Karnataka D.El.Ed 2nd year Question Paper 2022 Maths

D.El.Ed
Question
Paper

ÊÜÞ¨ÜÄ EñܤÃÜWÜÙÜá
MODEL ANSWERS
ÓÜíPæàñÜ ÓÜíTæ : 460-EK
Code No. : 460-EK
w.GÇ….Cw. ©Ì£à¿á ÊÜÐÜì
D.El.Ed. SECOND YEAR

IbfPÜ ËÐÜ¿á ― WÜ~ñÜ
Optional Subject ― MATHEMATICS
PܮܰvÜ ÊÜáñÜᤠCíXÉÐ… ÊÜÞ«ÜÂÊÜá
Kannada & English Medium
©®ÝíPÜ : 14. 07. 2022 ] [ WÜÄÐÜu AíPÜWÜÙÜá : 60
Date : 14. 07. 2022 ] [ Max. Marks : 60
PܮܰvÜ ÊÜÞ«ÜÂÊÜá
»ÝWÜ I
ŸÖÜá&BÁáR¿á EñܤÃÜWÜÙÜá ±ÜÅ£Áãí¨Üá ÓÜÄ EñܤÃÜPæR Jí¨æãí¨Üá AíPܨÜíñæ
1 × 10 = 10
1. (B) ÄÊæá୿ᬅ ÃæàTÝWÜ~ñÜ

2. (D) I ÊÜáñÜᤠIV

3. (A) ÓÜÊÜá®ÝXÃÜáñܤ¨æ

4. (A) MáOÝñܾPÜÊÝXÃÜáñܤ¨æ

5. (C) 32  42

6. (D) –4
7. (A) ŸãÅÓ… hæãÓ… ÊÜáñÜᤠÊÜÞÐÜìÊæÇ…

8. (B) WÜ~ñÜ¨Ü aÜoáÊÜqPæWÜÙÜá ÊÜáñÜᤠÊÜáãñÜì A®Üá»ÜÊÜWÜÚí¨Ü WÜ~ñÜ ±ÜÄPÜƳ®æWÜÙÜ®Üá°
ÓÜíÃÜbÓÜÆá ®æÃÜÊÝWÜáñܤ¨æ

D.El.Ed. – II [ Turn over

460-EK 2

9. (D) PÝ¿áìûÜÊÜáñæ ±ÜÄàûæ

10. (D) CÊÜâWÜÙæÆÉÊÜä

»ÝWÜ II
D PæÙÜX®ÜÊÜâWÜÙÜÈÉ ¿ÞÊÜâ¨Ý¨ÜÃÜã I¨Üá ±ÜÅÍæ°WÜÚWæ ÓÜíü±Ü¤ÊÝX EñܤÄÔ. ±ÜÅ£Áãí¨Üá EñܤÃÜÊÜâ
A«Üì ±ÜâoPæR ËáàÃÜ©ÃÜÈ 2 × 5 = 10

11.  F×ÓÜáPæãÙÜáÛÊÜâ¨Üá WÜ~ñÜ ÓÜíŸí«ÜWÜÙܬÜá® ÊÜáñÜᤠÓÜñÜÂÊæ­ÓÜáÊÜ ËaÝÃÜWÜÙÜ
ÊÝÂTæÂWÜÙÜ®Üá° ÃÜbÓÜáÊÜâ¨Üá ÊÜáñÜᤠÓÜñÜ AíÍÜWÜÙÜ®Üá° AÄ¿á¨æà CÃÜáÊÜâ¨Ü®Üá°
F×ÓÜáËPæ G®Üá°ÊÜÃÜá. C¨Ü®Üá° A±Üä|ìWæãívÜ ÊÜÞ×£¿á ÊÜáãÆ©í¨Ü
ÊÜÞvÜÇÝX¨æ.

E¨ÝÖÜÃÜOæ GÃÜvÜá ÓÜÊÜáÓÜíTæÂWÜÙÜ®Üá° PÜãw¨ÝWÜ ÊæãñܤÊÜâ ÓÜÊÜáÓÜíTæÂÁáà
BXÃÜáñܤ¨æ.

2+4=6

10 + 12 = 22

8 + 6 = 14

10 + 20 = 30

Êæáà騆 E¨ÝÖÜÃÜOæ¿á®Üá° ×àWæ F×ÔPæãÙÜáÛÊÜâ¨Üá. GÃÜvÜá
ÓÜÊÜáÓÜíTæÂWÜÙÜ®Üá° PÜãw¨ÝWÜ ÊæãñܤÊÜâ ÓÜÊÜáÓÜíTæÂÁáà BXÃÜáñܤ¨æ. 2

12. A) x + 10

B) x–5 1+1=2

13. ÓÜÖÜPÝÃÜ PÜÈPÝ ñÜíñÜÅ¨Ü E±ÜÁãàWÜWÜÙÜá
a) ˨ݦìWÜÙÜá Jí¨Üá WÜáÄ¿á¬Üá® ÓݘÓÜÆá JpÝrX PæÆÓÜ ÊÜÞvÜÆá
ÓÜÖÝ¿áÊÝX¨æ.
b) ÊæçË«ÜÂÊÜá¿á PÜÈPÝ aÜoáÊÜqPæWÜÙܬÜá® ŸÙÜÔ ËÐÜ¿á WÜÅ×Pæ¿á¬Üá®
ÓÜá«ÝÄÓÜÆá ÓÜÖÝ¿áÊÝX¨æ.

D.El.Ed. – II

3 460-EK

c) ˼®Ü° ÓÝÊÜá¥ÜÂìËÃÜáÊÜ ÊÜáPÜRÙÜ WÜáí²®æãí©Wæ ˨ݦì
PÝ¿áì¯ÊÜìÖÜOæ ÊÜÞw «Ü®ÝñܾPÜ ÓÜíÊÜÖÜ®ÜÊÜ®Üá° ¸æÙæÓÜáñܤ¨æ.
d) PÜÈPÝ ±ÜÅQÅÁá¿áÈÉ ÊÜáPÜRÙÜ®Üá° ÓÜQÅ¿áÊÝX ñæãvÜXÔPæãÙÜÛÆá
ÓÜÖÝ¿áPÜÊÝX¨æ.
¿ÞÊÜâ¨Ý¨ÜÃÜã ¬ÝÆáR ÓÜãPܤ AíÍÜWÜÙܬÜá® ¯ÜÄWÜ~ÓÜŸÖÜá¨Üá ½ × 4 = 2

‘4’ ÃÜ »ÝgÂñæ¿á ¯¿áÊÜá GÃÜvÜá AíQWÜÚXíñÜ ÖæaÝcXÃÜáÊÜ ÓÜíTæÂWÜÙÜ ¹w
14.
ÊÜáñÜᤠÖÜñܤÃÜ Óݧ®Ü©í¨Ý¨Ü ÓÜíTæ¿áá 4 Äí¨Ü »ÝWÜÊݨÜÃæ ¨Üñܤ ÓÜíTæ¿áá
4 Äí¨Ü »ÝWÜÊÝWÜáñܤ¨æ.

E¨ÝÖÜÃÜOæ

Th H T U
4 9 2 8 ÓÜíTæ BXÃÜÈ

28  4 = 7

 4928 ÓÜíTæ¿áá 4 Äí¨Ü »ÝWÜÊÝWÜáñܤ¨æ.

4928  4 = 1232 1+1=2

15.
WÜ~ñÜ A»ÝÂÓܱÜâÓܤPÜWÜÙÜ®Üá° ÓÜÌ¿áí PÜÈPÝ ±ÜâÓܤPÜ Gí¨Üá PÜÃæ¿Æá PÝÃÜ|

i) A»ÝÂÓܱÜâÓܤPÜWÜÙÜá ±ÜsܱÜâÓܤPÜPæR ±ÜäÃÜPÜÊÝX¨Ü᪠AWÜñÜÂÊÝ¨Ü ÓÜáÚÊÜâ ÊÜáñÜá¤
ÓÜãaÜ®æWÜÙÜ®Üá° J¨ÜXÓÜáñܤ¨æ.

ii) A»ÝÂÓܱÜâÓܤPÜWÜÙÜá ÓÜÌ¿áí ¯¨æàìÎñÜ PÜÈPÝ ÖÜíñÜWÜÙÜ®Üá° ÊÜáñÜᤠÓÜÊÜáÓÝÂ
±ÜÄÖÝÃÜPæR ÓÝPÜÐÜár ÓܧÙÝÊÜPÝÍÜÊÜ®Üá° J¨ÜXÓÜáñܤ¨æ. 1×2=2

16.
AíñÜhÝìÆ¨Ü WÜ~ñÜ ÓÜí±Ü®Üã¾ÆWÜÙÜá

i) WÜ~ñÜ ñÜÃÜWÜ£¿áÈÉ E¯ÜÁãàXÓÜƳvÜáÊÜ A¬æàPÜ hÝÆñÝ|WÜÙܬÜá® ŸÙÜÓÜÆá
ÓÜÖÝ¿áPÜÊÝX¨æ.

D.El.Ed. – II [ Turn over

460-EK 4

ii) ÍæçûÜ~PÜ ËÐÜ¿á ÊÜáñÜᤠÊÜÞ×£WÜÙÜ®Üá° ±ÜűÜíaܨݨÜÂíñÜ ±ÜÃÜÓܳÃÜ
ÖÜíbPæãÙÜÛÆá ÓÜÖÝ¿áÊÝX¨æ.

iii) ICT ¸æãà«Ü®Ý ˫ݮܩí¨Ü ÓÜáÆ»ÜÊÝ¨Ü Áãàg®Ý«ÝÄñÜ PÜÈPæ¿á®Üá°
¯ÜÄaÜÀáÓÜŸÖÜá¨ÝX¨æ.

iv) AíñÜÃ…hÝƨÜÈÉ ËË«Ü Äࣿá BoWÜÙܬÜá® ÊÜáñÜᤠaÜoáÊÜqPæWÜÙܬÜá®
¯ÜÄOÝÊÜáPÝÄ PÜÈPæãà¯ÜPÜÃÜ|WÜÙÝX ÎûÜPÜÃÜá ŸÙÜÓÜŸÖÜá¨Üá.

¿ÞÊÜâ¨Ý¨ÜÃÜã ®ÝÆáR ÓÜãPܤ AíÍÜWÜÙܬÜá® ¯ÜÄWÜ~ÓÜŸÖÜá¨Üá ½ × 4 = 2

±ÜÄàûÝíPÜWÜÙÜ ËÍæÉàÐÜOæ¿áÈÉ WÜÊÜá¯ÓÜáÊÜ AíÍÜWÜÙÜá
17.

i) ¨æãàÐÜWÜÙÜ ÓÜÌ»ÝÊÜ&WÜÊÜá®ÜËÆÉ¨Ü ñܱÜâ³WÜÚí¨Ý¨Ü ÇæPÝRaÝÃÜ A¥ÜÊÝ
±ÜÃÜPÜƳ®ÝñܾPÜ ¨æãàÐÜ.

ii) ËÐÜ¿á A¥ÜÊÝ E±ÜËÐÜ¿áÊÜâ ¨æãàÐÜWÜÙÜ®Üá° Öæãí©ÃÜáÊÜâ¨Üá.

iii) ˨ݦìWÜÚWæ PÜÐÜrÊæí¨Üá »ÝËÔÃÜáÊÜ ÓÜÊÜáÓæÂWÜÙÜá ŸÖÜáË«Ü¨Ü BÁáR,
ÇæPÝRaÝÃÜWÜÙÜá, ÍݹªPÜ ÓÜÊÜáÓæÂWÜÙÜá, ±ÜÄÖÜÃÝñܾPÜ ÓÜÊÜáÓæÂWÜÙÜá

iv) ËÐÜ¿á ÊÜáívÜ®æ & AaÜácPÜoárñÜ®Ü, ±ÝÃܨÜÍÜìPÜñæ. ½×4=2

EñܤÃÜ QàÈ¿á®Üá° WÜáÃÜá£ÓÜáÊÜ AíÍÜWÜÙÜá
18.

i) ˨ݦì¿áá ÓÜÊÜáÓæ¿á¬Üá® ¯ÜÄÖÜÄÓÜÆá ŸÙÜÓÜáÊÜ ¯ÜÅQÅÁá A¥ÜÊÝ
Ë«Ý®ÜÊÜ®Üá° ±ÜÄWÜ~ÓÜáÊÜâ¨Üá.

ii) ˨ݦìWÜÙÜá ±ÜÄÖÜÄst ±ÜÅ£Áãí¨Üá ÖÜíñÜÊÜ®Üá° ËÓܤÄÓÜáÊÜâ¨Ü®Üá° ÊÜáñÜá¤
ÓÜÊÜá¦ìÓÜáÊÜâ¨Ü®Üá° ±ÜÄWÜ~ÓÜáÊÜâ¨Üá.

iii) ÓÜÊÜáÓæÂWÜÙܬÜá® ¹wÓÜÆá A¬æàPÜ Ë«Ý¬ÜWÜÙܬÜá® ŸÙÜÓÜáÊÜâ¨Ü¬Üá®
WÜáÃÜá£ÓÜáÊÜâ¨Üá.

iv) ÊÜáPÜRÙÜ ¯ÆìüñÜ ñܱÜâ³WÜÙÜ®Üá° G~PÝ ¨æãàÐÜWÜÙÜá ÖæaÝcX ¨ÜíwÓܸÝÃܨÜá.
½×4=2

D.El.Ed. – II

5 460-EK

»ÝWÜ III
D PæÙÜX®Ü ±ÜÅ£ ±ÜÅÍæ°WÜã (a) A¥ÜÊÝ (b) BÁáRWÜÙÜ®Üá° ¯àvÜÇÝX¨æ. AÊÜâWÜÙÜÈÉ Jí¨ÜPæR EñܤÄÔ.
±ÜÅ£Áãí¨Üá EñܤÃÜÊÜâ Jí¨Üá ±Üâo ËáàÃܨÜíñæ EñܤÄÔ 5 × 8 = 40

19. a) ià®… ²¿Þhæ¿á ±ÜÅPÝÃÜ WÜ~ñÜ ÓÜíÃÜaÜ®æ¿á WÜá|ÆûÜ|WÜÙÜá

i) ÃÜaÜ®æ¿áá ±Üä|ìñæ¿á®Üá° Öæãí©¨æ

ii) ÃÜã±ÝíñÜÃÜ ¯¿áÊÜá

iii) ÃÜaÜ®æ¿áá ÓÜÌ¿áí ¯¿áíñÜÅPÜÊÝX¨æ.

±ÜÅ£Áãí¨Üá WÜá|ÆûÜ|¨Ü ËÊÜÃÜOæWæ 1×3=3

E¨ÝÖÜÃÜOæ
¯Üä|ì ÓÜíTæÂWÜÙÜ ¯Ü¨Üœ£

W = { 0, 1, 2, 3, 4, 5 ........ }

 { 0, 1, 2, 3, 4........ } W

 1, 2, 3, 4, 5, 6, 7 W

 GÃÜvÜá ÓÜíTæÂWÜÙÜ®Üá° PÜãw¨ÝWÜ,

2 + 3 = 5 W

7 + 5 = 12 W

 ¬ÝÊÜâ GÃÜvÜá ¯Üä|ì ÓÜíTæÂWÜÙܬÜá® PÜãw¨ÝWÜ ŸÃÜáÊÜ EñܤÃÜÊÜâ ¯Üä|ì
ÓÜíTæÂÁáà BXÃÜáñܤ¨æ.

WÜ~ñÜ ÃÜaܬæWæ ÓÜíŸí˜Ô¨Üíñæ ¿ÞÊÜâ¨Ý¨ÜÃÜã ÓÜãPܤ E¨ÝÖÜÃÜOæ¿á¬Üá®
¯ÜÄWÜ~ÓÜŸÖÜá¨Üá 2

A¥ÜÊÝ

D.El.Ed. – II [ Turn over

460-EK 6

b) ñÜÃÜWÜ£ PæãàOæ¿áÈÉ WÜ~ñÜ ñÝQìPÜñæ¿á®Üá° ±æäÅàñÝÕ×ÓÜáÊÜíñÜ ÖÜ ±ÜÄÓÜÃÜÊÜ®Üá°

ÓÜêÑrÓÜáÊÜ PèÍÜÆWÜÙÜá

i) ÎûÜPÜÃÜá ÊÜáPÜRÚWæ ñÜÃÜWÜ£ PæãàOæ¿áÈÉ Öæbc®Ü ±ÜÅÍæ°WÜÙÜ®Üá° PÜvÝx¿áÊÝX

PæàÙÜÆá ±æäÅàñÝÕ×ÓܸæàPÜá.

ii) ñÜÃÜWÜ£ PæãàOæ¿áÈÉ WÜ~ñÜ¨Ü ŸWæY aÜbìÓÜÆá ÊÜáñÜᤠÊÜÞñܬÝvÜÆá

Öæbc®Ü ÊæàÙæ¿á®Üá° ÊÜÂÀáÓܸæàPÜá.

iii) ÎûÜPÜÃÜá ñܯÜâ³ EñܤÃÜWÜÙܬÜá® ÊÜhÝWæãÚÓÜáÊÜí£ÆÉ A¨ÜÃÜ Ÿ¨ÜÈWæ ÊÜáPÜRÙÜá

ñÜÊÜá¾ EñܤÃÜÊÜ®Üá° ÓÜÊÜá¦ìÔPæãÙÜÛÆá ÖæàÙÜáÊÜâ¨Üá.

iv) ˨ݦìWÜÙÜá ÊÜÞñÜ®ÝvÜáÊÜ »ÝÐæÁãí©Wæ WÜ~ñÜ¨Ü bÖæ° ÊÜáñÜá¤
»ÝÐæ ÖÝWÜã ¯giàÊܮܨÜÈÉ WÜ~ñܨæãí©Wæ Pæãíw ¸æÓæ¿ááÊÜâ¨Üá.

v) ÓÜãPܤÊÝ¨Ü JWÜoáWÜÙÜ®Üá° ¯àw ¹wÓÜÆá ±æäÅàñÝÕ×ÓÜáÊÜâ¨Üá.

vi) ÊÜáPÜRÚWæ ÓÜÌñÜ@ ÓÜÊÜáÓæÂWÜÙÜ®Üá° ÓÜêÑrÔ A¨Ü®Üá° ±ÜÄÖÜÄÓÜÆá
±æäÅàñÝÕ×ÓÜáÊÜâ¨Üá.

vii) ÎûÜPÜÃÜá ÊÜáPÜRÙÜÈÉ FÖæWÜÙÜá ÊÜáñÜᤠÓÝÊÜÞ¯ÂàPÜÃÜ|WÜÙÜ®Üá° ÊÜÞvÜÆá
ÓÜÊÜáê¨Üœ ¯ÜÅÍæ®WÜÙܬÜá® ÃÜbÓÜÆá AÊÜPÝÍÜ ­àvÜáÊÜâ¨Üá.

¿ÞÊÜâ¨Ý¨ÜÃÜã I¨Üá ÓÜãPܤ AíÍÜWÜÙæãí©Wæ ËÊÜÄÓÜáÊÜâ¨Üá 1×5=5

D.El.Ed. – II

7 460-EK

20. a) £Å»Üág¨Ü ÊÜáãÃÜá Pæãà¬ÜWÜÙÜ ÊæãñܤÊÜâ GÃÜvÜá ÆíŸPæãà¬Ü £Å»ÜágPæR
ÓÜÊÜá®ÝXÃÜáñܤ¨æ.

¨Üñܤ ABC £Å»ÜágÊÝX¨æ.
Óݫܯà¿á
BAC + ABC + ACB = 180°
ÃÜaÜ®æ AC ÓÜÊÜÞíñÜÃÜÊÝX DE ÓÜÃÜÙÜÃæàTæ¿á®Üá° ÖݨÜá ÖæãàWÜáÊÜíñæ
GÙæ¿ááÊÜâ¨Üá.
ÖæàÚPæ PÝÃÜ|
£Å»Üág ABC ¿áÈÉ A, B ÊÜáñÜᤠC
Pæãà®ÜWÜÚÊæ.

DBA = BAC ¯àwÃÜáÊÜ AíÍÜ DE ÊÜáñÜᤠAC ÃæàTæWÜÙÜá
±ÜÃÜÓܳÃÜ ÓÜÊÜÞíñÜÃÜÊÝX¨æ ÊÜáñÜᤠGÃÜvÜá
ÓÜÊÜÞíñÜÃÜ ÓÜÃÜÙÜÃæàTæWÜÙÜ®Üá° dæà¨ÜPÜ
ÃæàTæ¿áá dæà©Ô¨ÝWÜ EípÝWÜáÊÜ
±Ü¿Þì¿á Pæãà®ÜWÜÙÜá ÓÜÊÜá®ÝXÃÜáñܤ¨æ.

EBC = BCA ±Ü¿Þì¿á Pæãà®ÜWÜÙÜá ÓÜÊÜá®ÝXÃÜáñܤÊæ.

DBA + ABC + EBC = 180° ±ÜäÃÜPÜ Pæãà®ÜWÜÙÜá

 BAC + ABC + ACB = ÓæàÄÓÜáËPæÀáí¨Ü / ±Ü¿Þì¿á
180°

 £Å»Üág¨Ü ÊÜáãÃÜá Pæãà®ÜWÜÙÜ ÆíŸPæãà¬Ü £Å»Üág = 90°
ÊæãñܤÊÜâ GÃÜvÜá ÆíŸPæãà¬Ü
£Å»ÜágPæR ÓÜÊÜá®ÝXÃÜáñܤ¨æ. 5
A¥ÜÊÝ

D.El.Ed. – II [ Turn over

460-EK 8

b) ¯ñÜÂÓÜËáàPÜÃÜ|ÊÜ®Üá° PÜÅÊÜ៨ܜWæãÚÓÜáÊÜâ¨Üá
( a  b )2  a 2  b 2  2ab

ABCD ËÔ¤à|ì = AEFG ËÔ¤à|ì + EBHF ËÔ¤à|ì + FHCI ËÔ¤à|ì
+ DGFI ËÔ¤à|ì

B¨Üá¨ÜÄí¨Ü ( a + b ) ( a + b ) = ( a  a ) + ( a  b ) + ( b  b ) + ( a  b )
( a  b )2  a 2  ab  b 2  ab
( a  b )2  a 2  b 2  2ab 5
21. a) 0 ÊÜáñÜᤠ1 ÃÜ ®ÜvÜá訆 ¼®Ü°ÃÝÎWÜÙÜá

0 ÊÜáñÜᤠ1 ÃÜ ®ÜvÜá訆 ¿ÞÊÜâ¨Ý¨ÜÃÜã ÓÜãPܤ 5 ¼®Ü°ÃÝÎWÜÙÜá 1×5=5

A¥ÜÊÝ

D.El.Ed. – II

9 460-EK

b) B¿áñÜ¨Ü ÓÜáñܤÙÜñæ

BÊÜêñܤ BPÜꣿá ÔàÊÜÞÃæàTæ¿á ÊæáàÇæ Jí¨Üá ÓÜáñÜᤠŸÃÜáÊÝWÜ PÜÅËáÔ¨Ü
ÖÝ©¿á ¨ÜãÃÜÊæà ÓÜáñܤÙÜñæ.

 B¿áñÜ¨Ü ÓÜáñܤÙÜñæ ABCD = AB + BC + CD + DA

= 7 cm + 3 cm + 7 cm + 3 cm

= 20 cm

B¿áñÜ¨Ü ÓÜáñܤÙÜñæ = E¨Üª + AWÜÆ + E¨Üª + AWÜÆ

 B¿áñÜ¨Ü ÓÜáñܤÙÜñæ = a + b + a + b = 2a + 2b = 2 ( a + b )

=2 E¨Üª + AWÜÆ

 B¿áñÜ¨Ü ÓÜáñܤÙÜñæ = 2 ( a + b ) = 2 ( 7 + 3 ) = 2 ( 10 )

= 20 cm

B¿áñÜ¨Ü ÓÜáñܤÙÜñæ = 20 cm

D.El.Ed. – II [ Turn over

460-EK 10

Jí¨Üá BÊÜêñܤ BPÜꣿáá BÊÜÄÔPæãíwÃÜáÊÜ ÊæáàÇ澄 ¯ÜŨæàÍÜÊܬÜá® A¨ÜÃÜ ËÔ¤à|ì
G®Üá°ÊÜÃÜá.
B¿áñÜ¨Ü ËÔ¤à|ì = 6 sq.cm
C¨Ü®Üá° 3  2 sq.cm Gí¨Üá ŸÃæ¿ááÊÜâ¨Üá
= E¨Üª  AWÜÆ

B¿áñÜ¨Ü ËÔ¤à|ì = E¨Üª  AWÜÆ

= 3 cm  2 cm

= 6 sq.cm 2½ + 2½ = 5

22. a) i) PÝWÜ¨Ü ÊÜáwbÓÜáËPæ¿á aÜoáÊÜqPæ
ii) bñÜÅ¨Ü ÊÜáãÆPÜ ¯ÜÅ£­˜ÓÜáËPæ
GÃÜvÜá ÓÜÊÜáËᣠÃæàTæWÜÚÃÜáÊÜ bñÜÅWÜÙÜá

B¿áñÜÊÜâ 2 ÓÜÊÜáËᣠÃæàTæWÜÙÜ®Üá° Öæãí©¨æ.
GÃÜvÜQRíñÜ ÖæaÜác ÓÜÊÜáËᣠÃæàTæWÜÙÜ®Üá° Öæãí©ÃÜáÊÜ bñÜÅWÜÙÜá

aèPÜÊÜâ 4 ÓÜÊÜáËᣠÃæàTæWÜÙÜ®Üá° Öæãí©¨æ.
¿ÞÊÜâ¨Ý¨ÜÃÜã ÓÜãPܤ aÜoáÊÜqPæ¿á¬Üá® ¯ÜÄWÜ~ÓÜŸÖÜá¨Üá 2½ + 2½ = 5

A¥ÜÊÝ

D.El.Ed. – II

11 460-EK

b) A®Üá±ÝñÜ
GÃÜvÜá ±ÜÄÊÜÞ|WÜÙÜ®Üá° GÐÜrÃÜÑrÊæ GíŸá¨Ü¬Üá® ÖæãàÈPæ ÊÜÞvÜáÊÜâ¨Ü¬Üá®
A®Üá±ÝñÜ G®Üá°ÊÜÃÜá. C¨Ü®Üá° ‘ :’ bÖæ® ŸÙÜÔ ÓÜãbÓÜáñæ¤àÊæ.
Jí¨æà ÊÜáãÆÊÜޮܨÜÈÉ C¨ÝªWÜ ÊÜÞñÜÅ GÃÜvÜá ±ÜÄOÝÊÜáWÜÙÜ®Üá°
ÖæãàÈÓÜŸÖÜá¨Üá.
aÜoáÊÜqPæWÜÙÜá
¨æç¬Üí©¬Ü ÓÜ­®ÊæàÍÜWÜÙܬÜá® ŸÙÜÔ A¬Üá¯ÝñÜÊܬÜá® PÜívÜá×w¿ááÊÜâ¨Üá
i) Jí¨Üá ñÜÃÜWÜ£¿á 30 ˨ݦìWÜÙÜÈÉ ¶Üâp…¸ÝÇ… CÐÜr¯ÜvÜáÊÜÊÜÃÜ
ÓÜíTæÂWÜã QÅPæp… CÐÜr¯ÜvÜáÊÜÊÜÃÜ ÓÜíTæÂWÜã CÃÜáÊÜ A¬Üá¯ÝñÜ.
ii) Jí¨Üá ñÜÃÜWÜ£¿á 35 ˨ݦìWÜÙÜÈÉ ÖÜávÜáX¿áÃÜ ÓÜíTæÂWÜã
ÖÜávÜáWÜÃÜ ÓÜíTæÂWÜã CÃÜáÊÜ A®Üá±ÝñÜ.
¿ÞÊÜâ¨Ý¨ÜÃÜã ÓÜãPܤ aÜoáÊÜqPæ¿á¬Üá® ¯ÜÄWÜ~ÓÜŸÖÜá¨Üá 1+2+2=5

23.
a) PÝ¿áì ËÍæÉàÐÜOæ¿á ±ÝÅÊÜááSÂñæ
i) ÎûÜPÜÄWæ ±ÝsÜÁãàg®æ¿á®Üá° PÜÅÊÜáÊÝX ÃÜã²ÓÜÆá ÓÜÖÝ¿áÊÝX¨æ.
ii) C¨Üá ÎûÜPÜ®Ü EñܤÊÜá ¸æãà«Ü®æWÝX ÊÜáñÜᤠEñܤÊÜá PÜÈPæ¿á®Üá°
A®ÜáPÜãÈÓÜÆá ®æÃÜÊÝX¨æ.
iii) ˨ݦìWÜÙÜ ±ÜäÊÜìþÝ®ÜÊÜ®Üá° ÓܾÄÓÜÆá ÓÜÖÝ¿áÊÝX¨æ.
iv) ¸æãà«Ü¬Ý E¨æªàÍÜÊܬÜá® ÓݘÓÜÆá ÓÜÖÝ¿áPÜÊÝX¨æ.
˨ݦìWÜÙÜá ÊÜáÖÜñܤÃÜ ÓÝÊÜޮܠA±ÜÊÜñÜì®ÜÊÜ®Üá° ÊÝÂTݯÓÜáÊÜÃÜá.
PÝ¿áìËÍæÉàÐÜOæ
 ˨ݦì¿áá ¿ÞÊÜâ¨Ý¨ÜÃÜã GÃÜvÜá ÓÜíTæÂWÜÙÜ®Üá° BÄÓÜáÊÜÃÜá.
 ¨Üñܤ ÓÜíTæÂWÜÙÜ®Üá° »ÝXÓÜáÊÜ WÜáOÝPÝÃÜ PæãàÐÜrPÜWÜÙÜ®Üá° ÓܾÄÓÜáÊÜÃÜá.
 GÃÜvÜá ÓÜíTæÂWÜÙÜ A±ÜÊÜñÜì®ÜWÜÙÜ®Üá° WÜáÃÜá£ÓÜáÊÜÃÜá.
 GÃÜvÜá ÓÜíTæÂWÜÙÜ ÓÝÊÜޮܠA±ÜÊÜñÜì®ÜWÜÙÜ®Üá° ÖÜávÜáPÜáÊÜÃÜá.

D.El.Ed. – II [ Turn over

460-EK 12

 ÊÜáÖÜñܤÃÜ ÓÝÊÜޮܠA±ÜÊÜñÜì®ÜÊÜ®Üá° WÜáÃÜá£ÓÜáÊÜÃÜá.
 ÊÜáÖÜñܤÃÜ ÓÝÊÜޮܠA±ÜÊÜñÜì®Ü¨Ü A¥ÜìÊÜ®Üá° ËÊÜÄÓÜáÊÜÃÜá.
 ÊÜáÖÜñܤÃÜ ÓÝÊÜޮܠA±ÜÊÜñÜì®ÜÊÜ®Üá° ÊÝÂTݯÓÜáÊÜÃÜá. 2+3=5

A¥ÜÊÝ
b) A®ÜáWÜÊÜá®Ü ÁãàaÜ®æ ÊÜÞ¨ÜÄ
±ÜÄPÜƳ®æ — ÓÜÃÜÙÜ¿ááWܾ
 hæãàwWÜÙÜá ±ÝÍÜÌìPæãà®ÜWÜÙÝXÃÜáñܤÊæ.
 D Pæãà®ÜWÜÙÜ ÊæãñܤÊÜâ 180° AÙÜñæ¿ÞXÃÜáñܤ¨æ.
 ÓÜÃÜÙÜ¿ááWܾWÜÙÜ ÊÝÂTÝ娆 ±ÜÃÜÓܳÃÜ ±ÝÍÜÌìÊÝXÃÜáÊÜ ÊÜáñÜᤠAÊÜâWÜÙÜ
AÙÜñæ 180° BXÃÜáÊÜ hæãàw Pæãà®ÜWÜÙÜá.
ÖÜíñÜWÜÙÜá
 ¨ÜñݤíÍÜWÜÙÜ®Üá° WÜáÃÜá£ÓÜáÊÜâ¨Üá ÊÜáñÜᤠ±Üqr ÊÜÞvÜáÊÜâ¨Üá.
 ÓÝÊÜޮܠWÜá|WÜÚÃÜáÊÜ AíÍÜWÜÙÜ®Üá° WÜáí±Üâ ÊÜÞvÜáÊÜâ¨Üá.
 ÊÜWÜìWÜÙܬÜá® A¼ÊÜê©œWæãÚÔ WÜáí¯ÜâWÜÙܬÜá® ÖæÓÜÄÓÜáÊÜâ¨Üá.
¯ÜÄPÜƳ¬æ¿á¬Üá® ŸÙÜÔ ¯ÜÅ£Áãí¨Üá ÖÜíñÜÊܬÜá® ËÊÜÄÓÜáÊÜâ¨Üá 5
24. a) ±ÜÄPÜƳ®æ ÊÜêñܤ¨Ü ÊÝÂÓÜ
±ÜÄPÜƳ®æ ÊÜáãwÓÜáËPæ ÊÜÞ¨ÜÄ

ÊÜágÆá I E¨ÝÖÜÃÜOæWÜÙÜ®Üá° ÊÜáíwÔ ÊÜáñÜᤠ±ÜÄPÜƳ®æ¿á®Üá° WÜáÃÜá£Ô.

ÖÜíñÜ 1 : Öè¨Üá / CÆÉ E¨ÝÖÜÃÜOæWÜÙÜ®Üá° ÊÜáíwÓÜáÊÜâ¨Üá.

ÖÜíñÜ 2 : Öè¨Üá / CÆÉ E¨ÝÖÜÃÜOæWÜÙÜ®Üá° ÖæãàÈÓÜáÊÜâ¨Üá.

ÖÜíñÜ 3 : ˨ݦìWÜÙÜá FÖÝPÜƳ®æ¿á®Üá° ±ÜÄÎàÈÓÜáÊÜÃÜá.

ÖÜíñÜ 4 : ˨ݦìWÜÙÜá ±ÜÄPÜƳ®æ¿á®Üá° ¯ÃÜã²ÓÜáÊÜÃÜá.

D.El.Ed. – II

13 460-EK

ÊÜágÆá II : ±ÜÄPÜƳ®æ ÊÜáãwÓÜáËPæ¿á®Üá° ±ÜÄàüÓÜáÊÜâ¨Üá.
ÖÜíñÜ 1 : ˨ݦìWÜÙÜá ÊæáÆáY£ìÓÜ¨Ü E¨ÝÖÜÃÜOæWÜÙÜ®Üá°
WÜáÃÜá£ÓÜáÊÜÃÜá.
ÖÜíñÜ 2 : ÓÜÄ¿Þ¨Ü FÖÝPÜƳ®æ¿á®Üá° ¨Üê{àPÜÄÓÜáÊÜÃÜá,
ÖæÓÜÄÓÜáÊÜÃÜá, ÊÝÂTݯÓÜáÊÜÃÜá.
ÖÜíñÜ 3 : ˨ݦìWÜÙÜá E¨ÝÖÜÃÜOæWÜÙÜ®Üá° ÓÜêiÓÜáÊÜÃÜá.
ÊÜágÆá III : bíñÜ®Ý PÜÅÊÜáWÜÙÜ ËÍæÉàÐÜOæ
ÖÜíñÜ 1 : ÎûÜPÜÃÜá bíñÜ®Ý PÜÅÊÜáWÜÙÜ®Üá° ËÍæÉàÐÜOæ ÊÜÞvÜáÊÜÃÜá.
ÖÜíñÜ 2 : ˨ݦìWÜÙÜá FÖÝPÜƳ®æWÜÙÜ ±ÝñÜÅÊÜ®Üá° ÊÜáñÜá¤
WÜá|«ÜÊÜáìWÜÙÜ®Üá° aÜbìÓÜáÊÜÃÜá.
ÖÜíñÜ 3 : ˨ݦìWÜÙÜá FÖÝPÜƳ®æWÜÙÜ ±ÝñÜÅÊÜ®Üá° ÊÜáñÜá¤
WÜá|«ÜÊÜáìWÜÙÜ®Üá° aÜbìÓÜáÊÜÃÜá.
±ÜÄPÜƳ®æ¿á ÊÜáãÆPÜ ±ÜÅ£ ÖÜíñÜÊÜ®Üá° ËÊÜÄÓÜáÊÜâ¨Üá
5
A¥ÜÊÝ
b) iW…&ÓÝ PÜÈPÝ ñÜíñÜÅ¨Ü ÖÜíñÜWÜÙÜá
i) ËÐÜ¿á ÊÜáñÜᤠ¯©ìÐÜr E¨æªàÍÜWÜÙÜ®Üá° ¯«ÜìÄÓÜáÊÜâ¨Üá.
ËÐÜ¿á ¸ÝÖÜáWÜÙÜ B«ÝÃÜ¨Ü Êæáà騆 £Å»Üág¨Ü Ë«ÜWÜÙÜá
ÓÜÊÜá¸ÝÖÜá £Å»Üág, ÓÜÊÜá©Ì¸ÝÖÜá £Å»Üág,
AÓÜÊÜá¸ÝÖÜá £Å»Üág.
E¨æªàÍÜWÜÙÜá ˨ݦìWÜÚWæ CÊÜâWÜÙÜ®Üá° A®ÜáPÜãÈÓÜÆá
 ±ÜÅ£Áãí¨Üá £Å»Üág¨Ü Ë«ÜÊÜ®Üá° ËÊÜÄÓÜáÊÜÃÜá
 ±ÜÅ£Áãí¨Üá £Å»Üág¨Ü Ë«ÜÊÜ®Üá° ÊÝÂTݯÓÜáÊÜÃÜá.
 ±ÜÅ£Áãí¨Üá £Å»Üág¨Ü Ë«ÜÊÜ®Üá° WÜáÃÜá£ÓÜáÊÜÃÜá.
 PæãqrÃÜáÊÜ £Å»ÜágWÜÙÜ®Üá° ¸æàÃæ ¸æàÃæ Ë«ÜWÜÙÝX ÊÜXàìPÜÄÓÜáÊÜâ¨Üá .

D.El.Ed. – II [ Turn over

460-EK 14

ii) PÜÈPæ¿á E±ÜPÜÃÜ|WÜÙÜ®Üá° ÁãàiÓÜáÊÜâ¨Üá.

iii) ˨ݦìWÜÙÜ®Üá° WÜáí±Üâ ÊÜÞvÜáÊÜâ¨Üá.

iv) ñÜýÃÜ WÜáí±ÜâWÜÙÝX ËíWÜwÓÜáÊÜâ¨Üá.

v) ñÜýÃÜá ñÜÊÜá¾ ÓÜÌíñÜ WÜáí±ÜâWÜÚWæ ñæWÜÙÜáÊÜÃÜá.

vi) WÜáí¯ÜâWÜÙÜÈÉ ÓÜÖÜPÝÃÜ ÊÜáñÜᤠ«Ü¬ÝñܾPÜ AÊÜÆퟬæ¿á¬Üá® ÓÜêiÓÜáÊÜâ¨Üá.

vii) WÜáí²®Ü ÓÝ«Ü®æ¿á®Üá° ÊÜåèÆÂÊÜÞ±Ü®Ü ÊÜÞvÜáÊÜâ¨Üá.

viii) Êæç¿áQ¤PÜÊÝX ˨ݦìWÜÙÜ WÜáOÝñܾPÜ PÜÈPæ¿á®Üá° ÊÜåèÆÂÊÜޱܮÜ
ÊÜÞvÜáÊÜâ¨Üá.
±ÜÄPÜƳ®æ¿á ÊÜáãÆPÜ ±ÜÅ£ ÖÜíñÜÊÜ®Üá° ËÊÜÄÓÜáÊÜâ¨Üá 5
25.
a) ËÐÜ¿á ÊÜáñÜᤠÊÜáívÜ®æWæ ÓÜíŸí˜Ô¨Üíñæ WÜ~ñÜ ±ÜâÓܤPÜ¨Ü ËÍæàÐÜ WÜá|WÜÙÜá

 ±ÜsܱÜâÓܤPܨÜÈÉÃÜáÊÜ ËÐÜ¿á ÊÜáívÜ®æ ¯SÃÜÊÝXÃܸæàPÜá ÊÜáñÜá¤
ÓÜí±Üä|ìÊÝXÃܸæàPÜá.

 WÜ~ñÜ ËÐÜ¿áPæR ÓÜíŸí˜Ô¨Üíñæ ¯ÜÅÓÜá¤ñÜ ¸æÙÜÊÜ~Wæ¿á¬Üá®
£ÚÓÜáÊÜí£ÃܸæàPÜá.

 WÜ~ñÜ ¯ÜsܯÜâÓܤPÜ¨Ü ËÐÜ¿áÊÜâ ˨ݦìWÜÙÜÈÉ ÎÓÜ᤟¨ÜœÊݨÜ
ÊÜåèÆÂWÜÙÜ®Üá° ¸æÙæÓܸæàPÜá.

 ËÐÜ¿á ÊÜáívÜ®æ¿áá ËÐÜ¿á¨Ü ±ÝÅÁãàXPÜ ÊÜáñÜᤠE±ÜÁãàWÜ
ÊÜåèÆÂÊÜ®Üá° ±ÝÅÊÜááSÂWæãÚÓܸæàPÜá.

 ËÐÜ¿á¨Ü ÓÝÊÜÞiPÜ ÊÜáñÜᤠÊÜá®ÜÃÜíg®ÝñܾPÜ ÊÜåèÆÂWÜÙÜ ÊæáàÇæ¿áã ÓÜÖÜ
¸æÙÜPÜá aæÆɸæàPÜá.

 WÜ~ñÜ ¯ÜâÓܤPÜPæR ÓÜíŸí˜Ô¨Ü bÖæ®WÜÙÜá, ÓÜãPܤ ¯Ü¨ÜWÜÙÜá ÊÜáñÜá¤
ÓÝÊÜÞ¯ÂàPÜÃÜ|WÜÙÜá ÓܳÐÜrÊÝX¿áã ¯SÃÜÊÝX¿áã ÊÝÂTݯÓܸæàPÜá
ÖÝWÜã ¯ÃÜã²ÓܸæàPÜá.

D.El.Ed. – II

15 460-EK

 bñÜÅWÜÙÜá ÊÜáñÜᤠ¬ÜûæWÜÙÜá ­SÃÜÊÝX¨Ü᪠A¬Üá¯Ý£à¿áÊÝX¨Ü᪠ÍÜá¨Üœ
ÊÜáñÜᤠJ±Ü³ÊÝX¨Ü᪠PÜÈ¿ááÊÜÊÜÃÜ »è£PÜ ÊÜáñÜᤠÓÝÊÜÞiPÜ ±ÜÄÓÜÃÜPæR
Öæãí¨Ü¸æàPÜá.
 ˨ݦìWÜÚWæ ÓÜÊÜáÓæ ±ÜÄÖÝÃÜ PèÍÜÆWÜÙÜ®Üá° ¸æÙæÓÜÆá ÓÝPÜÐÜár
ÓÜíTæ¿áÈÉ ÓÜÊÜáÓæÂWÜÙÜ®Üá° ¹wÔÃܸæàPÜá.
 ±ÜsܱÜâÓܤPܨÜÈÉÃÜáÊÜ »ÝÐæ¿áá ÓÜÃÜÙÜÊÝX¨Üáª, PÜÈ¿ááÊÜÊÜÃÜá
ÓÜáÆ»ÜÊÝX ÓÜí»ÝÑÓÜáÊÜí£ÃܸæàPÜá.
 GÇÝÉ ÖÜíñÜ¨Ü Ë¨Ý¦ìWÜÙÜ AÊÜÍÜÂPÜñæWÜÚWÜ®ÜáWÜá|ÊÝX ±ÜÅ£Áãí¨Üá
ZoPÜÊÜä ÓÝPÜÐÜár ÍæÅà~àPÜêñÜ A»ÝÂÓÜWÜÙÜ®Üá° Öæãí©ÃܸæàPÜá.
¿ÞÊÜâ¨Ý¨ÜÃÜã ÖÜñÜᤠÓÜãPܤ AíÍÜWÜÙÜá ½ × 10 = 5
A¥ÜÊÝ
b) WÜ~ñÜ ±ÜÅÁãàWÝÆ¿á¨Ü PÜÈPæãà±ÜPÜÃÜ|WÜÙÜá
 PÜaÝcÊÜÓÜá¤WÜÙÜá
 CñÜÃæ Ô§ÃÜÊÜÓÜá¤WÜÙÜá
 bñÜůÜoWÜÙÜá ÊÜáñÜᤠaÝp…ìWÜÙÜá
 PܱÜâ³ÖÜÆWæ / ËË«Ü ¸æãàv…ìWÜÙÜá
 ÊÜÞ¨ÜÄWÜÙÜá
 ñÜãPÜ ÊÜáñÜᤠAÙÜñæ ÊÜÞvÜáÊÜ ÓÝ«Ü®ÜWÜÙÜá
 WÜ~ñÜ bñÜÅPÜÇæ ÓÝ«Ü®ÜWÜÙÜá
 ÇæPÜRÓÜáÊÜ E±ÜPÜÃÜ|WÜÙÜá
¿ÞÊÜâ¨Ý¨ÜÃÜã 5 ÓÜãPܤ PÜÈPæãà±ÜPÜÃÜ|WÜÙÜ®Üá° E¨ÝÖÜÃÜOæ ÓÜ×ñÜ ËÊÜÄÓÜáÊÜâ¨Üá 1×5=5

26. a) ÃÜã±ÜOÝñܾPÜ ÊÜåèÆÂÊÜޱܮÜPæR E¨ÝÃÜOæWÜÙÜá
 ±ÜÅÍæ°WÜÙÜ®Üá° Ë®ÝÂÓÜWæãÚÓÜáÊÜâ¨Üá, WÜ~ñÜ¨Ü ¯Ü¨ÜŸí«ÜWÜÙܬÜá®
¹wÓÜáÊÜâ¨Üá CñÝ©.
 WÜáí±Üâ aÜoáÊÜqPæWÜÙÜ®Üá° ÁãàiÓÜáÊÜâ¨Üá

D.El.Ed. – II [ Turn over

460-EK 16

 WÜ~ñÜ ÖæàÚPæWÜÙÜ®Üá° aÜbìÓÜáÊÜâ¨Üá
 ±ÜÅÍÝ° ñÜíñÜÅ
 KWL ¬Üûæ ŸÙÜÓÜáÊÜâ¨Üá

 ÓÜÊÜáÓÝ ±ÜÄÖÝÃÜ ñÜíñÜÅ
¿ÞÊÜâ¨Ý¨ÜÃÜã 2 ÓÜãPܤ ÊÜåèÆÂÊÜÞ±Ü®Ü ñÜíñÜÅÊÜ®Üá° ËÊÜÄÓÜáÊÜâ¨Üá ÊÜáñÜá¤
ÓÜÊÜá¦ìÔPæãÙÜáÛÊÜâ¨Üá 2+2+1=5

A¥ÜÊÝ
b) ñܱÜâ³WÜÅ×PæWÜÙÜ®Üá° ¯ÊÝÄÓÜáÊÜ ÊÜÞWÜìWÜÙÜá
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ÓÜãPܤ ÊÜÞWÜìÊÜ®Üá° ËÊÜÄÓÜáÊÜâ¨Üá 1×4=4

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D.El.Ed. – II

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( English Version )

PART – I

Multiple choice questions. All the questions are compulsory : 1  10 = 10
1. (B) Riemannian Geometry
2. (D) I and IV
3. (A) equal
4. (A) negative
5. (C) 32  42
6. (D) –4
7. (A) Bruce Joyce and Marsha Weil
8. (B) To construct the mathematical concepts through activities and
concrete experience
9. (D) Performance Test
10. (D) All of these
PART – II
Answer any five of the following questions, each answer not exceeding
half a page. 2  5 = 10

11.  Conjecturing : A conjecture is a reasoning about
mathematical relationships and making a statement that is
thought to be true, but is not known to be true. It is usually
made on the basis of incomplete information.
Example : When two even numbers are added, the sum is
also an even number :
2+4=6
10 + 12 = 22
8 + 6 = 14
10 + 20 = 30
The conjecture for the above example "When we add two even
numbers, the sum is also an even number". 2
12. A) x + 10
B) x–5 1+1=2

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13. Advantages of co-operative learning technique :
a) It helps the students to work together to accomplish shared
goals.
b) It helps to use variety of learning activities to improve
students' understanding of a subject.
c) It provides students to work in mixed ability group and
promote positive interaction in the group
d) It helps active involvement of students in the learning
process.
( Any four related points can be considered ) ½4= 2

14. Rule for divisibility by 4 : "If the number is having more than
2 digits and the number formed by the digits from ten's and unit
place is divisible by 4, then that number is divisible by 4".
Example : Let the number be
Th H T U
4 9 2 8
28  4 = 7
 4928 is divisible by 4.
4928  4 = 1232 1+1= 2

15. "Workbooks in mathematics are called self-learning materials" :
i) Workbook is supplementary to the textbook that provides
required clues and hints
ii) Workbooks are provided with self-directing steps and
sufficient space for solving the problems. 12= 2
16. Online learning resources in Mathematics :
i) It helps the teachers to find many websites useful in
mathematics classroom
ii) It helps to share educational content and ideas worldwide
iii) It is easy to introduce project based learning using ICT
facilities
iv) Teachers can use different types of mathematics online
learning games and activities. ½4=2
( Any 4 suitable points )

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17. Points to analyze test scores in mathematics :
i) The nature of the errors — computational or conceptual
ii) The topic or sub-topic that had the most errors
iii) The type of problems that the student found difficult
(Multiple choice, computational, word problems, problem
solving etc.)
iv) The presentation of work — neatness and clarity. ½4= 2

18. The process of marking scheme according to NCF-2005 :
i) Consider the process or steps that the child used to solve the
problem.
ii) Consider the elaboration and justification of each step of the
solution.
iii) Multiple ways of representing and doing the problems.
iv) Careless mistakes (computational errors) should not be
penalized heavily. ½4
PART – III
There is an alternate question for each of the following. Answer either
(a) or (b) of them, each answer not exceeding one page. 5  8 = 40
19. a) Characteristics of structure of mathematics according to Jean
Piaget :
i) A structure is a totality
ii) Laws of transformation
iii) A structure is self-regulating.
( Explanation of each characteristic ) 13= 3
Example :
System of whole numbers.
i) W = { 0, 1, 2, 3, 4, 5 ........ }
 { 0, 1, 2, 3, 4........ }  W
 1, 2, 3, 4, 5, 6, 7  W , then
  If we add 2 numbers 2 + 3 = 5  W
7 + 5 = 12  W
 If we add two whole numbers we get another whole number.
( Any related example for structure can be considered ) 2
OR

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b) Ways to create a mathematics classroom environment that

encourages mathematical reasoning.

i) The teacher must encourage children to ask plenty of

questions.

ii) A significant amount of time must be spent discussing

and talking mathematics in class.

iii) Teachers should not dismiss wrong answers, but

encourage the children to justify the answers.

iv) Connect the language students speak with the symbols

and language of mathematics and make connections

with the real world and mathematics.

v) Make use of appropriate puzzles for students to solve

the problems.

vi) Encourage children to create their own word problems

vii) Teachers must frame rich questions and help students

to make conjectures and generalizations.

( Explain any five suitable points ) 15=5

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20. a) 'Sum of three angles of a triangle is equal to two right angles.

Data : ABC is a triangle

Prove :

BAC + ABC + ACB = 180°

Construction : Draw a line DE parallel to side AC

passing through the vertex B.

Statement Reason

Triangle ABC with angles A, B, C Given

DBA = BAC Lines DE and AC are parallel

and angles formed by 2

parallel lines and a transversal

— alternate angles are equal

EBC = BCA Alternate angles are equal

DBA + ABC + EBC = 180° Supplementary angles

 BAC + ABC + ACB = 180° By substitution

 Sum of 3 angles of a triangle is A right angle = 90°
equal to two right angles

OR 5

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b) Validate the identity :
( a  b )2  a 2  b 2  2ab

Area of ABCD = Area of AEFG + Area of EBHF + Area of FHCI +
Area of DGFI
Therefore ( a + b ) ( a + b ) = ( a  a ) + ( a  b ) + ( b  b ) +

(ab)
( a  b )  a  ab  b 2  ab
2 2

( a  b )2  a 2  b 2  2ab 5
21. a) Fractions between 0 and 1 :

( Representation of fractions on any five number lines) 15
OR

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b) Perimeter of a rectangle :

Perimeter is the distance covered along the boundary forming
a closed figure when you go round the figure once.

 Perimeter of a rectangle ABCD = AB + BC + CD + DA

= 7 cm + 3 cm + 7 cm + 3 cm = 20 cm

Perimeter of a rectangle = l + b + l + b

 In a rectangle, opposite sides are equal

 Perimeter of a rectangle = a + b + a + b = 2a + 2b = 2 ( a + b )

= 2 ( length + breadth)

 Perimeter of a rectangle = 2 ( a + b ) = 2 ( 7 + 3 ) = 2 ( 10 )

= 20 cm

Area of a rectangle : =

Amount of surface enclosed by a closed figure is called its area

Area of the rectangle = 6 sq.cm which can be written as

3  2 sq.cm = length  breadth

Area of the rectangle = length  breadth

= 3 cm  2 cm = 6 sq.cm 2½ + 2½ = 5

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22. a) i) Paper folding activity
ii) Pictorial/Diagrammatical representation
Figures with 2 lines of symmetry

Rectangle has 2 lines of symmetry
Figures with more than 2 lines of symmetry

Square has 4 lines of symmetry
( Any suitable activity can be considered ) 2½ + 2½ = 5
OR
b) Ratio :
Comparison of two quantities in terms of 'how many times' is
called as ratio. It is denoted by the symbol ' : '
Two quantities can be compared only if they are in same unit.
Activities :
Finding ratio by using real life situation :
i) Out of 30 students in a class, number of students liking
football to number of students liking cricket.
ii) Out of 35 students in a class, number of girls to the
number of boys.
( Any two suitable activities to be considered ) 1+2+2=5

23. a) Importance of task analysis in mathematics :

i) It helps the teacher to plan the lesson in order

ii) Helps the teacher to provide better instruction and
facilitate better learning

iii) Helps to recall previous knowledge of students

iv) Helps to achieve the instructional objective.

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Task Analysis :

Instructional objective :

"The pupil defines the highest common factor' (HCF)

Task analysis :

The pupil :

 selects any two numbers

 recalls multiplication tables of which divides those two
numbers

 recognizes the factors of the two numbers

 finds the common factors of two numbers

 identifies the highest common factor

 explains the meaning of highest common factor

 defines the highest common factor. 2+3=5

OR

b) Inductive Thinking Model :

Concept — Linear pair :

  The pair of angles is adjacent angles

 The sum of measurement of these angles is 180°

 Definition — A pair of angles which are adjacent to each

other and whose sum of their measurement is 180°

called linear pair

Stages :

 Identifying and listing the data

 Grouping the items on the basis of similar

characteristics

 Developing categories and labelling the groups.

( Explanation of each stage with the concept ) 5

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24. a) Concept : Diameter of a circle
Concept Attainment Model :
Phase I : Presentation of examples and identifying the
concept
Step 1 : Presentation of labelled examples as Yes/No
Step 2 : Comparison of yes or no examples
Step 3 : Students frame hypotheses and verify.
Step 4 : Students state the concept.

Phase II : Testing the concept Attainment :
Step 1 : Students identify the unlabelled examples
Step 2 : Teacher confirms the hypotheses, names the
concept and defines.
Step 3 : Students generate examples.
Phase III : Analysis of Thinking Strategies :
Step 1 : Teacher discusses the thinking strategies
Step 2 : Students discuss role of hypothesis and
attributes.
Step 3 : Students discuss types of hypotheses and
attributes.
( Explanation of each stage with the concept ) 5
OR
b) Steps of Jigsaw technique :
i) Determining the content and instructional objectives
Content : Types of triangle based on its sides —
Equilateral triangle, Isosceles triangle, Scalene triangle.
Objectives : To facilitate students to
 explain the meaning of each type of triangle
 define each type of triangle
 classify the given triangles into different types
ii) Planning of learning materials :
iii) Dividing the students into groups
iv) Breaking into expert groups
v) Experts return to their home group
vi) Creating the positive dependence and co-operation
among the groups
vii) Evaluating the performance of the group
viii) Evaluation of individual student's quantitative and
qualitative learning.
( Explanation of each stage with the concept ) 5

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25. a) Special qualities of textbook in mathematics :
 The content presented in the textbook should be accurate
and up to date.
 It should include the recent developments in mathematics
relating to the content
 The content of the textbook should enable the learners to
develop disciplinary values of the subject
 The content presented should highlight the practical and
utility value of the subject
 It should also cater to the social and recreational value of
the subject
 The terms, concepts and generalizations given in the
textbook should be clearly and accurately defined and
stated.
 The figures and the diagrams should be accurate,
proportionate, neat and clear with physical and social
environment of the learners.
 It should include adequate number of worked problems
for students to develop problem solving skills
 The language used in the textbook should be simple and
easily comprehended by the learners
 Each topic in the textbook should have sufficient graded
exercises to satisfy needs of all levels of learners.
( Any suitable ten points ) ½  10 = 5
OR
b) Learning materials for a mathematics laboratory :
 Raw materials
 Concrete materials
 Pictures and charts
 Boards
 Models
 Measuring and weighing instruments
 Drawing instruments
 Computing devices.
(Explanation of any five suitable materials with examples) 1  5 = 5
26. a) Examples for formative assessment :
 Designing questions, puzzles etc.
 Planning group activities
 Discussion on mathematical statements
 Questioning technique
 Using KWL chart
 Problem solving technique
( Explanation of any 2 suitable assessment techniques can
be considered ) 2+2+1=5
OR

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b) Ways to rectify misconceptions in mathematics :
 To make students justify or explain how students solved
the problem
 Identify or diagnose the exact nature of the
misconception
 Give counter examples that will challenge the
misconception
 Make students to solve tasks that create a conflict in
their mind
 Make students think in a way that they are able to
reconsider their misconception.
(Explanation for each suitable way 1  4 = 4 )
Justifying the remedial measure for any one
misconception. 1

D.El.Ed.-II