿâ¶ýÆóÿQË f™èþ HMîü¿¶ýÑõÜ¢, ‘a’ Ð@þ$ÇĶý¬ ‘b’ Ë ÑË$Ð@þË$ ¶MæüÐ@þ$…V>
(A) 2 Ð@þ$ÇĶý¬ 3 (B) 3 Ð@þ$ÇĶý¬ 2

(C) 4 Ð@þ$ÇĶý¬ 6 (D) 1 Ð@þ$ÇĶý¬ 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. ¶Mìü…¨ ÐésìýÌZ H¨ A…Mæü¶ÔóýÉìþ AÐ@þ#™èþ$…¨ ?
(A) 1, – 1, – 2, ..........

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

(C) 2, – 2, 2, – 2, ............

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

1113 () 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, ....... A…Mæü¶ÔóýÉìþ Äñý¬MæüP 11 Ð@þ 糧æþ…

(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. JMæü A…Mæü¶ÔóýÉìþÄñý¬MæüP Ððþ¬§æþsìý 10 糧éË Ððþ¬™èþ¢… 155 Ð@þ$ÇĶý¬ Ððþ¬§æþsìý 9 糧éË Ððþ¬™èþ¢…
126 AÆÿ¬¯@þ B ¶ÔóýÉìþÄñý¬MæüP 10 Ð@þ 糧æþ…

(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

1113 () 9 of 32

10. 2 x 2 + ax + 6 = 0 Ð@þÆæÿYçÜÒ$MæüÆæÿ×ý… Äñý¬MæüP JMæü Ð@þÊË… 2 AÆÿ¬™óþ ‘a’ ÑË$Ð@þ

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 Ð@þÆæÿçYÜÒ$MæüÆæÿ×ýÐ@þ¬ Äñý¬MæüP Ñ^èþ„æü×ìý

(A) q 2  4 pr (B) q 2  4 pr

(C) p 2  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

(C) p 2  4 pr (D) p 2  4qr

12. 4, x, 10 Ë$ A…Mæü¶ÔóýÉìþÌZ E…sôý x ÑË$Ð@þ

(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

1113 () 10 of 32

13. ax 2 + bx + c = 0 Ð@þÆæÿYçÜÒ$MæüÆæÿ×ý… Äñý¬MæüP Ð@þÊÌêË$

b  b 2  4ac b  b 2  4ac
(A) x = (B) x =
2a 2a

b  b 2  4c b  b 2  4ac
(C) x = (D) x =
2a 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 çÜÒ$MæüÆæÿ×ý… Äñý¬MæüP Ð@þÊÌêË$

(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

1113 () 11 of 32

15. Æðÿ…yæþ$ ¶MæüÐ@þ$ ³ç NÆæÿç~Ü…QÅË Ððþ¬™èþ¢… 27 AÆÿ¬¯@þ B ç³NÆæÿ~çÜ…QÅË$
(A) 7 Ð@þ$ÇĶý¬ 20 (B) 13 Ð@þ$ÇĶý¬ 14

(C) 1 Ð@þ$ÇĶý¬ 26 (D) – 13 Ð@þ$ÇĶý¬ – 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. CÐ@þÓºyìþ¯@þ _{™èþ…ÌZ sin  ÑË$Ð@þ

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

In the figure, the value of sin  is

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

1113 () 12 of 32

17. ( sin 30° + cos 60° – tan 45° ) ÑË$Ð@þ

(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  =

(A) 4 + tan 2  (B) 4 + cot 2 

(C) 2 + cot 2  (D) 3 + cot 2 

3 + sec 2  is equal to

(A) 4 + tan 2  (B) 4 + cot 2 

(C) 2 + cot 2  (D) 3 + cot 2 
19. JMæü Vøç³#Ææÿ… Äñý¬MæüP ´ë§æþ… ¯@þ$…yìþ 30 Ò$. §æþ*Ææÿ…ÌZ ¯óþË Ò$§æþ JMæü ¼…§æþ$Ð@þ# ¯@þ$…yìþ Vøç³#Ææÿ…
Äñý¬MæüP ÖÆæÿÛÐ@þ¬¯@þ$ ^èþ*_¯@þç³#yæþ$ HÆæÿµyæþ$ FÆæÿ®Ó Mø×ý… 30° AÆÿ¬™óþ B Vøç³#Ææÿ… G™èþ$¢

(A) 10 m (B) 30 m

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

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

1113 () 13 of 32

20. ( sin   cosec  ) Äñý¬MæüP ÑË$Ð@þ

(A) 2 (B) 1

1 3
(C) – (D)
2 2

The value of ( sin   cosec  ) is

(A) 2 (B) 1

1 3
(C) – (D)
2 2

21. A ( x1 , y1 ) Ð@þ$ÇĶý¬ B ( x 2 , y 2 ) ¼…§æþ$Ð@þ#˯@þ$ MæüË$ç³# çÜÆæÿâ¶ýÆóÿQ Ð@þ$«§æþż…§æþ$Ð@þ#¯@þ$
Mæü¯@þ$Vö¯@þ$rMæü$ çÜ*{™èþ…
 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 
   

1113 () 14 of 32

22. ( x1 , y1 ) Ð@þ$ÇĶý¬ ( x 2 , y 2 ) ¼…§æþ$Ð@þ#Ë Ð@þ$«§æþŧæþ*Ææÿ…

(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. HO§ðþ¯é §æþ™é¢…Ô¶ý…ÌZ CÐ@þÓºyìþ¯@þ A…Ô>ËÌZ GMæü$PÐ@þ ÝëÆæÿ$Ï VæüÇçÙx… ç³#¯@þÆ>Ð@þÆæÿ¢¯þ@ … ^ðþ…§æþ$ A…Ô¶ý…
(A) çÜVæür$ AÐ@þ#™èþ$…¨ (B) ºçßý$â¶ýMæü… AÐ@þ#™èþ$…¨
(C) Ð@þ$«§æþÅVæü™èþ… AÐ@þ#™èþ$…¨ (D) ÐéÅí³¢ AÐ@þ#™èþ$…¨
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. ¶Mìü…¨ §é¢™èþ…Ô¶ýÐ@þ¬ Äñý¬MæüP çÜVæür$
Ð@þ*Ææÿ$PË$ 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

1113 () 15 of 32

25. HO§ðþ¯é §æþ™é¢…Ô¶ýÐ@þ¬ Äñý¬MæüP Üç Væür$, ºçßý$â¶ýMæü… Ð@þ$ÇĶý¬ Ð@þ$«§æþÅVæü™éË Ð@þ$«§æþÅçÜ…º…«§æþ…
(A) 3 Ð@þ$«§æþÅVæü™èþ… = 2 çÜVæür$ + ºçßý$â¶ýMæü…

(B) 3 çÜVæür$ = 2 Ð@þ$«§æþÅVæü™èþ… + ºçßý$â¶ýMæü…

(C) çÜVæür$ = 3 Ð@þ$«§æþÅVæü™èþ… + ºçßý$â¶ýMæü…
(D) ºçßý$â¶ýMæü… = 3 çÜVæür$ + 2 Ð@þ$«§æþÅVæü™èþ…

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. OÐðþ$¯@þÐ@þ¬™ø ™èþĶý*Ææÿ$^óþĶý$ºyìþ¯@þ JMæü çÜ*¦ç³Ð@þ¬¯@þ$ MæüÇW…_ B Ððþ¬™èþ¢… {§æþÐ@þ…™ø JMæü Vøâ¶ýÐ@þ¬¯@þ$
™èþĶý*Ææÿ$^óþíܯ@þç³#yæþ$ B Vøâ¶ý… çœ$¯@þç³ÇÐ@þ*×ý…
(A) çÜ*¦ç³… çœ$¯@þç³ÇÐ@þ*×ê°Mìü Æðÿ…yæþ$Æðÿr$Ï E…r$…¨
(B) çÜ*¦ç³… çœ$¯@þç³ÇÐ@þ*×ý…ÌZ çÜVæü… E…r$…¨
(C) çÜ*¦ç³… çœ$¯@þç³ÇÐ@þ*×ê°Mìü Ð@þÊyæþ$ Æðÿr$Ï E…r$…¨
(D) çÜ*¦ç³… çœ$¯@þç³ÇÐ@þ*×ý…™ø çÜÐ@þ*¯@þ…V> E…r$…¨

A cylinder made of wax is melted and recast completely 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

1113 () 16 of 32

27. ™èþÆæÿVæü† A…™èþÆæÿ… Ð@þ$«§æþż…§æþ$Ð@þ#¯@þ$ Mæü¯@þ$Vö¯@þ$rMæü$ çÜ*{™èþ…
GVæü$Ð@þçßý§æþ$ª – ¨Væü$Ð@þçßý§æþ$ª
(A) 2
GVæü$Ð@þçßý§æþ$ª × ¨Væü$Ð@þçßý§æþ$ª
(B)
3
GVæü$Ð@þçßý§æþ$ª + ¨Væü$Ð@þçßý§æþ$ª
(C) 2
GVæü$Ð@þçßý§æþ$ª + ¨Væü$Ð@þçßý§æþ$ª
(D) 3
The formula to find the mid-point of the class interval is
Upper limit  low er limit
(A)
2
Upper limit  low er limit
(B)
3
Upper limit  low er limit
(C)
2
Upper limit  low er limit
(D)
3
28.  ABC ÌZ XY || BC AÆÿ¬¯@þ

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

1113 () 17 of 32

In the  ABC, XY || BC then

AX AC
(A) 
AB AY
AX AY
(B) 
BX CY
AX XY
(C) 
BX AY
AB AC
(D) 
BX AY
29. ¶Mìü…§æþ CÐ@þÓºyìþ¯@þ Æðÿ…yæþ$ {†¿¶ý$gê˯@þ$ VæüÐ@þ$°…_ DF MöË™èþ¯@þ$ Væü$Ç¢…^èþ…yìþ

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

1113 () 18 of 32

30.  ABC ~  PQR,  ABC OÐðþÔ>ËÅ… = 64 cm 2 Ð@þ$ÇĶý¬  PQR OÐðþÔ>ËÅ… = 100 cm 2 ,

AB = 8 cm AÆÿ¬¯@þ PQ ´÷yæþÐ@þ#

(A) 12 cm (B) 15 cm

(C) 10 cm (D) 8 cm

 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 ÌZ B = 90° Ð@þ$ÇĶý¬ BD  AC, AB = 6 cm, BC = 8 cm AÆÿ¬¯@þ
CD ´÷yæþÐ@þ#

(A) 10 cm (B) 6.4 cm

(C) 4.8 cm (D) 3.6 cm

1113 () 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. _{™èþ…ÌZ ^èþ*í³¯@þr$Ï ‘O’ Móü…{§æþ…V> VæüË Ð@þ–™é¢°Mìü A ¼…§æþ$Ð@þ# Ð@þ§æþª AT çܵÆæÿØÆóÿQ XĶý$ºyìþ¯@þ¨.
OTA = 30° Ð@þ$ÇĶý¬ OT = 4 cm AÆÿ¬¯@þ AT ´÷yæþÐ@þ#

(A) 4 cm (B) 2 cm

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

1113 () 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. _{™èþ…ÌZ PA, PBC Ð@þ$ÇĶý¬ CD Ë$ ‘O’ Móü…{§æþ…V> VæüË Ð@þ–™é¢°Mìü XĶý$ºyìþ¯@þ çܵÆæÿØÆóÿQË$.
PC = 8 cm Ð@þ$ÇĶý¬ AP = 5 cm AÆÿ¬¯@þ, CD çܵÆæÿØÆóÿQ ´÷yæþÐ@þ#

(A) 5 cm

(B) 3 cm

(C) 8 cm

(D) 13 cm

1113 () 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. ¶Mìü…¨ ÐéM>ÅËÌZ ™èþ糚 ÐéMæüÅÐ@þ¬¯@þ$ Væü$Ç¢…^èþ…yìþ.
(A) Ð@þ–™é¢°Mìü XĶý$ºyìþ¯@þ çܵÆæÿØÆóÿQ Ð@þ–™èþ¢Ð@þ¬¯@þ$ JMóü¼…§æþ$Ð@þ# Ð@þ§æþª çܵÇØçÜ$¢…¨
(B) Ð@þ–™é¢°Mìü JMæü çÜÆæÿâ¶ýÆóÿQ¯@þ$ X_¯@þç³#yæþ$ A¨ Ð@þ–™èþ¢Ð@þ¬ Äñý¬MæüP JMæü ¼…§æþ$Ð@þ# Væü$…yé
´ù™èþ$…¨
(C) Ð@þ–™é¢°Mìü JMæü çÜÆæÿâ¶ýÆóÿQ¯@þ$ X_¯@þç³#yæþ$ A¨ Ð@þ–™èþ¢Ð@þ¬¯@þ$ çܵÇØ…^èþ$ ¼…§æþ$Ð@þ#¯@þ$ çܵÆæÿØ
¼…§æþ$Ð@þ# A…sêÆæÿ$
(D) çܵÆæÿؼ…§æþ$Ð@þ# ¯@þ$…yìþ Ð@þ–™é¢°Mìü XĶý$ºyìþ¯@þ ÐéÅÝëÆæÿ®… çܵÆæÿØÆóÿQMìü Ë…º…V> E…r$…¨
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

1113 () 22 of 32

35. C_a¯@þ Ð@þ–™é¢°Mìü »êçßýż…§æþ$Ð@þ# ‘T ’ ¯@þ$…yìþ Æðÿ…yæþ$ çܵÆæÿØÆóÿQ˯@þ$ XĶý¬¯@þç³#yæþ$ _{™èþ…ÌZ
^èþ*í³¯@þ °Æ>Ã×ý… ™èþÆ>Ó™èþ XĶý$Ð@þËíܯ@þ §é°° Væü$Ç¢…^èþ…yìþ.

(A)

(B)

(C)

(D)

1113 () 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)

1113 () 24 of 32

36. JMæü Vøâ¶ý… Eç³Ç™èþË OÐþð Ô>ËÅ… 616 cm 2 AÆÿ¬¯@þ B Vøâ¶ý… ÐéÅÝëÆæÿ®…
(A) 49 cm (B) 14 cm
(C) 21 cm (D) 7 cm
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. _{™èþ…ÌZ ^èþ*í³¯@þ Ñ«§þæ …V> E¯@þ² Ô¶ý…Q$Ð@þ# çœ$¯@þç³ÇÐ@þ*×ý

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

(A) r 2 h (B) r ( r + l )
1
(C) r 2 h (D) r l
3

1113 () 25 of 32

38. JMæü OÐðþç³# ™ðþÆæÿÐ@þºyìþ¯@þ Ë…ºÐ@þ–™èþ¢´ë§æþ çÜ*¦ç³… ´ë§æþÐéÅÝëÆæÿ®… r cm Ð@þ$ÇĶý¬ G™èþ$¢ h cm
AÆÿ¬¯@þ §é° çÜ…ç³NÆæÿ~™èþË OÐðþÔ>ËÅ… Mæü¯@þ$Vö¯@þ$rMæü$ çÜ*{™èþ…

(A) (  r 2 + 2rh ) cm 2 (B) 2rh cm 2

1
(C)  r 2h cm 3 (D) (  r 2 + h ) cm 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) (  r 2 + 2rh ) cm 2 (B) 2rh cm 2

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

39. _{™èþ…ÌZ CÐ@þÓºyìþ¯@þ Ô¶ý…Q$Ð@þ# bóþ§æþMæü… Ð@þ¶Mæü™èþË OÐðþÔ>ËÅ… Mæü¯@þ$Vö¯@þ$rMæü$ çÜ*{™èþ…

1
(A)  l ( r1  r2 )
3

1
(B)  h ( r12  r22  r1 r2 )
3

(C) l ( r1  r2 )

(D) l ( r1  r2 )

1113 () 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)  l ( r1  r2 ) (B)  h ( r12  r22  r1 r2 )
3 3
(C) l ( r1  r2 ) (D) l ( r1  r2 )

40. JMæü çœ$¯@þ AÆæÿ®Vøâ¶ý… Äñý¬MæüP çÜ…ç³NÆæÿ~™èþË OÐðþÔ>ËÅ… 462 cm 2 Ð@þ$ÇĶý¬ §é° Ð@þ¶Mæü™èþË OÐðþÔ>ËÅ…
308 cm 2 AÆÿ¬¯@þ §é° ´ë§æþ OÐðþÔ>ËÅ…

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

(C) 154 cm 2 (D) 1078 cm 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

1113 () 27 of 32

( SPACE FOR ROUGH WORK )

1113 () 28 of 32

( SPACE FOR ROUGH WORK )

1113 () 29 of 32

( SPACE FOR ROUGH WORK )

1113 () 30 of 32

( SPACE FOR ROUGH WORK )

1113 () 31 of 32

1113 () 32 of 32