56A$Math$ ! 1.!Consider!each!condition!and!try!to!deduce!the!final!answer! First!condition!is!All!drivers!who!live!in!Ohio!have!car!insurance.! Second!condition!is!that!Antonia!does!not!live!in!Ohio!which!might!mean!she!doesn’t!hold!a!car! insurance,!but!we!don’t!for!certain!about!that.!What!if!she!holds!a!car!insurance!living!in,!say,! Texas.! Thus,!option!B!could!be!true,!but!not!“must!be!true”! Third!Condition!is!that!Catherine!has!car!insurance,!which!might!mean!she!lives!in!Ohio!or!it!also! might!mean!she!lives!in!some!other!state!where!she!holds!car!insurance.!Eliminate!C.! Fourth!condition!is!that!Tai!has!car!insurance.!Just!like!Catherine,!he!could!be!in!Ohio!or!some!other! state.!Eliminate!D.! Fifth!condition:!Jorge!lives!in!Maine!which!could!mean!he!could!or!coul!not!hold!a!driver’s!license.! Thus,!eliminate!E.! We’re!left!with!option!A.! Ans:!(A)! !! 2.! =! !! !! !! !!! (!!)! !! !!!! !!! =! !! ! ! =!P4! Ans:!(F)! ! 3.!TakePhome!pay!=!320!–!(18%!of!320)! =!320(0.82)! =!$262.4! Ans:!(B)! ! 4.!∠XYZ = ∠XZY = 50°!! (XY!=!XZ)! Sum!of!all!measures!of!triangle!is!180°! ∠XYZ + ∠XZY + ∠YXZ = 180°!! 50°!+!50°!+!∠YXZ!=!180°! ∠YXZ = 80°!! Ans:!(J)! ! 5.!The!construction!paper!is!9!inches!by!12!inches,!thus!3!circles!of!diameter!3!inches!can!be!cut! along!the!width!of!9!inches!and!4!circles!along!the!length!of!12!inches.! Number!of!circles!can!be!cut!out!of!a!construction!paper!=!4!×!3! =!12! 4!construction!papers!will!give!48!circles! Thus,!in!order!to!have!50!circles,!the!minimum!number!of!construction!paper!needed!=!5!! Ans:!(E)! ! 6.!The!numbers!increase!in!value!from!left!to!right!meaning!A!is!less!than!B.! Ans:!(H)! ! 7.!2(x!–!5)!=!P11! 2x!–!10!=!P11! 2x!=!P1! x!=!P1/2! Ans:!(E)! ! 8.!x2!–!x!–!20!! =!x2!P!5x!+!4x!–!20! =!x(x!–!5)!+!4(x!–!5)! =!(x!–!5)(x!+!4)! Ans:!(F)! ! 9.!Line!parallel!to!xPaxis!is!y=k!(for!some!constant!k)!and!above!7!units!is!y!=!7! Ans:!(A)! ! 10.! =! !" + !" ! ! !"!!"# ! ! !" Ans:!(G)! ! 11.!Original!wingspan!=!41!feet! Wingspan!of!1/20th!scale!model!of!airplane!=!1/20!×!41! =!2!!" ! Ans:!(C)! !! 12.!Let!the!total!slice!of!pizza!be!‘x’! ! x!=!4!+!3!+!2!+!(1!+!!x)! ! x!=!10!+!!x! ! ! x!=!10! x!=!20! Ans:!(K)! ! 13.!Cost!to!ship!the!furniture!after!hike!in!shipping!cost!=!80!+!(60%!of!80)! =!80(1.6)! =!$128! Ans:!(D)! ! 14.!! ! Area!of!figure!=!Area!of!rectangle!ABCD!+!Area!of!rectangle!DHFE! =!(18!×!10)!+!(8!×!5)! =!180!+!40! =!220!square!feet.! Area!of!100!square!feet!requires!1Pquart;!thus,!220!square!feet!area!would!require!3!quarts! Ans:!(H)! ! 15.!x!=!P4! 3x2!–!15x!=!3(P4)2!–!15(P4)! =!3(16)!+!60! =!48!+!60! =!108! Ans:!(E)! ! 16.!! In!triangle!AMD,! ∠AMD + ∠MAD + ∠ADM = 180°!! 90°!+!35°!+!∠ADM!=!180°! ∠ADM = 55°!! Therefore,!∠ADC = 55° + 90° = 145°! Sum!of!all!measures!of!angles!in!a!trapezoid!is!360°! ∠A + ∠B + ∠C + ∠D = 360°!! 35°!+!80°!+!145°!+!∠C=360°! ! ∠C = 100°!! Ans:!(G)! ! 17.!(6a3!–!5ac2!+!14c)!–!(8c!–!3a3!–!2ac2)! =!9a3!–!3ac2!+!6c! Ans:!(D)! ! 18.!f(x)!=!3x2!+!5x!–!5! f(P2)!=!3(P2)2!+!5(P2)!–!5! f(P2)!=!12!–!10!–!5! f(P2)!=!P3! Ans:!(J)! ! 19.!Let!the!price!of!one!tire!be!‘x’! 4!tires!are!sold!at!price!=!3x!! If!all!tires!were!sold!at!original!price,!then!cost!of!4!tires!would!be!4x! !"!!" Discount!off!regular!price!=! ! =!! ×100! !"! ×100! =!25%! Ans:!(C)! ! 20.!If!you!toss!a!coin,!it!will!come!up!a!head!or!tail!irrespective!of!number!of!times!coin!is!tossed.! Thus,!probability!of!landing!head!is!50%!i.e.!½! Ans:!(J)! ! 21.!The!code!consist!of!2!alphabets!followed!by!2!digits!which!can!be!repeated! _!×!_!×!_!×!_! The!consists!of!2!alphabets!of!which!the!first!can’t!be!O!and!next!can!be!any!alphabet!from!A!to!Z!! 25!×!26!×!_!×!_! The!last!characrters!consist!of!2!digits!out!of!10!which!can!be!repeated! 25!×!26!×!10!×!10! Ans:!(C)! ! 22.!Length!=!Width!+!6! Perimeter!=!2(length!+!width)! 48!=!2(width!+!6!+!width)! 24!=!2(width)!+!6! 18!=!2(width)! Width!=!9! Ans:!(J)! ! 23.!Slope!=! !! !!! !! !!! ! Grid!lines!are!spaced!every!½!unit.! Coordinate!of!P!=!(0,!2)! Coordinate!of!Q!=!(3/2,!3)! !!! Slope!=! ! !! ! ! ! Slope!=! ! ! Ans:!(A)! ! 24.!∠FZE + ∠CZE = 180°!! 52°!+!∠CZE = 180°! ∠CZE = 128°!! Ans:!(K)! ! 25.!Area!of!parallelogram!=!base!×!height! Base!of!parallelogram!=!55!! (11!squares!of!5!mm!side)! Height!of!parallelogram!=!20! (4!squares!of!5!mm!side)! Area!=!20!×!55! =!1100!! Ans:!(C)! ! 26.!Put!the!value!of!‘t’!from!the!answer!options!in!the!equation!and!check!which!gives!d!=!0.! Option!F:!t!=!1! d!=!P5t2!+!18t!+!8! d!=!P5(1)!+!18!+!8!=!21! Option!G:!t!=!2! d!=!P5(2)2!+!18(2)!+!8! d!=!P20!+!36!+!8!! d!=!24! Option!H:!d!=!P5(4)2!+!18(4)!+!8! d!=!P80!+!72!+!8!=!0! We!don’t!need!to!check!the!other!two!options.! Ans:!(H)! ! 27.!Area!of!triangle!=!½!×!base!×!height! Base,!AB!=!3! Height,!AC!=!4! Area!=!½!×!3!×!4! =!6!! Ans:!(B)! ! 28.!–(P3a3)2!! =!P(9a6)! Ans:!(F)! ! 29.!120!=!5/12!×!x! x!=!24!×!12! x!=!288! Ans:!(E)! ! !!! ! 30.!!! !! ÷ !!!! !!! =!!! !! ×(b − 3)! !! !! =!!! !!! =!1! Ans:!(F)! ! 31.!Option!A:!P1/2!=!P0.5! Option!B:!P3/4!=!P0.75! Option!C:!P7/8!=!P0.875! Option!D:!P9/16!=!P0.5625! Option!E:!P5/32!=!P0.15625! Option!E!is!the!greatest!among!all! Ans:!(E)! ! ! 32.!! x > −1! =!4x!>!P3! =!x!>! !! ! ! Ans:!(F)! ! 33.!Area!of!rectangular!8Pfoot!by!10Pfoot!dance!floor!=!8!×!10!! =!80!square!feet! Area!is!doubled!by!increasing!width!by!2!feet!and!consider!increasing!length!by!‘x’!feet.! Area!=!(8!+!2)!×!(10!+!x)! 160!=!10!×!(10!+!x)! 16!=!10!+!x! x!=!6! Ans:!(C)! ! !""#$%&' 34.!tan!θ!=!!"#$%&'( = ! !! ! In!right!angled!triangle!ABC,! AB2!+!BC2!=!AC2! 22!+!( 77)2!=!AC2! 4!+!77!=!AC2! AC2!=!81! AC!=!9! !""#$%&' ! !" sin!θ = !"#$%&'()& = !"! ! sin!θ = !! Ans:!(F)! ! 35.!x!+!2y!=!6! Rearrange!the!above!equation!in!slope!intercept!form,! 2y!=!Px!+!6! ! y!=!P! x + 3! Slope!=!P1/2!! A!line!parallel!to!this!line!has!the!same!slope.! Ans:!(B)! ! 36.!A!=!P(1!+!rt)! ! ! ! = 1 + rt!! − 1 = rt!! ! !!! ! r= = rt!! !!! !" !! Ans:!(G)! ! 37.!Charles!contributed!half!the!amount!Tanya!contributed.! If!Tanya!contributed!‘x’,!then!Charles!contributed!‘x/2’! x!+!x/2!=!24! 3x/2!=!24! x!=!16! Ans:!(D)! ! 38.!|x!–!1|!>!5! x!–!1!>!5! or!! Px!+!1!>!5! x!>!6!! ! or!! Px!>!4! x!>!6!! ! or!! x!<!P4! Ans:!(F)! ! 39.!4!<! !!<!9! Taking!square!on!both!sides!of!the!inequality! 16!<!x!<!81! Ans:!(E)! ! 40.!sin!A!=! ! ! = !" !" !" !""#$%&' !"#$%&'()& !" = !"! !! AB!=! ! ! Ans:!(J)! ! 41.!The!parabola!cuts!xPaxis!at!P2!and!3;!thus,!xPintercept!is!P2!and!3! y!=!(x!+!2)(x!–!3)!has!xPintercepts!P2!and!3! Ans:!(A)! ! !"#!!"!!""!!"#$%!& 42.!Average!=!!"#$%&!!"!!"#$%!&! 4!=! !"#!!"!!""!!"#$%!& ! ! Sum!=!20! 3!+!Sum!of!4!entries!=!20! Sum!of!4!entries!=!17! Average!of!integers!=!17/4! =!4!! ! Ans:!(J)!! !! 43.!x2!–!kx!–!8!=!0! Since!k!is!a!solution,! k2!–!k(Pk)!–!8!=!0!! (x!=!Pk)! 2 2 k !+!k !–!8!=!0! 2k2!=!8! k2!=!4! k!=!±2! Ans:!(B)! ! 44.!Triangle!T1!is!congruent!to!Triangle!T2!i.e.!all!the!sides!and!angles!of!triangle!T1!and!T2!are!equal.! Triangle!T1!is!similar!to!Triangle!T3!i.e.!all!the!angles!of!triangle!T1!and!T3!are!equal!and!sides!are!in! proportion.! Angles!of!all!three!triangles!are!equal;!thus!all!are!similar!to!each!other.! Point!2!is!true.! Triangle!T1!and!triangle!T3!have!same!angles!but!we!don’t!know!if!their!sides!are!equal;!therefore,! both!triangles!can!be!congruent!but!not!necessarily.! Ans:!(J)! ! 45.!The!cubes!are!cut!that!will!be!of!size!1!½!inches!plus!an!allowance!of!1/8!inches!due!to!width!of! saw!blade.! Maximum!number!of!blocks!=! !" !" !!! !!! ! ! ! (6!feet!=!12!inches)! =!! !! ! ! ! !" =! !" ! ! =!44.30! ≈ 44!! Ans:!(B)! ! 46.!y!=!kx!+!1!and!y!=!x!–!2!intersect!at!(4,!2);!thus,!this!point!satisfies!both!the!equations! y!=!kx!+!1! 2!=!4k!+!1! 1!=!4k! k!=!¼! Ans:!(G)! ! ! 47.!!y!=!! x − 2!represents!the!equation!of!the!side!of!a!rectangle! The!adjacent!sides!in!rectangle!are!perpendicular!to!each!other.!The!product!of!slopes!of!two! perpendicular!lines!is!P1! Slope!of!given!equation!is!¾! Slope!of!adjacent!side!will!be!P4/3! Option!E!has!an!equation!with!slope!P4/3! Ans:!(E)! ! 48.!Factors!of!30!=!1,!2,!3,!5,!6,!10,!15,!and!30! There!are!8!positive!factors!of!30! Ans:!(J)! ! 49.!Diagonals!of!square!divide!it!into!four!equal!triangles;!thus!in!order!for!rectangle!ABCD!to!be!a! square,!triangle!ABE!must!be!congruent!to!triangle!ADE!or!BEC.! Ans:!(B)! ! 50.!Designate!pencil!as!‘P’!and!pen!as!‘Pn’! 2P!+!Pn!=!0.55! ! P!(1)! P!+!2Pn!=!0.95! ! P!(2)! Multiplying!equation!1!with!2,!we!get! 4P!+!2Pn!=!1.1!!! P!(3)! Subtracting!equation!3!from!equation!2,!we!get! P3P!=!P0.15! P!=!0.05! P!+!2Pn!=!0.95! 0.05!+!2Pn!=!0.95! 2Pn!=!0.9! Pn!=!0.45! P!+!Pn!=!0.05!+!0.45!=!0.5! Ans:!(H)! ! 51.!i"+!i2!+!i3!+………..+!i49! We!know,!i!=!i5!=!i" i2!=!i6!=!P1! i3!=!i7!=!Pi! i4!=!i8!=!1! =!(i"+!i2!+!i3!+!i4)!+!(i"+!i2!+!i3!+!i4)!+!……….+!i! =!0!+!0!+!0!+…..+!i" =!i! Ans:!(A)! ! 52.!sin!x!>!0! sine!is!positive!between!0!and!π! cos!x!<!0! ! !" cosine!is!negative!between! ! !and! ! ! ! Therefore,!x!lies!between! ! !and!π! Ans:!(G)! ! 53.!! Diameter!of!circle!=!6!cm! Area!=!πr2! Area!=!π(3)2! Area!=!9π! Ans:!(C)! !! ! 54.!Circle!is!divided!into!12!equal!pieces.! Measure!angle!of!arc!AB!=!360°/12!=!30°! ! Measure!of!arc!=!!"# ×2πr! !" =!!"# ×2!(6)! =!π! Ans:!(H)! ! 55.!Terminal!side!is!OC.! ∠!"# = 40!! Terminal!angle!will!be!(360°!–!40°)!=!P320°!(clockwise)! Others!terminal!will!be!(40°!+!360°)!=!400°!and!addition!of!360°!henceforth! Among!all!options,!680°!is!not!a!coPterminal!angle.! Ans:!(C)! ! 56.!! Speed!! Time!! Distance! X! 8! 8x! 325! 7! 2275! ! !"#$%!!"#$%&'( Average!airspeed!=! 350!=! !!!(!"#)(!) !" !"#$%!!"#$ ! ! 350(15)!=!8x!+!(325)(7)! Ans:!(K)! ! 57.!! ! Area!of!quadrilateral!ABCD!=!Area!of!triangle!AMD!+!Area!of!rectangle!BCDM! Triangle!AMD!is!right!angled!triangle!and!has!side!3!–!4!–!5!(Pythagorean!triplet)! Area!of!triangle!AMD!=!½!×!3!×!4!=!6! Area!of!rectangle!BCMD!=!2!×!4!=!8! Area!of!quadrilateral!=!6!+!8!=!14! ! Similarly,!area!of!quadrilateral!DCFE!=!Area!of!triangle!DEN!+!Area!of!rectangle!DCFN! =!(½!×!12!×!5)!+!(2!×!12)! =!30!+!24! =!54! ! Ratio!=!14:!54! =!7:!27! Ans:!(D)! ! 58.!If!line!p!and!q!were!parallel!lines,!then!the!sum!of!angle!‘a’!and!‘b’!would!be!180°! But!both!seem!to!be!opening!outwards;!thus!(a!+!b)!>!180! Ans:!(K)! ! 59.!Standard!equation!of!circle!with!center!(h,!k)!is! (x!–!h)2!+!(y!–!k)2!=!r2! Points!(1,!a)!and!(a,!3)!pass!through!a!circle;!thus!they!will!satisfy!the!equation!of!the!circle! (x!–!2)2!+!(y!+!3)2!=!r2! For!point!(1,!a):!(1!–!2)2!+!(a!+!3)2!=!r2! For!point!(a,!3):!(a!–!2)2!+!(3!+!3)2!=!r2! !(1!–!2)2!+!(a!+!3)2!=!(a!–!2)2!+!(3!+!3)2! 1!+!a2!+!6a!+!9!=!a2!–!4a!+!4!+!36! 10a!=!30! a!=!3! Ans:!(E)! ! 60.!|ax!+!b|!=!|bx!+!a|! ax!+!b!=!bx!+!a!!! or!! P(ax!+!b)!=!bx!+!a! ax!–!bx!=!a!–!b!!! or!! Pax!–!bx!=!a!+!b! (a!–!b)x!=!a!–!b!! or!! Px(a!+!b)!=!a!+!b! x!=!1!! ! ! or!! x!=!P1! Ans:!(G)! ! ! 56A$Science$ 1. From'the'graph'in'figure'3,'we'can'interpret'that'as'the'average'discharge'increases,'total' annual'precipitation'also'increases.'Answer$A' ' 2. From'tables'1'and'2,'we'can'conclude'that'as'the'erosion'rate'in'limestone'bedrock' catchments'increases,'total'annual'precipitation'also'increases.'Answer$F' Catchment$ Erosion$rate$cm/1000year$ Total$annual$precipitation$ cm$ A' 0.9' 67.5' B' 1.9' 142.5' C' 2.7' 202.0' ' 3. According'to'figure'1,'as'the'average'discharge'increases,'total'annual'precipitation'also' increases.'Therefore,'for'300'cm'of'annual'precipitation,'the'value'of'average'discharge' would'be'over'12'm3/sec.'Answer$D' ' 4. From'the'2nd'and'3rd'columns'of'table'2,'we'can'conclude'that'there'is'no'clear'relationship' between'average'annual'air'temperature'and'total'annual'precipitation.'Answer$J' ' 5. According'to'table'1,'dissolved'load'in'limestone'catchment'is'more'than'solid'load,'which' means'that'limestone'is'more'susceptible'to'chemical'weathering'than'mechanical' weathering.'Answer$B' ' 6. According'to'table'2,'for'F'catchment,'total'annual'precipitation'is'less'than'25cm'(22.5cm),' and'therefore'according'to'table'1,'the'catchments'fulfilling'this'criterion'will'have'erosion' rate'less'than'1cm/1000'yr.'Answer$F' ' ' 7. According'to'the'hydrothermal'vent'hypothesis,'temperatures'between'10'degree'Celsius' and'700'degree'Celsius'promote'the'chemical'reactions.'Answer$B' ' 8. 'The'organic'molecules'combined'to'first'form'polymers'and,'later,'living,'self'replicating' cells.'Their'first'characteristic'was'the'ability'to'replicate.'Answer$G' ' 9. According'to'both'the'hypotheses,'conditions'in'the'ocean'were'right'for'the'organic' molecules'to'combine'and'replicate.'Answer$B' 10. The'outer'space'hypothesis'makes'the'assumption'that,'billions'of'years'ago,'earth'was' bombarded'by'bodies'from'outer'space'which'eventually'brought'organic'molecules'to' earth.'Answer$H' ' 11. According'to'the'outer'space'hypothesis,'organic'molecules'were'brought'to'earth'by'bodies' from'space'while'the'hydrothermal'vent'hypothesis'claims'that'these'organic'molecules' were'first'discovered'in'the'ocean.'Therefore,'the'statement'made'in'the'question'is' consistent'with'the'outer'space'hypothesis'only.'Answer$A' ' 12. According'to'both'hypotheses,'in'order'to'replicate,'these'organic'molecules'first'had'to' form'polymers.'Answer$H' ' 13. According'to'outer'space'hypothesis,'microscopic'interplanetary'dust'particles'contain'up'to' 10%'organic'material'while'meteorites'contain'up'to'5%'organic'material.'This'indicates'that' most'of'the'organic'molecules'were'brought'to'earth'by'interplanetary'dust'particles.$ Answer$C' ' 14. According'to'the'graph'of'distance'versus'time'for'the'3rd'motorbike,'at't=0,'the'bike'has' already'covered'some'distance'and'it'is'not'on'the'starting'line.'Answer$G' ' 15. According'to'the'graph'of'velocity'versus'time'for'the'2nd'motorbike,'the'velocity'of'the' motorbike'increases'until'it'reaches'certain'point'and'then'it'immediately'starts'to'decrease.' The'description'made'in'option'D'about'the'velocity'of'a'car'is'identical'to'this'behaviour.' Answer$D' ' ' 16. Acceleration'of'a'motorbike'is'the'rate'of'change'of'velocity'with'time.'Therefore,'if'the' acceleration'of'a'vehicle'is'negative,'we'can'conclude'that'its'velocity'is'decreasing.'Answer$ H' ' 17. If'a'baseball'is'tossed'back'and'forth'between'two'people,'it'would'have'positive'velocity' while'travelling'forward'and'negative'velocity'while'travelling'backwards.'These'two' motions'will'have'both'positive'and'negative'values'of'velocity'on'a'velocity'versus'time' curve.'Answer$A' ' 18. According'to'the'acceleration'versus'time'graph'in'set'3,'the'motorbike’s'acceleration'is'zero' between'points'C'and'D'but'the'velocity'of'motorbike'is'not'zero'between'these'2'points.' Answer$J' ' 19. According'to'table'1,'the'melting'point'of'iron'is'1535'degrees'Celsius.'Therefore'at' temperatures'above'1535'degree'Celsius,'iron'will'be'in'its'molten'state'and'it'can'be' drained'or'poured'out'of'the'reactor.'Answer$D'' ' ' 20. According'to'2nd'and'3rd'columns'of'table'1,'there'is'no'clear'relationship'between'atomic' number'and'melting'point'of'the'metals.'Answer$J' ' 21. The'energy'required'to'form'a'+1'ion'is'the'first'ionization'energy'of'the'metal,'and' according'to'table'1,'for'‘Sc’'metal,'this'first'ionization'energy'is'minimum.'Answer$A' ' ' 22. According'to'the'last'2'columns'of'table'1,'we'can'conclude'that'atomic'radius'of'all'metals' is'greater'than'their'ionic'radius'which'means'that'an'atom'is'larger'than'its'positive'ion.' Answer$F' ' 23. To'change'a'neutral'metal'atom'into'its'+2'ion'form,'first'one'electron'must'be'removed' from'the'neutral'atom'to'form'+1'ion'at'the'expenditure'of'first'ionization'energy,'and' subsequently'one'more'electron'must'be'removed'from'+1'ion'at'the'expenditure'of'second' ionization'energy'to'form'+2'ion.'Zinc'metal'has'first'ionization'energy'equal'to'906'KJ/mole' and'second'ionization'energy'equal'to'1733'kJ/mole.'Answer$D' ' 24. According'to'figure'1,'the'largest'percentage'of'seeds'collected'had'masses'between'0.6mg' and'1mg.'Answer$G' ' 25. According'to'figure'2,'seeds'under'0.5mg'produced'plants'with'considerable'masses.'The' greenhouse'pots'and'the'cleared'meadow'sites'showed'growth'of'plants'from'seeds'that' weighed'under'0.5'mg.'Answer$C' ' 26. According'to'figure'3,$the'average'mass'of'the'seeds'produced'by'plants'grown'in'the' greenhouse'from'2'mg'parent'seeds'was'approximately'2.5mg.'Answer$H' ' ' 27. According'to'the'3'curves'shown'in'figure'2,'we'can'conclude'that'the'mass'of'the'plants'was' greatest'for'plants'grown'in'greenhouse'pots'and'smallest'for'the'plants'grown'in'the' unclear'meadow'sites.'Answer$C''' ' 28. According'to'figure'2,'if'a'0.9'mg'seed'of'wildflower'species'was'planted'in'an'unclear'site,' plant'growth'would'have'taken'place,'but'in'figure'3'we'can'see'that'these'produced'plants' would'not'be'able'to'produce'any'seeds'when'planted'in'an'unclear'site.'Answer$F' ' ' ' '' ' 29. 'According'to'figure'1,'the'rate'of'photosynthesis'is'almost'above'100%'for'the'plants' exposed'to'a'wavelength'of'440Z460nm.'Answer$A' ' 30. As'bacteria'require'oxygen'for'survival,'most'of'the'bacteria'were'clustered'around'the' region'of'algae'where'it'was'releasing'more'oxygen.'Answer$F' ' 31. Photosynthetic'activity'depends'on'the'wavelength'of'light'and'not'on'the'intensity'of'light.' Answer$D' ' 32. According'to'figure'2,'carotenoids'absorb'light'from'the'range'400'to'550'nm'and'it'absorbs' very'little'light'between'the'wavelengths'550'and'575'nm.'Carotenoids'do'not'absorb' wavelengths'greater'than'575nm.'Therefore,'according'to'the'statement'made'in'the' question,'the'colour'of'carotenoids'would'most'probably'be'yellow'and'orange.'Answer$G' ' 33. 'In'study'1,'it'is'mentioned'that'bacteria'are'found'in'red'and'orange'region'rather'than'blue' region'as'they'take'oxygen'from'these'regions'of'algae'for'living.'Therefore,'blue'or'blue' purple'grow'light'will'not'add'any'advantage'in'the'process'of'photosynthesis.'Answer$D' ' 34. In'study'1'and'2,'living'algae'and'bean'plants'were'exposed'to'sunlight'to'examine'the'role' of'light’s'wavelength'and'plant'pigments'on'photosynthesis.'For'this'to'happen,'it'was' mandatory'for'both'the'studies'to'expose'them'regularly'to'the'same'temperatures.'' Answer$J' ' 35. To'produce'the'test'solutions'for'experiment'2,'steel'pipe'samples'were'taken'and'they' were'dissolved'in'acid'followed'by'addition'of'an'oxidant.'After'this,'the'solutions'were' diluted'with'water.'Answer$D' ' 36. The'Mn'in'the'test'solutions'could'not'be'compared'with'the'standard'solutions'unless'it' was'converted'to'MnO4Z.'For'this'reason,'an'oxidant'was'added'to'chemically'react'with'the' Mn+2'and'convert'it'to'MnO4Z.'Answer$G' ' 37. As'there'will'be'less'concentration'of'MnO4Z'in'the'solution,'the'value'of'absorbance'would' be'less.'Answer$C' ' 38. According'to'figure'1,'we'can'conclude'that'as'the'Mn'content'increases,'absorbance'also' increases.'From'table'1,'it’s'clear'that'the'solution'of'pipe'A'has'the'highest'absorbance;' therefore,'it'must'be'having'the'highest'content'of'Mn.'Answer$F' ' ' ' 39. According'to'table'1,'absorbance'of'sample'D'is'0.500.'If'we'plot'this'value'of'absorbance'in' figure'1,'we'will'get'the'corresponding'value'of'concentration'of'Mn'in'between'1.0'and' 1.5mg/L.'Answer$C' ' ' ' 40. As'the'solution'of'copper'ions'absorb'light'at'675'nm,'the'spectrometer'must'be'set'at'675' nm'for'the'standard'solutions'containing'copper'ions'as'well'as'for'the'metal'pipe'test' solution.'Answer$J'' ' ' ' ' ' ' '
