July Month sslc Vac Key answer 15-16 123
Transcription
July Month sslc Vac Key answer 15-16 123
Bharati Nagar 3rd cross, Near Dasankoppa Circle, Dharwad. Ph: 0836-2770268 M: 9343100184 MATHEMATICS KEY ANSWERS SSLC State 2015-16 Vacation July-15- Test – Ist Marks - 50 Time : 90minutes Answer the following questions. 1x10=10 Ans 1. c) Both (a) & (b) Ans 2. Ans 3. Ans 4. Ans 5. b) xy √ √ d) Two Squares c) √20 c) Dividend = (Divisor x Quotient) + remainder Ans 6. a) A Ans 13. Ans 14. Ans 15. > C b) -6 d) Every composite number cannot be expressed as a product of primes. b) √7 b) Product of its prime factors II. Fill in the blanks Ans 11. Ans 12. B O D Ans 7. Ans 8. Ans 9. Ans 10. > 1x5=5 The rationalizing factor of binomial surd is also called Conjugate of binomial surd 180 = 2x x 3y x 5. The value of x and y is 2 The product of √ and √ is ab ab √ In two triangles ‘b’ if their corresponding sides are proportional they they are called Similar If ‘a’ and ‘b’ are any two positive intetgers if H is HCF of (a,b) and L is LCM of (a,b) Then the relation between them is expressed as a x b = H x L. III. Answer the following questions 2x9=9 A 16. Ans: 2cm 3.7cm D > E 3cm B > C In ∆ ABC DE II BC = = ∴ = . = . ∴ CE = 5.55 17. (7 x 11 x 13 + 13) is a composite number. Justify the statement. Ans: (7 x 11 x 13) + 13 = 13 [(7 x 11) + 1] = 13 [77 + 1] = 13 x 78 ∴ ! "#$"! % &'#%( 18. Express 120 as a product of prime factors 120 2 60 2 30 2 15 3 5 5 120 = 2 x 2 x 2 x 3 x 5 120 = 23 x 3 x 5 1 19. Classify the like surds. 2√16 √81, √128 √192 + + + + Ans: 2√16 = 2√8 2 = 2 x 2√2 = 4√2 + + + + √81 = √273 = 3 √3 + + + √128 = √64 2 = 4 √3 + + + √192 = √64 3 = 4 √3 + + + Like surds are 2√16 , √128 + + √81 , √192 + + 20. Simplify 3x √ + 3√ - 2√9 Ans : 3x √ + 3√ - 2√9 = 3x √x + 3√x x x = 2√9 x x x x =3x √ + 3 √ − 2 3 x √ = 3x √ + 3 √ − 6 √ = 6x √ - 6x √ =0 21. Ans: Find the sum of (√12 + √20), (3√3 + 2√5) (√45 − √90) (√12 + √20) + 53√3 + 2√56 + (√45 − √90) = 5√4 3 + √4 56 + 53√3 + 2√56 + (√9 5 − √9 10) = (2√3 + 2√5) + (3√3 + 2√5) + (3√5 − 3√10) = (2 + 3) √5 + (2 + 2 + 3) √5 - 3√10 = 5√3 + 7√5 − 3√10 22. Rationalise the denominator and simplify Ans : 8 9 : ; can be written as = √5 R.F is √5 8 √: √; x √; √; 8 √8<: = = ; √8<: 8< 8 √: √; 8 9 : ; 23. Ans. In ∆ ABC, PQ II BC ∴ = = A => 8 ? =@ 8 12x = 72 – 6x 18x = 72 X = 4cm G.; ∴ = Q x B 24. Ans : In ∆ ABC, DE II BC AB AD = DE BC ;. x P x R (∴Given) (∵ Thales Theorem) A AD @ AE = 5.7cm ;. H @ AE = 3.6cm D G.; > 6cm > IV. Answer the following questions 25. Ans: 1515 35 5 3 7 105 = 3 x 5 x 7 1515 = 3x 5 x 101 HCF of 105 and 1515 is 3 x 5 = 15 a x b = HCF x LCM 105 x 1515 = 15 x LCM 105 x 1515 = LCM 15 10,605 = LCM = Orders are 3 & 5 LCM of 3 and 5 is 15 √4 x √2 = 4 I 8; 8 4 J J8; 8J x x2 8J ; (index form of the surds) 8; 8 2 J; J8; 505 5 105 = 3 x 5 x 7 1515 = 3 x 5 x 101 + C 3x3=9 105 3 = E 9.5cm B 26. Ans: C 101 Each index is multiplied & divided by the LCM) = 4 ;J 8; x2 J 8; = √4; √2 (rewrite the index form in surd form) KI = √1024 8 KI KI = √8192 KI ⋀ ⋀ = = 900 (∵ Given) QPR UTS 27. ⋀ PRQ = ⋀ UST (∵ Alternate Angles) ∴ by A – A Criteria PQR ∽ TUS. 28. The areas of similar triangles are proportional to the squares of the Corresponding Sides. A D B L C Data : ∆ ABC ∽ ∆ DEF AC = BD CE DT To prove : = E M F AE BT AUVW XY ∆ ACE AUVW XY ∆ BDT = CEZ DTZ Construction : Draw AL ⊥ BC & DM ⊥ EF Proof : Statement Compare ALB & DME. ∟ABL = ∟DEM ∴ ⇒ ∟ALB ALB ∽ A] B^ But CE DT AL Data Construction Equiangular AA - Criteria = ∟DME = 900 DME AC = BD = AC BD CE ∴ DM = DT Data Axiom 1 ∴ AUVW XY ∆ ACE AUVW XY ∆ BDT Now, 8J c CE c A] 8J c DT c B^ = WU (∆ ACE) WU (∆ BDT) Area of ∆ 1J2 bh CE c A] = DT c B^ CE A] = d e xd e DT B^ CE = = CEZ DTZ = WU (∆ BDT) √ = = √ √@? √ x √8g h √G Z B^ CEZ √g h √ √@ h √ √@ h √ Z 5√@6 ? 5√6 - √G c h c @? - = c √ h 8 - 8 5√ h 86 - = 7 5√2 + 16 - = - √; √g? √ x √i< h √8< Z Z 5√g 6 ? 5√6 √i c 8< h √8< g? i√8< h √8< 2 √10 @ √8< 8 5√ h 86? √8< CE DT QED DTZ = = = 2 f5 - √@? √ A] DT WU (∆ ACE) ∴ 29. CE x DT @ √g ? √ √g ? √ is proved Bharati Nagar 3rd cross, Near Dasankoppa Circle, Dharwad. Ph: 0836-2770268 M: 9343100184 UÀtÂvÀ GvÀÛgÀUÀ¼À ¼ÄÀ SSLC State 2015-16 Vacation July-15- Test – Ist Marks - 50 Time : 90minutes 1x10=10 1.F PɼÀV£ÀªÀÅUÀ¼À£ÀÄß GvÀÛj¹j 1. GvÀÛgÀ. c) (a) ªÀÄvÀÄÛ (b) 2. GvÀÛgÀ. b) xy √ √ 3. GvÀÛgÀ. d) JgÀqÀÄ ZËPÀUÀ¼ÀÄ 4. GvÀÛgÀ. c) √20 5. GvÀÛgÀ. c) ¨sÁdå = (¨sÁdPÀ x ¨sÁUÀ®§ã) + ±ÉõÀ 6. GvÀÛgÀ. a) A > B 0 D > C 7. GvÀÛgÀ. b) -6 8. GvÀÛgÀ. d) ¸ÀAPÀ®£ÀªÁV ªÀåPÀÛ¥Àr¸À§ºÀÄzÀÄ. 9. GvÀÛgÀ. b) √7 10. GvÀÛgÀ. b) CzÀgÀ C«¨sÁdå C¥ÀªÀvÀð£ÀUÀ¼À UÀÄt®§ÞªÁV ªÀåPÀÛ¥Àr¸À§ºÀÄzÀÄ. 2.. ©lÖ ¸ÀܼÀ¼À vÀÄA©j A©j 11. GvÀÛgÀ. JgÀqÀÄ ¢é¥ÀzÀ PÀgÀtÂUÀ¼À CPÀgÀtÂÃPÁgÀPÀªÀ£ÀÄß ¸ÀAAiÀÄÄVä JAzÀÄ PÀgÉAiÀÄĪÀgÀÄ. 12. GvÀÛgÀ. 2 13. GvÀÛgÀ. ab √ 14. GvÀÛgÀ. ¸ÀªÀÄgÀÆ¥À wæ¨sÀÄdUÀ¼ÀÄ 15. GvÀÛgÀ. a x b = H x L 1x5=5 2x9=18 3. F PɼÀV£ÀªÀÅUÀ¼À£ÀÄß GvÀÛj¹j A 16. GvÀÛgÀ. 2cm 3.7cm D > E 3cm B > C In ∆ ABC DE II BC = = . = ∴ = . ∴ CE = 5.55cm 17. GvÀÛgÀ. (7 x 11 x 13 + 13) EzÀÄ MAzÀÄ ¸ÀAAiÀÄÄPÀÛ ¸ÀASÉå. F ºÉýPÉAiÀÄ£ÀÄß ¸ÀªÀÄwð¹. (7 x 11 x 13) + 13 = 13 [(7 x 11) + 1] = 13 [77 + 1] = 13 x 78 ∴ EzÀÄ MAzÀÄ ¸ÀAAiÀÄÄPÀÛ ¸ÀASÉå 120 18. GvÀÛgÀ. 2 60 2 30 2 15 3 5 5 1 120 = 2 x 2 x 2 x 3 x 5 120 = 23 x 3 x 5 19. GvÀÛgÀ. 2 √16 = 2√8 2 = 2 x 2√2 = 4√2 + + + + √81 = √273 = 2 3 √3 = 6 √3 + + + √128 = √64 2 = 4 √2 + + + √192 = √64 3 = 4 √3 + 2 √16 + + + + ¸ÀªÀÄgÀÆ¥À PÀgÀtÂUÀ¼ÀÄ 2 √16 , √128 + + √81 , √192 + + 20. GvÀÛgÀ. 3x √ + 3√ - 2√9 = 3x √ + 3√ = 2√9 =3x √ + 3 √ − 2 3 √ = 3x √ + 3 √ − 6 √ = 6x √ - 6x √ =0 21. GvÀÛgÀ. (√12 + √20) + 53√3 + 2√56 + (√45 − √90) = 5√4 3 + √4 56 + 53√3 + 2√56 + (√9 5 − √9 10) = (2√3 + 2√5) + (3√3 + 2√5) + (3√5 − 3√10) = (2 + 3) √5 + (2 + 2 + 3) √5 - 3√10 = 5√3 + 7√3 − 3√10 22. GvÀÛgÀ. 8 √: √; 8 9 : ; √5 CPÀgÀtÂPÁgÀPÀ √5 8 √: √; x √; √; 8 √8<: = = ; 8<< √ 8< 23. GvÀÛgÀ. PɼÀV£À DPÀÈwAiÀÄ°è AB = 12cm ªÀÄvÀÄÛ BC = 6cm DzÀgÉ x PÀAqÀÄ»r¬Äj A x P x B ∆ ABC, PQ II BC Q x x R C ∴ = 8 ? => = 8 = @ 12x = 72 – 6x 18x = 72 X = 4cm 24. GvÀÛgÀ. In ∆ ABC, DE II BC (∴Given) AB BC ;. G.; ∴ = AD = AD (∵ Thales Theorem) DE A @ E= 5.7cm ;. H @ D G.; AE = 3.6cm > E 9.5cm 6cm B > C 3x3=9 4. F PɼÀV£ÀªÀÅUÀ¼À£ÀÄß GvÀÛj¹j. 105 25. GvÀÛgÀ. 5 1515 21 3 5 7 105 = 5 x 3 x 7 303 3 101 1515 = 5 x 3 x 501 105 = 3 x 5 x 7 1515 = 3 x 5 x 101 HCF of 105 and 1515 is 3 x 5 = 15, a x b = HCF x LCM 105 x 1515 = 15 x LCM 105 x 1515 = LCM 15 10,605 = LCM 26. GvÀÛgÀ. √4 ªÀÄvÀÄÛ √2 UÀ¼À PÀgÀtÂAiÀÄ PÀæªÀÄUÀ¼ÀÄ 3 ªÀÄvÀÄÛ 5 3 ªÀÄvÀÄÛ 5gÀ ®.¸Á.C 15 + = I √4 x √2 = 4 + I = 24 = 4 ;J 8; 8J x2 8J ; 8J 8;J 8; x2 J 8; x2 8J 8;J ; 8; = = = √4; √2 (rewrite the index form in surd form) KI √1024 8 KI KI √8192 KI ⋀ ⋀ ªÀÄvÀÄÛ UÀ¼À°è PQR UTS 27. GvÀÛgÀ. ⋀ QRQ = ⋀ = 900 (∴ zÀvÀÛ) UST ∴ A – A (PÉÆÃPÉÆà ¹zÁÞAvÀ) PQR TUS 4x2=8 5. F PɼÀV£ÀªÀÅUÀ¼À£ÀÄß GvÀÛj¹j 28. GvÀÛgÀ. zÀvÀÛ ∆ ABC ∽ ∆ DEF AC BD CE = DT = AE BT A ∆ ABC «¹ÛÃtð ¸ÁzsÀ¤ÃAiÀÄ : = ∆ DEF «¹ÛÃtð gÀZÀ£É : AL ⊥ BC ªÀÄvÀÄÛ DM ⊥ EF gÀa¹. ¸ÁzsÀ£É : ºÉýPÉUÀ¼ÀÄ ALB ªÀÄvÀÄÛ ⇒ A] B^ DzÀgÉ AL DME AC CE = DT BD B [∴ zÀvÀÛ]è = DME = 900 = EF DME UÀ¼À°è ALB ∴ ALB BC PÁgÀtUÀ¼ÀÄ ∟ABL = ∟DEM AC BD CE ∴ DM = DT [∴ gÀZÀ£É] [∴ PÉÆÃPÉÆà ¤zsÁðgÀPÀ UÀÄt] [∴ zÀvÀÛ]è [∴ UÀÄt®PÀët] ∆ ABC «¹ÛÃtð = 1J2 x BC x AL ∆ DEF «¹ÛÃtð = 1J2 x EF x DM D L C E M F 1J x BC x AL 2 ∴ ∆ ABC «¹ÛÃtð 1J x EF x DM 2 ∆ DEF «¹ÛÃtð FUÀ ∴ ∆ ABC «¹ÛÃtð BC x AL ∆ DEF «¹ÛÃtð CE EF x DM CE d A] B^ = = d e xd e DT = DT CEZ DTZ op DzÀgÉ, ∴ = qr CE DT = AE (∵ zÀvÀÛ) BT ∆ ABC «¹ÛÃtð = AB = BC = AC s ∆ DEF «¹ÛÃtð EF DF ¥ÀæªÉÄÃAiÀÄ ¸Á¢ü¹zÉ. √ 29. GvÀÛgÀ : = = √ √@? √ √@? √ x √8g h √G Z Z 5√@6 ? 5√6 - √G c h c @? - = c √ h 8 - 8 5√ h 86 - = 7 5√2 + 16 - = - √; √g? √ x √i< h √8< Z Z 5√g 6 ? 5√6 √i c 8< h √8< g? i√8< h √8< 2 √10 @ √8< 8 5√ h 86? √8< √g h √ √@ h √ √@ h √ = = 2 f5 - @ √g ? √ √g ? √ CE DT ¸Á¢ü¹zÉe Bharati Nagar 3rd cross, Near Dasankoppa Circle, Dharwad. Ph: 0836-2770268 M: 9343100184 «eÁÕ£À GvÀÛgÀUÀ¼ÀÄ SSLC State 2015-16 Vacation July-15- Test – Ist Marks - 50 Time : 90minutes 1. §ºÀÄ DAiÉÄÌAiÀÄ GvÀÛgÀUÀ¼ÀÄ1. 1. – (c) 2. – (d) 3. –(a) 4. –(b) 5. (d) 2. ºÉÆA¢¹ §gɬÄj 1.– (b) 2. – (a) 3. –(d) 4. –(c) 2. MAzÀÄ CAPÀzÀ GvÀÛgÀUÀ¼ÀÄ 1. ¸À¸Àå ºÁUÀÆ ¸À¹åÃvÀà£ÀßUÀ¼À eÉÊ«PÀ ªÀ¸ÀÄÛUÀ¼À ±ÀQÛAiÀÄ ªÀÄÆ®UÀ¼ÀÄ gÀÆ¥ÁAvÀgÀUÉƼÀÄî«PÉUÉ ±ÀQÛ J£ÀÄßvÉÛêÉ. 2. ¨sÀÆ«ÄAiÀÄ ªÀÄvÀÄÛ ¸ÁUÀgÀUÀ¼À ¸ÀgÁ¸Àj vÁ¥ÀªÀiÁ£ÀzÀ°è PÀ¼ÉzÀ 200 ªÀµÀðUÀ¼À°è GAmÁVgÀĪÀ KjPÉAiÀÄ£ÀÄß eÁUÀwPÀ vÁ¥ÀªÀiÁ£À KjPÉ JAzÀÄ PÀgÉAiÀÄÄvÁÛgÉ. 3. CtıÀQÛ ¸ÁܪÀgÀUÀ¼ÀÄ, CtÄ DzsÁjvÀ ±ÀÛçUÀ¼ÀÄ 4. ¥ÀgÀªÀiÁtÄ«£À PÉÃAzsÀæ ªÀÄvÀÄÛ CzÀgÀ CvÀåAvÀ ºÉÆgÀ PÀªÀZÀzÀ CAvÀgÀ 5. ±ÀQÛAiÀÄ ¸ÀtÚ ¸ÀtÚ ¥ÁåPÉlUÀ¼ÀÄ. 3. JgÀqÀÄ CAPÀzÀ GvÀÛUÀ¼ÀÄ 1. C) zsÁvÀÄUÀ¼À «ªÀgÀUÀ¼ÀÄ ¸ÀÄ®¨sÀªÁV MAzÉà PÀqÉ zÉÆgÉAiÀÄĪÀAvÉ ªÀiÁqÀÄvÀÛzÉ. D) E£ÀÆß D«µÁÌgÀUÉƼÀîzÀ zsÁvÀÄUÀ¼À ¥ÀgÀªÀiÁtÄgÁ² UÀÄtUÀ¼ÀÄ, ªÀÄwÛvÀgÉ «ªÀgÀUÀ¼À£ÀÄß G»¸À®Ä ¸ÁzsÀåªÁ¬ÄvÀÄ 2. ªÀiÁ£ÀªÀ£À £ÀgÀªÀÄAqÀ® ªÀåªÀ¸ÉÜAiÀÄ £ÉÃgÀªÁV ¥ÀgÀuÁªÀÄ ©ÃgÀÄvÀÛzÉ. QªÀÅqÀÄvÀ£À, vÀ¯É±ÀÆ¯É C¢üPÀ gÀPÀÛ MvÀÛqÀ. 3. ªÀļÉAiÀÄ PH ªÀiË®å 5.6 EzÀQÌAvÀ PÀrªÉÄ DzÀ°è CzÀPÉÌ DªÀÄè ªÀÄ¼É J£ÀÄߪÀgÀÄ. d®ªÁ¹ fëUÀ½UÉ ªÀiÁgÀPÀ. ¥ÁætÂUÀ¼À fë ¸ÀAzÀtÂAiÀÄ C£ÀÄ¥ÁvÀ KgÀÄ¥ÉÃgÀÄ, ¥ÉÆõÀPÁA±ÀUÀ¼À ªÀĵÀÖ, ZÀªÀÄðgÉÆÃUÀ, G¹gÁlzÀ vÉÆêÀÄzÀgÉ PÁgÀt. 4. C) CvÀåAvÀ ºÉaÑ£À vÁ¥ÀªÀiÁ£ÀzÀ°è ¤ªÁðvÀ CxÀªÁ PÀrªÉÄ UÁ½AiÀÄ G¥ÀAiÉÆÃUÀ¢AzÀ £ÀqɸÀ¯ÁUÀĪÀ gÁ¸ÁAiÀĤPÀ «WÀl£ÉAiÀÄ QæAiÉÄ – zÀºÀ£À ¥ÉÊgÉÆð¹¸ï D) fêÀ gÁ²AiÀÄÄ CªÁAiÀÄÄ«PÉ fÃtð QæAiÉÄAiÀÄ°è eÉÊ«PÀ C¤®ªÀ£ÀÄß GAlĪÀiÁqÀĪÀ QæAiÉÄ §AiÉÆÃUÁ幦üPÉñÀ£ï 5. PÀ½î¸À¸ÀåzÀ ©ÃdUÀ½AzÀ ¥ÀqÉAiÀÄĪÀ vÉÊ®ªÀ£ÀÄß røɯïUÉ ¥ÀAiÀiÁðAiÀĪÁV ¥ÀjªÀwð¸À¯ÁUÀÄwÛzÉ. PÀZÁÑ vÉÊ®ªÀ£ÀÄß ªÁtÂdå §¼ÀPÉAiÀÄ EAzsÀ£ÀªÁV ¥ÀjªÀwð¸ÀĪÀ ¥ÀæQæAiÉÄ. 6. C) «PÉÃA¢æPÀÈvÀ ªÀåªÀ¸ÉÜAiÀÄ°è F ±ÀQÛAiÀÄ£ÀÄß ¥ÀÆgÉʸÀ§ºÀÄzÀÄ. D) ±ÀÄzsÀÞ ¸ÀĹÜgÀ ¥Àj¸ÀgÀªÀ£ÀÄß C©üªÀÈ¢Þ¥Àr¹PÉƼÀÀÄzÀÄ 7. ªÉÄAqÀ°Ãªï£À DªÀvÀð PÉÆõÀÖPÀ ¤AiÀĪÀÄzÀ°è zsÁvÀÄUÀ¼À ®PÀëtUÀ¼ÀÄ CªÀÅUÀ¼À ¥ÀgÀªÀiÁtÄ gÁ²AiÉÆA¢UÉ DªÀvÀðUÉƼÀÄîvÀÛzÉ. “DzsÀĤPÀ DªÀvÀðPÉÆõÀÖPÀzÀ°è, zsÁvÀÄUÀ¼À ®PÀëtUÀ¼ÀÄ, CªÀÅUÀ¼À ¥ÀgÀªÀiÁtÄ ¸ÀASÉåAiÉÆA¢UÉ DªÀvÀðªÁUÀÄvÀÛzÉ. 8. C) ¨ÁèPï£À°ègÀĪÀ zsÁvÀÄUÀ¼À°è CqÀظÁ°£À ¸ÁªÀÄåQÌAvÀ PÀA§¸Á°£À ¸ÁªÀÄåQÌAvÀ ºÉZÁÑVgÀÄvÀÛzÉ. G¥ÀAiÉÆÃUÀzÀ zÀȶ֬ÄAzÀ ¥ÀævÉåÃPÀªÁV EqÀ¯ÁVzÉ. D) MAzÀÄ ªÀUÀðzÀ PɼÀUÉ §AzÀ ºÁUÉ ¥ÀgÀªÀiÁtÄ«UÉ ºÉZÀÄÑ PÀªÀZÀUÀ¼ÀÄ ¸ÉÃjPÉƼÀÄîvÀÛªÉ DzÀgÉ DªÀvÀðzÀ°è PÀªÀZÀUÀ¼À ¸ÀASÉå ºÉZÁÑUÀĪÀÅ¢®è. CzÉà PÀªÀZÀPÉÌ E¯ÉPÁÖç£ïUÀ¼ÀÄ ¸ÉÃjPÉƼÀÄîvÀÛªÉ & ¥ÀgÀªÀiÁtÄ«£À £ÀÆåQèAiÀÄgï E¯ÉPÁÖç£ÀUÀ¼À ªÉÄÃ¯É M¼ÀªÀÄÄR ¸É¼ÉvÀªÀ£ÀÄßAlÄ ªÀiÁqÀÄvÀÛzÉ. Bharati Nagar 3rd cross, Near Dasankoppa Circle, Dharwad. Ph: 0836-2770268 M: 9343100184 ENGLISH – Ist Key Answer SSLC State 2015-16 Vacation July-15- Test – Ist I. Answer the following in a word or sentence. Marks - 50 Time : 90minutes 3x1=3 1. The writer. 2. They are happy and proud to be so busy. 3. She was interested only in her work. II. Answer the following 5 x 2 = 10 1. The torrent in workers’ paradise is silent as it didn’t like to waste its energy singing while filling the girl’s pitcher. 2. Just as princess feels sorry for a poor beggar, the busy girl of the torrent felt sorry for a lovely man without work. 3. In workers’ paradise, work had begun to suffer. Many active people were now idle and were wasting their time on useless things such as painting and sculpture. This made the elders anxious. 4. The cranes pair for life. Having seen its mate shot dead by the hunters and being taken away, the female saurus is heartbroken with the beak she kissed the blood stained feathers of her mate which the wind had not carried away and sat down to hatch them in hope she could bring him back to life. This shows the agony and the despair of the bird. 5. The callousness of the hunters is brought out in the way the poet describes the killing of a male saurus bird. The hunters were not touched by the beauty of the bird. They picked up the dead bird by hands and jaws and flung it into a coarse bag as if it were a piece a dirty lines and the bag a washing bag. III. Using the following phrasal verbs, make your own sentences. 1. Run down -- The teacher’s health had run down as a result of overwork. 5 x 2 = 10 2. Run into -- The Government’s plan to introduce Ian Lokpal bill has run into strong apposition from anther parties. 3. Run out -- The children began to run out of the class room when a snake entered there. 4. Go about -- the driver used to go about his work looking unwell. 5. Go through -- she is going through a difficult time. IV. Using the following idioms and phrases, make your own sentences. 5 x 2 = 10 1. Take charge of - The older elephants took charge of a colt that had list its mother. 2. Shake off - We should shake off our bad habits to be successful in life. 3. Laught at – The girls laughed at the funny way the man was walking up the stairs. 4. Take to task - The teacher refused to take the boys to task for having her class. 5. Come to pass – People refused to believe that what the old man had said would come to pass. V. Use suitable articles [a, an, the] 5 x 2 = 10 1. There is a fly in the coffee. 2. The book you wanted is not available in the library? 3. The cow is an useful animal. 4. The Cauvery merges into Bay of Bengal. 5. I like the stories of the spics. VI. Give antonyms of the following. Words X Antonyms 1. Defense X Offense 2. Particular X General 3. Complex X Simple 4. Abstract X Concrete 5. Confidents X Diffident 6. Confuse X Enlighten 7. Neglect X Cherish 7x1=7 Bharati Nagar 3rd cross, Near Dasankoppa Circle, Dharwad. Ph: 0836-2770268 M: 9343100184 ENGLISH – IInd Key Answer SSLC State 2015-16 Vacation July-15- Test – Ist I. Answer the following in a word or sentence. 1. – c II. Marks - 50 Time : 90minutes 2. - a 3. - b 4. - b 5x1=5 5. - b Answer the following 5 x 2 = 10 6. It was disgraceful to swami’s father that his son used to sleep by the side of his granny though he was a student of second standard. Father also didn’t like the way that swami was brought up by his granny as well as his mother. 7. Swami concluded that his father’s proposition was frightful because it was great fearful for him to sleep alone. But his father forced him to sleep alone in his office without a light. 8. Swami was honored by his classmates as they looked at him with respect. The teachers patted his back with pleasure. The head master said that he was a true scout. Thus the congratulations were showered on him. 9. Swami felt relieved at the end because his mother supported him to sleep by the side of granny. She didn’t want that her son should risk his life. Father also kept quiet for the proposal to sleep by the side of granny. 10. Though there was absolute silence in the room, swami heard the ticking of a clock. Some insects were humming. The rustle of trees and snoring sounds were hoard by Swami. III. R. C 5 x 2 = 10 11. A) Father of Swami said this b) ‘You’ is referred to swami’s mother It was indirectly spoken to granny. Father thought that swami was spoiled by granny. 12. A) Swami asked like this. b) The answer was that the door might be kept open. But he shouldn’t roll up his bed and lie beside granny. POETRY SECTION IV. Answer the following in a word or sentence. 13. – c 14. – b 5 x 2 = 10 15 - a 16. One day granny had terrible fall while they were in the town. She climbed a tree and she couldn’t come down. 17. The doctor had recommended that granny should be in a bed for a week. She could not do any other activities. 18. Granny asked her son that he had to build to fulfill her demand. My father built a house on the top of a tree with a window and a door with my assistance. 19. a) Granny was sixty b) It reflects that granny being old was interested to climb a tree. 20. a) ‘Me’ is referred to speaker or a poet. b) Granny sits in a house who is on a tree top. She enjoys her life drinking sherry with her grandson. 21. Simile. 22. a) tragedy b) Fragment c) Sympathetic Bharati Nagar 3rd cross, Near Dasankoppa Circle, Dharwad. Ph: 0836-2770268 M+: 9343100184 SCIENCE KEY ANSWER SSLC State 2015-16 Vacation July-15- Test – Ist Marks - 50 Time : 90minutes PHYSICS I. M.C.Q Answers 1. a) T = 2&9 t u 3x1=3 2.b. Nature of medium II. Fill in the blanks 3. c. Fuel injunction pump 3x1=3 1. Transverse & Non-mechanical 2. Carburetor 3. Heat Engine III. Match the following 1. Steam Engine 2. Petrol Engine 3. Diesel Engine 3x1=3 i) External combustion engine ii) Spark plug iii) Injection pump IV. Answer the following 1. The disturbance produced in the medium is ‘wave’ 2. V = nv w v wvelocity of wave n w frequency of wave vw wavelength 2x1=2 V. Answer the following Transverse wave 1. The vibration of particle perpendicular to the direction of propagation 2. Propagate in the form of crest & triangles 3. These can be polarized 2x2=4 Longitudinal wave i) The vibration of the partite along the direction of propagation ii) Propagate in the form of compression & Rarefaction iii) These cannot be polarized VI. 1. Any two differences Internal Combustion Engine 1. It is smaller in size 2. It has higher Efficiency VII. Answer the following 1. Steam Engine External Combustion Engine 1. It is larger in size 2. It has lesser efficiency 3x1=3 CHEMISTRY I. Multiple choice question Answers 1. 2. 3. 4. 1x4=4 a. Energy source are very less d. 1300oC a. Fluorine c. Atomic Number II. Match the following Answer 1. Solar cell 2. Group 17 3. Group 2 1x3=3 a. Photovoltaic effect b. Halogens c. Alkaline Earth Metals III. Answers 1x3=3 1. Transesterification 2. “The properties of the elements are the periodic functions of their atomic mass” 3. Atomic size increases down the group. IV. 1. Ans: Elements in which electrons in an atom occupy ultimate shell leaving penultimate shells partially. Vacant are called transitional elements. 2. Two advantages i) Reduce CO2 content in atmosphere ii) Pollution free iii) Increase fertility of soil iv) Reduce water retention capacity of soil V. Answer 1x3=3 1. Abundant 2. Renewable 3. Pollution free & ecofriendly Supplied to urban rural & remote areas BIOLOGY I. Multiple choice questions 1. b) Hippocampus 2. a) On the right side 3. c) Green house gases II. Answer the following 1. Ans: Sound which causes unpleasant effects and discomfort to human ears is called noise. 2. Ans: Rise in average temperature of the earth’s atmosphere and oceans are called global warming. III. Answer the following 1. Gymnosperms consists of two of cones • Male cones contain numerous microsporophyll’s, which produce microspores that have male gametes. • Female cones contain megasporophylls, which produce ovule containing female gametes 2. Unique features of phylum chordates are • Presence of a solid supporting structure on the dorsal side of the body called notochord. • Presence of a dorsal, hollow, tubular nerve cord. • Presence of openings in the pharynx called gill slits, atleast in embryonic stage. IV. Answer the following 1. Flight adaptations of birds are i) Streamlined body ii) Forelimbs modified into wings iii) Special arrangement of features on the wings to provide the lift. iv) Presence of flight muscles. v) Long bones are pneumatic, filled with air vi) Many bones in the body are fused vii) Absence of teeth, replaced by a beak.