2018년 좋은책신사고 쎈 ( SSEN ) 초등 수학 5 - 2 답지

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2018년 좋은책신사고 쎈 ( SSEN ) 초등 수학 5 - 2.pdf 다운로드 | 답지저장소

초등수학`5-21소수의곱셈0142합동과대칭0313분수의나눗셈0504소수의나눗셈0625여러가지단위0786자료의표현088●학업성취도평가100●경시대비평가105단계기본다잡기는빠른정답찾기에만정답이있습니다.단계부터는빠른정답찾기와자세한풀이에정답과풀이가있습니다.쉽게이해되고문제해결력을길러주는~정답및풀이자세한풀이014~109빠른정답찾기002~013※빠른정답찾기의서술형평가유형은풀이과정을제외한정답만제시하였습니다. 15초등쎈수학5-2정답(001~013) 2015.6.2 7:35 PM 페이지001 SinsagoHitec 002쎈수학5－2008쪽1소수의곱셈01;1¡0;, 0.102;1¶0;03;1¶0;045, 5, 0.5054, 8, 0.080625, 275, 0.275072, 2, 2, 1.2085, 5, 45, 2.45098, 8, 272, 3.272108, 8, 2, 4, 5114, 4, 2, 25124, 4, 1135, 5, 1, 1, 21475, 75, 25, 3, 3, 415125, 125, 1, 8162, 2, 6, 0.6；<1725, 25, 75, 0.75；>188, 9；<1965, 64；>단계⑴010쪽012쪽014쪽01;10∞00;, 0.005；0.00502(cid:9066), 0.5703;1¶0;, 0.70405=06;1£0ª0;, 0.3907분수: 7;10¶00;, 소수: 7.007; 07(cid:9066)7;10¶00;과7.007은크기가같습니다.080.6096.5102.2km110.8121.2cm130.06145.3515㉡, ㉢16;2!5$;, 0.56170.75180.325191.624203.195m21;4!0(;에◯표220.314230.224kg240.62525⑴5, 1, 2(cid:100)⑵5.5263, 6, 10；3, 3, 527;1¢0;, ;5@;285;1∞0;(=;1%0%;), 5;2!;(=;;¡2¡;;)29㉡301;5!;L, 2;2!;L31;5$;32;1ª0§0;, ;2@5$;333;1¢0∞0;(=;1#0$0%;), 3;2ª0;(=;2^0(;)34㉡359;2@5#;초36;1$0^0%;, 4;1§0∞0;, 4;2!0#;(=;2(0#;)154-010.80단계⑴016쪽374;5¡0;38;1™0ª0§0;, ;1£2¶5;395;1£0¶0∞0;(=;1%0#0&0%;), 5;8#;(=;;¢8£;;)4050!0;, ;12!5;411;2§0ª0;km420.62543109개448;2!;450.754, ;5#0&0&;461;2¶0;47>48<492.650수산시장51;8(;52;4#;, 0.78, ;5$;, 0.81353주스54은우55우체국, 영화관, 문구점560, 1, 2, 3574개015;1™0•0;, 5.2802⑴125 ⑵;12^5;⑶0.048030.3204;9!6@;051;4!0!;m06분수: ;1£0º0¶0;km, 소수: 0.307km07123.75cm081.05kg09;2!0&;105;4#;11㉢, ㉠, ㉡018쪽단계⑴020쪽022쪽01⑴1.6(cid:100)⑵1.6(cid:100)⑶1.6024, 4, 16, 1.60316, 1.604⑴4.5(cid:100)⑵4.5(cid:100)⑶4.50515, 15, 45, 4.50645, 4.507(cid:9066)；2.7089, 9, 27, 2.70927, 2.71021, 21；273, 27.31113, 26, 273；27.31245, 45；450, 4.51345, 45；100, 451445, 45；45000, 450152.8, 28, 2801634, 340, 34001710, 230；230, 2318100, 100；100, 2.3191000, 230；230, 0.2320300, 30, 32116, 1.6, 0.16220.01, 63；0.63239, 7, 63, 0.632463, 0.632523, 42, 966, 9.6626966；10, 100；9.662746, 92, 966；9.6603단계⑵024쪽010.8, 2.402;1£0;_9==;1@0&;=2.73_910단계⑵ 15초등쎈수학5-2정답(001~013) 2015.6.8 3:48 PM 페이지002 SinsagoHitec 빠른정답빠른정답찾기003026쪽028쪽030쪽03;1¡0¢0;_8==;1!0!0@;=1.12041.5052.16064.9076.76082.8, 3.9209㉠, ㉡, ㉢, ㉣10(cid:9066)；3.911㉢1220.8136.951437.81527.361621.48175.26_26=_26==175.26_26=136.76184.5kg1918.2L2010시간2138_;1¶0;===26.62231.82337.12247.2259.122620.527④282.7m¤29④30543145.883223.83321.843417.5, 51.1235㉡36㉡, ㉢374.8m3880kg39기성, 0.45L400.75_9=6.75411.1242⑴㉡(cid:100)⑵㉢(cid:100)⑶㉠430.027_540에◯표44(cid:9066)곱해지는수가;10!0;배가되면곱의결과도;10!0;배(cid:100) 가됩니다.452.8, 28, 2804632.9, 329047㉢48650L4945, 4.5, 0.45500.38451③52147원, 14.7원, 1.47원53파란색끈, 0.78m540.1550.00156㉣575687058㉯590.21600.0576610.32620.10263(위에서부터) 0.231, 0.0252；0.294, 0.019864<650.04_0.16에◯표660.1867;1!0*;_;1$0@;=;1&0%0^;=7.56684.766915.0787031.8976713, 1, 272㉣735.4.9.2_5.21.0.9.8.42.7.4.6.0.82.8.5.5.8.42661038_71013676100526_2610052610014_8100032쪽034쪽740.054L758.75kg76⑴0.1시간(cid:100)⑵0.007m(cid:100)⑶0.183m7718.33780.4779(cid:9066)0.325_27과32.5_0.27에서소수를분수로나타내면325_27의곱의결과에;10¡00;배를한것과같습니다. 따라서0.325_27과32.5_0.27의값은8.775로같습니다.80100배817;2!;cm8249.5cm83336.3원, 3363원, 33630원840.16kg851034개86㉢8740.96883.3L8919.152900.001910.64m¤92701227.5cm0231.5kg03680kg04983.04cm¤05⑴2.75분(cid:100)⑵2.31km(cid:100)⑶2.16km060.09, 0.4070.5kg0843.2cm0925.2100.0011132.0418km122036쪽단계⑵0137, 100；0.37028.803;1£2¡5;041;2!0!;m05㉡, ㉢, ㉠, ㉣06③074；4, 2.80817.28097.2m107, 7；98, 9.8112.712318.21316.5m¤1495.4kg15<16⑴㉢(cid:100)⑵㉠(cid:100)⑶㉡175.2, 0.052180.48721913.3420정사각형모양의종이038쪽단원마무리1회012;1∞0;(=;1@0%;), 2.502;5@0&;kg031;2!5$;04혜란, 재중, 미소05;1•0;_9==;1&0@;=7.2061.20723.1L0814.82cm0956, 5.6；1010011③1221450원13=14㉠, ㉡, ㉣, ㉢155520kg16①17(위에서부터) 3.78, 24.75；33.75, 2.7721813.975km190.342098_910040쪽단원마무리2회 15초등쎈수학5-2정답(001~013) 2015.6.8 3:48 PM 페이지003 SinsagoHitec 004쎈수학5－2044쪽046쪽048쪽2합동과대칭01㉮02㉯, ㉱03합동04[] [] [] []05점ㄹ, 점ㄹ06변ㄹㅁ, 변ㄹㅁ07각ㄹㅁㅂ, 각ㄹㅁㅂ08변ㅁㅂ09각ㅁㅂㅅ10같습니다117, 6, 반지름, 원12(위에서부터) 3, 4, 2, 11310, 각도기, 8, ㄱ, ㄷ14(위에서부터) 4, 2, 1, 3158, 60, 70, 각16(위에서부터) 3, 1, 4, 2◯단계⑴056쪽39(cid:9066)40(cid:9066)41(cid:9066)42(cid:9066)43(cid:9066)44(cid:9066)45(cid:9066)465cm47(cid:9066)48(cid:9066)4930˘, (cid:9066)50현중51㉢52(cid:9066)[방법1]세변의길이를알아야합니다.[방법2]두변의길이와그사이에있는각의크기를알아야합니다.(cid:100) [방법3]한변의길이와그양끝각의크기를알아야합니다.53각ㄴㄱㄷ(또는각ㄷㄱㄴ)54한변의길이55㉢56③57㉡50˘30˘100˘2`cm60˘70˘4`cm70æ40æ3`cm3`cm60æ3`cm2.5cm2cm80˘3`cm1.5`cm60æ2cm2cm2cm1.5`cm1.5`cm2`cm3cm3.5cm4cm3`cm3`cm3`cm049쪽050쪽052쪽054쪽01라에◯표02가와사, 라와바03(cid:9066)모양은같지만크기가다르므로두도형을포개었을때완전히겹쳐지지않습니다.따라서두도형은합동이아닙니다.04(cid:9066)오른쪽도형의아래쪽변을위쪽변과평행하게만듭니다.05가와다, 나와라062쌍07가, 나, 라, 마08㉢09(cid:9066)10(cid:9066)1112(cid:9066)13(cid:9066)14㉡15점ㄹ, 점ㅂ, 점ㅁ16변ㄹㅂ, 변ㅂㅁ, 변ㄱㄷ17각ㄹㅂㅁ, 각ㄱㄷㄴ, 각ㄷㄱㄴ18각ㅇㅁㅂ194쌍씩20점ㄹ, 변ㄹㄷ21(위에서부터) ㄹ; ㅂ, ㅁ22(위에서부터) 4, 7237cm245cm2510cm2613cm2711cm2824cm29135cm¤308cm3110cm3240˘3390˘3450˘351353665˘3770˘38⑴30˘⑵30˘⑶120˘단계⑴01㉣024cm0398cm04⑴75˘(cid:100)⑵75˘, 40˘(cid:100)⑶115˘05(cid:9066)06가와라, 나와다0713cm08168cm¤0930˘10100˘4`cm4`cm4`cm058쪽단계⑴ 15초등쎈수학5-2정답(001~013) 2015.6.8 3:48 PM 페이지004 SinsagoHitec 빠른정답빠른정답찾기00511(cid:9066)4`cm3`cm2`cm060쪽062쪽01가02선대칭도형03대칭축04[] [] []05점ㅂ06변ㅁㄹ07각ㄱㅂㅁ08변ㄱㅂ09각ㄱㅂㅁ109011선분ㅂㅈ1213가14점대칭도형15대칭의중심16ㅇ17점ㄹ18변ㅁㅂ19각ㄱㄴㄷ20변ㅁㅂ21각ㅂㄱㄴ22점ㅇ23선분ㅂㅇ24◯단계⑵064쪽066쪽01[] [] [] []02②, ⑤03ㄷ, ㅂ040506⑤07㉢, ㉠, ㉡08(cid:9066)직선ㄱㄴ을따라접었을때글자가완전히겹쳐지지않습니다.따라서직선ㄱㄴ은대칭축이아닙니다.09Y,1개10(cid:9066)113개12점ㄴ, 변ㄷㄹ, 각ㅂㅁㄹ13(왼쪽에서부터) 점ㄱ, 점ㅁ；변ㄱㅁ, 변ㅁㄹ；각ㄴㄱㅁ, 각ㄱㅁㄹ14(왼쪽에서부터) 점ㄱ, 점ㄹ；변ㄴㄱ, 변ㄷㄹ；각ㄴㄱㅁ, 각ㄱㅁㄹ15(위에서부터) 60,416(위에서부터) 6,1301711cm1865˘19120˘20선분ㄴㅂ, 선분ㄷㅁ21(왼쪽에서부터) 20, 14225cm, 100˘◯단계⑵068쪽070쪽072쪽2364cm24⑴2배⑵168cm¤⑶336cm¤2548cm¤2610cm2728293031(cid:9066)3233HIDE34I3035[] [] []363개37다38다39㉠, ㉣404142ㄷ43가, 나, 라, 바；가나라바44점ㄹ, 변ㅁㅂ, 각ㄹㄷㄴ45㉡46(위에서부터) 점ㅁ, 점ㄹ；변ㅁㄹ, 변ㄹㅁ；각ㅁㄹㄷ, 각ㄹㅁㅂ473cm48(왼쪽에서부터) 10, 4049120, 850100˘51110˘5212cm539cm5413cm5518cm565cm5732cm58322cm¤5940cm¤60대칭의중심, 대응점6162636465(cid:9066)◯◯ 15초등쎈수학5-2정답(001~013) 2015.6.2 7:35 PM 페이지005 SinsagoHitec 006쎈수학5－2074쪽66, 24cm¤6711일, 26일683가지69㉠, ㉤, ㉥; ㉠, ㉦70(cid:9066)713쌍7212cm7345˘74그릴수있습니다.754cm7625˘77성윤7892cmㅇ1`cm1`cm016002120cm0350˘04⑴7cm, 3cm(cid:100)⑵5cm(cid:100)⑶16cm05100˘0612cm0734cm0816cm¤097개10155˘1162cm076쪽단계⑵01나, 라02㉣03정사각형0418cm05122˘0684cm¤07(cid:9066)08(cid:9066)09(cid:9066)10, 에◯표11셀수없이많이그릴수있습니다.12(위에서부터) 9, 80, 1001342cm14(cid:9066)15①, ⑤16점ㄹ, 점ㅁ, 점ㅂ1723cm1855˘1900, 60˘3`cm2.5`cm3`cm2`cm2`cm50æ85æ1.5`cm080쪽단원마무리2회086쪽084쪽3분수의나눗셈016；6026；6036044；4051, 4；1, 4061, 4071, 8081, 8091, 8101, 7, 1, 7111, 12, 1, 12121, 9134, 9144, 9155, 6, 5, 6163, 11, 3, 11171, 3, 4, 15181, 7191, 10, 3, 70201, 2, 1, 4, 11211, 5, 5, 15(1, 3)225, 2, 3231, 7, 11, 35241, 8, 1, 4, 9251, 4, 8, 20(2, 5)266, 92720, 1, 8283, 4, 3, 8295, 2, 7, 5, 14302, 18, 5, 1, 2, 53110, 30, 7, 1, 3, 1, 10, 7, 1, 3, 7088쪽011, 1, 9021_;2!;031_;2¡3;048_;2¡1;0512_;1¡7;단계단계01나02(cid:9066)나의아래쪽변의길이를한눈금길게그립니다.03(cid:9066)04점ㅁ05(왼쪽에서부터) 7; 8, 60670˘07(cid:9066)08(cid:9066)09변ㄴㄷ의길이10(cid:9066)8>3+4로가장긴변의길이가다른두변의길이의합보다깁니다.따라서다른두변이서로만나지않기때문에삼각형을그릴수없습니다.11④124개1310cm1426cm15120˘16③1718135˘1914cm20ㅇ 3cm60˘40˘80˘3`cm2`cm3`cm078쪽단원마무리1회 15초등쎈수학5-2정답(001~013) 2015.6.8 3:48 PM 페이지006 SinsagoHitec 빠른정답빠른정답찾기007090쪽092쪽06④07350809;8!;10;3¡2;11;7$;12;2!4#;13<14㉡15;1¡3;L16;1§1;m17㉯빵집18;1¡6;(=;4£8;)19;2™1;(=;6§3;)20;1£4;, ;7¡2;(=;14@4;)21㉡22;1¶5;23;9@;(=;3•6;)24;1£4;253426종현27;1™7;L28;8&;m29;3∞6;kg301;2!1^;31;5#;(=;3@5!;)322;1£0;÷6=;1@0#;_;6!;=;6@0#;33㉢, ㉣34;3!5*;L35;6!;km36;5#6!;km37;9%;_4/_=;1∞8;38;4#;392;2!;40㉣411;4!;kg42⑴;5!;kg⑵;7!;kg⑶;3!5@;kg43④44>45>46㉯47[] [] []48㉡, ㉣493；;3!;50;1•7;÷12, ;5™1;513개527, 3, 5；;4!0(; 53;3£2;54;7$;55;1¡0;56;4!;573개586592, 3, 4605;3@;(=:¡3¶:)cm611;1¡0;m625;5@;(=:™5¶:)m¤635;5#;cm¤641;6!;(=;6&;)65;8¡4;66;4#;67;7@;kg68미란694개70;1!7^;71;1¢5;÷2=_=;1™5;727개73;8#;kg74㉡, ㉢, ㉠75;4!;761;2!;cm¤77;9*;배12/124/15◯18/21094쪽096쪽098쪽01;7@;, 3;4!;(=:¡4£:)；1, 2, 302;1∞8;m032;1ª3;km04⑴2;5$;L⑵;1!5$;L⑶1;1!5#;L⑷;3¶0;L057, 2；7;2!;0623;5!;(=;;;!5!;§;;)cm07;1£9;배083;6!;09;4!;kg1018cm¤112;2ª8;cm100쪽단계011, 1, 402③03;1¡1;04;2ª8;053;6!;(=:¡6ª:)m06;7@;(=;1¢4;)07준호08;22@5;09;1§1;÷4, ;2£2;10㉡11㉠12㉢13<14;2∞1;L15;1!2!;(=;2@4@;)16;7%;(=;2@8);)173개182;7%;kg192;3@;(=;3*;), ;3@;, ;3!;2013;4#;cm¤102쪽단원마무리1회011_;1¡7;02;2∞1;_;9!;031;7%;(=:¡7™:)04;2∞4;053;6%;(=:™6£:)0615;9%;(=;;;!9$;º;;)L07;4™5;(=;18*0;)08;2£8;(=;1¡6•8;)09;8∞8;10①11>12>13;2•1;14;2∞4;m15_=;9@;16㉢171, 2, 3, 418;1¢5;kg19;2™7;2012;3@;cm¤129?1258?9104쪽단원마무리2회••••••••••••108쪽4소수의나눗셈0152, 156, 3, 1；52, 5.20252, 5.20310, 15, 315；15, 1.50415, 1.505100, 126, 504；126, 1.26062486, 226, 2486, 11, 1；226, 2.26071.25단계⑴ 15초등쎈수학5-2정답(001~013) 2015.6.2 7:35 PM 페이지007 SinsagoHitec 008쎈수학5－2110쪽116쪽112쪽114쪽01;1(0*;÷7=_=;1!0$;=1.402:£1™0∞:÷13=_=;1@0%;=2.50319.2, 16, 1.60426051.3061.9073.2085.3096.9, 2.31020.8115.312(cid:9066)[방법1]273÷7=39→27.3÷7=3.912(cid:9066)[방법2]139.1m142.1kg15768, 128, 12.8162.7L17㉢, ㉡, ㉣, ㉠18:¡1¢0∞0™:÷6=_=;1@0$0@;=2.42197.02÷3=;1&0)0@;÷3=_=;1@0#0$;=2.3420212.38223.56232.42244.37252.31, 3.6426㉡27284.36m293.26L30고은311.32kg32⑤33;1@0$;÷4=_=;1§0;=0.614/1624/10403.788<“30.244244062405640.06440.06440.00029.85÷518.42÷313/1234702÷10016/12421452÷1003.97<“27.3216363 0113/125325÷1017/11498/1034㉢35；0.8360.46370.76380.63390.4240②4142100배430.28L440.28kg45⑴4.68m⑵0.36m140.5417<“9.18148.5140.6814.068140.000.51.52.53.54.5035241081, 8, 9；17；18；0092, 3, 8；6；9；24101, 9, 4；75；32；0111, 5, 7；15；75；1051225, 5, 25, 5, 1；5, 0.513184, 23, 184；23, 0.2314119, 17, 119, 7, 1；17, 0.17150.72160.31170, 7；35180, 1, 4；12190, 6；54200, 4, 9；135단계⑴111쪽01:¡1ª0∞:÷6=_=;2^0%;=;1#0@0%;=3.2502하리；(cid:9066)7.4÷4=;1&0$0);÷4=_02하리；(cid:9066)7.4÷4=;1!0*0%;=1.85031.35042.15054.25065.15073.5608③090.26cm102.15분111.15kg12㉢, ㉡, ㉠, ㉣13:¡1•0¡:÷2=:¡1•0¡:_;2!;=:¡2•0¡:=;1(0)0%;=9.0513또는:¡1•0¡0º:÷2=_=;1(0)0%;=9.0514868152.04163.05179.05184.05196.05, 15.08, 12.0620㉢2144.05g2211.03L238.02g12/19051810/10014/1185740÷10016/265195÷10단계⑵120쪽122쪽117쪽118쪽01108, 27, 108；27, 135, 1.35021080, 135, 1080；135, 1.35030, 8；0, 8, 5；0；20042, 2, 5；15；0；30051, 8, 8；40；0；4006203, 203, 5；203, 406, 4.0607847, 121, 847；121, 605, 6.05082440, 305, 2440；305, 3.05091.03102.08115, 0, 9；45；81124, 0, 2；60；301325, 50, 2, 1, 25, 2.5142, 5, 25, 2.5152, 5；0162.4172.43182.43192.42단계⑵[방법3]27.3÷7=:™1¶0£:÷7=_27.3÷7=;1#0(;=3.917/139273÷10 15초등쎈수학5-2정답(001~013) 2015.6.8 3:49 PM 페이지008 SinsagoHitec 빠른정답빠른정답찾기009128쪽124쪽126쪽130쪽132쪽24:™1¶0º0º:÷12=_=;1@0@0%;=2.2525:¡1º0º0º:÷8=_=;1!0@0%;=1.2526③272.8280.25292.5301.37531지우개3296.25km330.25kg341.25kg352.1364.3372.7381.839진수40㉢412.81422.44433.33442.2345㉠, 1.4461.83470.07480.4L493.1kg504.33m5170.89초521.325354㉠551, 2, 356㉢576개584.41m¤594.45cm¤6012.5m¤617.5cm¤623.12cm636.05644.565⑴49m¤⑵5.44m662.75671.5m680.13690.99703.04712.91720.86730.6743.64cm750.12km76677㉡780795, 4, 3, 2；27.15807, 6, 5, 3；25.58134÷76；0.44823, 4, 3.48328.1m¤84대현85866.1876.4885, 6, 7, 88925.64kg9017.35m¤91920.18km938.51802.0613<“26.7818264000.784000.78400.0001234549220326572327264.4÷830.6÷1554.9÷1812318/11251000/100112/12252700/1000112.8kg02(위에서부터) 7；2, 9, 7, 8；2, 0, 3033.34cm0410시46분48초134쪽단계0196, 12, 96；12, 1.2026.4, 1.6038, 6, 4, 2；43.2043.11054.81cm06㉡070.7km081.23093.1510(cid:9066)[방법1]41.2÷5=:¢1¡0™:÷5=:¢1¡0™:_;5!;10(cid:9066)[방법1]41.2÷5=:¢5¡0™:=;1*0@0$;=8.2410(cid:9066)[방법2]4120÷5=824→41.2÷5=8.2410(cid:9066)[방법3]111.75배12㉡13㉠143.05cm¤156.05m167.08m17:¢1§0º:÷4=_=:¡1¡0∞:=11.5184.5196.32059.79g14/1115460/108.245<“41.204012102020 005⑴14장⑵10장⑶140장0668kg070.16m0813.74095.05107.2110.02136쪽단원마무리1회01÷27=_=;1^0*;=6.80231.2032.14048.2L053.2806(위에서부터) 4.29, 1.95, 7.15075.28m08②094개100.6611㉮124.15131412.05m155.1161.61715.6분1878÷24에◯표1962063장4.0514<“56.705670700127/1681836/10183610138쪽단원마무리2회142쪽5여러가지단위0110, 10021a, 1아르031000405300066077001a단계 15초등쎈수학5-2정답(001~013) 2015.6.8 3:49 PM 페이지009 SinsagoHitec 010쎈수학5－25320명548대553대5612a57(cid:9066)58강서구591.492kg6094.8a61(cid:9066)야구장의넓이는약1.1ha입니다.626ha63100km¤6470개65㉡, ㉢, ㉠664.8t673.6t거리100m08400926001081111ha, 1헥타르121km¤, 1제곱킬로미터131001415165001721842001930020621900022(위에서부터) 10000；100, 100；;100!00;(0.0001),22;10!0;(0.01)235700, 5724260000, 26000000257000269270.228t, 톤1km@1ha01144a02270m03324km¤0430kg05⑴30000m¤⑵10000그루⑶700t0629.25a07360m084ha091km¤101.35t116시간40분156쪽단계011a, 1아르0250380000000428a0515개06>07=08100, 100092000m10900m1116121500130.1km¤14㉣15570000, 57001617㉡182.4t197대200.3t3000`m@=3`a1200`a=12`km@5`ha=50000`m@158쪽단원마무리1회010.8026200003㉡0481a052a0648ha0720000개0814209;10!0;(0.01)배1012500ha110.44km¤1210513a14t15125m16㉡, ㉠, ㉢17㉮188배1915명2016.2t160쪽단원마무리2회154쪽6자료의표현164쪽01파란, 2022, 9, 90342, 48, 172044일05172, 4, 43；430642078, 2082, 44094, 4, 141010, 14, 14, 4；56, 4, 14단계⑴144쪽145쪽146쪽148쪽150쪽152쪽0140264a03(cid:9066)1a=(한변이10m인정사각형의넓이)1a=10_10=100(m¤)1a=100m¤이므로1a는1m¤의100배입니다.03(cid:9066)1a는1m¤에비해가로가10배, 세로가10배더크므로1a는1m¤의100배입니다.0488a0554a06(cid:9066)한변이100m인정사각형의넓이를1ha라씁니다.0712ha0813ha0979810미소1149121.05km¤136400ha140.42km¤15①16km¤17ha18(cid:9066)인천의넓이는약1002km¤입니다.1990020㉢21㉠2230000, ;10¡00;(0.001)23㉢2452ha25208a2670273km283029500m3080m31<32>33㉡342.2ha35나36800개37600개3875개3910.893t403배41680.274220명43㉢44소영45<46㉠, ㉡, ㉢474t48900493.6t503t514.6t52⑴2;2!;(2.5)시간(cid:100)⑵1125kg(cid:100)⑶1.125t단계 15초등쎈수학5-2정답(001~013) 2015.6.8 3:49 PM 페이지010 SinsagoHitec 빠른정답빠른정답찾기011114, 10, 2, 6；7, 6, 3, 712혜주133, 39；39, 151421, 841584, 16, 12167쪽166쪽168쪽170쪽01240분025명0348분0417m0538살0638.4kg070.4kg0836분0920, 22, 91014, 27, 22, 20, 9; 36, 361130012(cid:9066)2월과3월을비교하면3월의7000원에서500원을6000원에주면민정이가두달동안저금한평균금액은6500원이됩니다. 4월과5월을비교하면4월의7000원에서500원을6000원에주면민정이가두달동안저금한평균금액은6500원이됩니다. 따라서민정이가네달동안저금한평균금액은6500원입니다.13(cid:9066)[방법1]기준수를19로정하고합이38이되도록실내온도를2개씩묶으면(22, 16), (24, 14)입니다.따라서실내온도는평균19˘C입니다.13(cid:100) [방법2](평균실내온도)=13(cid:100) [방법2](평균실내온도)=;;ª5∞;;=19(˘C)1410명15느린편입니다.16‘높은’에◯표173명182266마리, 2638마리19(cid:9066)2013년보다2015년의호수의물고기수가늘어났으므로호수가깨끗해져서물고기수가늘어났다고예상할수있습니다.20민진이네모둠, 1권21(위에서부터) 92, 11.5；84, 1222할수없습니다.23재민이네자동차2420800개259600026143.6cm2736.2kg287022992점30⑴17살⑵12살⑶12살3187점22+24+19+14+165단계⑴174쪽176쪽178쪽01㉡02㉢03㉠04㉢05(cid:9066)해는동쪽에서떠서서쪽으로집니다.따라서해는동쪽에서뜰사건이일어날가능성은확실합니다.06(위에서부터) ㉢, ㉡, ㉣, ㉠07080910011㉡12113;2!;14;4!;15;4!;16⑴반⑵;2!;17;4#;183500kg19라마을, 가마을2017000kg21(왼쪽에서부터) 15；27, 20, 27；40, 23, 122228만마리23(cid:9066)그림그래프는표에비해도별소의마릿수를한눈에비교하기쉽습니다.2416명2525266570000, 8270000278, 2; 7, 4, 3;278, 2; 7, 4, 3;28라지역, 8270000명29꺾은선그래프30막대31꺾은선3232(cid:9066)33장미, 무궁화348명0510좋아하는꽃별학생수101001관광객수110123456학생수412143104121431010412143단계⑵172쪽01(cid:9066)02(cid:9066)03[(cid:100)] [(cid:100)] [◯]04105220006다07그림그래프08나09가10다단계⑵◯◯ 15초등쎈수학5-2정답(001~013) 2015.6.2 7:35 PM 페이지011 SinsagoHitec 012쎈수학5－201101명024일0323, 28, 190428, 19, 23; 40, 40, 400528.2cm06㉯학교079508㉠09㉢10㉡11㉠1213;4!;14;2!;15가지역, 다지역161410t17꺾은선그래프1818(cid:9066)051056789수출액41214310186쪽단원마무리1회188쪽단원마무리2회0191점02800원03(cid:9066)[방법1]기준수를33으로정하고합이66이되도록기록을2개씩묶으면(25, 41), (39, 27)입니다.따라서공던지기기록의평균은33m입니다.07(cid:9066)[방법2](평균기록)=07(cid:9066)[방법2](평균기록)==33(m)165525+39+33+27+4150110.5022.9m단계184쪽182쪽180쪽3510000, 100362537(cid:9066)[장점]호박생산량을나타낸그림그래프는표에비해생산량의많고적음을쉽게비교할수있습니다.37(cid:9066)[단점]표에비해도별호박생산량을한눈에바로알수없습니다.38꺾은선그래프39(cid:9066)39(cid:9066)4029일4111cm보다클것입니다.42(cid:9066)우리넷의평균키가150cm이지만계곡의평균깊이인145cm보다더작은사람이있을수있으니건너가면안돼!4344;4!;45가, 다, 나4632명471720kg4818개4912개50㉢, ㉠, ㉡5151525153⑴(cid:9066)청학마을의배생산량이가장많습니다. ⑵(cid:9066)세마마을이중앙마을보다배생산량이324-216=108(t) 더많습니다.중앙 세마 청학 100`t10`t1`t배생산량02040ABC강아지의수051815222910cm양파싹의키10000kg100kg1000`kg10a당호박생산량03⑴85점⑵89점, 356점⑶(cid:9066)88, 92, 86, 9004;4!;05050612개07어머니: 39살, 오빠: 17살, 아버지: 43살081090910(cid:9066)09(cid:9066)010203456방문객수10010자동차보유수10`t1`t사과생산량 15초등쎈수학5-2정답(001~013) 2015.6.8 3:49 PM 페이지012 SinsagoHitec 빠른정답빠른정답찾기0131쪽01④028, 8；128, 13.12803 <04 14, 14；364, 3.64059.1m06⑴㉠(cid:100)⑵㉡076.5608④090.82241079.46cm¤11㉯, ㉱12변ㅂㄹ1351˘14(cid:9066)15㉢, ㉣16⑤1718cm18⑤1972, 52050æ3cm2cm3쪽017, 1, 702;1∞3;03;1£4;, ;4¡2;04 <051;6!;cm065207;1¢1;L08;2!1);09;4!9#;kg1017;1!6#;km114, 0.7123.1cm136.6m14152.34cm¤160.4817㉢18나19820516.327<“44.2442.212.2212.1212.1412.1412.105쪽01⑴500⑵30220a03900m04 3개0543.2km¤06300ha079km¤087000000, 70000, 700, 709⑴900⑵8105t11호준이네모둠, 1번122점131, 014151516가마을17260kg18꺾은선그래프19209æ보다낮을1920것입니다.{æ`C}101520502468온도 시간 (분 후)물의온도100`kg10`kg보리판매량141-21-43-03쪽01;5@0#0(;0259.45L03105˘04 36cm¤0512061;1¡8;m070.34m083.06km09④, ⑤101.16117종류12700m130.92ha14㉠1531.5t1637번1737.5kg18㉠19;4#;20쌀생산량1`kg5`kg1`kg1쪽014개023;5!0&;032.98m04 6개052.5206100배0727.47cm¤0854cm0940˘104개116cm1264˘1380cm¤14;6•5;159시간1610cm174；;1!5$;18;8#;kg193;6%;km20㉠04민경이네집0591점064840kg07④08(cid:9066)숟가락과포크중에서숟가락을고를가능성은반반입니다.09㉢10011;2!;1224000명138000명14(cid:9066)막대그래프1516(cid:9066)165반, 3반, 1반, 4반, 2반051012345참가한학생수 15초등쎈수학5-2정답(001~013) 2015.6.8 3:49 PM 페이지013 SinsagoHitec 014쎈수학5－21소수의곱셈✽A단계기본다잡기⑴정답은‘정답002쪽’에있습니다.010쪽~017쪽01수직선한칸의크기는;10¡00;또는0.001입니다.0에서오른쪽으로5칸간수→;10∞00;=0.005(cid:8951);10∞00;, 0.005；0.00502;1∞0¶0;: 모눈한칸의크기는;10!0;이므로(cid:100)(cid:100)(cid:100)57칸을색칠합니다.→색칠한칸수가같으므로;1∞0¶0;=0.57(cid:8951)(cid:9066), 0.5703전체를똑같이10으로나눈것중의1은;1¡0;입니다.담긴물의양은전체를똑같이10으로나눈것중의7이므로분수로나타내면;1¶0;, 소수로나타내면0.7입니다.(cid:8951);1¶0;,0.704;5$;: 전체를똑같이5로나눈것중의4에표시합니다.0.8: 전체를똑같이10으로나눈것중의8에표시합니다.(cid:8951)05표시한부분이같으므로;5$;=0.8(cid:8951)=154-010.8006젤리판한칸의크기는;10!0;또는0.01입니다.남아있는젤리는100으로나눈것중의39이므로분수로나타내면;1£0ª0;, 소수로나타내면0.39입니다.(cid:8951);1£0ª0;, 0.3907➊7과7;10!0;사이를똑같이10으로나누었으므로눈금한칸의크기는;10¡00;또는0.001입니다.㉠`은7에서눈금7칸간곳이므로㉠`에알맞은분수는7;10¶00;이고, 소수는7.007입니다.➋두수7;10¶00;과7.007은크기가같습니다.08;5#;==;1§0;=0.6(cid:8951)0.6096;2!;=6+;2!;=6+=6+;1∞0;=6.5(cid:8951)6.5분모가2, 5인분수는분모를10으로고친후소수한자리수로나타냅니다.10➊2;5!;=2+;5!;=2+=2+;1™0;=2.2▶3점➋따라서원준이네집에서공원까지의거리를소수로나타내면2.2km입니다.▶2점11➊진분수는분자가분모보다작은분수이므로분모가5인가장큰진분수는;5$;입니다.▶3점➋;5$;==;1•0;=0.8▶3점4_25_21_25_21_52_53_25_2틀리는이유│전체가몇인지몰라서비커에담긴물의양이전체의얼마인지구하지못하는경우해결방안│비커에담긴물의양은전체1을똑같이10으로나눈것중의몇칸인지알아봅니다.채점기준➊㉠`에알맞은분수와소수를각각구한경우3점6점➋두수의관계를설명한경우3점채점기준➊거리를소수로나타내는과정을쓴경우3점5점➋거리를소수로나타낸경우2점채점기준➊분모가5인진분수중에서가장큰수를구한경우3점6점➋분모가5인진분수중에서가장큰수를소수로나타낸경우3점▶3점▶3점 15초등쎈수학5-2정답(014~030) 2015.6.2 7:38 PM 페이지014 SinsagoHitec 1. 소수의곱셈015본문|010쪽-013쪽1단원12(정삼각형의둘레)=;5@;_3=;5^;=1;5!;(cm)→1;5!;=1+=1+;1™0;=1.2따라서정삼각형의둘레는1.2cm입니다.(cid:8951)1.2cm13;5£0;===0.06(cid:8951)0.06145;2¶0;=5+;2¶0;=5+=5+=5.35(cid:8951)5.3515분수의분모가10또는100의약수인수는분모를10또는100으로고칠수있습니다.㉠;2!;==㉣;5¡0;==(cid:8951)㉡, ㉢;2!;을분모가100인분수로고치면분자가50이되고이를소수로나타내면0.50=0.5로분모가10인분수로고쳐소수로나타내는것과같게됩니다.따라서가장간단한방법은분모가10인분수로나타내는것입니다.16전체25칸중에서14칸에색칠되어있으므로;2!5$;===0.56(cid:8951);2!5$;, 0.5617➊(남은호두파이의양)=1-;4!;=;4#;▶3점➋;4#;을소수로나타내면;4#;==;1¶0∞0;=0.75따라서남은호두파이는전체의0.75입니다.▶3점18;4!0#;==;1£0™0∞0;=0.325(cid:8951)0.32513_2540_253_254_255610014_425_421001_250_25101_52_5351007_520_561003_250_21_25_2틀리는이유│정삼각형의둘레를분수로답한경우해결방안│정삼각형의둘레를분수로구한후분모가10, 100, 1000……인분수로고쳐서소수로나타냅니다.틀리는이유│;4!;을소수로고쳐서틀리는경우해결방안│전체를1로생각하여1-;4!;로남은부분을먼저구합니다.채점기준➊남은호두파이는전체의얼마인지분수로나타낸경우3점6점➋남은호두파이는전체의얼마인지소수로나타낸경우3점191;1¶2•5;=1+;1¶2•5;=1+1;1¶2•5;=1+;1§0™0¢0;=1.624(cid:8951)1.62420(액자의둘레)=3;2£0ª0;=3+;2£0ª0;=3+(액자의둘레)=3+;1¡0ª0∞0;=3.195(m)(cid:8951)3.195m21;5#;==;1§0;=0.6;2¶0;==;1£0∞0;=0.35;4!0(;==;1¢0¶0∞0;=0.475;5@0!;==;1¢0™0;=0.42따라서분수를소수로나타낼때소수점아래자릿수가가장많은것은;4!0(;입니다.(cid:8951);4!0(;에◯표22➊;1¡0;이3개인수는;1£0;, ;50!0;이7개인수는;50&0;이므로;1£0;+;50&0;=;5!0%0);+;50&0;=;5!0%0&;▶3점➋;5!0%0&;==;1£0¡0¢0;=0.314▶3점238/_=;1™2•5;==;1™0™0¢0;=0.224(kg)(cid:8951)0.224kg24㉠>㉡>㉢일때가장작은진분수는입니다.8>7>5이므로만들수있는가장작은진분수: ;8%;분수를소수로나타내면;8%;==;1§0™0∞0;=0.625(cid:8951)0.6255_1258_125㉢㉠28_8125_87250?1254157_2500_221_250_219_2540_257_520_53_25_239_5200_578_8125_8틀리는이유│가장작은진분수를어떻게만드는지모르는경우해결방안│가장작은진분수는분모에가장큰수를, 분자에가장작은수를놓아만듭니다.채점기준➊;1¡0;이3개, ;50!0;이7개인수를분수로구한경우3점6점➋;1¡0;이3개, ;50!0;이7개인수를소수로나타낸경우3점 15초등쎈수학5-2정답(014~030) 2015.6.2 7:38 PM 페이지015 SinsagoHitec 016쎈수학5－21소수의곱셈25⑴㉠>㉡>㉢일때가장큰대분수는㉠입니다.(cid:100) 5>2>1이므로만들수있는가장큰대분수: 5;2!;⑵분수를소수로나타내면⑵5;2!;=5+=5+;1∞0;=5.5(cid:8951)⑴5, 1, 2(cid:100)⑵5.5263.6=3;1§0;=3+=3;5#;(cid:8951)3, 6, 10；3, 3, 5소수한자리수는분모가10인분수로고친후약분이되면약분하여기약분수로나타냅니다.270.4=;1¢0;==;5@;(cid:8951);1¢0;, ;5@;285.5=5;1∞0;=5+=5;2!;(cid:8951)5;1∞0;(=;1%0%;), 5;2!;(=;;¡2¡;;)29➊기약분수로각각나타내면㉠0.6=;1§0;=;5#;㉡4.5=4;1∞0;=4;2!;㉢10.2=10;1™0;=10;5!;▶3점➋따라서기약분수로나타낼때분모가다른하나는㉡㉡`입니다.▶2점30(우유의양)=1.2=1;1™0;=1;5!;(L)(물의양)=2.5=2;1∞0;=2;2!;(L)(cid:8951)1;5!;L, 2;2!;L31➊0.1이8개인수: 0.8▶3점➋0.8=;1•0;=;5$;따라서기약분수로나타내면;5$;입니다.▶3점5÷510÷54÷210÷26÷210÷21_52_5㉢㉡320.96=;1ª0§0;==;2@5$;(cid:8951);1ª0§0;, ;2@5$;333.45=3;1¢0∞0;=3+=3;2ª0;(cid:8951)3;1¢0∞0;(=;1#0$0%;), 3;2ª0;(=;2^0(;)기약분수로나타내려면분모와분자를그들의최대공약수로나눕니다.34㉠0.44=;1¢0¢0;==;2!5!;㉡0.35=;1£0∞0;==;2¶0;㉢2.25=2;1™0∞0;=2+=2;4!;따라서기약분수로나타낼때분모가20이되는소수는㉡입니다.(cid:8951)㉡35(선수의기록)=9.92=9;1ª0™0;(선수의기록)=9+=9;2@5#;(초)(cid:8951)9;2@5#;초36(4.6보다0.05큰수)=4.6+0.05=4.654.65=;1$0^0%;=4;1§0∞0;=4+=4;2!0#;(cid:8951);1$0^0%;, 4;1§0∞0;, 4;2!0#;(=;2(0#;)37➊4.02를기약분수로나타내면4.02=4;10@0;=4+=4;5¡0;▶4점➋따라서조건을모두만족하는분수는4;5¡0;입니다.▶2점380.296=;1™0ª0§0;==;1£2¶5;(cid:8951);1™0ª0§0;, ;1£2¶5;395.375=5;1£0¶0∞0;=5+=5;8#;(cid:8951)5;1£0¶0∞0;(=;1%0#0&0%;), 5;8#;(=;;¢8£;;)375÷1251000÷125296÷81000÷82÷2100÷265÷5100÷592÷4100÷425÷25100÷2535÷5100÷544÷4100÷445÷5100÷596÷4100÷4채점기준➊기약분수로각각나타낸경우3점5점➋기약분수로나타낼때분모가다른하나를찾아기호를쓴경우2점채점기준➊0.1이8개인수를소수로나타낸경우3점6점➋0.1이8개인수를기약분수로나타낸경우3점틀리는이유│4.02를분모가100인분수로만나타낸경우해결방안│4.02를분모가100인분수로나타낸후기약분수로나타냅니다.채점기준➊4.02를기약분수로나타내는과정을쓴경우4점6점➋조건을모두만족하는분수를구한경우2점 15초등쎈수학5-2정답(014~030) 2015.6.2 7:38 PM 페이지016 SinsagoHitec 1. 소수의곱셈017본문|013쪽-016쪽1단원400과0.01사이를똑같이10으로나누었으므로눈금한칸의크기는0.001입니다.㉠` 0.002=;10™00;==;50!0;㉡` 0.008=;10•00;==;12!5;(cid:8951);50!0;, ;12!5;411.345=1;1£0¢0∞0;=1;2§0ª0;(km)(cid:8951)1;2§0ª0;km42➊0.625=;1§0™0∞0;=;8%;0.008=;10•00;=;12!5; 3.108=3;1¡0º0•0;=3;2™5¶0;0.006=;10§00;=;50#0;▶4점➋따라서소수를기약분수로나타낼때분모가가장작은소수는0.625입니다.▶2점430.218=;1™0¡0•0;==;5!0)0(;;5!0)0(;는;50!0;이109개인수입니다.(cid:8951)109개44㉠>㉡>㉢>㉣>㉤일때가장큰소수한자리수는㉠.㉡입니다.8>5>3>2>0이므로만들수있는가장큰소수한자리수: 8.5→8.5=8;1∞0;=8+=8;2!;(cid:8951)8;2!;45㉠>㉡>㉢일때자연수부분이0인가장큰소수세자리수는0.㉠`㉡`㉢입니다.7>5>4이므로만들수있는가장큰소수세자리수: 0.754→0.754=;1¶0∞0¢0;==;5#0&0&;(cid:8951)0.754, ;5#0&0&;754÷21000÷25÷510÷5218÷21000÷28÷81000÷82÷21000÷246➊㉠<㉡<㉢일때가장작은소수두자리수는㉠.㉡`㉢입니다.1<3<5이므로만들수있는가장작은소수두자리수: 1.35▶3점➋1.35=1;1£0∞0;=1;2¶0;따라서만들수있는가장작은소수두자리수를기약분수로나타내면1;2¶0;입니다.▶3점470.856=;1•0∞0§0;, ;5#;=;1§0º0º0;이므로;1•0∞0§0;>;1§0º0º0;→0.856>;5#;;5#;=;1§0;=0.6이므로0.856>0.6→0.856>;5#;(cid:8951)>484;2£0;=4;1¡0∞0;=4.15이므로4.15<4.25→4;2£0;<4.254;2£0;=4;1¡0∞0;, 4.25=4;1™0∞0;이므로4;1¡0∞0;<4;1™0∞0;→4;2£0;<4.25(cid:8951)<분수와소수의크기비교[방법1] 분수를소수로나타내어크기를비교합니다.[방법2] 소수를분수로나타내어크기를비교합니다.492.6=2;1§0;=2;1§0º0;, 2;5#0#;=2;1§0§0;이므로2;1§0º0;<2;1§0§0;→2.6<2;5#0#;(cid:8951)2.650➊분수를소수로나타내어크기를비교합니다.5;4¶0;=5;1¡0¶0∞0;=5.175이므로5.175>5.172→5;4¶0;>5.172▶3점➋따라서현우네집에서더먼곳은수산시장입니다.▶2점➊소수를분수로나타내어크기를비교합니다.5;4¶0;=5;1¡0¶0∞0;, 5.172=5;1¡0¶0™0;이므로5;1¡0¶0∞0;>5;1¡0¶0™0;→5;4¶0;>5.172▶3점➋따라서현우네집에서더먼곳은수산시장입니다.▶2점틀리는이유│소수가작을수록분모도작다고생각하여0.006이라고답하는경우해결방안│각수를분모가1000인분수로나타낸후기약분수로나타내어분모의크기를비교합니다.채점기준➊소수를각각기약분수로나타낸경우4점6점➋소수를기약분수로나타낼때분모가가장작은소수를찾아쓴경우2점채점기준➊가장작은소수두자리수를만든경우3점6점➋가장작은소수두자리수를기약분수로나타낸경우3점채점기준➊두수의크기를비교한경우3점5점➋현우네집에서더먼곳을찾아쓴경우2점 15초등쎈수학5-2정답(014~030) 2015.6.8 3:50 PM 페이지017 SinsagoHitec 018쎈수학5－21소수의곱셈018쪽~019쪽015.2와5.4사이를똑같이10으로나누었으므로눈금한칸의크기는;10@0;=0.02입니다.㉮`는5.2에서;10@0;=0.02씩4칸갔으므로분수로5;1™0•0;, 소수로5.28입니다.(cid:8951)5;1™0•0;, 5.2851;1!0)0(;=;1!0)0(0);, 1.003=;1!0)0)0#;, ;8(;=;1!0!0@0%;이므로;1!0!0@0%;>;1!0)0(0);>;1!0)0)0#;→;8(;>;1!0)0(;>1.003;1!0)0(;=1.09, ;8(;=;1!0!0@0%;=1.125이므로1.125>1.09>1.003→;8(;>;1!0)0(;>1.003(cid:8951);8(;52분수를소수로나타내어크기를비교하면;4#;=;1¶0∞0;=0.75, ;5$;=;1•0;=0.8이므로0.75<0.78<0.8<0.813→;4#;<0.78<;5$;<0.813(cid:8951);4#;, 0.78, ;5$;, 0.813530.65=;1§0∞0;, ;2!5^;=;1§0¢0;이므로;1§0∞0;>;1§0¢0;→0.65>;2!5^;따라서양이더적은것은주스입니다.(cid:8951)주스54➊(진아의몸무게)=35;2!0#;=35+➊(진아의몸무게)=35+;1§0∞0;=35.65(kg)35.7>35.65→35.7>35;2!0#;이므로은우의몸무게가더무겁습니다.▶3점➋따라서시소는은우가탄쪽으로내려갑니다.▶2점➊(은우의몸무게)=35.7=35;1¶0;➊(은우의몸무게)=35;1¶0º0;(kg)35;2!0#;=35+=35+;1§0∞0;=35;1§0∞0;(kg)35;1¶0º0;>35;1§0∞0;→35.7>35;2!0#;이므로은우의몸무게가더무겁습니다.▶3점➋따라서시소는은우가탄쪽으로내려갑니다.▶2점13_520_513_520_5틀리는이유│35.7과35;2!0#;에서자연수부분이같으므로소수점아래7과분자13의크기를비교해35;2!0#;이더크다고생각하여틀리는경우해결방안│분수를소수로나타내거나소수를분수로나타내어크기를비교합니다.55(도서관에서우체국까지의거리)=1;4£0;=1;10&0%0;=1.075(km)(도서관에서문구점까지의거리)=;4%;=1;4!;=1;1™0∞0;=1.25(km)1.075<1.125<1.25→1;4£0;<1.125<;4%;따라서도서관에서가까운곳부터차례로쓰면우체국, 영화관, 문구점입니다.(cid:8951)우체국, 영화관, 문구점56;5@;=0.4이므로0.(cid:8641)6<0.4→(cid:8641)<4따라서(cid:8641)안에들어갈수있는숫자는0, 1, 2, 3입니다.(cid:8951)0, 1, 2, 3분수를소수로나타내어소수의크기를비교할때에는자연수부분을먼저비교하고자연수부분이같으면소수첫째자리, 소수둘째자리숫자를차례로비교합니다.57➊2;8&;과2.88을분모가1000인분수로나타내면2;8&;=2;1•0¶0∞0;, 2.88=2;1•0•0º0;▶2점➋2;1•0¶0∞0;<(cid:8641)<2;1•0•0º0;이므로(cid:8641)` 안에들어갈수있는분수는자연수부분이2이고분모가1000일때분자는876부터879까지로모두4개입니다.▶2점➌따라서(cid:8641)안에들어갈수있는분모가1000인분수는모두4개입니다.▶2점채점기준➊2;8&;, 2.88을분모가1000인분수로각각나타낸경우2점6점➋(cid:8641)안에들어갈수있는분모가1000인분수의분자가될수있는수는모두몇개인지구한경우2점➌(cid:8641)안에들어갈수있는분모가1000인분수는모두몇개인지구한경우2점먼저수직선에서눈금한칸의크기를구합니다.전략틀리는이유│두분수가가분수임을생각하지않고소수만1보다크다고생각하여가장큰수가1.003이라고생각하여틀리는경우해결방안│분수를소수로나타내거나소수를분수로나타내어크기를비교하여가장큰수를찾습니다.채점기준➊분수와소수의크기를비교하여몸무게가더무거운사람을찾은경우3점5점➋시소는누가탄쪽으로내려가는지구한경우2점 15초등쎈수학5-2정답(014~030) 2015.6.2 7:38 PM 페이지018 SinsagoHitec 1. 소수의곱셈019본문|016쪽-019쪽1단원02⑴분수를소수세자리수로나타내려면분모가1000인분수로만들수있어야합니다.숫자카드로만들수있는세자리수중에서소수세자리수로나타낼수있는분수의분모는125입니다.⑵분자가될수있는수는분모에사용한숫자1, 2, 5를제외한나머지숫자3, 4, 6입니다. 따라서만들수있는분수는분모가125, 분자가각각3, 4, 6이므로`;12#5;, ;12$5;, ;12^5;입니다.→;12#5;, ;12$5;, ;12^5; 중에서가장큰분수: ;12^5;⑶;12^5;을소수로나타내면(cid:100) ;12^5;==;10$0*0;=0.048(cid:8951)⑴125 ⑵;12^5;⑶0.04803➊0.064=;10^0$0;=;12*5;이므로기약분수의분자는8입니다.➋0.12=;1¡0™0;=;2£5;이므로기약분수의분모는25입니다.➌→기약분수: ;2•5;;2•5;을소수로나타내면;2•5;==;1£0™0;=0.32(cid:8951)0.32040.125를기약분수로나타내면0.125=;1¡0™0∞0;=;8!;;8!;은분모와분자의최대공약수12로약분하여나타낸기약분수이므로약분하기전의분수를구합니다.;8!;==;9!6@;(cid:8951);9!6@;1_128_128_425_46_8125_8➊기약분수의분자구하기→➋기약분수의분모구하기→➌기약분수를소수로나타내기푸는순서먼저소수를기약분수로나타낸후기약분수로나타내기전의분수를구합니다.전략05벽돌의높이를분모가1000인분수로나타내면;8%;=;1§0™0∞0;, 0.608=;1§0º0•0;, ;2!0#;=;1§0∞0º0;, ;5@0^0(;=;1∞0£0•0;;1§0∞0º0;>;1§0™0∞0;>;1§0º0•0;>;1∞0£0•0;이므로가장높게쌓으려면높이가;1§0∞0º0;m(=;2!0#;m),;1§0™0∞0;m(=;8%;m)인두벽돌을쌓아야합니다.(가장높게쌓을수있는높이) =;1§0∞0º0;+;1§0™0∞0;=;1!0@0&0%;=1;1™0¶0∞0;=1;4!0!;(m)벽돌의높이를소수로나타내면;8%;=;1§0™0∞0;=0.625, ;2!0#;=;1§0∞0;=0.65,;5@0^0(;=;1∞0£0•0;=0.5380.65>0.625>0.608>0.538이므로가장높게쌓으려면높이가0.65m(=;2!0#;m), 0.625m(=;8%;m)인두벽돌을쌓아야합니다.(가장높게쌓을수있는높이)=0.65+0.625=1.275=1;1™0¶0∞0;=1;4!0!;(m)(cid:8951)1;4!0!;m06➊;1¡0¶0;=0.17이므로(공원입구`~`분수대`~`팔각정까지의거리)=0.17+0.137=0.307(km)▶4점➋0.307=;1£0º0¶0;이므로거리를분수로나타내면;1£0º0¶0;km,소수로나타내면0.307km입니다.▶3점높이가다음과같은벽돌이한개씩있습니다. 이중에서2개를골라높게쌓으려고합니다. 가장높게쌓을수있는높이는몇m인지기약분수로나타내시오.;8%;m 0.608m ;2!0#;m ;5@0^0(;m(가장높은벽돌의높이)+(두번째로높은벽돌의높이)[크기를비교하는방법][방법1] 분수와소수를분모가같은분수로나타내어비교하기[방법2] 분수를소수로나타내어비교하기▶▶ 15초등쎈수학5-2정답(014~030) 2015.6.8 3:50 PM 페이지019 SinsagoHitec 020쎈수학5－21소수의곱셈➊;1¡0¶0;=;1¡0¶0º0;, 0.137=;1¡¡0£0¶0;이므로(공원입구`~`분수대`~`팔각정까지의거리)=;1¡0¶0º0;+;1¡¡0£0¶0;=;1£0º0¶0;(km)▶4점➋;1£0º0¶0;=0.307이므로거리를분수로나타내면;1£0º0¶0;km,소수로나타내면0.307km입니다.▶3점07➊(진우의키)=165?_➊(진우의키)==148;2!;(cm)▶2점➋(동생의키)=148;2!;_;6%;=_(동생의키)==123;4#;(cm)▶3점➌123;4#;=123;1¶0∞0;=123.75이므로동생의키를소수로나타내면123.75cm입니다.▶3점08➊(떡을만드는데사용한쌀의무게)=11_;4#;=:£4£:=8;4!;=8;1™0∞0;=8.25(kg)▶4점➋(남아있는쌀의무게)=(처음에있던쌀의무게)=-(떡을만드는데사용한쌀의무게)=-(뻥튀기를만드는데사용한쌀의무게)=11-8.25-1.7=2.75-1.7=1.05(kg)▶4점495456/299297/22972910/233(cid:8778)`의→(cid:8778)_▲(cid:8774)▲(cid:8774)전략채점기준➊떡을만드는데사용한쌀의무게를구한경우4점8점➋남아있는쌀은몇kg인지소수로나타낸경우4점09➊홀수번째의규칙은;10!0;씩커지는것이고,짝수번째의규칙은0.03씩작아지는것입니다.▶3점➋8번째수와10번째수는짝수번째이므로(8번째수)=(6번째수)-0.03=0.91-0.03=0.88(10번째수)=(8번째수)-0.03=0.88-0.03=0.85▶2점➌0.85를기약분수로나타내면0.85=;1•0∞0;=;2!0&;따라서10번째에올수를기약분수로나타내면;2!0&;입니다.▶3점10➊가장큰대분수: 5;5$;둘째로큰대분수: 5;4#;가장큰소수두자리수: 5.54▶3점➋5;4#;=5;1¶0∞0;=5.75이므로5;4#;>5.54▶3점➌따라서만든두수중에서더큰수는5;4#;입니다.▶2점11➊주어진수를각각소수로나타내면㉠3;2!0!;=3;1∞0∞0;=3.55㉡0.001이3549개인수: 3.549㉢6의;5#;→6_;5#;=;;¡5•;;=;1#0^;=3.6→3.6>3.55>3.549➋따라서큰수부터차례로기호를쓰면㉢㉢,, ㉠㉠,, ㉡㉡``입니다.채점기준➊둘째로큰대분수와가장큰소수두자리수를각각만든경우3점8점➋만든두수의크기를비교한경우3점➌만든두수중에서더큰수를구한경우2점채점기준➊규칙을찾은경우3점8점➋10번째에올수를소수로구한경우2점➌10번째에올수를기약분수로나타낸경우3점채점기준➊분수를소수로또는소수를분수로나타내어공원입구에서분수대를거쳐팔각정까지의거리를구한경우4점7점➋이거리를분수와소수로각각나타낸경우3점채점기준➊진우의키는몇cm인지분수로구한경우2점8점➋동생의키는몇cm인지분수로구한경우3점➌동생의키는몇cm인지소수로나타낸경우3점분수와소수를규칙에따라늘어놓았을때10번째에올수를기약분수로나타내려고합니다. 풀이과정을쓰고, 답을구하시오.;10!0;, 0.97, ;10@0;, 0.94, ;10#0;, 0.91yy짝수번째의규칙으로구합니다.▼+;10!0;▲+;10!0;▲-0.03▲-0.03▲▶5점▶2점(cid:8778)`는▲의(cid:8774)배→(cid:8778)=▲_(cid:8774)전략채점기준➊크기를비교한경우5점7점➋큰수부터차례로기호를쓴경우2점 15초등쎈수학5-2정답(014~030) 2015.6.8 3:50 PM 페이지020 SinsagoHitec 1. 소수의곱셈021본문|019쪽-026쪽1단원✽A단계기본다잡기⑵정답은‘정답002쪽’에있습니다.024쪽~035쪽010.8씩3번색칠하였으므로2.4가됩니다.(cid:8951)0.8, 2.4020.3을;1£0;으로고쳐서분수의곱셈을한후분수를소수로나타냅니다.(cid:8951);1£0;_9==;1@0&;=2.7소수한자리수는분모가10인분수로고칩니다.030.14를;1¡0¢0;로고쳐서분수의곱셈을한후분수를소수로나타냅니다.(cid:8951);1¡0¢0;_8==;1!0!0@;=1.1204050607080.7_4=2.8, 0.98_4=3.92(cid:8951)2.8, 3.9209➊㉠0.2_8=1.6㉡0.6_3=1.8㉢0.5_6=3.0=3㉣0.8_5=4.0=4➋1.6<1.8<3<4이므로곱이작은것부터차례로기호를쓰면㉠㉠, ㉡㉡, ㉢㉢, ㉣㉣입니다.101.3씩3번을이어서색칠하면3.9가됩니다.(cid:8951)(cid:9066)；3.9(cid:8951)6.76_0.52_0.13_1.56_5.2_6.76(cid:8951)4.9_0.7_0.7_4.9(cid:8951)2.16_0.24_0.09_2.16(cid:8951)1.5_0.5_0.3_1.514_81003_91013200.51.52.50.80.80.811㉠1.5_4=;1!0%;_4==;1^0);=6㉡1.8_7=;1!0*;_7==:¡1™0§:=12.6(cid:8951)㉢12131415163.58_6=21.48(cid:8951)21.4817➊분수를소수로나타내는과정에서소수점의위치가틀렸습니다.➋따라서바르게계산하면5.26_26=;1%0@0^;_26==18(식빵을9개만드는데필요한밀가루의양)=(식빵을한개만드는데필요한밀가루의양)_(식빵수)=0.5_9=4.5(kg)(cid:8951)4.5kg192주일은14일입니다.(2주일동안마신물의양)=(하루에마신물의양)_(날수)=1.3_14=18.2(L)(cid:8951)18.2L20➊2시간30분=2.5시간수요일부터토요일까지는4일이므로(민혁이가수학공부를한시간)=2.5_4=➋10(시간)21소수한자리수를분모가10인분수로고쳐서분수의곱셈을한후분수를소수로나타냅니다.(cid:8951)38_;1¶0;===26.62661038_71013676100526_26100(cid:8951)27.36_3.42_0.0827.36(cid:8951)37.8_6.3_0.637.8(cid:8951)6.95_1.39_0.05_6.95(cid:8951)20.8_5.2_0.420.818_71015_410틀리는이유│어떤수에3을곱해야하는지모르는경우해결방안│그림을보고얼마씩3번이늘어났는지찾습니다.채점기준➊곱을각각구한경우3점5점➋곱이작은것부터차례로기호를쓴경우2점▶2점▶3점채점기준➊틀린부분을찾아이유를설명한경우3점5점➋바르게계산한경우2점▶3점▶2점틀리는이유│2시간30분을2.3시간이라고생각해서틀리는경우해결방안│30분=;6#0);시간=;2!;시간=0.5시간이므로2시간30분은2.5시간입니다.채점기준➊민혁이가수학공부를한시간은모두몇시간인지구하는과정을쓴경우4점6점➋민혁이가수학공부를한시간은모두몇시간인지구한경우2점=136.76 15초등쎈수학5-2정답(014~030) 2015.6.2 7:38 PM 페이지021 SinsagoHitec 022쎈수학5－21소수의곱셈35㉠9_1.9=17.1㉡4_1.25=5.00=5㉢13_2.76=35.88㉣5_5.3=26.5(cid:8951)㉡36➊계산결과가가보다큰수가되려면가에1보다큰수를곱해야합니다.곱하는수중1보다큰수는1.03, 10.2입니다.▶4점➋따라서계산결과가가보다큰것은㉡㉡, ㉢㉢`입니다.▶2점37(그림자의길이)=(막대의길이)_1.6=3_1.6=4.8(m)(cid:8951)4.8m38➊(아버지의몸무게)=(정인이의몸무게)_2.5=32_2.5▶3점➋=80(kg)▶2점39(종현이가마신음료수의양)=3_0.35=1.05(L)1.05<1.5이므로기성이가1.5-1.05=0.45(L) 더많이마셨습니다.(cid:8951)기성, 0.45L40곱의소수점의위치는곱해지는소수의소수점의위치와같으므로0.75_9=6.75(cid:8951)0.75_9=6.754128_4=112→28_0.04=1.12(cid:8951)1.1242⑴49_2.5=122.5→㉡⑵49_0.25=12.25→㉢⑶49_0.025=1.225→㉠(cid:8951)⑴㉡⑵㉢⑶㉠채점기준➊아버지의몸무게를구하는과정을쓴경우3점5점➋아버지의몸무게를구한경우2점;10!0;배▲;10!0;배▼222324252682_0.25=20.50=20.5(cid:8951)20.5소수점아래끝자리의0은생략하여나타냅니다.27①13_0.4=5.2②6_0.8=4.8③17_0.34=5.78④12_0.55=6.60=6.6⑤26_0.23=5.98→6.6>5.98>5.78>5.2>4.8(cid:8951)④28➊(직사각형의넓이)=3_0.9▶3점➋=2.7(m¤)▶2점298_4.6=8_;1$0^;==:£1§0•:=36.8→㉠=46, ㉡=36.8(cid:8951)④30313233345_3.5=17.5, 12_4.26=51.12(cid:8951)17.5, 51.12(cid:8951)21.84_0.13_1.68_1.04007.8013021.84(cid:8951)23.8_0.7_3.4_2.8021023.8(cid:8951)45.88_0.37_1.24_1.48007.4037045.88(cid:8951)54→54_1.2_4.5_6.0048054.08_4610(cid:8951)9.12_0.38_0.241.527.609.12(cid:8951)7.2_0.8_0.9_7.2(cid:8951)37.12_0.64_0.58_5.12032.0037.12(cid:8951)31.8_5.3_0.631.8틀리는이유│직사각형의넓이를구하는방법을몰라틀리는경우해결방안│(직사각형의넓이)=(가로)_(세로)로구합니다.채점기준➊직사각형의넓이를구하는과정을쓴경우3점5점➋직사각형의넓이를구한경우2점채점기준➊계산결과가가보다큰것을모두찾는과정을쓴경우4점6점➋계산결과가가보다큰것을모두찾아기호를쓴경우2점틀리는이유│가에어떤수를곱하면항상가보다크다고생각하는경우해결방안│가보다큰수가되려면가에1보다큰수를곱해야합니다.소수두자리수▲▲ 15초등쎈수학5-2정답(014~030) 2015.6.2 7:38 PM 페이지022 SinsagoHitec 1. 소수의곱셈023본문|026쪽-030쪽1단원4327_0.054=1.458, 0.027_540=14.580=14.58→1.458<14.58(cid:8951)0.027_540에◯표44➊165_3=495, 1.65_3=4.95▶2점➋따라서곱해지는수가;10!0;배가되면곱의결과도;10!0;배가됩니다.▶4점450.28_10=2.8, 0.28_100=28, 0.28_1000=280(cid:8951)2.8, 28, 280463.29_10=32.9, 32.9_100=3290(cid:8951)32.9, 329047㉠100_6.29=629㉡1000_0.629=629㉢10_629=6290따라서계산결과가다른하나는㉢입니다.(cid:8951)㉢48➊(0.5L짜리음료수의양)=0.5_1000=500(L)(1.5L짜리음료수의양)=1.5_100=150(L)▶4점➋(창고에있는음료수의양)=500+150=650(L)▶2점49450_0.1=45, 450_0.01=4.50=4.5,450_0.001=0.450=0.45(cid:8951)45, 4.5, 0.4550384_0.001=0.384(cid:8951)0.38451①5346_0.1=534.6②4504_0.01=45.04③1973_0.001=≥1.973④7500_0.001=7.500=7.5⑤920_0.001=0.920=0.92(cid:8951)③52➊(등유0.1L의값)=(등유1L의값)_0.1=1470_0.1=147(원)▶1점➋(등유0.01L의값)=(등유1L의값)_0.01=1470_0.01=14.7(원)▶2점➌(등유0.001L의값)=(등유1L의값)_0.001=1470_0.001=1.47(원)▶2점53(사용한파란색끈의길이)=9_0.1=0.9(m)(사용한분홍색끈의길이)=12_0.01=0.12(m)0.9>0.12이므로파란색끈을0.9-0.12=0.78(m)더많이사용했습니다.(cid:8951)파란색끈, 0.78m5487에서소수점을왼쪽으로한자리옮겨야8.7이되므로(cid:8641)안에알맞은수는0.1입니다.(cid:8951)0.155580에서소수점을왼쪽으로세자리옮겨야0.580=0.58이되므로(cid:8641)안에알맞은수는0.001입니다.(cid:8951)0.00156➊㉠, ㉡, ㉢은소수점이오른쪽으로두자리옮겨진것이므로100을곱한것입니다.㉣은소수점이오른쪽으로세자리옮겨진것이므로1000을곱한것입니다.▶4점➋따라서(cid:8641)안에알맞은수가다른하나는㉣㉣입니다.▶2점틀리는이유│곱해지는수인165와1.65가다르다는것만써서틀리는경우해결방안│곱해지는수가165에서1.65로;10!0;배가되면곱의결과도;10!0;배가된다는것을알고해결합니다.채점기준➊계산결과를각각구한경우2점6점➋계산결과를비교하여알게된점을쓴경우4점틀리는이유│③, ④, ⑤`를모두답이라고생각하는경우해결방안│④, ⑤`는자연수의끝에0이있으므로곱의소수끝자리에0이생깁니다. 0은생략할수있으므로곱의소수점아래자릿수가달라집니다.채점기준➊등유0.1L의값을구한경우1점5점➋등유0.01L의값을구한경우2점➌등유0.001L의값을구한경우2점채점기준➊0.5L짜리음료수의양과1.5L짜리음료수의양을각각구한경우4점6점➋창고에있는음료수의양은모두몇L인지구한경우2점틀리는이유│㉠, ㉡, ㉢, ㉣`의(cid:8641)안에알맞은수가모두같다고생각하는경우해결방안│곱해지는수(곱하는수)와계산결과의소수점의위치를보고(cid:8641)안에알맞은수를구합니다.채점기준➊(cid:8641)안에알맞은수가다른하나를찾는과정을쓴경우4점6점➋(cid:8641)안에알맞은수가다른하나를찾아기호를쓴경우2점 15초등쎈수학5-2정답(014~030) 2015.6.8 3:51 PM 페이지023 SinsagoHitec 024쎈수학5－21소수의곱셈6869(cid:8951)4.76(cid:8951)15.07870711.7_1.6=2.72, 9.3_0.9=8.37, 2.3_1.2=2.76→8.37>2.76>2.72(cid:8951)3, 1, 272㉠2.46_4.3=10.578(소수세자리수)㉡85.7_0.24=20.568(소수세자리수)㉢0.35_2.24=0.7840=0.784(소수세자리수)㉣≥5.72_0.5=2.860≥=2.86(≥소수두자리수)(cid:8951)㉣73➊곱해지는수가소수두자리수이고곱하는수가소수한자리수이므로계산결과는소수세자리수가되어야하는데소수점을잘못찍어틀렸습니다.▶3점741000m=1km이므로900m=0.9km(900m를달리는데필요한휘발유의양)=(1km를달리는데필요한휘발유의양)_(달리는거리)=0.06_0.9=0.054(L)(cid:8951)0.054L▶3점5.4.9.2_ .5.21.0.9.8.42.7.4.6.0.82.8.5.5.8.4➋(cid:8951)31.8976←소수두자리수←소수두자리수←소수네자리수_5.12_0.6.2315.361.02.4030.72.0031.89.76←소수두자리수←소수한자리수←소수세자리수_3.5.9_ 4.27.1.81.4.3.6.81.5.0.7.8←소수한자리수←소수한자리수←소수두자리수_1.7_2.81.3.63.4.64.7.6틀리는이유│곱하는두소수의소수점아래자릿수로생각하여㉢`이라고답하는경우해결방안│두소수의끝자리숫자가각각2의배수와5이면곱의끝자리숫자가0이므로소수점아래자릿수가달라집니다.575.687_10=56.87이므로㉠=56.870.8392_㉡=839.2에서839.2는0.8392에서소수점을오른쪽으로세자리옮겨야하므로㉡=1000㉠_㉡=56.87_1000=56870(cid:8951)5687058㉮분수를소수로나타내는과정에서소수점의위치가틀렸습니다.바르게계산하면(cid:100) `0.47_0.6=;1¢0¶0;_;1§0;=;1™0•0™0;=0.2825960(cid:8951)0.21(cid:8951)0.05766162(cid:8951)0.32(cid:8951)0.102630.7_0.33=0.231, 0.42_0.06=0.0252,0.7_0.42=0.294, 0.33_0.06=0.0198(cid:8951)(위에서부터) 0.231, 0.0252；0.294, 0.0198640.72_0.5=0.360=0.36,0.7_0.63=0.441→0.36<0.441(cid:8951)<650.08_0.08=0.00640.02_0.31=0.0062, 0.04_0.16=0.0064,0.13_0.05=0.0065(cid:8951)0.04_0.16에◯표66➊0.9>0.6>0.5>0.2이므로가장큰수: 0.9, 가장작은수: 0.2▶3점➋(가장큰수)_(가장작은수)=0.9_0.2=0.18▶3점671.8은;1!0*;, 4.2는;1$0@;로고쳐서분수의곱셈을한후분수를소수로나타냅니다.(cid:8951);1!0*;_;1$0@;=;1&0%0^;=7.56←소수두자리수←소수한자리수←소수세자리수_0.1.7_0.0.60.1.0.2←소수한자리수←소수한자리수←소수두자리수_0.8_0.40.3.2←소수두자리수←소수두자리수←소수네자리수_0.48_0.12964.880.05.76←소수한자리수←소수한자리수←소수두자리수_0.7_0.30.2.1▼소수세자리수채점기준➊가장큰수와가장작은수를각각찾은경우3점6점➋가장큰수와가장작은수의곱을구한경우3점채점기준➊틀린부분을찾아이유를설명한경우3점6점➋바르게계산한경우3점틀리는이유│900m를km로고치지않고0.06_900을계산하여틀리는경우해결방안│900m를km로고치면0.9km입니다.(cid:8951)㉯ 15초등쎈수학5-2정답(014~030) 2015.6.2 7:38 PM 페이지024 SinsagoHitec 1. 소수의곱셈025본문|030쪽-034쪽1단원75밀가루3봉지반은3;2!;봉지=3.5봉지이므로(밀가루전체의무게)=(한봉지에들어있는밀가루의무게)_(봉지수)=2.5_3.5=8.75(kg)(cid:8951)8.75kg76⑴6분=;6§0;시간`=;1¡0;시간`=0.1시간1시간=60분이므로(cid:8778)분=시간⑵(6분동안탄길이)=(한시간동안타는길이)_(양초가탄시간)=0.07_0.1=0.007(m)⑶(타고남은양초의길이)=0.19-0.007=0.183(m)(cid:8951)⑴0.1시간(cid:100)⑵0.007m(cid:100)⑶0.183m773.9_4.7=18.33(cid:8951)18.33781833이1.833이되려면소수점을왼쪽으로세자리옮겨야합니다.39의소수점이왼쪽으로한자리옮겨져3.9가되었으므로47의소수점을왼쪽으로두자리옮겨야합니다.→(cid:8641)=0.47(cid:8951)0.47790.325_27과32.5_0.27에서소수를분수로나타내면325_27의곱의결과에;10¡00;배를한것과같습니다.따라서0.325_27과32.5_0.27의값은8.775로같습니다.80㉠41_8=328→4.1_0.8=3.28㉡41_8=328→0.41_0.08=0.0328㉠은소수두자리수이고㉡은소수네자리수이므로㉠은㉡의100배입니다.(cid:8951)100배81㉡`이가장길기때문에자연수부분이7인두수를비교하여더큰수를찾습니다.7;2!;=7;1∞0º0;, 7.25=7;1™0∞0;이므로7;1∞0º0;>7;1™0∞0;→7;2!;>7.25따라서㉡`은가장긴7;2!;cm입니다.(cid:8951)7;2!;cm(cid:8778)6082(가로)=9.9_3=29.7(cm)(세로)=9.9_2=19.8(cm)(가로)+(세로)=29.7+19.8=49.5(cm)(cid:8951)49.5cm831바트(THB)=33.63원이므로10바트(THB)=33.63_10=336.3(원)100바트(THB)=33.63_100=3363(원)1000바트(THB)=33.63_1000=33630(원)(cid:8951)336.3원, 3363원, 33630원84(산철사의무게)=0.17_0.9=0.153(kg)(산노끈의무게)=0.014_0.5=0.007(kg)(산철사와노끈의무게)=0.153+0.007=0.16(kg)(cid:8951)0.16kg[[85~~92]]서술형평가유형의입니다.85⑴1;5¡0¶0;=1;10#0$0;=1.034(cid:100)；1.034▶2점⑵1.034는0.001이1034개인수입니다.▶2점⑶1034개▶1점86⑴㉠0.5=;1∞0;=;2!;(cid:100) ㉡0.62=;1§0™0;=;5#0!;(cid:100) ㉢0.788=;1¶0•0•0;=;2!5(0&;(cid:100) ; ;2!;, ;5#0!;, ;2!5(0&;⑵㉠;2!;에서분모와분자의차: 2-1=1(cid:100) ㉡;5#0!;에서분모와분자의차: 50-31=19(cid:100) ㉢;2!5(0&;에서분모와분자의차: 250-197=53(cid:100) 따라서분모와분자의차가가장큰소수는㉢입니다.⑶㉢87⑴어떤수를(cid:8641)라하면(cid:8641)÷8=0.64, (cid:8641)=0.64_8=5.12따라서어떤수는5.12입니다.; 5.12▶2점(cid:8641)를구할때에는곱셈과나눗셈의관계를이용합니다.(cid:8641)÷▲=(cid:8778)→(cid:8641)=(cid:8778)_▲또는(cid:8641)=▲_(cid:8778)⑵바르게계산하면5.12_8=40.96▶3점⑶40.96▶1점소수한자리수소수한자리수소수두자리수채점기준0.325_27과32.5_0.27의값이같은이유를설명한경우6점▶2점▶2점▶1점 15초등쎈수학5-2정답(014~030) 2015.6.8 3:51 PM 페이지025 SinsagoHitec 02(은정이의몸무게)=(어머니의몸무게)_0.8+2=50_0.8+2=42(kg)(아버지의몸무게)=(어머니의몸무게)_1.47=50_1.47=73.5(kg)(은정이와아버지의몸무게의차)=73.5-42=31.5(kg)(cid:8951)31.5kg030.25_4=1(m)이므로(쇠파이프1m의무게)=1.7_4=6.8(kg)(쇠파이프100m의무게)=6.8_100=680(kg)(cid:8951)680kg04(늘인정사각형의한변)=32+32_0.4=44.8(cm)(늘인정사각형의넓이)=44.8_44.8=2007.04(cm¤)(처음정사각형의넓이)=32_32=1024(cm¤)(늘어난부분의넓이)=(늘인정사각형의넓이)-(처음정사각형의넓이)=2007.04-1024=983.04(cm¤)(cid:8951)983.04cm¤05⑴2분45초=2;6$0%;분=2;4#;분=2;1¶0∞0;분=2.75분⑵(기차의길이)+(터널의길이)=0.84_2.75=2.31(km)⑶150m=0.15km이므로(터널의길이)=2.31-(기차의길이)=2.31-0.15=2.16(km)(cid:8951)⑴2.75분⑵2.31km⑶2.16km기차가터널을완전히통과한다는것은기차의뒷부분이터널을완전히빠져나온순간을말합니다.026쎈수학5－21소수의곱셈036쪽~037쪽01➊(색테이프16장의길이의합)=14.5_16=232(cm)➋(겹쳐진부분의길이의합)=0.3_15=4.5(cm)➌(이어붙인색테이프의전체길이)=232-4.5=227.5(cm)(cid:8951)227.5cm➊색테이프16장의길이의합구하기→➋겹쳐진부분의길이의합구하기→➌이은색테이프의전체길이구하기푸는순서쇠파이프1m의무게를구한다음쇠파이프100m의무게를구합니다.전략한변이32cm인정사각형이있습니다. 가로와세로를각각0.4배씩더늘인다면늘어난부분의넓이는몇cm¤입니까?늘인각변은(32+32_0.4)cm가됩니다.▼(늘인정사각형의넓이)-(처음정사각형의넓이)▼88⑴(서현이가마신우유의양)=(훈진이가마신우유의양)_1.5=2_1.5=3(L) ; 3L▶2점⑵(은민이가마신우유의양)=(서현이가마신우유의양)_1.1=3_1.1=3.3(L)▶3점⑶3.3L▶1점89⑴56_38=2128이므로56_0.38=21.28, 56_0.038=2.128; 21.28, 2.128▶3점⑵56_0.38-56_0.038=21.28-2.128=19.152▶2점⑶19.152▶1점90⑴어떤소수를(cid:8641)라하면2765.07_(cid:8641)=2.76507; 2765.07_(cid:8641)=2.76507▶2점⑵2765.07에서소수점을왼쪽으로세자리옮겨야2.76507이됩니다. →(cid:8641)=0.001따라서어떤소수는0.001입니다.▶3점⑶0.001▶1점91⑴창문의모양이정사각형이므로(붙인종이의넓이)=(한변)_(한변)=0.8_0.8=0.64(m¤)▶3점⑵0.64m¤▶2점92⑴4.28_1.52=6.5056；6.5056⑵6.5056<(cid:8641)이므로(cid:8641)안에들어갈수있는가장작은자연수는7입니다.⑶7▶2점▶2점▶1점(기차의길이)+(터널의길이)=(1분에달리는거리)_(터널을완전히통과하는데걸리는시간)전략 15초등쎈수학5-2정답(014~030) 2015.6.8 3:51 PM 페이지026 SinsagoHitec 11➊4바퀴반=4;2!;바퀴=4.5바퀴이고2주일은14일이므로(현수가2주일동안달린거리)=(운동장의둘레)_(운동장을달린바퀴수)_(날수)=254.3_4.5_14=16020.9(m)▶4점➋현수와아버지가2주일동안달린거리는같으므로(현수와아버지가2주일동안달린거리의합)=16020.9_2=32041.8(m) →32.0418km▶4점12➊0.8을(cid:8785)번곱했을때소수(cid:8785)째자리숫자는곱의소수점아래끝자리숫자이므로0.8을75번곱했을때소수75째자리숫자는곱의소수점아래끝자리숫자입니다.▶2점➋곱의소수점아래끝자리숫자는8, 4, 2, 6이반복됩니다.▶3점➌75÷4=18y3이므로소수75째자리숫자는8, 4, 2, 6이18번반복되고세번째숫자인2입니다.▶3점06㉠_2의곱의일의자리숫자가8이므로㉠=4또는9364_2=728, 369_2=738→㉠=99_㉡`의곱의일의자리숫자가6이므로9_4=36→㉡=4㉠`은소수둘째자리숫자이므로0.09,㉡`은소수첫째자리숫자이므로0.4를나타냅니다.(cid:8951)0.09, 0.407➊(식용유;4!;의무게)=2.7-2.15➊(식용유;4!;의무게)=0.55(kg)▶2점➋(식용유만의무게)=0.55_4=2.2(kg)▶3점➌(빈병의무게)=2.7-2.2=0.5(kg)▶3점08➊정삼각형끼리맞닿는변이많을수록만든무늬의둘레가짧습니다.맞닿는변이많도록정삼각형모양조각8개를놓으면만든무늬의둘레는정삼각형모양조각의한변의8배입니다.▶4점➋(만든무늬의둘레)=5.4_8=43.2(cm)▶3점09➊5★1.4=5_(5-1.4)_1.4=5_3.6_1.4=18_1.4=➋25.210➊가_30.27_나=≥30.27_(가_나)=≥0.03027소수두자리수소수다섯자리수30.27이0.03027이되려면소수점을왼쪽으로세자리옮겨야하므로(가_나)는0.001입니다.▶5점➋따라서가와나의곱은0.001입니다.▶2점1. 소수의곱셈027본문|034쪽-038쪽1단원채점기준➊식용유;4!;의무게를구한경우2점8점➋식용유만의무게를구한경우3점➌빈병의무게를구한경우3점채점기준➊만든무늬의둘레는정삼각형모양조각의한변의몇배인지구한경우4점7점➋만든무늬의둘레는몇cm인지구한경우3점채점기준➊가와나의곱이얼마인지구하는과정을쓴경우5점7점➋가와나의곱을소수로구한경우2점채점기준➊현수가2주일동안달린거리를구한경우4점8점➋현수와아버지가2주일동안달린거리의합을구한경우4점0.8을75번곱했을때소수75째자리숫자는무엇인지풀이과정을쓰고, 답을구하시오.곱의소수점아래끝자리숫자▼채점기준➊소수75째자리숫자와소수점아래끝자리숫자가같음을쓴경우2점8점➋곱의소수점아래끝자리의반복되는숫자를구한경우3점➌소수75째자리숫자를구한경우3점채점기준➊5★1.4를계산하는과정을쓴경우4점7점➋5★1.4를계산한경우3점곱하는횟수1번2번3번4번5번6번yy끝자리숫자842684yy038쪽~039쪽1회01색칠한부분은1을똑같이100으로나눈것중의37이므로;1£0¶0;=0.37입니다.(cid:8951)37, 100；0.37먼저0.8을여러번곱하여곱의소수점아래끝자리숫자의규칙을찾습니다.전략 15초등쎈수학5-2정답(014~030) 2015.6.2 7:38 PM 페이지027 SinsagoHitec 1148_0.3=14.4, 30_0.4=12, 21_0.7=14.714.7>14.4>12이므로가장큰곱: 21_0.7=14.7가장작은곱: 30_0.4=12→차: 14.7-12=2.7(cid:8951)2.71237_8.6=318.2(cid:8951)318.213(직사각형의넓이)=(가로)_(세로)=3_5.5=16.5(m¤)(cid:8951)16.5m¤14➊(경민이의몸무게)=(어머니의몸무게)_0.8=53_0.8=42.4(kg)▶3점➋(두사람의몸무게의합)=53+42.4=95.4(kg)▶2점150.043_27=1.161, 43_0.27=11.61→1.161<11.61두수를구성하고있는숫자가같으므로≥0.043_27<≥43_0.27소수세자리수소수두자리수(cid:8951)<16⑴28.62_10=286.2→㉢⑵286.2_100=28620→㉠⑶2.862_1000=2862→㉡(cid:8951)⑴㉢⑵㉠⑶㉡1752_0.1=5.2, 5.2_0.01=0.052(cid:8951)5.2, 0.052180.87_0.56=0.4872(cid:8951)0.487219➊어떤수를(cid:8641)라하면(cid:8641)+4.6=7.5이므로(cid:8641)=7.5-4.6=2.9▶3점➋바르게계산하면(cid:8641)_4.6=2.9_4.6=13.34▶2점028쎈수학5－21소수의곱셈028;5$;=8+;5$;=8+=8+;1•0;=8.8(cid:8951)8.8030.248=;1™0¢0•0;==;1£2¡5;(cid:8951);1£2¡5;04➊1.55=1;1∞0∞0;=1;2!0!;▶3점➋따라서선호가가지고있는색테이프의길이를기약분수로나타내면1;2!0!;m입니다.▶2점05➊분수를소수로나타내어크기를비교하면㉡;5$;=0.8(cid:100)㉣;8#;=0.3750.8>0.79>0.67>0.375→;5$;>0.79>0.67>;8#;▶3점➋따라서큰수부터차례로기호를쓰면㉡㉡, ㉢㉢, ㉠㉠, ㉣㉣입니다.▶2점06①, ②, ④, ⑤는2.4로계산결과가같습니다.(cid:8951)③070.7+0.7+0.7+0.7=0.7_4=2.8(cid:8951)4；4, 2.808(cid:8951)17.2809➊한시간=60분이므로한시간은10분의6배입니다.(한시간동안기어간거리)=1.2_6=➋7.2(m)1014_0.7=14_===9.8(cid:8951)7, 7；98, 9.8981014_710710_2.16_6.2817.28248÷81000÷84_25_2채점기준➊경민이의몸무게는몇kg인지구한경우3점5점➋두사람의몸무게의합은몇kg인지구한경우2점채점기준➊어떤수를구한경우3점5점➋바르게계산한값을구한경우2점채점기준➊선호가가지고있는색테이프의길이를기약분수로나타내는과정을쓴경우3점5점➋선호가가지고있는색테이프의길이를기약분수로나타낸경우2점채점기준➊분수와소수의크기를비교한경우3점5점➋큰수부터차례로기호를쓴경우2점채점기준➊한시간동안기어간거리는몇m인지구하는과정을쓴경우3점5점➋한시간동안기어간거리는몇m인지구한경우2점 15초등쎈수학5-2정답(014~030) 2015.6.2 7:38 PM 페이지028 SinsagoHitec 20➊(정사각형모양의종이의넓이)=0.9_0.9=0.81(m¤)(평행사변형모양의종이의넓이)=0.94_0.8=0.752(m¤)▶3점➋0.81>0.752이므로정사각형모양의종이의넓이가더넓습니다.▶2점1. 소수의곱셈029본문|038쪽-041쪽1단원040쪽~041쪽2회012;2!;=2+;2!;=2+=2+;1∞0;=2.5(cid:8951)2;1∞0;(=;1@0%;), 2.5020.54=;1∞0¢0;==;5@0&;→농구공의무게는;5@0&;kg입니다.(cid:8951);5@0&;kg03➊1<5<6이므로만들수있는가장작은소수두자리수: 1.56▶2점➋1.56=1;1∞0§0;=1;2!5$;따라서만들수있는가장작은소수두자리수를기약분수로나타내면1;2!5$;입니다.▶3점04재중: 8;2!0!;=8;1∞0∞0;=8.55(분)혜란: 8;12&5;=8;10%0^0;=8.056(분)8.056<8.55<8.6→8;12&5;<8;2!0!;<8.6따라서학교에빨리도착한사람부터차례로이름을쓰면혜란, 재중, 미소입니다.(cid:8951)혜란, 재중, 미소빨리도착한사람부터쓰는것이므로크기비교를하여시간이적게걸린사람부터차례로씁니다.54÷2100÷21_52_5채점기준➊정사각형과평행사변형모양의종이의넓이를각각구한경우3점5점➋어느종이의넓이가더넓은지구한경우2점050.8을;1•0;로고쳐서분수의곱셈을한후분수를소수로나타냅니다.(cid:8951);1•0;_9==;1&0@;=7.2060.2<0.4<0.7<3<6이므로가장작은수: 0.2가장큰수: 6(가장작은수)_(가장큰수)=0.2_6=1.2(cid:8951)1.2073주일은21일이므로(3주일동안마시는당근주스의양)=1.1_21=23.1(L)(cid:8951)23.1L08➊정육각형은길이가같은변이6개있으므로(정육각형의둘레)=(한변)_6=2.47_6=➋14.82(cm)098_7=56, 8_0.7=5.6곱하는수가;1¡0;배가되면곱의결과도;1¡0;배가됩니다.(cid:8951)56, 5.6；1010➊87_0.23=87_;1™0£0;==:™1º0º0¡:=20.01이므로㉠=2001, ㉡=20.01▶3점➋㉡_100-㉠=20.01_100-2001=2001-2001=0▶2점11계산결과가㉮보다작은수가되려면㉮와곱하는수가1보다작아야합니다. →③㉮_0.68(cid:8951)③12➊1000g=1kg이므로1300g=1.3kg▶1점➋(돼지고기1300g의값)=16500_1.3▶2점➌=21450(원)▶2점87_231008_910채점기준➊가장작은소수두자리수를만든경우2점5점➋가장작은소수두자리수를기약분수로나타낸경우3점채점기준➊정육각형의둘레는몇cm인지구하는과정을쓴경우3점5점➋정육각형의둘레는몇cm인지구한경우2점채점기준➊㉠과㉡`에알맞은수를각각구한경우3점5점➋㉡_100-㉠의값을구한경우2점채점기준➊1300g을kg으로고친경우1점5점➋돼지고기1300g의값을구하는식을세운경우2점➌돼지고기1300g의값을구한경우2점 15초등쎈수학5-2정답(014~030) 2015.6.2 7:38 PM 페이지029 SinsagoHitec 030쎈수학5－21소수의곱셈130.308_100=30.8308_0.1=30.8→30.8=30.8(cid:8951)=14㉠2.33_0.01=0.0233(소수네자리수)㉡16.5_0.01=0.165(소수세자리수)㉢2.56_10=25.6(소수한자리수)㉣0.2356_100=23.56(소수두자리수)(cid:8951)㉠, ㉡, ㉣, ㉢15(통나무1000m의무게)=(통나무1m의무게)_(길이)=5.52_1000=5520(kg)(cid:8951)5520kg16①32.16에서소수점을오른쪽으로두자리옮겨야3216이되므로(cid:8641)=100②146.5에서소수점을왼쪽으로두자리옮겨야1.465가되므로(cid:8641)=0.01③0.327에서소수점을오른쪽으로한자리옮겨야3.27이되므로(cid:8641)=10④0.07에서소수점을왼쪽으로두자리옮겨야0.0007이되므로(cid:8641)=0.01⑤28에서소수점을왼쪽으로세자리옮겨야0.028이되므로(cid:8641)=0.001(cid:8951)①175.4_0.7=3.78 , 6.25_3.96=24.75,5.4_6.25=33.75, 0.7_3.96=2.772(cid:8951)(위에서부터) 3.78, 24.75；33.75, 2.77218➊3시간15분=3;6!0%;시간=3;4!;시간(cid:9233)3시간15분=3;1™0∞0;시간=3.25시간▶2점➋(아버지께서걸은거리)=4.3_3.25=13.975(km)▶3점199350이0.935가되려면소수점을왼쪽으로네자리옮겨야합니다.275의소수점이왼쪽으로두자리옮겨져2.75가되었으므로34의소수점을왼쪽으로두자리옮겨야합니다.→(cid:8641)=0.34(cid:8951)0.3420➊0.7을34번곱했을때소수34째자리숫자는곱의소수점아래끝자리숫자입니다.▶1점➋곱의소수점아래끝자리숫자는7, 9, 3, 1이반복됩니다.▶2점➌34÷4=8` …`2이므로소수34째자리숫자는7, 9, 3, 1이8번반복되고두번째숫자인9입니다.▶2점채점기준➊3시간15분을3.25시간으로고친경우2점5점➋아버지께서걸은거리를구한경우3점채점기준➊소수34째자리숫자와소수점아래끝자리숫자가같음을쓴경우1점5점➋곱의소수점아래끝자리의반복되는숫자를구한경우2점➌소수34째자리숫자를구한경우2점 15초등쎈수학5-2정답(014~030) 2015.6.2 7:38 PM 페이지030 SinsagoHitec 2. 합동과대칭0312합동과대칭2단원✽A단계기본다잡기`⑴정답은‘정답004쪽’에있습니다.01가, 나, 다는모양과크기가같아서포개었을때완전히겹쳐집니다.(cid:8951)라에◯표02모양과크기가같은두도형을모두찾으면가와사, 라와바입니다.(cid:8951)가와사, 라와바03모양은같지만크기가다르므로두도형을포개었을때완전히겹쳐지지않습니다. 따라서두도형은합동이아닙니다.합동인두도형은모양과크기가같습니다.04오른쪽도형의아래쪽변을위쪽변과평행하게만듭니다.05직사각형을점선을따라자른후포개었을때완전히겹쳐지는도형은가와다, 나와라입니다.(cid:8951)가와다, 나와라06정사각형을점선을따라자르면와, 와→2쌍(cid:8951)2쌍07정사각형의한가운데를지나는점선을따라잘라야두도형이합동이됩니다. 정사각형의한가운데를지나는점선: 가, 나, 라, 마(cid:8951)가, 나, 라, 마049쪽~057쪽08➊㉠㉡➊㉢㉣네변의길이가모두같은마름모는두대각선을따라잘랐을때항상잘린도형이모두합동입니다.▶4점➋따라서항상잘린도형이모두합동이되는사각형은㉢㉢``입니다.▶2점[[09~~10]]모눈종이의선을이용하여주어진도형의꼭짓점과똑같은위치에점을찍은다음, 찍은점을연결하여합동인도형을그립니다.09(cid:8951)(cid:9066)10(cid:8951)(cid:9066)11오른쪽도형은왼쪽도형을돌리기한위치에있습니다.각꼭짓점에서모눈몇칸을움직였는지확인하여나머지부분을완성합니다.(cid:8951)12두도형이합동이되도록선을긋는방법은여러가지가있습니다.(cid:9066)(cid:8951)(cid:9066)13모양과크기가같은삼각형4개가(cid:8951)(cid:9066)되도록선을긋는방법은여러가지입니다.채점기준합동이아닌이유를설명한경우5점채점기준두도형이합동이되도록만들려면오른쪽도형을어떻게하면되는지설명한경우5점틀리는이유│합동인도형을모두찾지않고한가지만찾은경우해결방안│합동인도형을한가지찾은후남은도형이합동인지생각해봅니다.틀리는이유│합동인도형은왼쪽도형과똑같은방향으로똑같이그려야한다고생각하여나머지부분을완성하지못하는경우해결방안│합동인도형을그릴때밀기, 뒤집기, 돌리기한것도모양과크기가같은도형입니다.채점기준➊항상잘린도형이모두합동이되는사각형을찾는과정을쓴경우4점6점➋항상잘린도형이모두합동이되는사각형을찾아기호를쓴경우2점 15초등쎈수학5-2정답(031~049) 2015.6.8 3:52 PM 페이지031 SinsagoHitec 032쎈수학5－22합동과대칭22변ㄱㄹ의대응변은변ㅇㅁ이므로(변ㄱㄹ)=(변ㅇㅁ)=4cm변ㄴㄷ의대응변은변ㅅㅂ이므로(변ㄴㄷ)=(변ㅅㅂ)=7cm(cid:8951)(위에서부터) 4, 723변ㅁㅂ의대응변은변ㄷㄴ이므로(변ㅁㅂ)=(변ㄷㄴ)=7cm(cid:8951)7cm24변ㄹㅂ의대응변은변ㄱㄴ이므로(변ㄹㅂ)=(변ㄱㄴ)=5cm(cid:8951)5cm25변ㄱㄷ의대응변은변ㄹㅁ이므로(변ㄱㄷ)=(변ㄹㅁ)=10cm(cid:8951)10cm26변ㄱㄷ의대응변은변ㄷㅁ이므로(변ㄱㄷ)=(변ㄷㅁ)=13cm(cid:8951)13cm27➊변ㅂㄹ의대응변은변ㄷㄱ이므로(변ㅂㄹ)=(변ㄷㄱ)=4+7▶3점➋=11(cm)▶2점28변ㄴㄷ의대응변은변ㅂㅁ이므로(변ㄴㄷ)=(변ㅂㅁ)=6cm변ㄱㄷ의대응변은변ㄹㅁ이므로(변ㄱㄷ)=(변ㄹㅁ)=8cm(삼각형ㄱㄴㄷ의둘레)=10+6+8=24(cm)(cid:8951)24cm삼각형ㄱㄴㄷ과삼각형ㄹㅂㅁ은합동이므로두삼각형의둘레는같습니다.29대응변의길이는서로같으므로(변ㄹㄷ)=(변ㅁㅇ)=9cm(직사각형ㄱㄴㄷㄹ의넓이)=15_9=135(cm¤)(cid:8951)135cm¤14㉡변ㄴㄷ의대응변은변ㅁㅂ입니다.(cid:8951)㉡15합동인두도형을완전히포개었을때겹쳐지는점을대응점이라고합니다.(cid:8951)점ㄹ, 점ㅂ, 점ㅁ16합동인두도형을완전히포개었을때겹쳐지는변을대응변이라고합니다.(cid:8951)변ㄹㅂ, 변ㅂㅁ, 변ㄱㄷ17합동인두도형을완전히포개었을때겹쳐지는각을대응각이라고합니다.(cid:8951)각ㄹㅂㅁ, 각ㄱㄷㄴ, 각ㄷㄱㄴ대응변또는대응각을찾을때에는먼저대응점을찾은다음대응점의순서대로나열합니다.18점ㄴ, 점ㄷ, 점ㄹ의대응점은각각점ㅇ, 점ㅁ, 점ㅂ이므로각ㄴㄷㄹ의대응각은각ㅇㅁㅂ입니다.(cid:8951)각ㅇㅁㅂ사각형ㄱㄴㄷㄹ을180˘돌리면사각형ㅅㅇㅁㅂ과완전히겹쳐지므로두도형은합동입니다.19사각형은꼭짓점, 변, 각이각각4개씩있으므로대응점, 대응변, 대응각이각각4쌍씩있습니다.20합동인두삼각형이겹쳐져있는것입니다.점ㄱ의대응점은점ㄹ이고, 변ㄱㄴ의대응변은변ㄹㄷ입니다.(cid:8951)점ㄹ, 변ㄹㄷ21점ㄱ의대응점→점ㅂ점ㄴ의대응점→점ㄹ점ㄷ의대응점→점ㅁ(cid:8951)(위에서부터) ㄹ; ㅂ, ㅁㄱ ㄴ ㄹ ㅁ ㅂ ㄷ ㄱ ㄴ ㄷ ㄴ ㄷ ㄹ 채점기준대응점, 대응변, 대응각이각각몇쌍씩있는지설명한경우5점틀리는이유│두도형에서대응변을찾을수있지만기호를쓰는것을헷갈려틀리는경우해결방안│각점의대응점을찾아기호를써넣습니다.틀리는이유│두삼각형이겹쳐져있어서대응변을찾지못하는경우해결방안│먼저대응점을찾으면점ㅂ과점ㄷ, 점ㄹ과점ㄱ이므로변ㅂㄹ의대응변은변ㄷㄱ입니다.채점기준➊변ㅂㄹ의길이는몇cm인지구하는과정을쓴경우3점5점➋변ㅂㄹ의길이는몇cm인지구한경우2점 15초등쎈수학5-2정답(031~049) 2015.6.2 7:40 PM 페이지032 SinsagoHitec 2. 합동과대칭033본문|051쪽-054쪽2단원30➊두사각형은합동이므로둘레가같습니다.(사각형ㅁㅂㅅㅇ의둘레)=23cm두사각형이합동이므로대응변의길이가서로같습니다.(변ㅁㅂ)=(변ㄹㄱ)=4cm ▶3점➋(변ㅂㅅ)=23-(4+5+6)=8(cm)▶2점31➊대응변의길이가서로같으므로(변ㄱㄷ)=(변ㄹㄴ)=12cm➋(삼각형ㄱㄴㄷ의둘레)=(변ㄱㄴ)+(변ㄴㄷ)+(변ㄷㄱ)=5+(변ㄴㄷ)+12=27(cm)(변ㄴㄷ)=27-5-12=➌10(cm)32각ㅁㄹㅂ의대응각은각ㄷㄱㄴ이므로(각ㅁㄹㅂ)=(각ㄷㄱㄴ)=40˘(cid:8951)40˘33각ㄹㅂㅁ의대응각은각ㄱㄴㄷ이므로(각ㄹㅂㅁ)=(각ㄱㄴㄷ)=90˘(cid:8951)90˘34삼각형의세각의크기의합은180˘이고각ㄹㅁㅂ의대응각은각ㄱㄷㄴ이므로(각ㄹㅁㅂ)=(각ㄱㄷㄴ)=180˘-(40˘+90˘)=50˘(cid:8951)50˘35두사각형이합동이므로㉠`=45˘이고사각형의네각의크기의합은360˘이므로(cid:8641)=360˘-(45˘+90˘+90˘)=135˘(cid:8951)13536➊각ㅁㄹㄴ의대응각은각ㄱㄷㄴ이므로▶3점➋(각ㅁㄹㄴ)=(각ㄱㄷㄴ)=65˘▶2점45æ㉠æ37➊각ㄹㅂㅁ의대응각은각ㄴㄱㄷ이므로(각ㄹㅂㅁ)=(각ㄴㄱㄷ)=40˘▶2점➋이등변삼각형은두각의크기가같으므로(각ㅂㄹㅁ)=(180˘-40˘)÷2=70˘▶4점이등변삼각형의성질①두변의길이가같습니다.②두각의크기가같습니다.38⑴삼각형의세각의크기의합은180˘이므로(cid:100) (각ㄱㄷㄴ)=180˘-(100˘+50˘)=30˘⑵각ㅁㄷㄹ의대응각은각ㄱㄷㄴ이므로(cid:100) (각ㅁㄷㄹ)=(각ㄱㄷㄴ)=30˘⑶직선위의한점을꼭짓점으로하는각의크기는180˘이므로(cid:100) (각ㄱㄷㅁ)=180˘-(30˘+30˘)=120˘(cid:8951)⑴30˘⑵30˘⑶120˘39①②③①점ㄴ을중심으로반지름이3cm인원의일부분을그립니다.②점ㄷ을중심으로반지름이3cm인원의일부분을그립니다.③두원이만나는점ㄱ을찾아점ㄱ과점ㄴ, 점ㄱ과점ㄷ을각각잇습니다.(cid:8951)(cid:9066)3`cm3`cm3`cm3`cm3`cm3`cm3`cm3`cm3`cm틀리는이유│사각형ㅁㅂㅅㅇ의둘레를구하지못하는경우해결방안│합동인도형에서대응변의길이는서로같으므로두도형의둘레가같음을이용합니다.틀리는이유│두삼각형이겹쳐져있어서대응각을찾지못해틀리는경우해결방안│두삼각형을포개었을때를생각하여대응점을찾은후대응각을찾습니다.채점기준➊각ㄹㅂㅁ의크기를구한경우2점6점➋각ㅂㄹㅁ의크기를구한경우4점채점기준➊각ㅁㄹㄴ의대응각을찾은경우3점5점➋각ㅁㄹㄴ의크기는몇도인지구한경우2점채점기준➊변ㄱㄷ의길이를구한경우1점6점➋변ㄴㄷ의길이를구하는과정을쓴경우3점➌변ㄴㄷ의길이를구한경우2점채점기준➊변ㅂㅅ의길이는몇cm인지구하는과정을쓴경우3점5점➋변ㅂㅅ의길이는몇cm인지구한경우2점 15초등쎈수학5-2정답(031~049) 2015.6.8 3:53 PM 페이지033 SinsagoHitec 034쎈수학5－22합동과대칭40①②③④①길이가4cm인선분ㄴㄷ을긋습니다.②점ㄴ을중심으로반지름이3.5cm인원의일부분을그립니다.③점ㄷ을중심으로반지름이3cm인원의일부분을그립니다.④두원이만나는점ㄱ을찾아점ㄱ과점ㄴ, 점ㄱ과점ㄷ을각각잇습니다.(cid:8951)(cid:9066)41세변이각각2cm, 1.5cm, 1.5cm인삼각형을40번과같은방법으로차례로그립니다.(cid:8951)(cid:9066)42정삼각형은세변의길이가같으므로(한변)=6÷3=2(cm)따라서한변이2cm인정삼각형을그립니다.(cid:8951)(cid:9066)43두변이각각3cm, 1.5cm이고그사이에있는각이60˘인삼각형이되도록나머지부분을그립니다.(cid:8951)(cid:9066)3`cm1.5`cm60æ2cm2cm2cm1.5`cm1.5`cm2`cm3cm3.5cm4cm4`cm3.5`cm3`cm4`cm3`cm4`cm3.5`cm012344`cm44①②③④①길이가2.5cm인선분ㄴㄷ을긋습니다.②점ㄴ을꼭짓점으로하여각도기로80˘인각을그립니다.③점ㄴ에서2cm인곳에점ㄱ을찍습니다.④점ㄱ과점ㄷ을잇습니다.(cid:8951)(cid:9066)자와각도기를사용하면두변의길이와그사이에있는각의크기가주어진삼각형과합동인삼각형을그릴수있습니다.45두변이모두3cm이고그사이에있는각이60˘인삼각형을44번과같은방법으로차례로그립니다.(cid:8951)(cid:9066)46➊두변이각각4cm, 3cm이고그사이에있는각이90˘인삼각형을그리면다음과같습니다.▶3점➋삼각형에서자를사용하여나머지한변의길이를재면5cm입니다.따라서나머지한변은5cm입니다.▶3점90˘4cm3cm3`cm60æ3`cm2.5cm2cm80˘80æ2`cm2.5`cm01280æ2`cm2.5`cm08070605040302010090901001101201301401501601701802.5`cm0122.5`cm채점기준➊삼각형을그린경우3점6점➋나머지한변은몇cm인지구한경우3점틀리는이유│눈짐작으로변의길이를예상하여왼쪽삼각형과비슷한크기의삼각형을그린경우해결방안│왼쪽삼각형의세변의길이를재어합동인삼각형을그립니다. 15초등쎈수학5-2정답(031~049) 2015.6.8 3:53 PM 페이지034 SinsagoHitec 2. 합동과대칭035본문|054쪽-057쪽2단원47한변이3cm이고그양끝각이각각40˘, 70˘인삼각형이되도록나머지부분을그립니다.(cid:8951)(cid:9066)48①②③④①길이가4cm인선분ㄴㄷ을긋습니다.②점ㄴ을꼭짓점으로하여각도기로60˘인각을그립니다.③점ㄷ을꼭짓점으로하여각도기로70˘인각을그립니다.④두각의변이만나는점ㄱ을찾아삼각형ㄱㄴㄷ을완성합니다.(cid:8951)(cid:9066)49➊삼각형의세각의크기의합은180˘이므로(나머지한각의크기)=180˘-(100˘+50˘)=30˘ ▶3점➋▶3점50˘30˘100˘2`cm60˘70˘4`cm70æ60æ4`cm1008110120130140150160170180909080170605040302010060æ4`cm08070605040302010090901001101201301401501601701804cm012344cm70æ40æ3`cm50지유: 세각의크기가주어진삼각형은여러가지로그릴수있습니다.적어도한변의길이를더알아야합동인삼각형을그릴수있습니다.(cid:8951)현중(cid:9066)세각의크기가70˘, 50˘, 60˘로주어진삼각형그리기(cid:100) `→크기가다르므로합동이아닙니다.51㉢다른조건이더있어야합동인삼각형을그릴수있습니다.(cid:8951)㉢52[방법1]세변의길이를알아야합니다.[방법2]두변의길이와그사이에있는각의크기를알아야합니다.[방법3]한변의길이와그양끝각의크기를알아야합니다.53두변의길이를알고있으므로그사이에있는각ㄴㄱㄷ(또는각ㄷㄱㄴ)의크기를더알면합동인삼각형을그릴수있습니다.(cid:8951)각ㄴㄱㄷ(또는각ㄷㄱㄴ)54➊한변의길이와그양끝각의크기를알면합동인삼각형을그릴수있으므로▶3점➋세변중에서한변의길이만더알면합동인삼각형을그릴수있습니다.▶2점55㉠한변의길이와그양끝각의크기를알면합동인삼각형을그릴수있습니다.㉡두변의길이와그사이에있는각의크기를알면합동인삼각형을그릴수있습니다. ㉢세각의크기를알면모양은같으나크기가다른삼각형을그릴수있으므로합동인삼각형을그릴수없습니다.(cid:8951)㉢50æ60æ70æ50æ60æ70æ50æ60æ70æ틀리는이유│삼각형의세각의크기의합이180˘임을몰라서나머지한각의크기를구하지못하는경우해결방안│삼각형의세각의크기의합이180˘임을이용하여나머지한각의크기를구합니다.채점기준➊나머지한각의크기를구한경우3점6점➋삼각형을그린경우3점채점기준알아야할조건을3가지방법으로설명한경우6점6점알아야할조건을2가지방법으로설명한경우4점알아야할조건을1가지방법으로설명한경우2점채점기준➊적어도어느조건을더알아야하는지구하는과정을쓴경우3점5점➋적어도어느조건을더알아야하는지구한경우2점 15초등쎈수학5-2정답(031~049) 2015.6.8 3:53 PM 페이지035 SinsagoHitec 036쎈수학5－22합동과대칭56삼각형을그리려면(가장긴변의길이)<(다른두변의길이의합)①4<2+3②6<6+3③6=2+4④12<7+8⑤9<8+8(cid:8951)③(cid:9066)세변의길이가모두5cm로같은경우가장긴변의길이: 5cm다른두변의길이의합: 5+5=10(cm)이므로삼각형을그릴수있는조건을만족합니다.57삼각형의세각의크기의합은180˘입니다. ㉠한각이180˘이면나머지두각의크기의합이0˘가되므로삼각형을그릴수없습니다. ㉢한변의양끝각이각각100˘, 90˘이면양끝각의크기의합이180˘보다크므로삼각형을그릴수없습니다.(cid:8951)㉡한변의양끝각의크기의합이180˘인경우도나머지한각이0˘가되므로삼각형을그릴수없습니다.01㉣(cid:9066)(cid:100) (cid:100) (넓이)=8_3(넓이)=4_6(cid:100) (cid:100) (넓이)=24(cm¤)(넓이)=24(cm¤)위와같이넓이는같지만합동이아닌직사각형이있을수도있습니다.(cid:8951)㉣둘레나넓이가같아도모양과크기가다른도형은합동이아닙니다.02➊(처음종이의한변)=32÷4=8(cm)➋(만든정사각형의한변)=8÷2=4(cm)(cid:8951)4cm8`cm6cm4cm3cm8cm058쪽~059쪽➊처음종이의한변의길이구하기→➋만든정사각형의한변의길이구하기푸는순서03변ㄷㅁ의대응변은변ㄱㄷ이므로(변ㄷㅁ)=(변ㄱㄷ)=24cm변ㄴㄷ의대응변은변ㄹㅁ이므로(변ㄴㄷ)=(변ㄹㅁ)=25cm변ㄷㄹ의대응변은변ㄱㄴ이므로(변ㄷㄹ)=(변ㄱㄴ)=7cm(선분ㄴㄹ)=25-7=18(cm)(도형전체의둘레)=7+18+25+24+24=98(cm)(cid:8951)98cm04⑴삼각형ㄱㄴㄷ에서(cid:100) 삼각형의세각의크기의합은180˘이므로(cid:100) (각ㄴㄱㄷ)=180˘-(40˘+65˘)=75˘⑵삼각형ㄱㄹㅂ과삼각형ㅁㅂㄹ에서(cid:100) 각ㄹㅁㅂ의대응각은각ㅂㄱㄹ이므로(cid:100) (각ㄹㅁㅂ)=(각ㅂㄱㄹ)=75˘(cid:100) 삼각형ㄹㄴㅁ과삼각형ㅂㅁㄷ에서(cid:100) 각ㅂㅁㄷ의대응각은각ㄹㄴㅁ이므로(cid:100) (각ㅂㅁㄷ)=(각ㄹㄴㅁ)=40˘⑶(각ㄹㅁㄷ)=(각ㄹㅁㅂ)+(각ㅂㅁㄷ)=75˘+40˘=115˘(cid:8951)⑴75˘(cid:100)⑵75˘, 40˘(cid:100)⑶115˘05(직사각형의넓이)=(가로)_(세로)=8_2=16(cm¤)정사각형은네변의길이가모두같고16=4_4이므로정사각형의한변은4cm입니다.(cid:8951)(cid:9066)4`cm4`cm4`cm합동인직각삼각형2개를겹치지않게붙여놓은도형입니다. 이도형전체의둘레는몇cm입니까?24cm25cm7cmㄱ ㄴ ㄷ ㄹ ㅁ 겹쳐진부분의선분의길이는더하지않습니다.먼저직사각형의넓이를구합니다.전략틀리는이유│한변의길이가나머지두변의길이의합보다작으면삼각형을그릴수있다고생각하는경우해결방안│(가장긴변의길이)<(다른두변의길이의합)을이용하여해결합니다.▶(도형의둘레)=(변ㄱㄴ)+(선분ㄴㄹ)+(변ㄹㅁ)+(변ㅁㄷ)+(변ㄷㄱ)(선분ㄴㄹ)=(변ㄴㄷ)-(변ㄹㄷ)=(변ㄹㅁ)-(변ㄱㄴ)▶▶ 15초등쎈수학5-2정답(031~049) 2015.6.2 7:40 PM 페이지036 SinsagoHitec 2. 합동과대칭037본문|057쪽-059쪽2단원06➊가: 두변이각각12cm, 8cm이고그사이에있는각이70˘인삼각형나: (나머지한각)=180˘-(80˘+70˘)=30˘→한변이12cm이고그양끝각이각각80˘, 30˘인삼각형다: 한변이12cm이고그양끝각이각각80˘, 30˘인삼각형라: (나머지한각)=180˘-(70˘+40˘)=70˘→두변이각각12cm, 8cm이고그사이에있는각이70˘인삼각형▶5점➋따라서서로합동인삼각형끼리짝지으면가와라, 나와다입니다.▶2점07➊삼각형ㅁㄴㄷ과삼각형ㅂㄹㄷ에서변ㄴㄷ과변ㄹㄷ은정사각형ㄱㄴㄷㄹ의한변으로길이가같고,▶2점➋변ㅁㄷ과변ㅂㄷ은정사각형ㅁㄷㅂㅅ의한변으로길이가같습니다.▶2점➌각ㄴㄷㅁ과각ㄹㄷㅂ은모두직각이므로두삼각형은합동입니다.(선분ㅁㄴ)=(선분ㅂㄹ)=13cm▶4점08➊대응변의길이는서로같으므로(변ㄱㄴ)=(변ㅂㅁ)=6cm(변ㄱㅅ)=(변ㅂㅅ)=5cm(변ㅈㄹ)=(변ㅈㅁ)=13cm▶3점➋(변ㄱㄹ)=5+10+13=28(cm)▶2점➌(직사각형ㄱㄴㄷㄹ의넓이)=28_6=168(cm¤)▶3점09➊삼각형3개가서로합동인이등변삼각형이므로(각ㄴㅁㄱ)=(각ㄴㅁㄹ)=170˘÷2=85˘(각ㄴㄱㅁ)=(각ㄴㅁㄱ)=85˘▶3점➋삼각형의세각의크기의합은180˘이므로(각ㄱㄴㅁ)=180˘-(85˘+85˘)=10˘▶2점➌(각ㄱㄴㅁ)=(각ㅁㄴㄹ)=(각ㄹㄴㄷ)=10˘이므로(각ㄱㄴㄷ)=10˘_3=30˘▶2점10➊삼각형ㄱㅁㄹ과삼각형ㄷㅁㄴ은합동입니다.대응각의크기는서로같으므로(각ㅁㄱㄹ)=(각ㅁㄷㄴ)=40˘▶2점➋삼각형의세각의크기의합은180˘이므로(각ㄱㅁㄹ)=180˘-(40˘+60˘)=80˘▶3점➌직선위의한점을꼭짓점으로하는각의크기는180˘이므로(각ㄹㅁㄷ)=180˘-80˘=100˘▶2점11➊(가장긴변의길이)<(다른두변의길이의합)이므로(cid:8641)<3+2=5에서(cid:8641)=1, 2, 3, 4(cid:8641)cm가가장긴변이므로(cid:8641)=4이어야합니다.따라서세변이각각3cm, 4cm, 2cm인삼각형을그립니다.▶5점➋▶3점4`cm3`cm2`cm정사각형의네변의길이가같음을이용하여해결합니다.전략채점기준➊변ㄴㄷ과변ㄹㄷ의길이가같음을설명한경우2점8점➋변ㅁㄷ과변ㅂㄷ의길이가같음을설명한경우2점➌선분ㅁㄴ의길이를구한경우4점직사각형ㄱㄴㄷㄹ을사각형ㄱㄴㅇㅅ과사각형ㅂㅁㅈㅅ이합동이되도록접었습니다. 직사각형ㄱㄴㄷㄹ의넓이는몇cm¤인지풀이과정을쓰고, 답을구하시오.13`cm13`cm10`cm5cm5cm6`cm6`cm대응변의길이가서로같습니다.(직사각형의넓이)=(가로)_(세로)채점기준➊변ㄱㄴ, 변ㄱㅅ, 변ㅈㄹ의길이를각각구한경우3점8점➋변ㄱㄹ의길이를구한경우2점➌직사각형ㄱㄴㄷㄹ의넓이를구한경우3점채점기준➊각ㄴㄱㅁ의크기를구한경우3점7점➋각ㄱㄴㅁ의크기를구한경우2점➌각ㄱㄴㄷ의크기를구한경우2점채점기준➊각ㅁㄱㄹ의크기를구한경우2점7점➋각ㄱㅁㄹ의크기를구한경우3점➌각ㄹㅁㄷ의크기를구한경우2점세변의길이가주어졌을때, 삼각형을그릴수있는조건을이용하여식을세워서해결합니다.전략채점기준➊어떤삼각형을그려야하는지설명한경우5점8점➋삼각형을그린경우3점채점기준➊가, 나, 다, 라는어떤삼각형인지각각설명한경우5점7점➋서로합동인삼각형끼리각각짝지은경우2점▶▶ 15초등쎈수학5-2정답(031~049) 2015.6.2 7:40 PM 페이지037 SinsagoHitec 038쎈수학5－22합동과대칭✽A단계기본다잡기`⑵정답은‘정답005쪽’에있습니다.064쪽~075쪽01한직선을따라접어서완전히겹쳐지는도형을찾습니다.(cid:8951)[] [] [] [] 02①③④①, ③, ④`는직선을따라접으면완전히겹쳐지므로선대칭도형입니다.②, ⑤``는어떤직선을따라접어도완전히겹쳐지지않습니다.(cid:8951)②, ⑤03➊ㄷ, ㅂ；▶2점선대칭도형은한직선을따라접었을때양쪽의모양이같습니다.▶3점[[04~~05]]도형이완전히겹쳐지도록접었을때의접은직선을그립니다.04(cid:8951)05(cid:8951)선대칭도형에서대칭축의수는도형의모양에따라다릅니다.06원의중심을지나는직선으로원을접으면완전히겹쳐지는데원의중심을지나는직선은셀수없이많습니다.(cid:8951)⑤◯07➊대칭축의수를각각알아보면㉠㉡㉢2개1개5개▶3점➋5>2>1이므로대칭축의수가많은도형부터차례로기호를쓰면㉢㉢,, ㉠㉠,, ㉡㉡`입니다.▶2점08직선ㄱㄴ을따라접었을때글자가완전히겹쳐지지않습니다.따라서직선ㄱㄴ은대칭축이아닙니다.09한직선을따라접었을때완전히겹쳐지는알파벳을찾습니다.(cid:8951)Y,1개10한직선을따라접어서완전히겹쳐지는글자나그림을찾은다음완전히겹쳐지도록접었을때접은직선인대칭축을각각그립니다.(cid:8951)(cid:9066)11➊가나가의대칭축은4개이고, 나의대칭축은1개입니다.▶4점➋→차: 4-1=3(개)▶2점12선분ㅅㅇ을따라접었을때겹쳐지는점, 겹쳐지는변,겹쳐지는각을각각찾습니다.(cid:8951)점ㄴ, 변ㄷㄹ, 각ㅂㅁㄹ틀리는이유│대칭축을세로로만생각하여‘ㄷ’을답으로생각하지않는경우해결방안│여러방법으로대칭축을그려선대칭도형인자음을찾습니다.채점기준➊선대칭도형인자음을모두찾아쓴경우2점5점➋선대칭도형의공통점을설명한경우3점채점기준➊㉠, ㉡, ㉢`의대칭축의수를각각구한경우3점5점➋대칭축의수가많은도형부터차례로기호를쓴경우2점채점기준직선ㄱㄴ이대칭축이아닌이유를설명한경우5점틀리는이유│나의대칭축의수를도형안의선의수와같은5개로생각하는경우해결방안│대칭축을중심으로양쪽모양이같다는것을이용하여대칭축의수를구합니다.채점기준➊가와나의대칭축의수를각각구한경우4점6점➋가와나의대칭축의수의차는몇개인지구한경우2점➋ 15초등쎈수학5-2정답(031~049) 2015.6.2 7:40 PM 페이지038 SinsagoHitec 2. 합동과대칭039본문|064쪽-066쪽2단원13직선가를대칭축으로할때(cid:8951)(왼쪽에서부터) 점ㄱ, 점ㅁ；변ㄱㅁ, 변ㅁㄹ;각ㄴㄱㅁ, 각ㄱㅁㄹ14직선나를대칭축으로할때(cid:8951)(왼쪽에서부터) 점ㄱ, 점ㄹ；변ㄴㄱ, 변ㄷㄹ; 각ㄴㄱㅁ, 각ㄱㅁㄹ대응변과대응각을쓸때에는각점의대응점을찾은후순서대로쓰면됩니다.[[15~~16]]선대칭도형에서대응변의길이와대응각의크기는각각같습니다.15(cid:8951)(위에서부터) 60,416(cid:8951)(위에서부터)6,13017➊대응변의길이는서로같으므로(변ㄱㅇ)=(변ㄱㄴ)▶3점➋=11cm▶2점18직선위의한점을꼭짓점으로하는각의크기는180˘이므로(각ㄷㄹㅁ)=180˘-115˘=65˘대응각의크기는서로같으므로㉠=(각ㄷㄹㅁ)=65˘(cid:8951)65˘ㄱ ㄴㄷ ㄹ ㅁ ㉠ 115˘나 ㄱ ㄴ ㄷ ㄹ ㅁ 가 ㄱ ㄴ ㄷ ㄹ ㅁ 19직선위의한점을꼭짓점으로하는각의크기는180˘이므로(각ㅅㅂㅁ)=180˘-120˘=60˘대응각의크기는서로같고사각형의네각의크기의합은360˘이므로㉠=(각ㅇㅅㅂ)=360˘-(90˘+90˘+60˘)=120˘(cid:8951)120˘20대응점을이은선분은대칭축에의해이등분됩니다.(cid:8951)선분ㄴㅂ, 선분ㄷㅁ21대응점을이은선분은대칭축과수직으로만나고삼각형의세각의크기의합은180˘이므로(각ㅂㄱㅅ)=180˘-(70˘+90˘)=20˘대응점에서대칭축까지의거리는같으므로(선분ㄷㅂ)=(선분ㅅㅂ)_2=(선분ㄴㅁ)_2=7_2=14(cm)(cid:8951)(왼쪽에서부터) 20, 1422선분ㅅㅇ은변ㄴㄹ을이등분하므로(선분ㄴㄷ)=10÷2=5(cm)대응각의크기는서로같으므로(각ㄱㄴㄷ)=(각ㅁㄹㄷ)=80˘대응점을이은선분은대칭축과수직으로만나고사각형의네각의크기의합은360˘이므로(각ㄴㄱㅂ)=360˘-(80˘+90˘+90˘)=100˘(cid:8951)5cm, 100˘10`cm80æ70æcm1420æ7`cm대응점대응변대응각점ㄷ점ㄱ변ㄴㄷ변ㄴㄱ점ㅁ점ㄹ변ㄱㅁ변ㄷㄹ대응점대응변대응각틀리는이유│각ㄱㅂㄷ과각ㄴㄷㅂ의크기가90˘임을모르는경우해결방안│선대칭도형에서대응점을이은선분은대칭축과수직으로만납니다.120˘ㄷ ㉠ ㄹ ㅅ ㅂ ㄱ ㅇ ㄴ ㅁ 채점기준➊변ㄱㅇ의길이를구하는과정을쓴경우3점5점➋변ㄱㅇ의길이를구한경우2점각ㄱㄴㄷ각ㄴㄱㅁ각ㄴㄷㄹ각ㄱㅁㄹ점ㄴ점ㄱ변ㄴㄷ변ㄱㅁ점ㄷ점ㅁ변ㄷㄹ변ㅁㄹ각ㄴㄷㄹ각ㄴㄱㅁ각ㄷㄹㅁ각ㄱㅁㄹ 15초등쎈수학5-2정답(031~049) 2015.6.2 7:40 PM 페이지039 SinsagoHitec 040쎈수학5－22합동과대칭23대응점에서대칭축까지의거리는같으므로(선분ㄱㅅ)=(선분ㅂㅅ)=7cm(변ㄴㄷ)=(변ㅁㄹ)=12cm(선분ㄹㅇ)=(선분ㄷㅇ)=5cm(변ㅂㅁ)=(변ㄱㄴ)=8cm(도형의둘레)={(선분ㄱㅅ)+(변ㄱㄴ)+(변ㄴㄷ)+(선분ㄷㅇ)}_2=(7+8+12+5)_2=64(cm)(cid:8951)64cm선대칭도형은대칭축을중심으로양쪽의모양이같으므로도형의둘레는한쪽모양의길이의2배입니다.24⑴선대칭도형은대칭축을중심으로접었을때완전히겹쳐지므로완성한선대칭도형전체의넓이는주어진도형의넓이의2배입니다.⑵주어진도형은윗변이10cm, 아랫변이18cm,높이가12cm인사다리꼴이므로(cid:100) (주어진도형의넓이)=(10+18)_12÷2=168(cm¤)주어진도형의넓이를사각형과삼각형으로나누어구할수도있습니다.(주어진도형의넓이)=(12_10)+(12_8÷2)=168(cm¤)⑶(완성한선대칭도형전체의넓이)=168_2=336(cm¤)(cid:8951)⑴2배⑵168cm¤ ⑶336cm¤25➊선분ㄱㄷ은선분ㄴㄹ을이등분하므로(선분ㄴㅁ)=8÷2=4(cm)▶2점➋대응점을이은선분은대칭축과수직으로만나므로(삼각형ㄱㄴㄷ의넓이)=12_4÷2=24(cm¤)▶2점➌삼각형ㄱㄴㄷ과삼각형ㄱㄹㄷ의넓이는같으므로(사각형ㄱㄴㄷㄹ의넓이)=24_2=48(cm¤)▶2점선대칭도형에서대칭축에의해나누어진두도형은합동이므로넓이가같습니다.26➊선분ㄱㄹ은변ㄴㄷ을이등분하므로(변ㄴㄷ)=6_2=12(cm)➋삼각형ㄱㄴㄷ의둘레가32cm이므로(변ㄱㄴ)+(변ㄴㄷ)+(변ㄱㄷ)=32cm(변ㄱㄴ)+12cm+(변ㄱㄷ)=32cm(변ㄱㄴ)+(변ㄱㄷ)=32-12=20(cm)대응변의길이는서로같으므로(변ㄱㄴ)=(변ㄱㄷ)=20÷2▶3점➌=10(cm)▶2점[[27~~28]]각점의대응점을찾은후대응점과대응점을서로연결하여선대칭도형을완성합니다.27(cid:8951)28(cid:8951)[[29~~30]]대칭축을중심으로양쪽의모양이같도록그립니다.29(cid:8951)30(cid:8951)31도형의일부분을그린다음대칭축을긋고대칭축을중심으로각점의대응점을찾아표시합니다.표시한대응점과대응점을서로연결하여선대칭도형을그립니다.(cid:8951)(cid:9066)6cm6cm채점기준➊선분ㄴㅁ의길이를구한경우2점6점➋삼각형ㄱㄴㄷ의넓이를구한경우2점➌사각형ㄱㄴㄷㄹ의넓이를구한경우2점틀리는이유│변ㄱㄷ의길이가주어지지않아변ㄱㄴ의길이를구하지못한다고생각하는경우해결방안│대응변의길이는서로같으므로(변ㄱㄴ)=(변ㄱㄷ)이고대응점에서대칭축까지의거리는같으므로(선분ㄷㄹ)=(선분ㄴㄹ)임을이용합니다.채점기준➊변ㄴㄷ의길이를구한경우1점6점➋변ㄱㄴ의길이를구하는과정을쓴경우3점➌변ㄱㄴ의길이를구한경우2점▶1점 15초등쎈수학5-2정답(031~049) 2015.6.2 7:40 PM 페이지040 SinsagoHitec 2. 합동과대칭041본문|067쪽-070쪽2단원32각점의대응점을찾은후대응점과대응점을서로연결하여선대칭도형을완성합니다.(cid:8951)[[33~~34]]선대칭이되도록그림을완성한다음어떤단어와수가되는지확인합니다.33(cid:8951)HIDE34(cid:8951)I3035점ㅇ을중심으로180˘돌렸을때처음도형과완전히겹쳐지는도형을찾습니다.(cid:8951)[] [] []36㉠마름모㉡정삼각형㉢사다리꼴(cid:100) (cid:100) (cid:100) ㉣정육각형㉤정오각형㉥평행사변형(cid:100) (cid:100) (cid:100) 따라서점대칭도형은㉠, ㉣, ㉥`으로모두3개입니다.(cid:8951)3개37➊다；▶2점➋다도형은어떤점을중심으로180˘돌려도처음도형과완전히겹쳐지지않습니다.▶3점38어떤점을중심으로180˘돌렸을때처음알파벳과완전히겹쳐지는알파벳을찾습니다.가나다(cid:8951)다◯◯39➊㉠→㉡→㉢→㉣→▶4점➋따라서선대칭도형도되고점대칭도형도되는것은㉠㉠, ㉣㉣입니다.▶2점[[40~~41]]도형이완전히겹쳐지도록180˘돌렸을때중심이되는점을찾습니다.40(cid:8951)41(cid:8951)42➊대칭의중심은도형이완전히겹쳐지도록180˘돌렸을때중심이되는점을찾으면됩니다. ;▶3점➋ㄷ▶2점43어떤점을중심으로180˘돌렸을때처음도형과완전히겹쳐지는도형을찾으면가, 나, 라, 바입니다.(cid:8951)가,나,라,바；가`나`라`바`44점ㅇ을중심으로180˘돌렸을때겹쳐지는점, 겹쳐지는변, 겹쳐지는각을각각찾습니다.(cid:8951)점ㄹ, 변ㅁㅂ, 각ㄹㄷㄴ45㉡변ㄹㄷ의대응변은변ㅈㅇ입니다.(cid:8951)㉡선대칭도형, 점대칭도형점대칭도형선대칭도형선대칭도형, 점대칭도형채점기준➊점대칭도형이아닌것을찾아기호를쓴경우2점5점➋이유를설명한경우3점틀리는이유│선대칭도형이되면점대칭도형이될수없다고생각하여틀리는경우해결방안│한직선을따라접어서완전히겹쳐지는도형인선대칭도형을찾고, 한도형을어떤점을중심으로180˘돌렸을때처음도형과완전히겹쳐지는도형인점대칭도형을찾습니다.틀리는이유│각각의도형을정확히그리지못해틀리는경우해결방안│각각의도형을정확히그린후점대칭도형을찾습니다.채점기준➊선대칭도형과점대칭도형을각각찾은경우4점6점➋선대칭도형과점대칭도형이되는것을모두찾아기호를쓴경우2점채점기준➊대칭의중심을찾는방법을설명한경우3점5점➋대칭의중심을찾아기호를쓴경우2점 15초등쎈수학5-2정답(031~049) 2015.6.2 7:40 PM 페이지041 SinsagoHitec 042쎈수학5－22합동과대칭46(cid:8951)(위에서부터) 점ㅁ, 점ㄹ；변ㅁㄹ, 변ㄹㅁ；각ㅁㄹㄷ, 각ㄹㅁㅂ47대응변의길이는서로같으므로(변ㄱㄴ)=(변ㄹㅁ)=3cm(cid:8951)3cm48대응변의길이는서로같으므로(변ㄱㄴ)=(변ㄹㅁ)=10cm대응각의크기는서로같으므로(각ㄴㄱㅂ)=(각ㅁㄹㄷ)=40˘(cid:8951)(왼쪽에서부터) 10, 4049대응변의길이는서로같으므로(변ㅇㅅ)=(변ㄹㄷ)=8cm대응각의크기는서로같으므로(각ㄴㄱㅇ)=(각ㅂㅁㄹ)=120˘(cid:8951)120, 850➊대응각의크기는서로같고사각형의네각의크기의합은360˘이므로(각ㅁㅂㄱ)=(각ㄴㄷㄹ)=360˘-(75˘+105˘+80˘)=➋100˘51대응변의길이는서로같으므로(변ㄹㅇ)=(변ㄴㅇ)(변ㄱㅇ)=(변ㄹㅇ)이므로(변ㄱㅇ)=(변ㄴㅇ)이되어삼각형ㄱㄴㅇ은이등변삼각형입니다.(각ㄱㄴㅇ)=(각ㄴㄱㅇ)=35˘삼각형의세각의크기의합은180˘이므로(각ㄱㅇㄴ)=180˘-(35˘+35˘)=110˘(cid:8951)110˘52대응점에서대칭의중심까지의거리는같으므로(선분ㄴㅂ)=(선분ㄴㅈ)_2=6_2=12(cm)(cid:8951)12cm53➊대칭의중심은대응점을이은선분을이등분하므로(선분ㅁㅇ)=18÷2▶3점➋=9(cm)▶2점54대응점에서대칭의중심까지의거리는같으므로(선분ㄷㅇ)=(선분ㄱㅇ)=7cm두대각선의길이의합이40cm이므로(선분ㄴㄹ)=40-7_2=26(cm)(선분ㄴㅇ)=(선분ㄹㅇ)=26÷2=13(cm)(cid:8951)13cm다각형에서이웃하지않은두꼭짓점을이은선분을대각선이라고합니다.55➊대응점에서대칭의중심까지의거리는같으므로(선분ㅂㅇ)=(선분ㄷㅇ)=2cm ▶2점➋대응변의길이는서로같으므로(선분ㄹㄷ)=(선분ㄱㅂ)=11-2_2=7(cm)▶2점➌(선분ㄱㄹ)=11+7=18(cm)▶2점56대응변의길이는서로같으므로(변ㄷㄹ)=(변ㄱㄴ)=8cm(변ㄴㄷ)=(변ㄹㄱ)=(26-8-8)÷2=5(cm)(cid:8951)5cm57점대칭도형에서대칭의중심은대응점을이은선분을이등분하고직사각형에서두대각선의길이는같으므로(선분ㄴㅇ)=(선분ㄹㅇ)=(선분ㄱㅇ)=(선분ㄷㅇ)(삼각형ㄴㄷㅇ의둘레)=(변ㄴㄷ)+(선분ㄴㅇ)+(선분ㄷㅇ)=(변ㄴㄷ)+≥(선분ㄱㅇ)+≥(선분ㄷㅇ)=15+17=32(cm)(cid:8951)32cm채점기준➊각ㅁㅂㄱ의크기를구하는과정을쓴경우3점5점➋각ㅁㅂㄱ의크기를구한경우2점채점기준➊선분ㅁㅇ의길이를구하는과정을쓴경우3점5점➋선분ㅁㅇ의길이를구한경우2점틀리는이유│선분ㄴㅇ의길이를구하지못하는경우해결방안│직사각형은두대각선의길이가같다는것을이용합니다.대응하는것선대칭도형점대칭도형점ㅁ점ㄹ점ㄱ의대응점변ㅁㄹ변ㄹㅁ변ㄱㄴ의대응변각ㅁㄹㄷ각ㄹㅁㅂ각ㄱㄴㄷ의대응각틀리는이유│각ㅁㅂㄱ의대응각을각ㄷㄴㄱ이라고생각하여105˘라고답하는경우해결방안│점ㅇ을중심으로180˘돌렸을때겹쳐지는각을찾은후사각형의네각의크기의합을이용하여구합니다.채점기준➊선분ㅂㅇ의길이를구한경우2점6점➋선분ㄹㄷ의길이를구한경우2점➌선분ㄱㄹ의길이를구한경우2점 15초등쎈수학5-2정답(031~049) 2015.6.2 7:40 PM 페이지042 SinsagoHitec 2. 합동과대칭043본문|070쪽-073쪽2단원58➊(주어진도형의넓이)=23_14÷2=161(cm¤)▶3점➋완성한점대칭도형전체의넓이는주어진도형의넓이의2배이므로(완성한점대칭도형전체의넓이)=161_2=322(cm¤)▶3점59(점대칭도형전체의넓이)=(사다리꼴의넓이)_2={(5+8)_4÷2}_2=52(cm¤)(색칠하지않은부분의넓이)=4_3=12(cm¤)(색칠한부분의넓이)=52-12=40(cm¤)(cid:8951)40cm¤60[점대칭도형그리는순서]①각점에서대칭의중심을지나는직선을긋습니다.②각점에서대칭의중심까지의거리와같도록대응점을찍습니다.③각대응점을이어점대칭도형을완성합니다.(cid:8951)대칭의중심, 대응점[[61~~62]]대칭의중심에서거리가같고방향이반대인곳에대응점을찍은다음각대응점을이어점대칭도형을완성합니다.61(cid:8951)62(cid:8951)[[63~~64]]자를이용하여대칭의중심에서거리가같고방향이반대인곳에대응점을찍은다음각대응점을이어점대칭도형을완성합니다.63(cid:8951)64(cid:8951)23cm14cm65한도형을어떤점을중심으로180˘돌렸을때처음도형과완전히겹쳐지도록정사각형4개를변끼리이어점대칭도형을만듭니다. (cid:8951)(cid:9066)66대칭의중심에서거리가같고방향이반대인곳에대응점을찍은다음각대응점을이으면가로가6cm, 세로가4cm인직사각형이됩니다.(완성한점대칭도형의넓이)=6_4=24(cm¤)(cid:8951), 24cm¤67모양과크기가같아서포개었을때, 완전히겹쳐지는달의모양을찾으면11일과26일입니다.(cid:8951)11일, 26일68합동인사각형:, , 따라서합동인사각형은3가지종류입니다.(cid:8951)3가지69선대칭도형: ㉠(cid:100)㉤(cid:100)㉥점대칭도형: ㉠(cid:100)㉦(cid:8951)㉠, ㉤, ㉥；㉠, ㉦70대응점끼리연결한선분이만나는점이대칭의중심입니다.(cid:8951)(cid:9066)ㅇ26일11일1`cm1`cm틀리는이유│점대칭도형을완성하여도형전체의넓이를구하는과정에서실수하여틀리는경우해결방안│완성한도형전체의넓이가주어진도형의넓이의2배임을이용하여해결합니다.채점기준➊주어진도형의넓이를구한경우3점6점➋완성한점대칭도형전체의넓이를구한경우3점 15초등쎈수학5-2정답(031~049) 2015.6.8 3:54 PM 페이지043 SinsagoHitec 044쎈수학5－22합동과대칭[[71~~78]]서술형평가유형의입니다.71⑴삼각형1개로이루어진합동인삼각형:삼각형ㄱㄹㅁ과삼각형ㄱㅂㅁ,삼각형ㄹㄴㅁ과삼각형ㅂㄷㅁ→2쌍삼각형2개로이루어진합동인삼각형:삼각형ㄱㄴㅁ과삼각형ㄱㄷㅁ→1쌍(합동인삼각형의수)=2+1=3(쌍)▶4점⑵3쌍▶2점72⑴삼각형ㄴㅁㅂ과삼각형ㄹㄷㅂ은합동이므로대응변의길이가서로같습니다.변ㄹㅂ의대응변은변ㄴㅂ이므로(변ㄹㅂ)=(변ㄴㅂ)=5cm변ㄹㄷ의대응변은변ㄴㅁ이므로(변ㄹㄷ)=(변ㄴㅁ)=4cm; 5cm, 4cm▶3점⑵(삼각형ㄹㅂㄷ의둘레)=5+3+4=12(cm)▶2점⑶12cm▶1점그림에서찾을수있는합동인삼각형삼각형ㄱㄴㄹ과삼각형ㄷㄹㄴ, 삼각형ㄱㄴㄹ과삼각형ㅁㄴㄹ,삼각형ㅁㄴㄹ과삼각형ㄷㄹㄴ, 삼각형ㄴㅁㅂ과삼각형ㄹㄷㅂ73⑴삼각형ㄱㄴㅁ과삼각형ㄹㅁㄷ에서변ㄴㅁ의대응변은변ㅁㄷ이므로두변의길이는같습니다.▶2점⑵(변ㄴㅁ)=(변ㅁㄷ)이므로삼각형ㅁㄴㄷ은이등변삼각형입니다.㉠=(180˘-90˘)÷2=45˘▶3점⑶45˘▶1점74⑴그릴수있습니다.▶2점⑵삼각형의세각의크기의합은180˘이고,이등변삼각형은두각의크기가같으므로각ㄱㄴㄷ과각ㄱㄷㄴ의크기는같습니다.(각ㄱㄴㄷ)=(각ㄱㄷㄴ)=(180˘-50˘)÷2=65˘(cid:100) 따라서한변의길이와그양끝각의크기를알수있으므로주어진이등변삼각형과합동인삼각형을그릴수있습니다.▶3점75⑴(삼각형의넓이)=(변ㄴㄷ)_12÷2=48(변ㄴㄷ)=48_2÷12=8(cm) ; 8cm▶2점⑵선분ㄱㄹ은변ㄴㄷ을이등분하므로(선분ㄴㄹ)=8÷2=4(cm)▶2점⑶4cm▶1점076쪽~077쪽01➊㉠=60˘÷2=30˘➋대응점을이은선분은대칭축과수직으로만나므로㉡=90˘➌삼각형의세각의크기의합은180˘이므로(cid:8641)=180˘-(30˘+90˘)=60˘(cid:8951)6060˘60㉠ ㉡ ˘➊㉠`의크기구하기→➋㉡`의크기구하기→➌(cid:8641)`의각도구하기푸는순서76⑴(각ㄱㄴㄹ)=(각ㄷㄴㄹ)=35˘; 35˘▶2점⑵삼각형의세각의크기의합은180˘이므로(각ㄴㄱㄹ)=180˘-(35˘+120˘)=25˘▶2점⑶25˘▶1점77⑴도희가고른단어에서점대칭도형인알파벳: O, O→2개성윤이가고른단어에서점대칭도형인알파벳: S, H, O→3개2<3이므로게임에서이기는사람은성윤이입니다.▶3점⑵성윤▶2점78⑴(선분ㅇㅈ)=(선분ㄹㅈ)=3cm사각형ㄱㄴㄷㅇ은마름모이므로(선분ㄷㅇ)=(변ㄱㄴ)=13cm(변ㄷㄹ)=13-3-3=7(cm); 7cm▶2점⑵마름모는네변의길이가같고(변ㅅㅇ)=(변ㄷㄹ)=7cm이므로(cid:100) (도형의둘레)=13_6+7_2=78+14=92(cm)▶3점⑶92cm▶1점 15초등쎈수학5-2정답(031~049) 2015.6.8 3:54 PM 페이지044 SinsagoHitec 2. 합동과대칭045본문|074쪽-077쪽2단원02선대칭도형은대칭축을따라접었을때완전히겹쳐지므로완성한선대칭도형전체의둘레는주어진부분의길이의2배입니다. 모눈1칸은3cm이고주어진부분은모눈20칸이므로(완성한선대칭도형전체의둘레)=(3_20)_2=120(cm)변ㄱㄴ의길이는모눈2칸이고6cm이므로모눈1칸은3cm입니다.완성한선대칭도형전체의둘레는모눈40칸의길이와같으므로(완성한선대칭도형전체의둘레)=3_40=120(cm)(cid:8951)120cm03(각ㅂㄷㄹ)=(각ㅂㄷㄴ)=100˘(각ㄷㄹㅁ)=(각ㄷㄴㄱ)=115˘(각ㅁㅂㄷ)=(각ㄱㅂㄷ)=(360˘-170˘)÷2=95˘사각형의네각의크기의합은360˘이므로(각ㅂㅁㄹ)=360˘-(100˘+115˘+95˘)=50˘(cid:8951)50˘04⑴대응변의길이는서로같으므로(변ㄹㅁ)=(변ㅇㄱ)=7cm대응점에서대칭의중심까지의거리는같으므로(선분ㅂㅈ)=(선분ㄴㅈ)=3cm⑵(변ㄷㄴ)=29-(6+7+5+3+3)=5(cm)⑶(선분ㄷㅅ)=(선분ㄷㅈ)_2=(5+3)_2=16(cm)(cid:8951)⑴7cm, 3cm(cid:100)⑵5cm(cid:100)⑶16cm변ㄱㄴ은6cm입니다. 직선가를대칭축으로하는선대칭도형을완성했을때, 완성한선대칭도형전체의둘레는몇cm입니까?ㄱ ㄴ 가 모눈2칸: 6cm→모눈1칸: 3cm▼(한쪽의둘레)_2▼05➊선분ㄱㅇ, 선분ㄴㅇ은원의반지름이므로삼각형ㄱㄴㅇ은이등변삼각형입니다.➋→(각ㅇㄴㄱ)=(각ㅇㄱㄴ)=40˘삼각형의세각의크기의합은180˘이므로(각ㄱㅇㄴ)=180˘-(40˘+40˘)=100˘➌대응각의크기는서로같으므로(각ㄷㅇㄹ)=(각ㄱㅇㄴ)=100˘(cid:8951)100˘두변의길이가같은삼각형을이등변삼각형이라고합니다. 이등변삼각형은길이가같은두변과함께하는각의크기가같습니다.06➊대응변의길이는서로같으므로선분ㄱㄷ을대칭축으로하는선대칭도형에서㉠=㉣, ㉡=㉢선분ㄴㄹ을대칭축으로하는선대칭도형에서㉠=㉡, ㉣=㉢→㉠=㉡=㉢=㉣▶4점➋사각형ㄱㄴㄷㄹ의네변의길이가모두같으므로(변ㄱㄴ)=48÷4=12(cm)▶3점07➊대칭축과수직으로만나는선분은대응점을이은선분입니다.각각의대응점에서대칭축까지의거리는같으므로(선분ㅅㅇ)=(선분ㄱㅇ)=8cm(선분ㅂㅈ)=(선분ㄴㅈ)=3cm(선분ㄷㅊ)=(선분ㅁㅊ)=6cm▶5점➋(대칭축ㅋㅌ과수직으로만나는모든선분의길이의합)=(8+3+6)_2=34(cm)▶2점8cm8cm6cm6cm3cm3cmㄱㄴㄹㄷ㉠㉣㉢㉡➊삼각형ㄱㄴㅇ은어떤삼각형인지구하기→➋각ㄱㅇㄴ의크기구하기→➌각ㄷㅇㄹ의크기구하기푸는순서채점기준➊사각형ㄱㄴㄷㄹ의네변의길이가같음을구한경우4점7점➋변ㄱㄴ의길이를구한경우3점선대칭도형에서각각의대응점에서대칭축까지의거리는같습니다.전략채점기준➊선분ㅅㅇ, 선분ㅂㅈ, 선분ㄷㅊ의길이를각각구한경우5점7점➋대칭축ㅋㅌ과수직으로만나는모든선분의길이의합을구한경우2점 15초등쎈수학5-2정답(031~049) 2015.6.8 3:54 PM 페이지045 SinsagoHitec 046쎈수학5－22합동과대칭➋사각형의네각의크기의합은360˘이므로(각ㄱㄴㄷ)=360˘-(115˘+90˘)=155˘▶3점11➊대응점에서대칭의중심까지의거리는같으므로(선분ㅅㅈ)=(선분ㄷㅈ)=4cm▶2점➋대응변의길이는서로같으므로(변ㄴㄷ)=(변ㅂㅅ)=15-(4+4)=7(cm)▶3점➌도형의둘레는반쪽둘레의2배와같으므로(도형의둘레)={(변ㄱㄴ)+(변ㄴㄷ)+(변ㅅㅇ)+(변ㄱㅇ)}_2=(3+7+5+16)_2=62(cm)▶3점채점기준➊각ㄴㄱㄹ과각ㄷㄹㄱ의크기의합을구한경우4점7점➋각ㄱㄴㄷ의크기를구한경우3점채점기준➊선분ㅅㅈ의길이를구한경우2점8점➋변ㄴㄷ의길이를구한경우3점➌도형의둘레를구한경우3점078쪽~079쪽1회01가, 다, 라는모양과크기가같아서포개었을때완전히겹쳐집니다.(cid:8951)나02나의아래쪽변의길이를한눈금길게그립니다.가의아래쪽변의길이를한눈금짧게그립니다.가 나 가 나 08➊삼각형ㄱㄴㄷ의오른쪽에합동인삼각형ㄱㄷㄹ을만들면오른쪽과같습니다.➋사각형ㄱㄴㄷㄹ은선분ㄱㄷ을대칭축으로하는선대칭도형입니다.(각ㄴㄱㄹ)=30˘+30˘=60˘(각ㄱㄴㄹ)=(각ㄱㄹㄴ)=(180˘-60˘)÷2=60˘따라서삼각형ㄱㄴㄹ은정삼각형입니다.→(변ㄴㄹ)=8cm▶3점➌삼각형ㄱㄴㄷ에서변ㄱㄷ을밑변으로하면높이는8÷2=4(cm)이므로(삼각형ㄱㄴㄷ의넓이)=8_4÷2=16(cm¤)▶3점09➊8698보다큰점대칭이되는네자리수: 8888, 8968, 9006, 96,9696, 9886, 9966▶5점➋따라서만들수있는수는모두7개입니다.▶2점10➊점ㄱ과점ㄹ을선분으로이으면사각형ㄱㄴㄷㄹ과사각형ㄹㅁㅂㄱ은합동입니다.(각ㄷㄹㄱ)=(각ㅂㄱㄹ)이므로(각ㄴㄱㄹ)+(각ㄷㄹㄱ)=115˘▶4점115æ8`cm8`cm30æ30æ8`cm오른쪽삼각형ㄱㄴㄷ의넓이는몇cm¤인지구하려고합니다. 삼각형ㄱㄴㄷ의오른쪽에합동인삼각형ㄱㄷㄹ을각ㄷㄱㄹ의크기가30˘가되게만들어구하는풀이과정을쓰고, 답을구하시오.채점기준➊8698보다큰점대칭이되는네자리수를모두만든경우5점7점➋8698보다큰점대칭이되는네자리수는모두몇개인지구한경우2점8`cm8`cm60æ60æ30æ8`cm30æ8`cm채점기준➊삼각형ㄱㄴㄷ의오른쪽에합동인삼각형ㄱㄷㄹ을각ㄷㄱㄹ의크기가30˘가되게만든경우3점9점➋변ㄴㄹ의길이를구한경우3점➌삼각형ㄱㄴㄷ의넓이를구한경우3점채점기준두도형이합동이되도록만들려면어떻게하면되는지설명한경우5점0, , 5, 6, 8, 9중에서180˘돌려서처음숫자가되는숫자는0, , 8이고, 6과9는180˘돌리면각각9와6이되므로0, , 6,8, 9로네자리수를만듭니다.전략▶3점 15초등쎈수학5-2정답(031~049) 2015.6.2 7:40 PM 페이지046 SinsagoHitec 2. 합동과대칭047본문|077쪽-079쪽2단원03모눈의칸수를세어모양과크기가같게그립니다.(cid:8951)(cid:9066)04두삼각형을완전히포개었을때점ㄴ과겹쳐지는점은점ㅁ입니다.(cid:8951)점ㅁ05합동인두삼각형에서대응변의길이는서로같습니다.(cid:8951)(왼쪽에서부터) 7; 8, 606➊각ㅇㅁㅂ의대응각은각ㄴㄷㄹ이므로(각ㅇㅁㅂ)=(각ㄴㄷㄹ)=60˘▶1점➋각ㅂㅅㅇ의대응각은각ㄹㄱㄴ이므로(각ㅂㅅㅇ)=(각ㄹㄱㄴ)=120˘▶1점➌사각형의네각의크기의합은360˘이므로(각ㅁㅇㅅ)=360˘-(60˘+110˘+120˘)=70˘▶3점사각형의네각의크기의합은360˘입니다.07자와컴퍼스를사용하여나머지두변을완성합니다.(cid:8951)(cid:9066)08삼각형의세각의크기의합은180˘이므로(나머지한각)=180˘-(80˘+60˘)=40˘한변이3cm이고그양끝각이각각60˘, 40˘인삼각형을그리면(cid:8951)(cid:9066)3cm60˘40˘80˘080706050403020100909010011012013014015016017018010081101201301401501601701809090801706050403020100012360æ40æ3cm3cm3cm3`cm2`cm3`cmcm7cm8cm6cm7cm8cm6채점기준➊각ㅇㅁㅂ의크기를구한경우1점5점➋각ㅂㅅㅇ의크기를구한경우1점➌각ㅁㅇㅅ의크기를구한경우3점09변ㄴㄷ의길이를더알면두변의길이와그사이에있는각의크기를알게되어주어진삼각형과합동인삼각형을그릴수있습니다.(cid:8951)변ㄴㄷ의길이10가장긴변의길이가다른두변의길이의합보다짧아야합니다.8>3+4로가장긴변의길이가다른두변의길이의합보다깁니다.따라서다른두변이서로만나지않기때문에삼각형을그릴수없습니다.11④(cid:8951)④12→4개(cid:8951)4개13대응변의길이는서로같으므로(변ㅁㄹ)=(변ㄱㄴ)=10cm(cid:8951)10cm14대응점에서대칭축까지의거리는같으므로(선분ㄴㄹ)=(선분ㄹㄷ)_2=13_2=26(cm)(cid:8951)26cm15➊대응각의크기는서로같고사각형의네각의크기의합은360˘이므로(각ㅂㅁㄹ)=(각ㅂㄱㄴ)=360˘-(60˘+90˘+90˘)▶3점➋=120˘▶2점16의한가운데점을중심으로180˘` 돌리면처음문자와완전히겹쳐집니다.(cid:8951)③채점기준삼각형을그릴수없는이유를설명한경우5점채점기준➊각ㅂㅁㄹ의크기를구하는과정을쓴경우3점5점➋각ㅂㅁㄹ의크기를구한경우2점 15초등쎈수학5-2정답(031~049) 2015.6.2 7:40 PM 페이지047 SinsagoHitec 048쎈수학5－22합동과대칭03넓이가같아도모양과크기가다른도형은합동이아닙니다.(cid:8951)정사각형둘레나넓이가같은두정다각형(정삼각형, 정사각형, 정오각형……)은항상합동입니다.04대응변의길이는서로같으므로(변ㅂㅅ)=(변ㄷㄹ)=22cm(변ㅅㅇ)=(변ㄹㄱ)=13cm(변ㅁㅂ)=67-(14+13+22)=18(cm)(cid:8951)18cm두도형은합동이므로사각형ㄱㄴㄷㄹ의둘레도67cm입니다.05➊대응각의크기는서로같으므로(각ㅁㅇㅅ)=(각ㄴㄱㄹ)▶3점➋=122˘▶2점06➊변ㅁㅇ의대응변은변ㄴㄱ이므로(변ㅁㅇ)=(변ㄴㄱ)=12cm▶3점➋(직사각형ㅁㅂㅅㅇ의넓이)=12_7=84(cm¤)▶3점07한변이1.5cm이고그양끝각이각각85˘, 50˘인삼각형이되도록나머지부분을완성합니다.(cid:8951)(cid:9066)08(cid:8951)(cid:9066)3`cm2`cm2`cm2cm2cm3`cm3`cm2cm2cm01233`cm3`cm50æ85æ1.5`cm채점기준➊각ㅁㅇㅅ의크기를구하는과정을쓴경우3점5점➋각ㅁㅅㅇ의크기를구한경우2점채점기준➊직사각형ㅁㅂㅅㅇ의가로`(변ㅁㅇ또는변ㅂㅅ)를구한경우3점6점➋직사각형ㅁㅂㅅㅇ의넓이를구한경우3점080쪽~081쪽2회01모양과크기가같은두도형을찾으면나와라입니다.(cid:8951)나, 라02정육각형의한가운데를지나는점선을따라잘라야두도형이합동이됩니다.정육각형의한가운데를지나는점선은㉣입니다.(cid:8951)㉣17도형이완전히겹쳐지도록180˘돌렸을때중심이되는점을찾습니다.(cid:8951)대칭의중심을찾으려면대응점끼리연결한선분을적어도2개이상그려야합니다.18➊대응각의크기는서로같고사각형의네각의크기의합은360˘이므로(각ㄴㄷㄹ)=(각ㅁㅂㄱ)=360˘-(110˘+55˘+60˘)▶3점(각ㄴㄷ`➋=135˘▶2점19➊대응점에서대칭의중심까지의거리는같으므로(선분ㅁㅈ)=(선분ㄱㅈ)=7cm▶3점➋(선분ㄱㅁ)=7_2=14(cm)▶2점20대칭의중심에서거리가같고방향이반대인곳에대응점을찍은다음각대응점을이어점대칭도형을완성합니다.(cid:8951)ㅇ 채점기준➊각ㄴㄷㄹ의크기를구하는과정을쓴경우3점5점➋각ㄴㄷㄹ의크기를구한경우2점채점기준➊선분ㅁㅈ의길이를구한경우3점5점➋선분ㄱㅁ의길이를구한경우2점 15초등쎈수학5-2정답(031~049) 2015.6.8 3:54 PM 페이지048 SinsagoHitec 2. 합동과대칭049본문|079쪽-081쪽2단원09①②③④①길이가2.5cm인선분ㄴㄷ을긋습니다.②점ㄴ을꼭짓점으로하여각도기로60˘인각을그립니다.③점ㄴ에서3cm인곳에점ㄱ을찍습니다.④점ㄱ과점ㄷ을잇습니다.(cid:8951)(cid:9066)두변의길이와그사이에있는각의크기가주어졌으므로자와각도기를사용하여합동인삼각형을그립니다.이때3cm인변을먼저그려도관계없습니다.10한직선을따라접었을때모양이완전히겹쳐지는알파벳을찾습니다.(cid:8951), 에◯표, : 점대칭도형인알파벳11➊셀수없이많이그릴수있습니다.；▶2점➋원의중심을지나는직선으로원을접으면완전히겹쳐지는데원의중심을지나는직선은셀수없이많습니다. 따라서원은대칭축을셀수없이많이그릴수있습니다.▶3점12선대칭도형에서대응변의길이와대응각의크기는각각같습니다.(cid:8951)(위에서부터)9, 80, 10060˘3`cm2.5`cm60æ3cm2.5cm01233cm60æ2.5cm08070605040302010090901001101201301401501601701802.5cm0122.5cm13➊대응변의길이는서로같으므로(선분ㄱㅂ)=(선분ㅁㅂ)=5cm(선분ㄷㄹ)=(선분ㄷㄴ)=7cm(변ㅁㄹ)=(변ㄱㄴ)=9cm(선대칭도형의둘레)=(5+9+7)_2▶3점➋=42(cm)▶2점14생활주변에서선대칭도형인그림을찾습니다.(cid:8951)(cid:9066)15정사각형, 원은선대칭도형이면서점대칭도형입니다.①⑤(cid:8951)①, ⑤②정삼각형은선대칭도형이지만점대칭도형은아닙니다.16점ㅇ을중심으로180˘돌렸을때겹쳐지는점을각각찾습니다.(cid:8951)점ㄹ, 점ㅁ, 점ㅂ17➊대응변의길이는서로같으므로(변ㄱㅂ)=(변ㄹㄷ)=8cm(변ㅁㅂ)=(변ㄴㄷ)=6cm▶2점➋대응점에서대칭의중심까지의거리는같으므로(선분ㄴㅁ)=2_2=4(cm)▶1점➌(사각형ㄱㄴㅁㅂ의둘레)=5+4+6+8=23(cm)▶2점18대응각의크기는서로같고삼각형의세각의크기의합은180˘이므로(각ㄷㄹㄱ)=(각ㄱㄴㄷ)=180˘-(45˘+80˘)=55˘(cid:8951)55˘19어떤점을중심으로180˘돌려서처음숫자가되는숫자는, 0이고, 6과9를180˘돌리면각각9와6이됩니다., 0, 6, 9를사용하여69보다작은점대칭이되는네자리수를만들면00, 입니다.(cid:8951)00, 채점기준➊변ㄱㅂ, 변ㅁㅂ의길이를각각구한경우2점5점➋선분ㄴㅁ의길이를구한경우1점➌사각형ㄱㄴㅁㅂ의둘레를구한경우2점채점기준➊대칭축을셀수없이많이그릴수있다고쓴경우2점5점➋이유를설명한경우3점채점기준➊선대칭도형의둘레를구하는과정을쓴경우3점5점➋선대칭도형의둘레를구한경우2점 15초등쎈수학5-2정답(031~049) 2015.6.2 7:40 PM 페이지049 SinsagoHitec 050쎈수학5－23분수의나눗셈▶▶▶▶✽A단계기본다잡기정답은‘정답006쪽’에있습니다.088쪽~099쪽01색칠한부분은1을똑같이9로나눈것중의한칸입니다.→1÷9색칠한부분의크기는1의;9!;입니다. →1_;9!;(cid:8837)1÷9=1_;9!;(cid:8951)1, 1, 9021÷2=1_;2!;(cid:8951)1_;2!;031÷23=1_;2¡3;(cid:8951)1_;2¡3;048÷21=8_;2¡1;(cid:8951)8_;2¡1;0512÷17=12_;1¡7;(cid:8951)12_;1¡7;06④7÷3=7_;3!;(cid:8951)④07➊5÷㉠=5_=5_;3¡5;▶3점➋따라서㉠`에알맞은자연수는35입니다.▶2점081÷4=1_;4!;=;4!;5÷4=5_;4!;=;4%;3÷5=3_;5!;=;5#;(cid:8951)091÷8=(cid:8951);8!;101÷32=(cid:8951);3¡2;114÷7=(cid:8951);7$;47132181㉠1213÷24=(cid:8951);2!4#;(cid:8784)÷(cid:8774)=13두나눗셈의몫을분수로각각나타내면3÷8=;8#;, 5÷12=;1∞2;(;8#;, ;1∞2;) →(;2ª4;, ;2!4);) (cid:8837);8#;<;1∞2;(cid:8951)<두분수의크기를비교할때에는통분하여분모를같게한다음분자의크기를비교합니다.14➊㉠1÷21=;2¡1; ㉡9÷2=;2(;=4;2!;➊㉢13÷15=;1!5#;▶3점➋따라서나눗셈의몫이1보다큰것은㉡㉡`입니다.▶2점(cid:8784)÷(cid:8774)에서(cid:8784)>(cid:8774)이면나눗셈의몫이1보다큽니다.15(한컵에담을주스의양)=(전체주스의양)÷(컵수)=1÷13=;1¡3;(L)(cid:8951);1¡3;L16➊(한명에게나누어주는색테이프의길이)=(전체색테이프의길이)÷(학생수)➊=6÷11=➋;1§1;(m)17쿠키한개를만드는데사용한밀가루의무게를각각알아보면㉮빵집: 2÷60=;6™0;=;3¡0;(kg)㉯빵집: 5÷100=;10%0;=;2¡0;(kg)단위분수는분모가작을수록큰수이므로;3¡0;<;2¡0;따라서쿠키한개를만드는데더많은밀가루를사용한빵집은㉯빵집입니다.(cid:8951)㉯빵집(cid:8784)(cid:8774)1324틀리는이유│;3¡0;kg과;2¡0;kg중분모가큰;3¡0;kg이더크다고생각하는경우해결방안│단위분수는분모가작을수록큰수입니다.채점기준➊㉠`에알맞은자연수를구하는과정을쓴경우3점5점➋㉠`에알맞은자연수를구한경우2점채점기준➊㉠, ㉡, ㉢`의몫을분수로나타낸경우3점5점➋나눗셈의몫이1보다큰것을찾아기호를쓴경우2점채점기준➊학생한명에게나누어주는색테이프의길이를구하는과정을쓴경우3점5점➋학생한명에게나누어주는색테이프의길이를구한경우2점틀리는이유│5÷㉠을곱셈으로나타내지못해서틀리는경우해결방안│÷와_의앞의수가5로같고÷가_로바뀌었으므로나눗셈을곱셈으로고쳐서㉠을구합니다.••••••••••••▼▼ 15초등쎈수학5-2정답(050~061) 2015.6.8 3:55 PM 페이지050 SinsagoHitec 3. 분수의나눗셈051본문|088쪽-091쪽3단원18;8#;÷6=_=;1¡6;(cid:8951);1¡6;(=;4£8;)19;7^;÷9=_=;2™1;(cid:8951);2™1;(=;6§3;)20;7#;÷2=;7#;_;2!;=;1£4;;9@;÷16=_=;7¡2;(cid:8951);1£4;, ;7¡2;(=;14@4;)21;1£0;÷12=_=;4¡0;㉠;1!6%;÷5=_=;1£6;㉡;2¶0;÷14=_=;4¡0;㉢;2ª0;÷6=_=;4£0;(cid:8951)㉡22;5&;÷3=;5&;_;3!;=;1¶5;(cid:8951);1¶5;23;3*;÷12=_=;9@;(cid:8951);9@;(=;3•6;)24;;™7¢;;÷16=_=;1£4;(cid:8951);1£4;문제에‘기약분수로나타내시오.’라고되어있을경우반드시답을약분하여기약분수로나타냅니다.25;;¢7•;;÷18=_=;2•1;→㉠=8, ㉡=18, ㉢=8㉠+㉡+㉢=8+18+8=34(cid:8951)3426➊종현: ;;¡9º;;÷12=_=;5∞4;선경: :£6∞:÷14=_=;1∞2;▶3점114?2535?6112?6510?9118?3848?7116?2324?7112?328/316/239/20114?217/2015/1315?16112?413/10116?812/919/326/716/213/8➋따라서계산을잘못한학생은종현이입니다.▶2점27(비커한개에담은식초의양)=(전체식초의양)÷(비커수)=;1!7);÷5=_=;1™7;(L)(cid:8951);1™7;L28➊(자른노끈한도막의길이)=(전체노끈의길이)÷(도막수)➊=;;¢8ª;;÷7=_▶3점➋=;8&;(m)▶2점29(전체밥의무게)=(쌀의무게)+(늘어난무게)=;1∞2;+;9%;=;3!6%;+;3@6);=;3#6%;(kg)(한사람이먹을수있는밥의무게)=(전체밥의무게)÷(사람수)=;3#6%;÷7=_=;3∞6;(kg)(cid:8951);3∞6;kg통분할때에는두분모의최소공배수를이용하면편리합니다.305;7@;÷3=:£7¶:_;3!;=;2#1&;=1;2!1^;(cid:8951)1;2!1^;314;5!;÷7=_=;5#;(cid:8951);5#;(=;3@5!;)17/1321?517/1535?3617/1749?815/1210?17틀리는이유│48과18(㉡)을2또는3으로약분하여㉠을24또는16으로구한경우해결방안│48과18(㉡)을약분하여18(㉡)이3이되었으므로두수를6으로약분한것입니다.채점기준➊나눗셈의몫을각각구한경우3점5점➋계산을잘못한학생의이름을쓴경우2점채점기준➊자른노끈한도막의길이를구하는과정을쓴경우3점5점➋자른노끈한도막의길이를구한경우2점틀리는이유│전체밥의무게를;9%;kg으로생각하여;9%;÷7로계산한경우해결방안│전체밥의무게는쌀의무게보다;9%;kg만큼늘어났으므로;1∞2;+;9%;=;3#6%;(kg)입니다. 15초등쎈수학5-2정답(050~061) 2015.6.8 3:55 PM 페이지051 SinsagoHitec 052쎈수학5－23분수의나눗셈37세수를한꺼번에계산합니다.;9%;_4÷8=;9%;_4/_=;1∞8;(cid:8951);9%;_4/_=;1∞8;보기의방법대로계산해야하므로왼쪽에서부터순서대로계산하지않고세수를한꺼번에계산해야함에주의합니다.38;3@;_9÷8=_9/_=;4#;_9/÷8=6÷8=;8^;=;4#;(cid:8951);4#;393;4#;÷3_2=__2/=;2%;=2;2!;3;4#;÷3_2=__23;4#;÷3_2=_2/=;2%;=2;2!;(cid:8951)2;2!;40➊㉠5;6%;÷7_2=__2/=;3%;=1;3@;➊㉡5;6%;_2÷7=_2/_=;3%;=1;3@;➊㉢3;3!;÷2=_=;3%;=1;3@;➊㉣1;3@;÷4=;3%;_;4!;=;1∞2;▶3점➋따라서계산결과가다른것은㉣㉣입니다.▶2점41(한사람이가진소금의무게)=(한봉지에들어있는소금의무게)_(봉지수)÷(사람수)=1;2!;_5÷6=_5_=;4%;=1;4!;(kg)16/213/212/1510?317/11535?6/3117/1535?6/3154/213/1515?4113/1515?4/2323/118/4312/3/118/2118/2132➊대분수를가분수로고치지않고계산하여잘못되었습니다.▶3점➋[바른계산]2;1£0;÷6=;1@0#;_;6!;=;6@0#;▶2점33㉠3;4#;÷10=_=;8#;㉡5;3!;÷4=_=;3$;=1;3!;㉢5;5#;÷7=_=;5$;㉣8;5$;÷11=_=;5$;(cid:8951)㉢, ㉣34(하루에마신식혜의양)=(전체식혜의양)÷(날수)=3;5#;÷7=;;¡5•;;_;7!;=;3!5*;(L)(cid:8951);3!5*;L일주일은7일입니다.35(1분동안간거리)=(집에서도서관까지의거리)÷(걸린시간)=3;2!;÷21=_=;6!;(km)(cid:8951);6!;km36➊원모양의연못이므로(나무와나무사이의간격수)=(심으려는나무수)=8구간(나무와나무사이의간격)=(연못의둘레)÷(나무와나무사이의간격수)=4;7#;÷8=;;£7¡;;_;8!;▶4점➋=;5#6!;(km)▶2점121?317/2111?1444?517/1428?514/1416?3110?2315?4채점기준➊계산이잘못된이유를설명한경우3점5점➋바르게계산한경우2점채점기준➊나무와나무사이의간격을구하는과정을쓴경우4점6점➋나무와나무사이의간격을구한경우2점틀리는이유│연못의둘레를따라나무8그루를심으면나무와나무사이의간격수는7구간이라고생각하는경우해결방안│원모양의연못의둘레를따라나무를심으면나무와나무사이의간격수는심으려는나무수와같습니다.채점기준➊계산결과를각각구한경우3점5점➋계산결과가다른것을찾아기호를쓴경우2점 15초등쎈수학5-2정답(050~061) 2015.6.2 7:54 PM 페이지052 SinsagoHitec 3. 분수의나눗셈053본문|091쪽-094쪽3단원451;9@;÷11=_=;9!;;4!;÷6=;4!;_;6!;=;2¡4;단위분수는분모가클수록작은수이므로;9!;>;2¡4;입니다.(cid:8951)>46➊㉮:¢9º:÷8=_=;9%;➊㉯3;7#;÷3=_=;7*;=1;7!;▶3점➋;9%;<1;7!;이므로나눗셈의몫이더큰것은㉯㉯`입니다.▶2점47;1§1;÷2=_=;1£1;;;¡8∞;;÷5=_=;8#;2;5@;÷4=_=;5#;세나눗셈의몫의분자가모두같으므로분모가클수록작은수입니다. →;1£1;<;8#;<;5#;따라서;1§1;÷2의풍선이가장높이올라갑니다.(cid:8951)[ (cid:8776)] [ (cid:100) ] [ (cid:100) ]48㉠2;5#;÷6=;;¡5£;;_;6!;=;3!0#;→;3!0#;<;2!;㉡1;5$;÷3=_=;5#;→;5#;>;2!;㉢1;4!;÷5=_=;4!;→;4!;<;2!;㉣5;5#;÷7=_=;5$;→;5$;>;2!;(cid:8951)㉡, ㉣분수가;2!;보다크려면(분자)_2>(분모)이어야합니다.(cid:9066);5#;6>5→;5#;>;2!;17/1428?515/115/413/139/514/1312?515/1315?812/136/1113/1824?718/1540?9111?1111?9(5봉지에들어있는소금의무게)=1;2!;_5=;2#;_5=:¡2∞:=7;2!;(kg)(한사람이가진소금의무게)=7;2!;÷6=_=;4%;=1;4!;(kg)(cid:8951)1;4!;kg42⑴(배한개의무게)=(배4상자의무게)÷(상자수)÷(한상자에들어있는배의수)(cid:100) =12;5$;÷4÷16(cid:100) =__=;5!;(kg)⑵(사과한개의무게)=(사과3상자의무게)÷(상자수)÷(한상자에들어있는사과의수)(cid:100) =6;7#;÷3÷15(cid:100) =__=;7!;(kg)⑶(배한개의무게)+(사과한개의무게)(cid:100) =;5!;+;7!;=;3¶5;+;3∞5;=;3!5@;(kg)(cid:8951)⑴;5!;kg⑵;7!;kg⑶;3!5@;kg43나뉠수가;6%;로모두같고나누는수는9가가장크므로④;6%;÷9의몫이가장작습니다.(cid:8951)④44;1@6%;÷5=_=;1∞6;;;™8¡;;÷28=_=;3£2;(;1∞6;, ;3£2;) →(;3!2);, ;3£2;) (cid:8837);1∞6;>;3£2;(cid:8951)>128?4321?815/1525?16115?113/1115?45?7116?114/1116?64?516/2515?2틀리는이유│각각을계산하여찾으려고하다가계산실수로틀리는경우해결방안│나뉠수가같을때나누는수가클수록몫이작아집니다.채점기준➊㉮`와㉯`의몫을각각구한경우3점5점➋나눗셈의몫이더큰것을찾아기호를쓴경우2점_2 15초등쎈수학5-2정답(050~061) 2015.6.2 7:54 PM 페이지053 SinsagoHitec 054쎈수학5－23분수의나눗셈545_(cid:8641)=:™7º:→(cid:8641)=:™7º:÷5=_=;7$;(cid:8951);7$;55(cid:8641)_24=2;5@;→(cid:8641)=2;5@;÷24=_=;1¡0;(cid:8951);1¡0;56➊_18?=22▶2점➋5;2!;÷㉮=22→㉮=5;2!;÷22=_=;4!;▶4점57:¢6¡:÷2=:¢6¡:_;2!;=;1$2!;=3;1∞2;(cid:8641)<3;1∞2;→(cid:8641)=1, 2, 3(3개)(cid:8951)3개58➊16;2!;÷3=_=:¡2¡:=5;2!;▶2점➋5;2!;<(cid:8641)→(cid:8641)안에들어갈수있는자연수: 6, 7, 8……따라서(cid:8641)안에들어갈수있는자연수중에서가장작은수는6입니다.▶3점592;5!;÷6_5=_;6!;_5/=:¡6¡:=1;6%;3;4#;_6÷5=_6/_=;2(;=4;2!;1;6%;와4;2!;사이에있는자연수: 2, 3, 4(cid:8951)2, 3, 460정삼각형은세변의길이가모두같으므로(한변)=(정삼각형의둘레)÷(변의수)=17÷3=;;¡3¶;;=5;3@;(cm)(cid:8951)5;3@;(=:¡3¶:)cm(정삼각형의둘레)=(한변)_315/13315?4/2 1115/113/11133?2122?2111?22119/1124?2112?515/1420?749나뉠수가1로일정하므로(cid:8641)안의수가가장작을때나눗셈의몫이가장큽니다.3<4<6<7<8이므로(cid:8641)=3→1÷3=;3!;(cid:8951)3；;3!;50➊나뉠수가작을수록나누는수가클수록나눗셈의몫이작아집니다.따라서몫이더작게되는나눗셈식은;1•7;÷12이고➋몫은;5™1;입니다. 나뉠수의분모가17로일정하므로분자가작을수록나뉠수는작아집니다.➊만들수있는나눗셈식: ;1•7;÷12, ;1!7@;÷8;1•7;÷12=_=;5™1;, ;1!7@;÷8=_=;3£4;(;5™1;, ;3£4;) →(;10$2;, ;10(2;) (cid:8837);5™1;<;3£4;따라서몫이더작게되는나눗셈식은;1•7;÷12이고➋몫은;5™1;입니다.51만들수있는가분수: ;2%;, ;2&;, ;5&;만들수있는나눗셈식: ;2%;÷7, ;2&;÷5, ;5&;÷2→3개(cid:8951)3개52만들수있는대분수: 3;7%;, 5;7#;, 7;5#;나누는수가16으로일정하므로나뉠수가가장클때나눗셈의몫이가장큽니다.→7;5#;÷16=_=;4!0(;(cid:8951)7, 3, 5；;4!0(;53;1!6%;÷(cid:8641)=10→(cid:8641)=;1!6%;÷10=_=;3£2;(cid:8951);3£2;110?2315?16116?81938?518/2312?17112?328/17틀리는이유│곱셈식과나눗셈식의관계를모르는경우해결방안│(cid:8784)÷(cid:8641)=▲→(cid:8641)=(cid:8784)÷▲를이용하여식을만든후계산합니다.채점기준➊분수의곱셈을계산한경우2점6점➋㉮`에알맞은수를구한경우4점채점기준➊나눗셈의몫을구한경우2점5점➋(cid:8641)안에들어갈수있는자연수중에서가장작은수를구한경우3점채점기준➊몫이더작게되는나눗셈식을만든경우3점5점➋몫이더작게되는나눗셈의몫을구한경우2점▶3점▶2점▶3점▶2점틀리는이유│(진분수)÷(자연수)와(가분수)÷(자연수)의개수를모두구해6개라고답한경우해결방안│분모에넣을숫자를정한뒤분모보다큰수를분자에넣어가분수를만들고나머지한장을나누는수에놓아(가분수)÷(자연수)를만듭니다. 15초등쎈수학5-2정답(050~061) 2015.6.8 3:55 PM 페이지054 SinsagoHitec 3. 분수의나눗셈055본문|094쪽-097쪽3단원(노란색삼각형의넓이)=(빨간색삼각형의넓이)÷4(노란색삼각형의넓이)=3÷4=;4#;(cm¤)(cid:8951);4#;67멜론한개의무게는참외9개의무게와같으므로(멜론한개의무게)=(참외한개의무게)_9→(참외한개의무게)=(멜론한개의무게)÷9→=2;7$;÷9=_=;7@;(kg)(cid:8951);7@;kg68첫번째갈림길에서나눗셈의몫을각각구하면2÷3=;3@;, 5÷4=;4%;=1;4!;(cid:8837);3@;<1;4!;두번째갈림길에서나눗셈의몫을각각구하면1;1ª6;÷5=_=;1∞6;;9*;÷4=_=;9@;(;1∞6;, ;9@;) →(;1¢4∞4;, ;1£4™4;) (cid:8837);1∞6;>;9@;따라서주희가도착하는곳은미란이네집입니다.(cid:8951)미란69;3%;÷2=;3%;_;2!;=;6%;, 2;3@;÷2=_;=;3$;=1;3!;,;3$;÷2=_=;3@;, ;3!;÷3=;3!;_;3!;=;9!;,2;3@;÷3=;3*;_;3!;=;9*;, ;3%;÷3=;3%;_;3!;=;9%;,;3!;÷2=;3!;_;2!;=;6!;, ;3@;÷3=;3@;_;3!;=;9@;나눗셈의몫중에서;3!;이상1미만인수는;6%;, ;3@;, ;9*;, ;9%;이므로남는풍선은4개입니다.(cid:8951)4개(cid:8778)이상(cid:8774)미만인수는(cid:8778)보다크거나같고(cid:8774)보다작은수입니다.12/124/312/148/314/128/915/1525?1619/1218?761➊(높이)=(평행사변형의넓이)÷(밑변)➊(높이)=7;1¶0;÷7=_➊(높이)=;1!0!;=➋1;1¡0;(m)62(색칠된부분의넓이)=27÷5=;;™5¶;;=5;5@;(m¤)(cid:8951)5;5@;(=:™5¶:)m¤63(정육각형을6등분한것중의한칸의넓이)=8;5@;÷6=_=;5&;=1;5@;(cm¤)(색칠된부분의넓이)=1;5@;_4=;5&;_4(색칠한부분의넓이)=:™5•:=5;5#;(cm¤)(cid:8951)5;5#;cm¤64어떤자연수를(cid:8641)라하여잘못계산한식을세우면(cid:8641)_6=42→(cid:8641)=42÷6=7[바른계산](cid:8641)÷6=7÷6=;6&;=1;6!;(cid:8951)1;6!;(=;6&;)`65➊어떤분수를(cid:8641)라하여잘못계산한식을세우면(cid:8641)_9=;2@8&;→(cid:8641)=;2@8&;÷9=_=;2£8;▶3점➋[바른계산](cid:8641)÷9=;2£8;÷9=_=;8¡4;▶3점66(주황색삼각형의넓이)=(빨간색삼각형의넓이)÷2(주황색삼각형의넓이)=3÷2=;2#;=1;2!;(cm¤)(노란색삼각형의넓이)=(주황색삼각형의넓이)÷2(노란색삼각형의넓이)=1;2!;÷2=;2#;_;2!;=;4#;(cm¤)19/313/2819/1327?2816/1742?517/11177?10채점기준➊어떤분수를구한경우3점6점➋바르게계산한값을구한경우3점틀리는이유│색칠된부분의넓이를8;5@;÷6으로구한경우해결방안│정육각형을6등분하였으므로÷6을한후, 6등분한도형중4칸을색칠했으므로_4를하여구합니다.빨간색 삼각형 넓이 주황색 삼각형 넓이 노란색 삼각형 넓이 ÷2÷4÷2채점기준➊평행사변형의높이를구하는과정을쓴경우3점5점➋평행사변형의높이를구한경우2점 15초등쎈수학5-2정답(050~061) 2015.6.8 3:55 PM 페이지055 SinsagoHitec 056쎈수학5－23분수의나눗셈⑵(;8#;, ;5$;) →(;4!0%;, ;4#0@;) →;8#;<;5$;⑴(;5$;, ;7%;) →(;3@5*;, ;3@5%;) →;5$;>;7%;⑴(;8#;, ;7%;) →(;5@6!;, ;5$6);) →;8#;<;7%;⑴(cid:8837);5$;>;7%;>;8#;▶2점⑶㉡, ㉢, ㉠▶1점75⑴가: ;1!5$;÷8=_=;6¶0;⑴나: 3;1£0;÷9=_=;3!0!;⑴；;6¶0;(=;1¡2¢0;), ;3!0!;(=;9#0#;)▶2점⑵(;6¶0;, ;3!0!;) →(;6¶0;, ;6@0@;) (cid:8837);6¶0;<;3!0!;⑴(몫의차)=;3!0!;-;6¶0;=;6@0@;-;6¶0;⑴(몫의차)=;6!0%;=;4!;▶2점⑶;4!;▶1점76⑴(도형을똑같이8개로자른것중의한개의넓이)⑵=2;5@;÷8=_=;1£0;(cm¤)⑵(이어붙인도형의넓이)⑵=_5/=;2#;=1;2!;(cm¤)▶4점⑵1;2!;cm¤▶2점77⑴(마름모의넓이)=4;9$;_6÷2=_6/_⑴(마름모의넓이)=;;¢3º;;=13;3!;(cm¤)⑴(사다리꼴의넓이)=(4+6)_3÷2=10_3÷2=30÷2=15(cm¤)⑴；13;3!;cm¤, 15cm¤▶3점⑵(마름모의넓이)÷(사다리꼴의넓이)⑵=13;3!;÷15=_=;9*;(배)▶2점⑶;9*;배▶1점115?3840?312/112/409/31310?218/2312?519/31133?1018/4714?15[[70~~77]]서술형평가유형의입니다.70⑴㉠=12÷17=;1!7@;⑴㉡=㉠÷3=;1!7@;÷3=_=;1¢7;⑴；;1!7@;, ;1¢7;(=;5!1@;)▶2점⑵㉠`+㉡`=;1!7@;+;1¢7;=;1!7^;▶2점⑶;1!7^;▶1점71⑴성호▶1점⑵나눗셈을곱셈으로나타낼때⑵;1¢5;÷2를;1¢5;_;2!;로나타내야하는데⑵;1¢5;_2로나타내었습니다.▶2점⑶;1¢5;÷2=_=;1™5;▶2점72⑴:¡5•:÷12=_=;1£0;, :¡5¶:÷6=:¡5¶:_;6!;=;3!0&;⑴；;1£0;(=;6!0*;), ;3!0&;▶3점⑵;1£0;<<;3!0&;→;3ª0;<<;3!0&;⑴→9<(cid:8641)<17⑴(cid:8641)안에들어갈수있는자연수: 10, 11, 12……, 16→7개▶2점⑶7개▶1점73⑴(쌀통한개에담을쌀의양)=(전체쌀의양)÷(쌀통수)⑵=3;8#;÷9=_=;8#;(kg)▶3점⑵;8#;kg▶2점74⑴㉠;4#;÷2=;4#;_;2!;=;8#;⑴㉡;;¡5™;;÷3=_=;5$;⑴㉢3;7$;÷5=_=;7%;⑴；;8#;, ;5$;(=;1!5@;), ;7%;(=;3@5%;)▶2점15/1525?713/1412?519/1327?8(cid:8641)30(cid:8641)30112?2318?512/124/1513/1412?17 15초등쎈수학5-2정답(050~061) 2015.6.2 7:54 PM 페이지056 SinsagoHitec 3. 분수의나눗셈057본문|098쪽-101쪽3단원100쪽~101쪽012÷7=;7@;, 13÷4=;;¡4£;;=3;4!;;7@;와3;4!;사이에있는자연수: 1, 2, 3(cid:8951);7@;, 3;4!;(=:¡4£:)；1, 2, 302세로를(cid:8641)m라하면가로는세로의2배이므로((cid:8641)_2)m입니다.(직사각형의둘레)={(가로)+(세로)}_2이므로((cid:8641)_2+(cid:8641))_2=;6%;, (cid:8641)_3_2=;6%;, (cid:8641)_6=;6%;,(cid:8641)=;6%;÷6=;6%;_;6!;=;3∞6;(m)(가로)=(cid:8641)_2=_2/=;1∞8;``(m)(cid:8951);1∞8;m031시간(60분)은10분의6배이므로(거북이가10분동안간거리)=;1$3@;÷6=_=;1¶3;(km)(토끼가10분동안간거리)--(거북이가10분동안간거리)=;1$3@;-;1¶3;=;1#3%;=2;1ª3;(km)(cid:8951)2;1ª3;km04⑴(처음식용유의양)=_=:¡5¢:=2;5$;(L)⑵(8명에게나누어주고남은식용유의양)⑵=(처음식용유의양)_;3!;=2;5$;_;3!;⑵=:¡5¢:_;3!;=;1!5$;(L)⑶(8명에게나누어준식용유의양)(cid:100) =(처음식용유의양)-(8명에게나누어주고남은식용유의양)(cid:100)=2;5$;-;1!5$;=2;1!5@;-;1!5$;=1;1@5&;-;1!5$;=1;1!5#;(L)24/572/116/1742?131536?18둘레가;6%;m이고가로는세로의2배인직사각형이있습니다. 이직사각형의가로는몇m입니까?{(가로)+(세로)}_2▼(cid:8641)m▼((cid:8641)_2)m((cid:8641)2개+(cid:8641)1개)(cid:8641)3개▼⑷(한사람에게나누어준식용유의양)⑷=(8명에게나누어준식용유의양)÷8(cid:100) =1;1!5#;÷8=_(cid:100) =;3¶0;(L)(cid:8951)⑴2;5$;L⑵;1!5$;L⑶1;1!5#;L⑷;3¶0;L052<3<5<7이므로2;7!;_7÷2=_7/_;2!;=;;¡2∞;;=7;2!;따라서몫이가장크게되는나눗셈식은2;7!;_7÷2이고몫은7;2!;입니다.(cid:8951)7, 2；7;2!;06➊그림과같이판화와판화사이의간격은판화한변의길이와같고간격수는12구간이므로전체길이는판화13+12=25(장)을나란히붙인것과같습니다.▶2점➋(판화한변)=580÷25=:∞2•5º:▶3점➌=;;;!5!;§;;=23;5!;(cm)▶2점판화를나란히벽에붙이면판화와판화사이의간격수는판화수보다1작습니다.(간격수)=(판화수)-107➊(cid:8784)÷(cid:8785)=;1ª9;÷3=_=;1£9;(배)▶5점➋따라서(cid:8784)는(cid:8785)의;1£9;배입니다.▶2점13/139/19580`cm25장 1157/118/2728?15곱하는수가클수록나누는수가작을수록계산결과가커집니다.전략토끼가10분동안간거리에서거북이가10분동안간거리를뺍니다.전략채점기준➊전체길이는판화몇장의길이와같은지구한경우2점7점➋판화한변의길이를구하는과정을쓴경우3점➌판화한변의길이를구한경우2점(cid:8784)÷(cid:8785)를이용하여(cid:8784)는(cid:8785)의몇배인지구합니다.전략채점기준➊(cid:8784)는(cid:8785)의몇배인지구하는과정을쓴경우5점7점➋(cid:8784)는(cid:8785)의몇배인지구한경우2점 15초등쎈수학5-2정답(050~061) 2015.6.8 3:56 PM 페이지057 SinsagoHitec 058쎈수학5－23분수의나눗셈08➊(;2%;와:£6∞:사이의크기)➊=:£6∞:-;2%;=:£6∞:-:¡6∞:➊=:™6º:=:¡3º:=3;3!;▶2점➋(수직선한칸의크기)=3;3!;÷5=_=;3@;▶3점➌(㉠`이나타내는수)=;2%;+;3@;=:¡6∞:+;6$;(㉠`이나타내는수)=:¡6ª:=3;6!;▶2점09➊(음료수한상자의무게)=(음료수5상자의무게)÷5➊=19;4#;÷5=;;¶4ª;;_;5!;➊=;2&0(;=3;2!0(;(kg)▶3점➋(음료수15개의무게)=(음료수한상자의무게)-(빈상자한개의무게)➊=3;2!0(;-;5!=3;2!0(;-;2¢0;=3;2!0%;=3;4#;(kg)▶3점➌(음료수한개의무게)=3;4#;÷15=_(음료수한개의무게)=;4!;(kg)▶2점10➊(색칠된부분의밑변)➊=10;5$;÷3=_=:¡5•:=3;5#;(cm)(색칠된부분의넓이)=3;5#;_10÷2(색칠한부분의넓이)=_10?_=➋18(cm¤)12/112/185/113/11854?5115?1115?415/1210?3채점기준➊;2%;와:£6∞:사이의크기를구한경우2점7점➋수직선한칸의크기를구한경우3점➌㉠`이나타내는수를구한경우2점;2%;와:£6∞:사이의크기를5등분하여수직선한칸의크기를구합니다.전략➊(삼각형ㄱㄴㄷ의넓이)➊=10;5$;_10÷2➊=_10?_=54(cm¤)작은삼각형3개의밑변과높이가각각같으므로작은삼각형3개의넓이는모두같습니다.(색칠된부분의넓이)=(삼각형ㄱㄴㄷ의넓이)÷3➋(색칠한부분의넓이)=54÷3=➋18(cm¤)11➊정오각형은5개의변의길이가모두같으므로(전체철사의길이)=(정오각형의한변)_(변의수)=3;7%;_5=:™7§:_5=:¡7#;º:=18;7$;(cm)▶3점➋이철사를모두사용하여모양과크기가같은정사각형을2개만들었으므로(정사각형의둘레)=(전체철사의길이)÷2(정사각형의둘레)=18;7$;÷2=_(정사각형의둘레)=:§7∞:=9;7@;(cm)▶3점➌정사각형은네변의길이가모두같으므로(정사각형의한변)=9;7@;÷4=:§7∞:_;4!;(정사각형의한변)=;2^8%;=2;2ª8;(cm)▶2점(정사각형의한변)=18;7$;÷2÷4로구해도됩니다.→(정사각형의한변)=18;7$;÷2÷4=__;4!;→(정사각형의한변)=;2^8%;=2;2ª8;(cm)12/165130÷712/165130÷712/122754?5/1음료수한상자의무게에서빈상자한개의무게를빼서음료수15개의무게를구합니다.전략채점기준➊음료수한상자의무게를구한경우3점8점➋음료수15개의무게를구한경우3점➌음료수한개의무게를구한경우2점채점기준➊색칠된부분의넓이를구하는과정을쓴경우4점7점➋색칠된부분의넓이를구한경우3점철사를구부려서한변이3;7%;cm인정오각형을만들었습니다. 이철사를겹치지않게모두사용하여모양과크기가같은정사각형을2개만들었다면만든정사각형의한변은몇cm가되겠는지풀이과정을쓰고, 답을구하시오.▼전체철사의길이: (3;7%;_5)cm(전체철사의길이)÷2▼(정사각형의둘레)÷4▼채점기준➊전체철사의길이를구한경우3점8점➋정사각형의둘레를구한경우3점➌정사각형의한변의길이를구한경우2점 15초등쎈수학5-2정답(050~061) 2015.6.8 3:56 PM 페이지058 SinsagoHitec 3. 분수의나눗셈059본문|101쪽-103쪽3단원01색칠한부분은1을똑같이4로나눈것중의하나입니다.→1÷4색칠한부분의크기는1의;4!;입니다. →1_;4!;(cid:8837)1÷4=1_;4!;(cid:8951)1, 1, 402①5÷6=5_;6!;②2÷5=2_;5!;④7÷6=7_;6!;⑤13÷15=13_;1¡5;(cid:8951)③031÷11=(cid:8951);1¡1;049÷28=(cid:8951);2ª8;05➊(한도막의길이)➊=(전체끈의길이)÷(도막수)➊=19÷6=➋:¡6ª:=3;6!;(m)06;7$;÷2=_=;7@;(cid:8951);7@;(=;1¢4;)07아라: ;1∞2;÷15=_=;3¡6;준호: ;1§3;÷3=_=;1™3;따라서나눗셈의몫을바르게구한학생은준호입니다.(cid:8951)준호08➊어떤분수를(cid:8641)라하여잘못계산한식을세우면(cid:8641)_10=;9*;→(cid:8641)=;9*;÷10=_=;4¢5;▶3점➋[바른계산](cid:8641)÷10=;4¢5;÷10=_=;22@5;▶2점110?524/45110?548/913/126/13115?315/1212/124/7928111102쪽~103쪽1회09➊나뉠수가클수록나누는수가작을수록나눗셈의몫이커집니다.따라서몫이가장크게되는나눗셈식은;1§1;÷4이고▶3점➋몫은;2£2;입니다.▶2점10;;£8£;;÷2=;;£8£;;_;2!;=;1#6#;=2;1¡6;(cid:8951)㉡11;;¡4∞;;÷3=_=;4%;=1;4!;(cid:8951)㉠12;;¡5§;;÷4=_=;5$;(cid:8951)㉢13:™5¡:÷7=_=;5#;:™3º:÷8=_=;6%;(;5#;, ;6%;) →(;3!0*;, ;3@0%;) (cid:8837);5#;<;6%;(cid:8951)<14➊(한사람이마신수정과의양)➊=(전체수정과의양)÷(사람수)➊=;3%;÷7=;3%;_;7!;=➋;2∞1;(L)157;3!;÷8=_=;1!2!;(cid:8951);1!2!;(=;2@4@;)162;7^;÷4=_=;7%;(cid:8951);7%;(=;2@8);)175;5@;÷9=_=;5#;9;1ª1;÷3=_=;1#1^;=3;1£1;;5#;<(cid:8641)<3;1£1;→(cid:8641)=1, 2, 3(3개)(cid:8951)3개13/136108÷1119/1327?514/1520?718/41122?318/2520?317/1321?514/1416?513/1515?4채점기준➊한도막의길이를구하는과정을쓴경우3점5점➋한도막의길이를구한경우2점채점기준➊어떤분수를구한경우3점5점➋바르게계산한값을구한경우2점채점기준➊몫이가장크게되는나눗셈식을만든경우3점5점➋몫이가장크게되는나눗셈의몫을구한경우2점채점기준➊한사람이마신수정과의양을구하는과정을쓴경우3점5점➋한사람이마신수정과의양을구한경우2점 15초등쎈수학5-2정답(050~061) 2015.6.2 7:55 PM 페이지059 SinsagoHitec 060쎈수학5－23분수의나눗셈18➊(한모둠이가지는콩의무게)=(전체콩의무게)÷(모둠수)➊=13;7$;÷5=_=;;¡7ª;;=➋2;7%;(kg)198÷3=;3*;=2;3@;2;3@;÷4=_=;3@;;3@;÷2=_=;3!;(cid:8951)2;3@;(=;3*;), ;3@;, ;3!;20➊(색칠된부분의가로)➊=8;4!;÷3=_➊=:¡4¡:=2;4#;(cm)▶3점➋(색칠된부분의넓이)=(가로)_(세로)➋(색칠한부분의넓이)=2;4#;_5=:¡4¡:_5➋(색칠한부분의넓이)=:∞4∞:=13;4#;(cm¤)▶2점(색칠된부분의넓이)=(가장큰직사각형의넓이)÷3으로구해도됩니다.13/11133?412/112/314/128/315/11995?7채점기준➊한모둠이가지는콩의무게를구하는과정을쓴경우3점5점➋한모둠이가지는콩의무게를구한경우2점채점기준➊색칠된부분의가로를구한경우3점5점➋색칠된부분의넓이를구한경우2점011÷17=1_;1¡7;(cid:8951)1_;1¡7;02;2∞1;÷9=;2∞1;_;9!;(cid:8951);2∞1;_;9!;0312÷7=;;¡7™;;=1;7%;(cid:8951)1;7%;(=:¡7™:)104쪽~105쪽2회04➊가장작은수: 5, 가장큰수: 24▶2점➋→5÷24=;2∞4;▶3점05➊(cid:8641)_6=23→(cid:8641)=23÷6=:™6£:=3;6%;▶3점➋따라서(cid:8641)안에알맞은분수는3;6%;(=:™6£:)입니다.▶2점06(한통에담은물의양)=140÷9==15;9%;(L)(cid:8951)15;9%;(=)L07;1•5;÷12=_=;4™5;(cid:8951);4™5;(=;18*0;)082;7$;÷24=_=;2£8;(cid:8951);2£8;(=;1¡6•8;)09;1∞1;÷8=;1∞1;_;8!;=;8∞8;(cid:8951);8∞8;10①;3$;÷2=_=;3@;②:¡9¢:÷7=_=;9@;③;1•1;÷4=_=;1™1;④:¡7º:÷5=_=;7@;⑤;1§3;÷3=_=;1™3;분자가같은분수는분모가작을수록더큰분수이므로몫이가장큰것은①입니다.(cid:8951)①13/126/1315/1210?714/128/1117/1214?912/124/3124?4318?7112?328/1514091409채점기준➊(cid:8641)안에알맞은분수를구하는과정을쓴경우3점5점➋(cid:8641)안에알맞은분수를구한경우2점채점기준➊가장작은수와가장큰수를각각구한경우2점5점➋가장작은수를가장큰수로나눈몫을분수로나타낸경우3점 15초등쎈수학5-2정답(050~061) 2015.6.2 7:55 PM 페이지060 SinsagoHitec 3. 분수의나눗셈061본문|103쪽-105쪽3단원(나뉠수)<(나누는수) (cid:8837)(몫)<1㉠4;5#;÷4에서4;5#;>4이므로(몫)>1㉡10;7!;÷10에서10;7!;>10이므로(몫)>1㉢6;2!;÷7에서6;2!;<7이므로(몫)<1따라서몫이진분수인것은㉢입니다.(cid:8951)㉢진분수는분자가분모보다작은분수이므로1보다작습니다.17➊2;3@;÷9=;3*;_;9!;=;2•7;9;1¶5;÷2=_=;1&5!;=4;1!5!;▶3점➋;2•7;과4;1!5!;사이에있는자연수: 1, 2, 3, 4`▶2점18➊빈바구니의무게가;5!;kg이므로(복숭아9개의무게)=2;5#;-;5!;=2;5@;(kg)▶2점➋(복숭아한개의무게)=(복숭아9개의무게)÷9=2;5@;÷9=_=;1¢5;(kg)▶3점19;9$;÷2÷3=__;3!;=;2™7;(cid:8951);2™7;세수의나눗셈또는혼합계산은반드시앞에서부터순서대로계산하거나세수를한꺼번에계산합니다.20➊(직사각형을8등분한것중의한칸의넓이)➊=50;3@;÷8=_➊=:¡3ª:=6;3!;(cm¤)▶3점➋(색칠된부분의넓이)=6;3!;_2=:¡3ª:_2➋(색칠된부분의넓이)=:£3•:=12;3@;(cm¤)▶2점18/119152÷312/124/919/3412?512/171142÷1511;1ª0;÷4=;1ª0;_;4!;=;4ª0;;5$;÷6=_=;1™5;(;4ª0;, ;1™5;) →(;1™2¶0;, ;1¡2§0;) (cid:8837);4ª0;>;1™5;(cid:8951)>12;1!3$;÷2=_=;1¶3;, ;4(;÷6=_=;8#;(;1¶3;, ;8#;) →(;1∞0§4;, ;1£0ª4;) (cid:8837);1¶3;>;8#;(cid:8951)>13➊㉠=;;™7¢;;÷8=_=;7#;➊㉡=㉠÷9=;7#;÷9=_=;2¡1;▶2점➋(;7#;, ;2¡1;) →(;2ª1;, ;2¡1;) (cid:8837);7#;>;2¡1;▶1점➌㉠`-㉡=;7#;-;2¡1;=;2ª1;-;2¡1;=;2•1;▶2점14정육각형은6개의변의길이가모두같으므로(정육각형의한변)=(전체철사의길이)÷(변의수)(정육각형의한변)=;4%;÷6=;4%;_;6!;=;2∞4;(m)(cid:8951);2∞4;m(정(cid:8774)각형의변의수)=(cid:8774)개156;9$;÷29=_=;9@;(cid:8951)_=;9@;16㉠4;5#;÷4=;;™5£;;_;4!;=;2@0#;=1;2£0;㉡10;7!;÷10=;;¶7¡;;_;1¡0;=;7&0!;=1;7¡0;㉢6;2!;÷7=;;¡2£;;_;7!;=;1!4#;따라서몫이진분수인것은㉢입니다.129?1258?9129?1258?919/313/718/1324?716/239/412/1714?1316/324/5채점기준➊㉠`과㉡`에알맞은수를각각구한경우2점5점➋㉠`과㉡`에알맞은수의크기를비교한경우1점➌㉠`과㉡`에알맞은수의차를구한경우2점채점기준➊복숭아9개의무게를구한경우2점5점➋복숭아한개의무게를구한경우3점채점기준➊두나눗셈의몫을각각구한경우3점5점➋두나눗셈의몫사이에있는자연수를모두구한경우2점채점기준➊직사각형을8등분한것중의한칸의넓이를구한경우3점5점➋색칠된부분의넓이를구한경우2점 15초등쎈수학5-2정답(050~061) 2015.6.8 3:56 PM 페이지061 SinsagoHitec 062쎈수학5－24소수의나눗셈✽A단계기본다잡기⑴정답은‘정답007쪽’에있습니다.111쪽~116쪽019.8÷7=;1(0*;÷7=_=;1!0$;=1.4(cid:8951);1(0*;÷7=_=;1!0$;=1.4소수한자리수는분모가10인분수로나타냅니다.0232.5÷13=:£1™0∞:÷13=_=;1@0%;=2.5(cid:8951):£1™0∞:÷13=_=;1@0%;=2.50319.2÷12=:¡1ª0™:÷12=_=;1!0^;=1.6(cid:8951)19.2, 16, 1.604➊10.5÷5=:¡1º0∞:÷5=_=;1@0!;=2.1→㉠=5, ㉡=21▶3점➋㉠+㉡=5+21=26▶2점0506070809(cid:8951)6.9, 2.342.33<“6.94643.943.940.0406.92<“13.8412401.8401.840.00(cid:8951)5.318005.324<“127.21812040007.240007.2400.000(cid:8951)3.2403.24<“12.8412405.8405.840.00(cid:8951)1.941.95<“9.54544.544.540.0(cid:8951)1.341.33<“3.94343.943.940.015/121105/10112/116192÷10113/125325÷10113/125325÷1017/11498/1017/11498/10채점기준➊㉠과㉡에알맞은수를각각구한경우3점5점➋㉠과㉡에알맞은수의합을구한경우2점10➊나누는수가8로같고몫이26에서2.6으로;1¡0;배되었으므로나뉠수도;1¡0;배되어야합니다.208→20.8X▶3점➋따라서(cid:8641)안에알맞은수는20.8입니다.▶2점어떤수를;1¡0;배하면소수점이왼쪽으로한칸옮겨집니다.11140.8<156.8<163.2<169.6→가장큰수: 169.6(가장큰수)÷32=169.6÷32=5.3(cid:8951)5.312[방법1]273÷7=39→27.3÷7=3.9[방법2][방법3]27.3÷7=:™1¶0£:÷7=_[방법3]27.3÷7=;1#0(;=3.913(서우가가지고있는리본끈의길이)=(동원이가가지고있는리본끈의길이)÷6=54.6÷6=9.1(m)(cid:8951)9.1m14(한사람당모은헌종이의무게)=(이번달에모은헌종이의무게)÷(학생수)=31.5÷15=2.1(kg)(cid:8951)2.1kg17/139273/103.97<“27.3216363 0채점기준➊(cid:8641)안에알맞은수를구하는과정을쓴경우3점5점➋(cid:8641)안에알맞은수를구한경우2점채점기준두가지방법으로계산한경우5점5점한가지방법으로계산한경우2점틀리는이유│곱셈식과나눗셈식의관계를이용하여(cid:8641)÷8=2.6→(cid:8641)=2.6_8=20.8로구한경우해결방안│몫이;1¡0;배가되면나뉠수도;1¡0;배가됩니다. 15초등쎈수학5-2정답(062~077) 2015.6.2 7:56 PM 페이지062 SinsagoHitec 4. 소수의나눗셈063본문|111쪽-114쪽4단원1576.8kg은0.1kg이768개입니다. →㉠=768768개를6명이똑같이나누어가지면768÷6=128(개) →㉡=1280.1kg짜리무게추128개의무게는12.8kg입니다.→㉢=12.8(cid:8951)768, 128, 12.816➊1시간=60분이므로1시간3분=63분▶2점➋(1분동안나온물의양)=(전체나온물의양)÷(걸린시간)=170.1÷63=2.7(L)▶4점1721.93÷17=:™1¡0ª0£:÷17=_21.93÷17=;1!0@0(;=1.29(cid:8951)㉢, ㉡, ㉣, ㉠1814.52÷6=:¡1¢0∞0™:÷6=_=;1@0$0@;=2.42(cid:8951):¡1¢0∞0™:÷6=_=;1@0$0@;=2.42소수두자리수는분모가100인분수로나타냅니다.19➊7.02는소수두자리수이므로분모가100인분수로고쳐야하는데분모가10인분수로고쳐서잘못되었습니다.▶3점➋[바른계산]7.02÷3=;1&0)0@;÷3=_[바른계산]7.02÷3=;1@0#0$;=2.34▶2점13/1234702/10016/12421452/10016/12421452/100117/11292193/1002029.85를30으로어림하면30÷5=629.85<30이므로29.85÷5<618.42를18로어림하면18÷3=618.42>18이므로18.42÷3>6(cid:8951)2122(cid:8951)2.38(cid:8951)3.56몫의소수점은나뉠수의소수점자리에맞추어찍습니다.23242526㉠㉡㉢㉣(cid:8951)㉡14005.6332<“180.161416014020.114019.214000.9614000.9614000.00505.359<“48.15545503.1502.750.04550.04550.00014005.3623<“123.281411514008.214006.914001.3814001.3814000.001405.2613<“68.3814651403.31402.61400.781400.78140.000(cid:8951)2.31, 3.641403.6412<“43.6814361407.61407.2140.048140.048140.000502.315<“11.55510501.5501.550.00550.00550.000(cid:8951)4.3714004.3762<“270.941424814022.914018.614004.3414004.34140.0000(cid:8951)2.421402.4219<“45.9814381407.91407.6140.038140.038140.000403.566<“21.36418403.3403.040.03640.03640.00042.384<“9.524841.541.240.3240.3240.0029.85÷518.42÷3틀리는이유│1시간3분을63분으로고치지않고170.1÷13으로계산한경우해결방안│걸린시간을분으로나타내어계산합니다.채점기준➊물이나오는데걸린시간을분으로나타낸경우2점6점➋1분동안나온물의양을구한경우4점㉢㉡㉣㉠틀리는이유│계산중간과정만보고틀린부분이없다고생각하는경우해결방안│소수첫째자리숫자가0이어도소수점아래자리수가두자리이므로7.02는소수두자리수입니다.채점기준➊계산이잘못된이유를설명한경우3점5점➋바르게계산한경우2점 15초등쎈수학5-2정답(062~077) 2015.6.2 7:56 PM 페이지063 SinsagoHitec 064쎈수학5－24소수의나눗셈27➊몫의소수점을나뉠수의소수점자리에맞추어찍어야하는데잘못찍었습니다.▶3점➋[바른계산]▶2점28(한사람이가지게되는철사의길이)=(전체철사의길이)÷(사람수)=30.52÷7=4.36(`m)(cid:8951)4.36m292주일은14일이므로(하루에마신생수의양)=(2주일동안마신생수의양)÷(날수)=45.64÷14=3.26(L)(cid:8951)3.26L일주일은7일입니다.300.01m가115개이므로한사람당1.15m씩가지게됩니다. 따라서잘못말한학생은고은이입니다.(cid:8951)고은31➊(가습기7대의무게)=(가습기한대의무게)_7=0.8_7=5.6(`kg)▶2점➋(라디오8대의무게)=(전체무게)-(가습기7대의무게)=16.16-5.6=10.56(`kg)▶2점➌(라디오한대의무게)=(라디오8대의무게)÷8=10.56÷8=1.32(kg)▶2점403.788<“30.244244062405640.06440.06440.000327.68÷8=;1&0^0*;÷8=_7.68÷8=;1ª0§0;=0.96(cid:8951)⑤332.4÷4=;1@0$;÷4=_=;1§0;=0.6(cid:8951);1@0$;÷4=_=;1§0;=0.634➊3.42÷6=;1#0$0@;÷6=_➊3.42÷6=;1∞0¶0;=0.57→㉠=57, ㉡=6, ㉢=57, ㉣=0.57▶3점➋따라서㉠에알맞은수와같은것은㉢㉢``입니다.▶2점350부터4.8까지의작은눈금수: 48개작은눈금48개를6부분으로나누면(한부분의작은눈금수)=48÷6=8(개)작은눈금한칸의크기가0.1이므로8칸의크기는0.8입니다.→4.8÷6=0.8(cid:8951)；0.83637(cid:8951)0.46(cid:8951)0.763839(cid:8951)0.63(cid:8951)0.42140.4216<“6.72146.4140.3214.032140.0040.635<“3.1543.040.1540.1540.001400.7623<“17.48141611401.381401.38140.0040.468<“3.6843.240.4840.4840.000.51.52.53.54.503524116/157342/10014/1624/1014/1624/1018/196768/100틀리는이유│라디오8대의무게를16.16-0.8로계산하는경우해결방안│라디오8대의무게는전체무게에서가습기7대의무게를뺀것과같습니다.채점기준➊가습기7대의무게를구한경우2점6점➋라디오8대의무게를구한경우2점➌라디오한대의무게를구한경우2점채점기준➊㉠, ㉡, ㉢, ㉣에알맞은수를각각구한경우3점5점➋㉠에알맞은수와같은것을찾아기호를쓴경우2점채점기준➊계산이잘못된이유를설명한경우3점5점➋바르게계산한경우2점③④①② 15초등쎈수학5-2정답(062~077) 2015.6.2 7:56 PM 페이지064 SinsagoHitec 4. 소수의나눗셈065본문|114쪽-120쪽4단원40(나뉠수)<(나누는수) →(몫)<1①26.24>8→26.24÷8>1②5.74<7→5.74÷7<1③38.4>12→38.4÷12>1④55.2>6→55.2÷6>1⑤41.04>9→41.04÷9>1(cid:8951)②나눗셈의몫은다음과같습니다.①3.28②0.82③3.2④9.2⑤4.56419를17로나눌수없으므로몫의일의자리에0을쓰고소수점을찍어야하는데몫의자리를잘못맞추어써서잘못되었습니다.(cid:8951)5.4에◯`표；42➊나누는수가9로같고몫인62는0.62의100배이므로▶3점➋나뉠수인558도㉠의100배입니다.▶2점몫이100배가되면나뉠수도100배가됩니다.43(비커한개에담은알코올의양)=2.24÷8=0.28(L)(cid:8951)0.28L44➊(참치통조림한개의무게)➊=(참치통조림28개의무게)÷28=7.84÷28=➋0.28(kg)45⑴(리본13개를만드는데필요한색테이프의길이)=4.56+0.12=4.68(m)⑵(리본한개를만드는데필요한색테이프의길이)=4.68÷13=0.36(m)(cid:8951)⑴4.68m⑵0.36m140.5417<“9.18148.5140.6814.068140.00틀리는이유│나눗셈의몫을모두구하다계산을잘못하는경우해결방안│나뉠수와나누는수의크기를비교합니다.틀리는이유│두나눗셈식사이의관계를알아내지못한경우해결방안│나누는수가같으므로몫이몇배인지이용하여나뉠수가몇배인지구합니다.채점기준➊558은㉠`의몇배인지구하는과정을쓴경우3점5점➋558은㉠`의몇배인지구한경우2점✽A단계기본다잡기⑵정답은‘정답008쪽’에있습니다.120쪽~133쪽0119.5÷6=:¡1ª0∞:÷6=_=;2^0%;=;1#0@0%;=3.25(cid:8951):¡1ª0∞:÷6=_=;2^0%;=;1#0@0%;=3.2502➊하리；▶2점➋7.4를분자가740인분수로고치려면분모가100이어야하는데10으로써서잘못되었습니다.[바른계산]7.4÷4=;1&0$0);÷4=_=;1!0*0%;=1.85▶3점0304(cid:8951)1.35(cid:8951)2.150506(cid:8951)4.25(cid:8951)5.15몫이소수로나누어떨어질때까지나뉠수의오른쪽끝자리에0이계속있는것으로생각하고계산합니다.07(cid:8951)3.56503.565<“17.80515502.8402.5500.30400.3040.0005.158<“41.20401284040 04.256<“25.502415123030 01402.1516<“34.4014321402.41401.61408014080140.0041.354<“5.404441441240.2040.2040.0014/1185740÷10016/265195÷1016/265195÷10채점기준➊참지통조림한개의무게를구하는과정을쓴경우3점5점➋참지통조림한개의무게를구한경우2점채점기준➊계산을잘못한학생의이름을쓴경우2점5점➋바르게계산한경우3점 15초등쎈수학5-2정답(062~077) 2015.6.8 3:57 PM 페이지065 SinsagoHitec 066쎈수학5－24소수의나눗셈1318.1÷2=:¡1•0¡:÷2=:¡1•0¡:_;2!;=:¡2•0¡:=;1(0)0%;=9.0518.1÷2=:¡1•0¡0º:÷2=_=;1(0)0%;=9.05(cid:8951):¡1•0¡:÷2=:¡1•0¡:_;2!;=:¡2•0¡:=;1(0)0%;=9.05또는:¡1•0¡0º:÷2=_=;1(0)0%;=9.0514➊40.4÷5=:¢1º0¢:_;5!;=:¢5º0¢:(cid:9233)40.4÷5=;1*0)0*;=8.08→㉠=10, ㉡=50, ㉢=808▶3점➋㉠+㉡+㉢=10+50+808=868▶2점1516(cid:8951)2.04(cid:8951)3.051718(cid:8951)9.05(cid:8951)4.0519(cid:8951)6.05, 15.08, 12.06112.069<“108.5491818.54.54 015.085<“75.40525254040 06.054<“24.20242020 04.0518<“72.90729090 09.056<“54.30543030 01443.0514<“42.7014421400.701400.701400..042.045<“10.20410400.20400.2040.00012/19051810/10012/19051810/100틀리는이유│㉢의값을404라생각하여틀리는경우해결방안│㉢`이있는분수에서분모가100이므로분모가100인분수로고쳤다는것을생각하여해결합니다.채점기준➊㉠, ㉡, ㉢`에알맞은수를각각구한경우3점5점➋㉠, ㉡, ㉢`에알맞은수의합을구한경우2점08①②③④⑤(cid:8951)③091시간=60분이므로(양초가1분동안탄길이)=(양초가1시간동안탄길이)÷60=15.6÷60=0.26(`cm)(cid:8951)0.26cm10(하루에늦게가는시간)=(2주일에늦게가는시간)÷14=30.1÷14=2.15(분)(cid:8951)2.15분2주일은14일입니다.11➊(샴푸12개의무게)=(샴푸를담은상자의무게)-(빈상자의무게)=14.1-0.3=13.8(`kg)▶3점➋(샴푸한개의무게)=(샴푸12개의무게)÷12=13.8÷12=1.15(`kg)▶3점128.2÷4=;1*0@0);÷4=_=;1@0)0%;=2.05(cid:8951)㉢, ㉡, ㉠, ㉣82÷4의몫은자연수로나누어떨어지지않으므로분모가100인분수;1*0@0);으로고친후계산합니다.14/1205820/1006.2534<“212.502048568170170 02.2522<“49.5044554411011008.55<“42.540252500.458<“3.6032404003.256<“19.501815123030 0틀리는이유│나뉠수의오른쪽끝자리에0을내려계산하지않는다는것의의미를모르는경우해결방안│나뉠수의오른쪽끝자리에0을내려계산하지않아도나누어떨어지는것을찾습니다.채점기준➊샴푸12개의무게를구한경우3점6점➋샴푸한개의무게를구한경우3점㉢㉡㉠㉣ 15초등쎈수학5-2정답(062~077) 2015.6.2 7:56 PM 페이지066 SinsagoHitec 4. 소수의나눗셈067본문|120쪽-124쪽4단원20➊㉠72.48÷24=3.02을㉡24.16÷8=3.02㉢91.5÷30=3.05을㉣45.3÷15=3.02▶3점➋따라서나눗셈의몫이다른것은㉢㉢입니다.▶2점21(컵한개에담아야하는부침개가루의무게)=528.6÷12=44.05(`g)(cid:8951)44.05g22(1km를달리는데사용하는기름의양)=99.27÷9=11.03(`L)(cid:8951)11.03L23➊연필1타는12자루이므로(연필2타)=12_2=24(자루)▶2점➋(연필한자루의무게)=192.48÷24=8.02(g)▶3점➊(연필1타의무게)=192.48÷2=96.24(g)▶2점➋연필1타는12자루이므로(연필한자루의무게)=96.24÷12=8.02(g)▶3점2427÷12=:™1¶0º0º:÷12=_=;1@0@0%;=2.25(cid:8951):™1¶0º0º:÷12=_=;1@0@0%;=2.252510÷8=:¡1º0º0º:÷8=_=;1!0@0%;=1.25(cid:8951):¡1º0º0º:÷8=_=;1!0@0%;=1.2518/11251000/10018/11251000/100112/12252700/100112/12252700/100틀리는이유│연필1타가몇자루인지모르는경우해결방안│연필1타는12자루이므로연필2타는12_2=24(자루)입니다.채점기준➊연필2타가몇자루인지구한경우2점5점➋연필한자루의무게를구한경우3점채점기준➊연필1타의무게를구한경우2점5점➋연필한자루의무게를구한경우3점269÷2=;1(0);÷2=_=;1$0%;=4.59÷2=9_;2!;=9_;1∞0;=;1$0%;=4.5(cid:8951)③27282930몫이소수로나누어떨어질때까지나뉠수의오른쪽끝자리에0이계속있는것으로생각하고계산합니다.31➊①②③➋나눗셈의몫을표에서찾으면①1.125→지지지②2.4→우지지③0.84→개따라서번호순서대로글자를읽으면지우개입니다.▶2점32(한시간동안달린거리)=(전체달린거리)÷(달린시간)=385÷4=96.25(km)(cid:8951)96.25km0.8425<“21.00200100100 02.45<“12.0102020 01.12516<“18.0001620164032 80 80 0(cid:8951)1.3751.3758<“11.00083024605640400(cid:8951)2.52.54<“10.0820200(cid:8951)0.251440.2548<“12.001409.61402.401402.401400.00(cid:8951)2.81442.815<“42.014301412.01412.01404.012/14590/10틀리는이유│분수로고쳐서계산하는과정을한가지만아는경우해결방안│9÷2에서9를;1(0);으로고쳐서계산하거나9_;2!;로고쳐서계산할수있습니다.채점기준➊나눗셈의몫을각각구한경우3점5점➋나눗셈의몫을표에서찾아번호순서대로글자를읽은경우2점채점기준➊나눗셈의몫을각각구한경우3점5점➋나눗셈의몫이다른것을찾아기호를쓴경우2점①②④⑤▶3점 15초등쎈수학5-2정답(062~077) 2015.6.8 3:57 PM 페이지067 SinsagoHitec 4142434445㉠8÷6=1.33yy→1.4㉡11÷5=2.2㉢68÷17=4(cid:8951)㉠, 1.446➊11>6이므로큰수: 11, 작은수: 6▶2점➋(큰수)÷(작은수)=11÷6=1.833……→1.83▶3점4741÷14=2.928yy몫을올림하여소수첫째자리까지나타내기: 41÷14=2.92yy→3(3.0)몫을올림하여소수둘째자리까지나타내기:41÷14=2.928yy→2.93→두수의차: 3-2.93=0.07(cid:8951)0.0748(컵한개에담은물의양)=(전체물의양)÷(컵수)=3÷7=0.4≤2yy→0.4L(cid:8951)0.4L2.23013<“29.00“0 263026140139 103.3336<“20.00“0 1820182018 2018 22.4449<“22.00“0 1840364036 4036 42.81811<“31.00“0 22.0009.0008.8002.202.1102.0902.0882.002068쎈수학5－24소수의나눗셈채점기준➊큰수와작은수를구한경우2점5점➋큰수를작은수로나눈몫을버림하여소수둘째자리까지구한경우3점틀리는이유│2.92……를올림하여소수첫째자리까지나타내지못하는경우해결방안│소수첫째자리미만을올림하여소수첫째자리숫자9에1을더하면3.0=3이됩니다.33➊8도막중한도막으로국을끓였으므로(국을끓이는데사용한쇠고기의무게)=(전체쇠고기의무게)÷(도막수)=2÷8=➋0.25(kg)34(진희가산사과의무게)=2.5_8=20(kg)(한봉지에들어있는사과의무게)=20÷16=1.25(kg)(cid:8951)1.25kg3536373839➊성희: 5÷9=0.5≤5yy→0.6진수: 13÷7=1.8≤5yy→1.9▶3점➋따라서몫을반올림하여소수첫째자리까지바르게나타낸사람은진수입니다.▶2점몫을반올림할때는구하려는자리바로아래자리의숫자가0, 1, 2, 3, 4이면버리고, 5, 6, 7, 8, 9이면올립니다.40㉠16÷7=2.28≤5……→2.29㉡22÷3=7.33≤3……→7.33㉢41÷9=4.55≤5……→4.56따라서소수둘째자리숫자가6인것은㉢`입니다.(cid:8951)㉢1.8312<“22.00“12.4610.0619.66.40.36. 42.669<“24.00“1860546054 64.336<“26.00“2420182018 22.147<“15.00“1410573028 2채점기준➊나눗셈의몫을반올림하여소수첫째자리까지각각나타낸경우3점5점➋몫을바르게나타낸사람을찾아이름을쓴경우2점(cid:8951)2.7(cid:8951)1.8(cid:8951)2.1(cid:8951)4.3(cid:8951)2.81(cid:8951)2.44(cid:8951)3.33(cid:8951)2.23틀리는이유│한봉지에들어있는사과를2.5÷16으로구한경우해결방안│(한상자에들어있는사과의무게)=(진희가산사과의무게)÷(봉지수)를이용하여해결합니다.yy→2.1yy→4.3yy→2.81yy→2.44yy→2.7yy→3.33yy→1.8yy→2.23채점기준➊국을끓이는데사용한쇠고기의무게를구하는과정을쓴경우3점5점➋국을끓이는데사용한쇠고기의무게를구한경우2점 15초등쎈수학5-2정답(062~077) 2015.6.2 7:56 PM 페이지068 SinsagoHitec 4. 소수의나눗셈069본문|124쪽-127쪽4단원49(포장김치한팩의무게)=(포장김치6팩의무게)÷6=19÷6=3.16……→3.1kg(cid:8951)3.1kg50➊(한도막의길이)=(전체나무막대의길이)÷(도막수)=13÷3=4.33≤3……→➋4.33m511분=60초이므로10분38초=638초(한번운행하는데걸린시간)=(9번운행하는데걸린시간)÷9=638÷9=70.888yy→70.89초(cid:8951)70.89초(cid:8784)분▲초=(60_(cid:8784)+▲)초52→1.32>1.3>1.29(cid:8951)1.3253→8.05>3.05>2.04(cid:8951)64.4÷830.6÷1554.9÷181234123.0518<“54.9041544111.904100.90410.0001522.0415<“30.601530151 .60151 .60150.00048.058<“64.404644.404.404.004121.2941<“52.8941414111.84108.24103.694103.69410.0001521.3215<“19.8015151514.81514.51500.301500.30150.00041.37<“9.14742.142.140.054➊9÷13=0.692yy㉠몫을반올림하여소수첫째자리까지나타내기: ㉠9÷13=0.6≤9yy→0.7▶2점➋㉡몫을반올림하여소수둘째자리까지나타내기: 9÷13=0.69≤2yy→0.69▶2점➌0.7>0.69이므로더큰수는㉠㉠``입니다.▶2점5543.4÷14=3.1이므로3.1>(cid:8641)→(cid:8641)=1, 2, 3(cid:8951)1, 2, 35614÷8=1.75, 7.4÷4=1.851.6<1.7<1.75<1.82<1.85<1.91따라서14÷8의몫보다크고7.4÷4의몫보다작은것은㉢입니다.(cid:8951)㉢수직선위에나타내면수의범위를더정확히알수있습니다.57➊51.45÷15=3.43이므로▶2점➋3.43<3.(cid:8641)6→(cid:8641)=4, 5, 6, 7, 8, 9▶3점(cid:8456)따라서1부터9까지의자연수중에서(cid:8641)안에들어갈수있는수는모두6개입니다.▶1점58(색칠된부분의넓이)=39.69÷9=4.41(m¤)(cid:8951)4.41m¤59색칠된부분은전체를2등분한것중의하나이므로(색칠된부분의넓이)=8.9÷2=4.45(cm¤)(cid:8951)4.45cm¤1.61.71.81.91.751.85채점기준➊한도막은몇m인지구하는과정을쓴경우3점5점➋한도막은몇m인지반올림하여소수둘째자리까지나타낸경우2점틀리는이유│3.1>(cid:8641)에서(cid:8641)=1, 2로구한경우해결방안│3도3.1보다작은자연수이므로(cid:8641)안에들어갈수있습니다.채점기준➊몫을반올림하여소수첫째자리까지나타낸경우2점6점➋몫을반올림하여소수둘째자리까지나타낸경우2점➌더큰수를찾아기호를쓴경우2점틀리는이유│(cid:8641)=4일때를구하지않은경우해결방안│(cid:8641)=4일때, 3.43<3.46이므로4도(cid:8641)안에들어갈수있습니다.채점기준➊나눗셈의몫을구한경우2점6점➋(cid:8641)안에들어갈수있는수를모두구한경우3점➌(cid:8641)안에들어갈수있는수의개수를구한경우1점㉠㉡㉢㉣ 15초등쎈수학5-2정답(062~077) 2015.6.2 7:56 PM 페이지069 SinsagoHitec 070쎈수학5－24소수의나눗셈66정오각형은5개의변의길이가모두같으므로(cid:8641)=(정오각형의둘레)÷(변의수)=13.75÷5=2.75(cm)(cid:8951)2.7567(정삼각형의둘레)=(사용한전체철사의길이)÷(만든정삼각형의수)=18÷4=4.5(`m)(정삼각형의한변)=(정삼각형의둘레)÷(변의수)=4.5÷3=1.5(`m)(cid:8951)1.5m합동인정삼각형은대응변의길이가모두같으므로둘레가같습니다.68곱셈식과나눗셈식의관계를이용하여곱셈식을나눗셈식으로나타내면0.13_6=0.78→0.78÷6=0.13따라서(cid:8641)안에알맞은수는0.13입니다.(cid:8951)0.136912_(cid:8641)=11.88→(cid:8641)=11.88÷12=0.99(cid:8951)0.99♥_(cid:8641)=▲→(cid:8641)=▲÷♥70➊76÷(cid:8641)=25→(cid:8641)=76÷25▶3점➋=3.04▶2점71어떤소수를(cid:8641)`라하여잘못계산한식을세우면(cid:8641)_15=43.65→(cid:8641)=43.65÷15=2.91(cid:8951)2.9172어떤소수를(cid:8641)`라하여잘못계산한식을세우면(cid:8641)_12=123.84→(cid:8641)=123.84÷12=10.32[바른계산](cid:8641)÷12=10.32÷12=0.86(cid:8951)0.8673➊어떤자연수를(cid:8641)`라하여잘못계산한식을세우면(cid:8641)_14=70→(cid:8641)=70÷14=5▶3점➋(cid:8641)÷9=5÷9=0.5≤5yy→0.6▶3점틀리는이유│(cid:8641)를구하는식을세우지못해틀리는경우해결방안│곱셈식과나눗셈식의관계를이용하여(cid:8641)안에알맞은소수를구하는식을만듭니다.채점기준➊어떤자연수를구한경우3점6점➋어떤자연수를9로나눈몫을반올림하여소수첫째자리까지구한경우3점60색칠된부분은전체를3등분한것중의하나이므로(색칠된부분의넓이)=37.5÷3=12.5(m¤)(cid:8951)12.5m¤61➊(전체테이프의넓이)=15_3=45(cm¤)▶3점➋(색칠된부분의넓이)=45÷6=7.5(cm¤)▶3점62(세로)=(직사각형의넓이)÷(가로)=21.84÷7=3.12(cm)(cid:8951)3.12cm63(밑변)=(평행사변형의넓이)÷(높이)=48.4÷8=6.05(m)(cid:8951)6.0564➊(㉮의넓이)➊=9_6=54(`cm¤)(㉯의넓이)=67.5-54=13.5(`cm¤)(cid:8641)_6÷2=13.5, (cid:8641)_6=13.5_2=27, (cid:8641)=27÷6=4.5(`cm)▶4점➋따라서(cid:8641)안에알맞은수는4.5입니다.▶2점➊도형은윗변이9cm, 아랫변이(9+(cid:8641))cm,높이가6cm인사다리꼴입니다.{9+(9+(cid:8641))}_6÷2=67.5, (18+(cid:8641))_6÷2=67.5,(18+(cid:8641))_6=67.5_2=135,18+(cid:8641)=135÷6=22.5,(cid:8641)=22.5-18=4.5(`cm)▶4점➋따라서(cid:8641)안에알맞은수는4.5입니다.▶2점65⑴(삼각형모양의꽃밭의넓이)=(밑변)_(높이)÷2=12.25_8÷2=98÷2=49(m¤)⑵(세로)=(직사각형모양의꽃밭의넓이)÷(가로)=(삼각형모양의꽃밭의넓이)÷(가로)⑵(세로)=49÷9=5.44≤4yy→5.44m(cid:8951)⑴49m¤(cid:100)⑵5.44m6`cm9cm㉮ ㉯ cm틀리는이유│도형의넓이를구하는방법을모르는경우해결방안│도형을직사각형과삼각형으로나누어넓이를구하거나도형이사다리꼴임을이용하여넓이를구합니다.채점기준➊(cid:8641)안에알맞은수를구하는과정을쓴경우4점6점➋(cid:8641)안에알맞은수를구한경우2점채점기준➊(cid:8641)안에알맞은소수를구하는과정을쓴경우3점5점➋(cid:8641)안에알맞은소수를구한경우2점채점기준➊전체테이프의넓이를구한경우3점6점➋색칠된부분의넓이를구한경우3점 15초등쎈수학5-2정답(062~077) 2015.6.8 3:58 PM 페이지070 SinsagoHitec 4. 소수의나눗셈071본문|127쪽-130쪽4단원74(점과점사이의간격수)=10-1=9(구간)(점과점사이의거리)=32.76÷9=3.64(cm)(cid:8951)3.64cm(점과점사이의간격수)=(점의수)-175➊(도로한쪽에설치하는조형물의수)=50÷2=25(개)▶2점➋(조형물과조형물사이의간격수)=25-1=24(구간)▶2점➌(조형물과조형물사이의거리)=2.88÷24=0.12(km)▶2점76소수점아래반복되는숫자: 6, 9(`2개)15÷2=7y1→몫의소수15째자리숫자: 6(cid:8951)677㉠30÷11=2.7272yy㉠소수점아래반복되는숫자: 7, 2(2개)40÷2=20→소수40째자리숫자: 2㉡25÷9=2.777yy소수점아래반복되는숫자: 7(1개)→소수40째자리숫자: 7(cid:8951)㉡㉠`에서소수점아래반복되는숫자가7, 2로2개이므로소수점아래홀수째자리숫자는7, 짝수째자리숫자는2입니다.78➊소수점아래반복되는숫자: 7,0,3(3개)▶3점➋101÷3=33y2이므로소수101째자리숫자: 0▶3점2.70370327<“73.00000“0 5419018910081 190 189 100 81 1979나뉠수가클수록, 나누는수가작을수록나눗셈의몫은커집니다.54.3÷2(cid:8837)(cid:8951)5, 4, 3, 2；27.15나뉠수가작을수록, 나누는수가클수록나눗셈의몫은작아집니다.80몫이가장크게되는나눗셈식을만들려면나뉠수는가장크게, 나누는수는가장작게해야합니다.76.5÷3(cid:8837)(cid:8951)7, 6, 5, 3；25.5㉠`<㉡`<㉢`<㉣`<㉤`일때몫이가장크게되는나눗셈식(cid:8641)`(cid:8641).(cid:8641)÷(cid:8641)→㉤`㉣.㉢÷㉠81➊몫이가장작게되는나눗셈식을만들려면나뉠수는가장작게, 나누는수는가장크게해야합니다.몫이가장작게되는나눗셈식: 34÷76▶3점➋▶3점㉠<㉡<㉢<㉣<㉤일때몫이가장작게되는나눗셈식(cid:8641)(cid:8641)÷(cid:8641)(cid:8641)→㉠㉡÷㉤㉣0.44776<“34.00“0 30.4360304 560 532 2825.53<“76.5616151515027.152<“54.30414141312110110 0채점기준➊소수점아래반복되는숫자를구한경우3점6점➋소수101째자리숫자를구한경우3점채점기준➊몫이가장작게되는나눗셈식을만든경우3점6점➋몫을소수둘째자리미만을버림하여나타낸경우3점틀리는이유│몫이가장크게되는나눗셈식을어떻게만들어야하는지모르는경우해결방안│몫이가장크게되는나눗셈식을만들려면나뉠수는가장크게, 나누는수는가장작게해야합니다.채점기준➊도로한쪽에설치하는조형물의수를구한경우2점6점➋조형물과조형물사이의간격수를구한경우2점➌조형물과조형물사이의거리를구한경우2점yyyy→0.44 15초등쎈수학5-2정답(062~077) 2015.6.2 7:56 PM 페이지071 SinsagoHitec 072쎈수학5－24소수의나눗셈[[86~~93]]서술형평가유형의입니다.86⑴(cid:8641)_6=36.6→(cid:8641)=36.6÷6；36.6÷6▶2점⑵▶2점⑶6.1▶1점87⑴(마름모의넓이)=(한대각선)_(다른대각선)÷2→7_(cid:8641)÷2=22.4,7_(cid:8641)=22.4_2=44.8,⑴→(cid:8641)=44.8÷7=6.4(cm)▶3점⑵6.4▶2점88⑴⑵; 7.54, 7.59▶2점⑵7.54<7.5(cid:8641)<7.59따라서1부터9까지의자연수중에서(cid:8641)안에들어갈수있는수는5, 6, 7, 8입니다.▶3점⑶5, 6, 7, 8▶1점89⑴첫째가갖게될쌀을(cid:8641)kg라하면(둘째가갖게될쌀)=(첫째가갖게될쌀)_2=(cid:8641)_2(막내가갖게될쌀)=(첫째가갖게될쌀)_3=(cid:8641)_3쌀을남김없이나누어가졌으므로세형제가갖게될쌀을합하면153.84kg입니다.(cid:8641)+(cid:8641)_2+(cid:8641)_3=(cid:8641)_6=153.84, (cid:8641)=153.84÷6=25.64(kg)▶4점⑵25.64kg▶2점90⑴상추를심은부분은전체밭을똑같이4부분으로나눈것중하나이므로(상추를심은부분의넓이)=(전체밭의넓이)÷4=69.4÷4=17.35(m¤)▶3점⑵17.35m¤▶2점807.598<“60.72856004.7004.0000.72000.7200.000807.544<“30.16828002.1002.0000.16000.1600.000606.16<“36.6636605.6605.660.0082무게추54개를16곳으로나누면54÷16=3`y`6→㉠=30.1kg짜리무게추64개를16곳으로나누면64÷16=4→㉡=4한곳에놓이는무게추는1kg짜리무게추3개와0.1kg짜리무게추4개이므로모두3.4kg입니다. →㉢=3.4(cid:8951)3, 4, 3.483네반의강의실의넓이가모두같으므로네반의강의실의가로는모두같습니다.(지혜반의강의실의가로)=22.48÷4=5.62(m)(지혜반의강의실의넓이)=5.62_5=28.1(m¤)네반의넓이가모두같으므로지혜반의강의실의넓이는전체반의강의실의넓이를4등분한것중의하나입니다.(전체반의강의실의넓이)=22.48_5=112.4(m¤)(지혜반의강의실의넓이)=112.4÷4=28.1(m¤)(cid:8951)28.1m¤84(소금을3명이똑같이나눈무게)=153.78÷3=51.26(g)51.26g과학생들이각각들고있는물건의무게의차가작을수록무게가더가까운물건입니다.준기`: 87.5-51.26=36.24`(g)미영`: 85.3-51.26=34.04(g)대현`: 51.26-33.46=17.8(g)17.8<34.04<36.24이므로소금을3명이똑같이나눈무게와가장가까운물건을들고있는학생은대현이입니다.(cid:8951)대현85<가로열쇠>❶98.4÷2=49.2❸1.92÷6=0.32❺36÷11=3.272……→3.27<세로열쇠>❶21.3÷5=4.26❷26.65÷13=2.05❹245÷9=27.222……→27.22(cid:8951)12345492203265723272 15초등쎈수학5-2정답(062~077) 2015.6.2 7:56 PM 페이지072 SinsagoHitec 4. 소수의나눗셈073본문|131쪽-134쪽4단원91⑴소수첫째자리계산에서7을13으로나눌수없으므로몫의소수첫째자리에0을써야하는데쓰지않아서잘못되었습니다.▶3점⑵▶2점92⑴(1분동안간거리)=(지영이네집~ 할머니댁)÷(걸린시간)=1.26÷7=0.18(km)▶3점⑵0.18km▶2점93⑴36÷6=6, 6÷6=1이므로규칙은(들어온수)÷6입니다. ▶3점⑵51÷6=8.5▶2점⑶8.5▶1점1802.0613<“26.7818264000.784000.78400.000134쪽~135쪽01(전체밀가루의무게)=22.4_4=89.6(kg)(하루에사용하는밀가루의무게)=(전체밀가루의무게)÷(날수)=89.6÷7=12.8(kg)(cid:8951)12.8kg일주일은7일입니다.02㉣`㉤-58=20이므로㉣`㉤=20+58=78→㉣=7, ㉤=8㉡`㉢_2=58이므로㉡`㉢=58÷2=29→㉡=2, ㉢=9203-㉥`㉦`㉧=0이므로㉥`㉦`㉧=203→㉥=2, ㉦=0, ㉧=329_㉠=203이므로㉠=203÷29=7(cid:8951)(위에서부터) 7；2, 9, 7, 8；2, 0, 30.2`㉠㉡`㉢<“㉣`.㉤“358`203㉥㉦㉧`003➊(처음직사각형의넓이)=6_16.7=100.2(cm¤)➋(줄인직사각형의가로)=6-1=5(cm)➌늘린직사각형의세로를(cid:8641)`cm라하면5_(cid:8641)=100.2→(cid:8641)=100.2÷5=20.04(cm)➍(늘려야하는세로)=20.04-16.7=3.34(cm)(cid:8951)3.34cm04일주일은7일이므로➊(하루에늦게가는시간)=(일주일에늦게가는시간)÷7=18.48÷7=2.64(분)하루에2.64분씩늦게가므로➋(5일동안늦게가는시간)=2.64_5=13.2(분)➌(5일후오전11시에가리키는시각)=오전11시-(5일동안늦게가는시간)=오전11시-13.2분=오전11시-(13분12초)=오전10시46분48초(cid:8951)10시46분48초1분=60초→;6¡0;분=1초0.2분=;1™0;분=;6!0@;분=12초05⑴40.5÷3=13.5→14장⑵28.32÷3=9.44→10장⑶(필요한색종이의수)=14_10=140(장)(cid:8951)⑴14장(cid:100)⑵10장(cid:100)⑶140장소수점이하를버림또는반올림으로계산하면도화지에빈공간이생길수있으므로딱맞게덮으려면올림하여자연수로색종이의수를구해야합니다.전체밀가루의무게를구한후날수로나눕니다.전략➊처음직사각형의넓이구하기→➋줄인직사각형의가로구하기→➌늘린직사각형의세로구하기→➍늘려야하는세로구하기푸는순서가로가6cm, 세로가16.7cm인직사각형이있습니다. 이직사각형의가로를1cm줄이면세로는몇cm늘려야처음넓이와같아집니까?(직사각형의넓이)=6_16.7=100.2(cm¤)▼6-1=5(cm)▼5_(cid:8641)=100.2▼(늘린직사각형의세로)=(cid:8641)cm▼➊하루에늦게가는시간구하기→➋5일동안늦게가는시간구하기→➌5일후오전11시에가리키는시각구하기푸는순서 15초등쎈수학5-2정답(062~077) 2015.6.8 3:58 PM 페이지073 SinsagoHitec 074쎈수학5－24소수의나눗셈09➊48.6★12=(48.6+12)÷12=60.6÷12▶4점➋=5.05▶3점10➊밑변12cm, 높이9cm로삼각형의넓이를구하면(삼각형의넓이)=12_9÷2=108÷2=54(cm¤)▶3점➋밑변15cm, 높이(cid:8641)cm로구한삼각형의넓이도54cm¤이므로15_(cid:8641)÷2=54, 15_(cid:8641)=54_2=108, (cid:8641)=108÷15=7.2(cm)따라서(cid:8641)안에알맞은소수는7.2입니다.▶4점(cid:8784)_▲÷2=(cid:8774)_♥÷2→(cid:8784)_▲=(cid:8774)_♥11➊52÷9=5.777yy에서52에가장작은소수를더하였을때나올수있는소수두자리수인몫은5.78입니다.▶4점➋어떤소수중에서가장작은수를(cid:8641)라하면(52+(cid:8641))÷9=5.78, 52+(cid:8641)=5.78_9=52.02,(cid:8641)=52.02-52=0.02▶4점채점기준➊나뉠수에어떤소수를더하였을때나올수있는소수두자리수인몫을구한경우4점8점➋어떤소수중에서가장작은수를구한경우4점06➊철326.4kg으로철근24개를만들었으므로(철근한개의무게)=326.4÷24=13.6(kg)▶4점➋(철근5개의무게)=(철근한개의무게)_5=13.6_5=68(kg)▶3점07➊45cm=0.45m이므로(그림의가로의길이의합)=0.45_7=3.15(m)(간격의길이의합)=(벽의길이)-(그림의가로의길이의합)=4.43-3.15=1.28(m)▶3점➋그림이7장이므로(간격수)=7+1=8(구간)▶2점➌각간격의길이는모두같으므로(그림과그림사이의간격)=1.28÷8=0.16(m)▶3점08➊(3에서20.9까지의크기)=20.9-3=17.9▶2점➋(한칸의크기)=(3에서20.9까지의크기)÷5=17.9÷5=3.58▶2점➌(cid:8641)안에알맞은수는3에서3.58씩3칸을간것이므로(cid:8641)=3+3.58_3=3+10.74=13.74▶3점채점기준➊간격의길이의합을구한경우3점8점➋간격수를구한경우2점➌그림과그림사이의간격을구한경우3점3에서20.9까지의크기인(20.9-3)을구한후5로나누면한칸의크기를구할수있습니다.전략가대신48.6을, 나대신12를넣어식을만듭니다.전략채점기준➊3에서20.9까지의크기를구한경우2점7점➋한칸의크기를구한경우2점➌(cid:8641)안에알맞은수를구한경우3점채점기준➊48.6★12를구하는과정을쓴경우4점7점➋48.6★12를구한경우3점채점기준➊삼각형의넓이를구한경우3점7점➋(cid:8641)` 안에알맞은소수를구한경우4점채점기준➊철근한개의무게를구한경우4점7점➋철근5개의무게를구한경우3점4.43m는그림7장의가로와벽과그림, 그림과그림사이8구간의길이의합입니다.전략52÷9의나뉠수에어떤소수를더하면몫이소수두자리수이고나머지가0이됩니다. 어떤소수중에서가장작은수는얼마인지풀이과정을쓰고, 답을구하시오.52÷9=5.777……▼{52+(어떤소수)}÷9=5.78▼52+(어떤소수)=5.78_9=52.02,(어떤소수)=52.02-52=0.02▶(cid:8784)cm(cid:8774)cm♥cm▲cm 15초등쎈수학5-2정답(062~077) 2015.6.8 3:58 PM 페이지074 SinsagoHitec 4. 소수의나눗셈075본문|135쪽-137쪽4단원136쪽~137쪽1회019.6÷8=;1(0^;÷8=_=;1!0@;=1.2(cid:8951)96, 12, 96；12, 1.202(cid:8951)6.4, 1.603몫이가장크게되는나눗셈식을만들려면나뉠수는가장크게, 나누는수는가장작게해야합니다.86.4÷2(cid:8837)(cid:8951)8, 6, 4, 2；43.204(cid:8951)3.1105➊(한도막의길이)=(전체대파의길이)÷(도막수)=43.29÷9▶3점➋=4.81(`cm)▶2점06㉠㉡㉢→0.32<0.34<0.35(cid:8951)㉡1400.3442<“14.281412.61401.681401.68140.000140.3223<“7.36146.9140.4614.046140.00140.3515<“5.25144.5140.7514.075140.0033.1111<“34.213312111111 0143.22<“86.47870670670647064700041.64<“6.44442.442.440.0106.48<“51.2748703.2703.270.0018/11296/1007(자동차가1분동안달리는거리)=(자동차가25분동안달리는거리)÷25=17.5÷25=0.7(km)(cid:8951)0.7km08➊▶3점➋(몫의합)=0.8+0.43=1.23▶2점0925.2>8이므로(큰수)÷(작은수)=25.2÷8=3.15(cid:8951)3.1510[방법1]41.2÷5=:¢1¡0™:÷5=:¢1¡0™:_;5!;[방법1]41.2÷5=:¢5¡0™:=;1*0@0$;=8.24[방법2]4120÷5=824→41.2÷5=8.24[방법3]41.2÷5=:¢1¡0™0º:÷5=:¢1¡0™0º:_;5!;=;1*0@0$;=8.24로계산하여도답으로인정합니다.11➊(아버지의몸무게)÷(미란이의몸무게)=66.5÷38▶3점➋=1.75(배)▶2점1213(cid:8951)㉠1801.0542<“44.1018424002.104002.10400.000(cid:8951)㉡1802.0435<“71.4018704001.404001.40400.0008.245<“41.204012102020 01400.4334<“14.621413.61401.021401.02140.0001400.817<“13.61413.6140.00채점기준➊한도막의길이를구하는과정을쓴경우3점5점➋한도막의길이를구한경우2점채점기준➊두나눗셈의몫을각각구한경우3점5점➋두나눗셈의몫의합을구한경우2점채점기준두가지방법으로계산한경우5점5점한가지방법으로계산한경우2점채점기준➊몇배인지구하는과정을쓴경우3점5점➋몇배인지구한경우2점8241 15초등쎈수학5-2정답(062~077) 2015.6.2 7:56 PM 페이지075 SinsagoHitec 076쎈수학5－24소수의나눗셈138쪽~139쪽2회01183.6÷27=÷27=_=;1^0*;=6.8(cid:8951)÷27=_=;1^0*;=6.80203(cid:8951)31.2(cid:8951)2.1404➊(1분동안받은물의양)=(13분동안받은물의양)÷13=106.6÷13▶3점➋=8.2(L)▶2점05나뉠수가;10!0;배가되면몫도;10!0;배가됩니다.1968÷6=328→19.68÷6=3.28(cid:8951)3.28나누는수가같을때나뉠수의소수점이왼쪽으로두칸옮겨지면몫의소수점도왼쪽으로두칸옮겨집니다.06(cid:8951)(위에서부터) 4.29, 1.95, 7.1507(한사람이가진털실의길이)=(전체털실의길이)÷(사람수)=31.68÷6=5.28(m)(cid:8951)5.28m307.153<“21.45321410.4419.3410.15419.1540.0001801.9511<“21.4518114010.44019.94010.554019.55400.000804.295<“21.45520401.4401.0400.45400.4540.0002.1445<“96.30906345180180 0331.23<“93.639303.4303.4303.6303.630.00127/1681836/10183610127/1681836/10183610채점기준➊성민이가1분동안받은물의양을구하는과정을쓴경우3점5점➋성민이가1분동안받은물의양을구한경우2점14색칠된부분은전체를4등분한것중의하나이므로(색칠된부분의넓이)=12.2÷4=3.05(cm¤)(cid:8951)3.05cm¤15정사각형은4개의변의길이가모두같으므로(한변)=(텃밭의둘레)÷(변의수)(한변)=24.2÷4=6.05(m)(cid:8951)6.05m16➊(나무와나무사이의간격수)=16-1=15(구간)▶2점➋(나무와나무사이의거리)=106.2÷15=7.08(`m)▶3점1746÷4=:¢1§0º:÷4=_=:¡1¡0∞:=11.5(cid:8951):¢1§0º:÷4=_=:¡1¡0∞:=11.5184_(cid:8641)=18→(cid:8641)=18÷4=4.5(cid:8951)4.519(cid:8951)6.3나눗셈의몫을소수첫째자리까지나타낼때몫의소수둘째자리숫자가5보다작으면버리고, 5보다크거나같으면올립니다.20➊(귤한개의무게)=(귤14개의무게)÷14=837÷14=59.785yy→➋59.79g6.339<“57.00“5430273027 314/1115460/1014/1115460/10채점기준➊나무와나무사이의간격수를구한경우2점5점➋나무와나무사이의거리를구한경우3점yy→6.3채점기준➊귤한개의무게를올림하여소수둘째자리까지나타내는과정을쓴경우3점5점➋귤한개의무게를올림하여소수둘째자리까지나타낸경우2점 15초등쎈수학5-2정답(062~077) 2015.6.8 6:3 PM 페이지076 SinsagoHitec 4. 소수의나눗셈077본문|137쪽-139쪽4단원14(가로)=(직사각형의넓이)÷(세로)=72.3÷6=12.05(m)(cid:8951)12.05m15➊㉠_16=32.8→㉠=32.8÷16=2.05▶2점➋24.4÷㉡=8→㉡=24.4÷8=3.05▶2점➌㉠+㉡=2.05+3.05=5.1▶1점16(cid:8951)1.6171시간18분=78분이므로(공원을한바퀴도는데걸린시간)=(공원을5바퀴도는데걸린시간)÷5=78÷5=15.6(분)(cid:8951)15.6분1826÷18=1.444yy→1.4443÷21=2.047yy→2.0478÷24=3.25(cid:8951)78÷24에◯표197÷11=0.6363yy소수점아래반복되는숫자: 6, 3(2개)49÷2=24y1→소수49째자리숫자: 6(cid:8951)6소수점아래규칙을찾을때까지나눗셈을합니다.20➊(가로한줄에필요한타일의수)=45÷7=6.42yy→7장▶2점➋(세로한줄에필요한타일의수)=60÷7=8.57yy→9장▶2점➌(필요한타일의수)=7_9=63(장)▶1점1801.620<“32.018204012.04012.0400.00채점기준➊㉠`에알맞은수를구한경우2점5점➋㉡`에알맞은수를구한경우2점➌㉠`과㉡`에알맞은수의합을구한경우1점채점기준➊가로한줄에필요한타일의수를구한경우2점5점➋세로한줄에필요한타일의수를구한경우2점➌필요한타일의수를구한경우1점08①4.2÷3=1.4②6.4÷8=0.8③17.04÷4=4.26④2.34÷2=1.17⑤78.72÷32=2.46(나뉠수)<(나누는수)이면나눗셈의몫이1보다작습니다.①4.2>3→4.2÷3>1②6.4<8→6.4÷8<1③17.04>4→17.04÷4>1④2.34>2→2.34÷2>1⑤78.72>32→78.72÷32>1(cid:8951)②0917.1÷18=0.950.95<0.9(cid:8641)이므로5<(cid:8641)→(cid:8641)=6, 7, 8, 9(4개)(cid:8951)4개10➊어떤소수를(cid:8641)라하면(cid:8641)÷3=1.32→(cid:8641)=1.32_3=3.96▶2점➋(cid:8641)÷6=3.96÷6=0.66▶3점11㉮37.2÷8=4.65㉯100.1÷22=4.55→4.65>4.55(cid:8951)㉮12➊정육각형은6개의변의길이가모두같으므로(cid:8641)=24.9÷6=4.15(cm)▶3점➋따라서(cid:8641)안에알맞은수는4.15입니다.▶2점(둘레가(cid:8784)인정▲각형의한변의길이)=(cid:8784)÷▲13➊소수첫째자리계산에서7을14로나눌수없으므로몫의소수첫째자리에0을써야하는데쓰지않아서잘못되었습니다.▶3점➋[바른계산]4.0514<“56.705670700채점기준➊어떤소수를구한경우2점5점➋어떤소수를6으로나눈몫을구한경우3점채점기준➊(cid:8641)안에알맞은수를구하는과정을쓴경우3점5점➋(cid:8641)안에알맞은수를구한경우2점채점기준➊계산이잘못된이유를설명한경우3점5점➋바르게계산한경우2점▶2점 15초등쎈수학5-2정답(062~077) 2015.6.2 7:56 PM 페이지077 SinsagoHitec 078쎈수학5－25여러가지단위✽A단계기본다잡기정답은‘정답009쪽’에있습니다.01(정사각형의넓이)=20_20=400(m¤)100m¤=1a이므로400m¤=4a(cid:8951)402100m¤=1a이므로6400m¤=64a(cid:8951)64a▲00m¤=▲a031a=(한변이10m인정사각형의넓이)=10_10=100(m¤)1a=100m¤이므로1a는1m¤의100배입니다.1a는1m¤에비해가로가10배, 세로가10배더크므로1a는1m¤의100배입니다.04200m¤=2a이므로(배추를심은부분의넓이)=90-2=88(a)(cid:8951)88a넓이의단위가다를경우하나의단위로통일하여계산합니다.05(평행사변형의넓이)=90_60=5400(m¤)=54(a)(cid:8951)54a(평행사변형의넓이)=(밑변)_(높이)061ha는한변이100m인정사각형의넓이입니다.(cid:8951)(cid:9066)한변이100m인정사각형의넓이를1ha라씁니다.‘한변이100cm인정사각형의넓이를1m¤라씁니다.’도답으로인정합니다.07➊(운동경기장의넓이)=400_300=120000(m¤)=1200(a)=➋12(ha)145쪽~155쪽08➊논의넓이를ha로나타내면200a=2ha이므로▶3점➋(논과산의넓이의합)=2+11=13(ha)▶2점09100a=1ha이므로㉠0.12ha=12a㉡260a=2.6ha㉢3.4ha=340a㉣780a=7.8ha→((cid:8641)안에알맞은수의합)=12+2.6+3.4+780=798(cid:8951)798ha를a로나타낼때에는소수점을오른쪽으로두칸, a를ha로나타낼때에는소수점을왼쪽으로두칸옮깁니다.101km¤=100ha이므로3km¤=300ha따라서넓이가3km¤와같은것을들고있는학생은미소입니다.(cid:8951)미소영현: 30ha=0.3km¤11(정사각형의넓이)=7×7=49(km¤)(cid:8951)49122000m=2km, 700m=0.7km이므로(사다리꼴의넓이)={(윗변)+(아랫변)}_(높이)÷2=(1+2)_0.7÷2=3_0.7÷2=2.1÷2=1.05(km¤)1km=1000m이므로(사다리꼴의넓이)={(윗변)+(아랫변)}_(높이)÷2=(1000+2000)_700÷2=3000_700÷2=2100000÷2=1050000(m¤)=10500(a)=105(ha)=1.05(km¤)(cid:8951)1.05km¤13정사각형은네변의길이가모두같으므로(아파트단지의한변)=32÷4=8(km)(아파트단지의넓이)=8_8=64(km¤)=6400(ha)(cid:8951)6400ha채점기준1a가1m¤의100배인이유를설명한경우5점틀리는이유│m¤를ha로나타내지못하는경우해결방안│m¤를a로나타낸후a를ha로나타냅니다.틀리는이유│길이의단위를통일하지않고넓이를구하는경우해결방안│m를km로또는km를m로나타낸후넓이를km¤로나타냅니다.채점기준➊논의넓이를ha로나타낸경우3점5점➋논과산의넓이의합이몇ha인지구한경우2점채점기준➊운동경기장의넓이가몇ha인지구하는과정을쓴경우3점5점➋운동경기장의넓이가몇ha인지구한경우2점 15초등쎈수학5-2정답(078-087) 2015.6.8 4:0 PM 페이지078 SinsagoHitec 5. 여러가지단위079본문|145쪽-148쪽5단원14➊1000m=1km이므로350m=0.35km▶2점➋1.2km_350m=1.2km_0.35km=0.42km¤▶4점km를m로나타내어계산해도됩니다.15①뒷산의넓이→ha②테니스경기장의넓이→a③교실바닥의넓이→m¤④경기도의넓이→km¤⑤책상의넓이→m¤(또는cm¤)(cid:8951)①16시, 도등은km¤로나타내는것이편리합니다.(cid:8951)km¤17어린이대공원의넓이는ha로나타내는것이편리합니다.(cid:8951)ha18방: m¤, 실내수영장: a, 염전: ha, 인천: km¤(cid:8951)(cid:9066)인천의넓이는약1002km¤입니다.19넓이의단위를a로통일하면0.2km¤=2000a, 60000m¤=600a, 35ha=3500a2000a+(cid:8641)a+600a=3500a, 2600a+(cid:8641)a=3500a→(cid:8641)=3500-2600=900(a)(cid:8951)9002061km¤=6100ha=610000a=61000000m¤(cid:8951)㉢1km¤=100ha=10000a=1000000m¤21610ha=61000a=6100000m¤(cid:8951)㉠22➊1km¤=10000a이므로3km¤=30000a따라서3km¤는1a의30000배입니다. →㉠=30000▶2점➋1ha=10000m¤이므로5ha=50000m¤따라서50m¤는5ha의;10¡00;(0.001)배입니다.→㉡=;10¡00;(0.001)▶3점23➊㉠26000a=260ha=2.6km¤→(cid:8641)=km¤㉡38ha=0.38km¤ →(cid:8641)=km¤㉢90000m¤=900a=9ha→(cid:8641)=ha㉣7000ha=70km¤ →(cid:8641)=km¤▶3점➋따라서넓이의단위가다른것은㉢㉢입니다.▶2점24진경이네아파트주차장의넓이를(cid:8641)ha라하면(우진이네아파트주차장의넓이)=(진경이네아파트주차장의넓이)+1300a→0.65km¤=(cid:8641)ha+1300a0.65km¤=65ha, 1300a=13ha65ha=(cid:8641)ha+13ha→(cid:8641)=65-13=52(ha)(cid:8951)52ha25➊한차선의도로폭이4m이고2개차선이므로(확장되는도로폭)=4_2=8(m)▶1점➋2.6km=2600m이므로(확장되는도로의넓이)=2600_8=20800(m¤)▶3점➌=208(a)▶2점2617.5a=1750m¤이므로(삼각형의넓이)=(밑변)_(높이)÷2→(밑변)=(삼각형의넓이)_2÷(높이)=1750_2÷50=70(m)(cid:8951)7027(평행사변형의넓이)=(밑변)_(높이)→(높이)=(평행사변형의넓이)÷(밑변)=12÷4=3(km)(cid:8951)3km(평행사변형의넓이)=(밑변)_(높이)이므로①(밑변)=(평행사변형의넓이)÷(높이)②(높이)=(평행사변형의넓이)÷(밑변)틀리는이유│(밑변)=(삼각형의넓이)÷(높이)로구하는경우해결방안│(삼각형의넓이)=(밑변)_(높이)÷2이므로(밑변)=(삼각형의넓이)_2÷(높이)입니다.채점기준➊㉠에알맞은수를구한경우2점5점➋㉡에알맞은수를구한경우3점채점기준➊m를km로나타낸경우2점6점➋계산결과를km¤로나타낸경우4점틀리는이유│무조건큰수를작은수로나누어몇배인지구한경우해결방안│(cid:8784)는▲의(cid:8641)배이면(cid:8641)=(cid:8784)÷▲입니다.채점기준➊확장되는도로폭을구한경우1점6점➋확장되는도로의넓이는몇a인지구하는과정을쓴경우3점➌확장되는도로의넓이는몇a인지구한경우2점채점기준➊(cid:8641)안에알맞은넓이의단위를각각찾은경우3점5점➋(cid:8641)안에알맞은넓이의단위가다른것을찾아기호를쓴경우2점 15초등쎈수학5-2정답(078-087) 2015.6.8 4:0 PM 페이지079 SinsagoHitec 080쎈수학5－25여러가지단위2818000ha=180km¤이므로(마름모의넓이)=(한대각선)_(다른대각선)÷2→(다른대각선)=(마름모의넓이)_2÷(한대각선)=180_2÷12=30(km)(cid:8951)3029➊150ha=1.5km¤이므로▶2점➋(직사각형의넓이)=(가로)_(세로)→(가로)=(직사각형의넓이)÷(세로)=1.5÷3=0.5(km)=500(m)▶3점30➊64a=6400m¤▶2점➋정사각형모양의과수원의한변을(cid:8641)m라하면(cid:8641)_(cid:8641)=6400, 80_80=6400이므로▶2점➌(cid:8641)=80m▶2점315.6km¤=560ha이므로560ha<5600ha→5.6km¤<5600ha5600ha=56km¤이므로5.6km¤<56km¤→5.6km¤<5600ha(cid:8951)<320.24ha=2400m¤이므로2400m¤>240m¤→0.24ha>240m¤240m¤=0.024ha이므로0.24ha>0.024ha→0.24ha>240m¤(cid:8951)>단위가다른넓이를비교할경우넓이의단위를통일한후비교합니다.33➊㉠2.35km¤=23500a이므로23500a>4670a →2.35km¤>4670a▶2점➋㉡15ha=150000m¤이므로 150000m¤>81000m¤→15ha>81000m¤▶2점➌따라서넓이를바르게비교한것은㉡㉡입니다.▶1점34넓이의단위를ha로통일하면0.23km¤=23ha, 2320a=23.2ha23.2ha>23ha>21ha이므로가장넓은밭: 고구마밭, 가장좁은밭: 무밭(고구마밭의넓이)-(무밭의넓이)=23.2-21=2.2(ha)(cid:8951)2.2ha35➊1000m=1km이므로4500m=4.5km, 2400m=2.4km(가의넓이)=4.5_2.4÷2=5.4(km¤)▶2점➋(나의넓이)=3_2=6(km¤)▶2점➌5.4km¤<6km¤이므로넓이가더넓은것은나입니다.▶2점1km=1000m임을이용하여가와나의넓이를m¤로나타낸후비교해도됩니다.368a=800m¤이므로8a는1m¤의800배입니다.따라서의자를800개까지놓을수있습니다.(cid:8951)800개37➊36ha=3600a▶2점➋(만들수있는텃밭의수)=3600÷6=600(개)▶3점38(전체논의넓이)=3_6=18(km¤)=1800(ha)(만들수있는논의수)=1800÷24=75(개)(cid:8951)75개391000kg=1t이므로10893kg=10.893t(cid:8951)10.893t401t=1000kg이므로3.9t=3900kg(트럭의무게)÷(버스의무게)=3900÷1300=3(배)(cid:8951)3배채점기준➊ha를a로나타낸경우2점5점➋텃밭을몇개까지만들수있는지구한경우3점채점기준➊과수원의넓이를m¤로나타낸경우2점6점➋같은두수의곱이6400인수를구한경우2점➌과수원의한변은몇m인지구한경우2점채점기준➊㉠에서넓이를비교한경우2점5점➋㉡에서넓이를비교한경우2점➌넓이를바르게비교한것을찾아기호를쓴경우1점채점기준➊가의넓이를구한경우2점6점➋나의넓이를구한경우2점➌더넓은것의기호를쓴경우2점채점기준➊직사각형의넓이를km¤로나타낸경우2점5점➋직사각형의가로는몇m인지구한경우3점틀리는이유│넓이의단위를통일하지않고계산한경우해결방안│넓이의단위를하나로통일하여계산합니다. 15초등쎈수학5-2정답(078-087) 2015.6.8 4:0 PM 페이지080 SinsagoHitec 5. 여러가지단위081본문|148쪽-152쪽5단원47(무게의합)=900+3100=4000(kg)=4(t)(cid:8951)4t48무게의단위를kg으로통일하면1t=1000kg이므로3.8t=3800kg2400kg+500kg+(cid:8641)kg=3800kg, 2900kg+(cid:8641)kg=3800kg→(cid:8641)=3800-2900=900(kg)(cid:8951)900무게의단위를t으로통일하여계산해도됩니다.492t=2000kg이므로(트럭의무게)+(쌀의무게)+(보리의무게)=2000+730+870=3600(kg)=3.6(t)(cid:8951)3.6t50➊(지영이네마을의무수확량)=796+836+729+639=3000(kg)▶3점➋1000kg=1t이므로3000kg=3t▶2점51(세제60상자의무게)=30_60=1800(kg)=1.8(t)(음료수70상자의무게)=40_70=2800(kg)=2.8(t)→(세제와음료수의무게)=1.8+2.8=4.6(t)(cid:8951)4.6t52⑴2시간30분=2;6#0);시간=2;2!;(2.5)시간⑵(2시간30분동안받는물의무게)⑵=450_2;2!;=450÷_=1125(kg)⑶1000kg=1t이므로1125kg=1.125t(cid:8951)⑴2;2!;(2.5)시간(cid:100)⑵1125kg(cid:100)⑶1.125t530.6t=600kg이므로(탈수있는어린이의수)=600÷30=20(명)(cid:8951)20명52/122541➊㉠270kg=0.27t→(cid:8641)=0.27▶2점➋㉡0.68t=680kg→(cid:8641)=680▶2점➌((cid:8641)안에알맞은수의합)=0.27+680=680.27▶1점1000kg=1t, 1kg=0.001t42(곰3마리의무게)=0.4_3=1.2(t)=1200(kg)시소의왼쪽에태우는사람을(cid:8641)명이라하면(시소왼쪽의무게)=((cid:8641)_60)kg시소가수평이되게하려면왼쪽과오른쪽의무게가같아야하므로(cid:8641)_60=1200→(cid:8641)=1200÷60=20(명)(cid:8951)20명43㉠딸기→g㉡오븐→kg㉢자동차→t(cid:8951)㉢44냉장고의무게는1t보다작으므로kg으로나타내어야합니다.→약125kg(cid:8951)소영452800kg=2.8t이므로2.8t<17t→2800kg<17t17t=17000kg이므로2800kg<17000kg→2800kg<17t(cid:8951)<46➊무게의단위를kg으로통일하면㉠24000g=24kg㉡250kg㉢12t=12000kg→24kg<250kg<12000kg▶3점➋따라서무게가가벼운것부터차례로기호를쓰면㉠㉠,,㉡㉡,,㉢㉢입니다.▶2점1000g=1kg채점기준➊무게의단위를통일하여무게를비교한경우3점5점➋무게가가벼운것부터차례로기호를쓴경우2점채점기준➊무수확량은모두몇kg인지구한경우3점5점➋무수확량은모두몇t인지구한경우2점틀리는이유│시소의양쪽의무게가같아야수평이된다는것을모르는경우해결방안│시소의왼쪽에60kg인사람을몇명태워야양쪽의무게가같아지는알아봅니다.채점기준➊㉠에서(cid:8641)안에알맞은수를구한경우2점5점➋㉡에서(cid:8641)안에알맞은수를구한경우2점➌(cid:8641)안에알맞은수의합을구한경우1점틀리는이유│쌀과보리의무게만더한경우해결방안│쌀, 보리, 트럭의무게를모두더합니다. 15초등쎈수학5-2정답(078-087) 2015.6.8 4:0 PM 페이지081 SinsagoHitec 082쎈수학5－25여러가지단위[[60~~67]]서술형평가유형의입니다.60⑴(일반주차공간한칸의넓이)=3_6=18(m¤)(장애인전용주차공간한칸의넓이)=4_6=24(m¤); 18m¤, 24m¤▶3점⑵(주차공간전체의넓이)=18_500+24_20=9000+480=9480(m¤)=94.8(a)▶2점⑶94.8a▶1점61⑴a는야구장의넓이를나타내기에너무작은단위이므로ha를사용하여나타내야합니다.▶3점⑵(cid:9066)야구장의넓이는약1.1ha입니다.▶2점62⑴(전체밭의넓이)=400_400=160000(m¤)=16(ha)▶2점⑵; 16ha⑵(고추를심은부분의넓이)⑵=(전체밭의넓이)_;8#;=16?_=6(ha)▶2점⑶6ha▶1점63⑴정사각형은네변의길이가모두같으므로(숲의한변)=40÷4=10(km); 10km▶2점⑵(숲의넓이)=10_10=100(km¤)▶2점⑶100km¤▶1점64⑴1400m=1.4km이므로(전체땅의넓이)=(2+3)_1.4÷2=3.5(km¤)=350(ha)⑵; 350ha▶2점⑵(만들수있는땅의수)=(전체땅의넓이)÷5=350÷5=70(개)▶3점⑶70개▶1점65⑴㉠(직사각형의넓이)=800_500=400000(m¤)=40(ha)⑵㉢(삼각형의넓이)=4_1÷2=2(km¤)=200(ha)⑵; 40ha, 200ha▶2점⑵250ha>200ha>40ha이므로넓이가넓은것부터차례로기호를쓰면㉡, ㉢, ㉠입니다.▶2점⑶㉡, ㉢, ㉠▶1점38/1254(쌀500가마니의무게)=80_500=40000(kg)=40(t)5t트럭에는5t까지실을수있으므로(필요한트럭수)=40÷5=8(대)(cid:8951)8대55➊(화물2000상자의무게)=10_2000=20000(kg)=20(t)▶3점➋20÷7=2…6이므로화물을7t씩비행기2대에실으면6t이남습니다.남은6t도옮겨야하므로비행기는적어도3대가필요합니다.▶3점남은6t도모두옮겨야함에주의합니다.56확장할부분은가로가20m, 세로가60m인직사각형모양이므로(확장할부분의넓이)=20_60=1200(m¤)=12(a)(cid:8951)12a57지도상에서한변이100m인정사각형을그립니다.(cid:8951)(cid:9066)58세지역의넓이의단위를ha로통일하면은평구: 297100a=2971ha강서구: 4140ha송파구: 33.88km¤=3388ha4140ha>3388ha>2971ha이므로가장넓은곳은강서구입니다.(cid:8951)강서구59(2006년음식물쓰레기총발생량)=1909t(2009년음식물쓰레기총발생량)=417t(총발생량의차)=1909-417=1492(t)=1.492(kg)(cid:8951)1.492kg거리100m채점기준➊화물2000상자의무게를구한경우3점6점➋비행기는적어도몇대가필요한지구한경우3점틀리는이유│쌀500가마니의무게를구하지않고계산하는경우해결방안│쌀500가마니의무게를t으로나타낸후5로나누면트럭이적어도몇대필요한지구할수있습니다. 15초등쎈수학5-2정답(078-087) 2015.6.2 7:59 PM 페이지082 SinsagoHitec 5. 여러가지단위083본문|152쪽-156쪽5단원01천을이어붙여서만들수있는가장작은정사각형의한변은천의가로(30m)와세로(40m)의최소공배수입니다.(가장작은정사각형의넓이)=120_120=14400(m¤)=144(a)(cid:8951)144a02➊9a=900m¤=30m_30m이므로넓이가9a인정사각형모양의땅의한변은30m입니다.➋9ha=90000m¤=300m_300m이므로넓이가9ha인정사각형모양의땅의한변은300m입니다.➌(더늘려야하는길이)=300-30=270(m)(cid:8951)270m2>≥30 405>≥15 205>13 24→(최소공배수)=2_5_3_4=120156쪽~157쪽03세로를(cid:8641)km라하면가로는세로의4배인((cid:8641)_4)km입니다.놀이공원의둘레가90km이므로((cid:8641)_4+(cid:8641))_2=90, (cid:8641)_5_2=90, (cid:8641)_10=90,(cid:8641)=90÷10=9(km)세로가9km이므로(가로)=9_4=36(km)(놀이공원의넓이)=36_9=324(km¤)(cid:8951)324km¤(cid:8641)_4+(cid:8641)=(cid:8641)_504(인형80상자의무게)=(인형한상자의무게)_80=25_80=2000(kg)4.1t=4100kg이므로(장난감70상자의무게)=(전체무게)-(인형80상자의무게)=4100-2000=2100(kg)(장난감한상자의무게)=(장난감70상자의무게)÷70=2100÷70=30(kg)(cid:8951)30kg05⑴1ha=10000m¤이므로3ha=30000m¤⑵(심을수있는배나무의수)=30000÷3=10000(그루)⑶(수확할수있는배의무게)=(한그루에서수확할수있는배의무게)=_(배나무의수)=70_10000=700000(kg)=700(t)(cid:8951)⑴30000m¤⑵10000그루⑶700t➊넓이가9a인땅의한변구하기→➋넓이가9ha인땅의한변구하기→➌한변을몇m씩더늘려야하는지구하기푸는순서천을이어붙여서만들수있는정사각형의한변은천의가로(30m)와세로(40m)의공배수입니다.전략민지네공장에서판매한인형80상자와장난감70상자의무게의합은4.1t입니다. 인형한상자의무게가25kg일때장난감한상자의무게는몇kg입니까?(인형80상자의무게)+(장난감70상자의무게)(장난감70상자의무게)÷7066⑴(배추밭의넓이)=75_32=2400(m¤)=24(a)▶2점(cid:100) ; 24a⑵1a당거름을200kg주므로(배추밭에주는거름의양)=200_24=4800(kg)=4.8(t)▶3점⑶4.8t▶1점67⑴(삼촌이옮긴보리의무게)=80_20=1600(kg)(cid:100) (아버지가옮긴보리의무게)=80_25=2000(kg)(cid:100) ; 1600kg, 2000kg▶2점⑵(삼촌이옮긴보리의무게)+(아버지가옮긴보리의무게)=1600+2000=3600(kg)=3.6(t)▶3점⑶3.6t▶1점(가로)(세로)놀이공원의세로를(cid:8641)km라하여가로를(cid:8641)를사용하여나타낸후(직사각형의둘레)={(가로)+(세로)}_2를이용합니다.전략 15초등쎈수학5-2정답(078-087) 2015.6.2 7:59 PM 페이지083 SinsagoHitec 084쎈수학5－25여러가지단위06➊(사다리꼴의넓이)=(50+25)_30÷2=1125(m¤)▶2점➋정사각형은두대각선의길이가같은마름모이므로(정사각형의넓이)=60_60÷2=1800(m¤)▶2점➌(도형의넓이)=(사다리꼴의넓이)+(정사각형의넓이)=1125+1800=2925(m¤)▶2점➍=29.25(a)▶2점07➊(㉮의세로)=260-180=80(m)(㉮의넓이)=240_80=19200(m¤)▶3점➋8.4ha=84000m¤이므로(㉯의넓이)=(도형의넓이)-(㉮의넓이)=84000-19200=64800(m¤)▶2점➌(㉯의넓이)=㉠_180=64800→㉠=64800÷180=360(m)▶2점08➊(숲의넓이)=400_300÷2=60000(m¤)=6(ha)▶2점➋(소나무를심은부분)=1-(단풍나무를심은부분)(소나무를심은부분)=1-;3!;=;3@;▶2점➌(소나무를심은부분의넓이)=6/_=4(ha)▶3점23/12㉮ ㉯ 260`m240`m80`m180`m㉠ 09➊(직사각형의둘레)=(정사각형의둘레)=3_4=12(km)▶1점➋직사각형의가로를(cid:8641)km라하면(직사각형의둘레)=((cid:8641)+2)_2=12, (cid:8641)+2=6, (cid:8641)=4(km)▶2점➌(직사각형의넓이)=4_2=8(km¤)(정사각형의넓이)=3_3=9(km¤)▶2점➍(두도형의넓이의차)=9-8=1(km¤)▶2점10➊1.5km=1500m(밭의넓이)=1500_2000=3000000(m¤)=3(km¤)▶3점➋(필요한씨의무게)=450_3=1350(kg)▶2점➌=1.35(t)▶2점2000m=2km이므로(밭의넓이)=1.5_2=3(km¤)로구해도됩니다.11➊(1분동안나오는물의무게)=10.5+14.5=25(kg)▶3점➋10t=10000kg이므로(걸리는시간)=(채워야할물의무게)÷(1분동안나오는물의무게)=10000÷25=400(분)▶3점➌60분=1시간이므로400분=6시간40분입니다.▶2점채점기준➊㉮`의넓이를구한경우3점7점➋㉯`의넓이를구한경우2점➌㉠은몇m인지구한경우2점채점기준➊숲의넓이를구한경우2점7점➋소나무를심은부분은전체의얼마인지구한경우2점➌소나무를심은부분의넓이를구한경우3점채점기준➊밭의넓이는몇km¤인지구한경우3점7점➋필요한씨의무게는몇kg인지구한경우2점➌필요한씨의무게는몇t인지구한경우2점채점기준➊1분동안나오는물의무게를구한경우3점8점➋걸리는시간은몇분인지구한경우3점➌걸리는시간은몇시간몇분인지구한경우2점도형을두직사각형으로나눈다음각각의넓이를이용하여㉠을구합니다.전략전체를1이라고생각하고소나무를심은부분은전체의얼마인지구합니다.전략정사각형은두대각선의길이가서로같은마름모입니다.전략채점기준➊사다리꼴의넓이를구한경우2점8점➋정사각형의넓이를구한경우2점➌도형의넓이는몇m¤인지구한경우2점➍도형의넓이는몇a인지구한경우2점채점기준➊직사각형의둘레를구한경우1점7점➋직사각형의가로를구한경우2점➌직사각형과정사각형의넓이를각각구한경우2점➍두도형의넓이의차를구한경우2점10t의물이들어가는빈수영장에1분에10.5kg의물이나오는수도와1분에14.5kg의물이나오는수도를동시에틀어물을채우려고합니다. 수영장에물을가득채우는데걸리는시간은몇시간몇분인지풀이과정을쓰고, 답을구하시오.(1분동안나오는물의무게)=10.5+14.5=25(kg)(채워야할물의무게)÷(1분동안나오는물의무게)=400(분) 15초등쎈수학5-2정답(078-087) 2015.6.8 4:0 PM 페이지084 SinsagoHitec 01한변이10m인정사각형의넓이를1a라쓰고1아르라고읽습니다.(cid:8951)1a, 1아르02100m¤=1a이므로500m¤=5a(cid:8951)5031km¤=1000000m¤이므로8km¤=8000000m¤(cid:8951)800000004➊(직사각형의넓이)=80_35=2800(m¤)▶3점➋=28(a)▶2점05(땅의넓이)=60_150÷2=4500(m¤)=45(a) (만들수있는텃밭의수)=45÷3=15(개)(cid:8951)15개06710a=7.1ha이므로7.1ha>7ha→710a>7ha7ha=700a이므로710a>700a→710a>7ha(cid:8951)>071km¤=100ha이므로32km¤=3200ha(cid:8951)=a를ha로나타낼때에는소수점을왼쪽으로두칸옮깁니다. 이때소수점아래마지막0은생략할수있습니다.→710a=7.10Yha=7.1ha081ha는한변이100m인정사각형의넓이입니다. (cid:8951)100, 10009➊(목장의넓이)=40ha=400000m¤▶2점➋(세로)=(목장의넓이)÷(가로)=400000÷200=2000(m)▶3점158쪽~159쪽1회10➊81ha=810000m¤▶2점➋정사각형의한변을(cid:8641)m라하면(cid:8641)_(cid:8641)=810000, 900_900=810000이므로▶2점➌(cid:8641)=900m▶1점11(평행사변형의넓이)=4_4=16(km¤)(cid:8951)1612(삼각형의넓이)=6_5÷2=15(km¤)=1500(ha)(cid:8951)150013➊(밭의넓이)=(300+500)_250÷2=100000(m¤)▶3점➋1000000m¤=1km¤이므로100000m¤=0.1km¤따라서밭의넓이는0.1km¤입니다.▶2점14나라, 도시등땅의넓이의단위는km¤로나타내는것이적당합니다.(cid:8951)㉣151000000m¤=10000a=100ha→57000000m¤=570000a=5700ha(cid:8951)570000, 5700163000m¤=30a, 1200a=0.12km¤(cid:8951)17➊무게의단위를kg으로통일하면㉠2900g=2.9kg㉡2kg㉢0.05t=50kg→2kg<2.9kg<50kg▶3점➋따라서무게가가장가벼운것은㉡㉡입니다.▶2점무게의단위를g또는t으로통일하여비교해도됩니다.3000`m@=3`a1200`a=12`km@5`ha=50000`m@5. 여러가지단위085본문|157쪽-159쪽5단원채점기준➊밭의넓이는몇m¤인지구한경우3점5점➋밭의넓이는몇km¤인지구한경우2점채점기준➊무게의단위를통일하여무게를비교한경우3점5점➋무게가가장가벼운것을찾아기호를쓴경우2점채점기준➊직사각형의넓이는몇m¤인지구한경우3점5점➋직사각형의넓이는몇a인지구한경우2점채점기준➊목장의넓이를m¤로나타낸경우2점5점➋목장의세로는몇m인지구한경우3점채점기준➊ha를m¤로나타낸경우2점5점➋같은두수의곱이810000인수를구한경우2점➌정사각형의한변을구한경우1점 15초등쎈수학5-2정답(078-087) 2015.6.8 4:1 PM 페이지085 SinsagoHitec 01100m¤=1a이므로80m¤=0.8a(cid:8951)0.8021km¤=100ha이므로620km¤=62000ha(cid:8951)6200003㉠1a=100m¤이므로154a=15400m¤㉡1ha=100a이므로9.3ha=930a따라서잘못나타낸것은㉡입니다.(cid:8951)㉡04➊(잔디밭의한변)=360÷4=90(m)▶2점➋(잔디밭의넓이)=90_90=8100(m¤)=81(a)▶3점05(직사각형의넓이)=25_24=600(m¤)(정사각형의넓이)=20_20=400(m¤)(넓이의차)=600-400=200(m¤)=2(a)(cid:8951)2a06(마름모의넓이)=(작은삼각형의넓이)_4=(600_400÷2)_4=480000(m¤)=48(ha)마름모의한대각선은다른대각선을반으로나누므로(마름모의넓이)=(600_2)_(400_2)÷2=1200_800÷2=480000(m¤)=48(ha)(cid:8951)48ha07➊6ha=60000m¤이므로▶2점➋(만들수있는땅의수)=60000÷3=20000(개)▶3점086쎈수학5－25여러가지단위160쪽~161쪽2회18(음료수50상자의무게)=18_50=900(kg)=0.9(t)(통조림100상자의무게)=15_100=1500(kg)=1.5(t)(음료수와통조림상자의무게)=0.9+1.5=2.4(t)(음료수50상자의무게)=18_50=900(kg)(통조림100상자의무게)=15_100=1500(kg)(음료수와통조림상자의무게)=900+1500=2400(kg)=2.4(t)(cid:8951)2.4t19➊(냉장고200대의무게)=125_200=25000(kg)=25(t)▶2점➋4t트럭에는4t까지실을수있고25÷4=6``…``1이므로냉장고를4t씩트럭6대에실으면1t이남습니다.남은1t도옮겨야하므로트럭은적어도7대가필요합니다.▶3점201시간15분=1;6!0%;시간=1;4!;(1.25)시간(1시간15분동안받는물의무게)=240_1;4!;=240÷_=300(kg)=0.3(t)(cid:8951)0.3t1시간=60분이므로1분=;6¡0;시간입니다.→(cid:8784)시간▲분=(cid:8784)시간▲6054/160채점기준➊냉장고200대의무게를구한경우2점5점➋트럭은적어도몇대가필요한지구한경우3점채점기준➊땅의넓이를m¤로나타낸경우2점5점➋만들수있는땅의수를구한경우3점채점기준➊잔디밭의한변을구한경우2점5점➋잔디밭의넓이는몇a인지구한경우3점 15초등쎈수학5-2정답(078-087) 2015.6.2 7:59 PM 페이지086 SinsagoHitec 08➊1km¤=100ha이므로0.78km¤=78ha, 6400ha=64km¤▶3점➋((cid:8641)안에알맞은수의합)=78+64=142▶2점0912km¤=1200ha이므로12ha는12km¤의;10!0;(0.01)배입니다.(cid:8951);10!0;(0.01)배101km¤=100ha이므로125km¤=12500ha(cid:8951)12500ha11➊1100m=1.1km이므로직사각형의세로를(cid:8641)km라하면(직사각형의둘레)=(1.1+(cid:8641))_2=3, 1.1+(cid:8641)=3÷2=1.5, (cid:8641)=1.5-1.1=0.4(km)▶3점➋(직사각형의넓이)=1.1_0.4=0.44(km¤)▶2점12넓이의단위를a로통일하면0.3km¤=3000a, 29ha=2900a, 500m¤=5a이므로3000a-2900a+5a=(cid:8641)a→(cid:8641)=3000-2900+5=105(a)(cid:8951)10513축구장의넓이→약83a(cid:8951)a14비행기의무게→약225t(cid:8951)t15가로와세로가각각300m, 200m인직사각형의넓이에서가로와세로가각각80m, ㉠`인직사각형의넓이를뺍니다.5ha=50000m¤이므로(도형의넓이)=300_200-80_㉠=50000, 60000-80_㉠=50000, 80_㉠=60000-50000=10000, ㉠=10000÷80=125(m)(cid:8951)125m16➊넓이의단위를ha로통일하면㉠0.25km¤=25ha㉡84000a=840ha㉢13ha→840ha>25ha>13ha▶3점➋따라서넓이가넓은것부터차례로기호를쓰면㉡㉡,, ㉠㉠,, ㉢㉢입니다.▶2점넓이의단위를a또는km¤로통일하여비교해도됩니다.171.3t=1300kg이므로1300kg>250kg→1.3t>250kg따라서무게가더무거운것은㉮입니다.250kg=0.25t이므로1.3t>0.25t→1.3t>250kg따라서무게가더무거운것은㉮입니다.(cid:8951)㉮184t=4000kg이므로(코끼리의무게)÷(물소의무게)=4000÷500=8(배)(cid:8951)8배190.9t=900kg이므로(어른5명의무게)=70_5=350(kg)(더태울수있는무게)=900-350=550(kg)550÷35=15`y`25이므로어린이를15명까지더태울수있습니다.(cid:8951)15명사람수는자연수이므로몫을자연수까지만구합니다.20➊0.18km=180m이므로(밭의넓이)=60_180÷2=5400(m¤)=54(a)▶2점➋(필요한거름의양)=300_54=16200(kg)=16.2(t)▶3점5. 여러가지단위087본문|159쪽-161쪽5단원채점기준➊직사각형의세로를구한경우3점5점➋직사각형의넓이는몇km¤인지구한경우2점채점기준➊(cid:8641)안에알맞은수를각각구한경우3점5점➋(cid:8641)안에알맞은수의합을구한경우2점넓이의단위구하려는넓이m¤ahakm¤방, 거실등체육관, 운동장등공원, 논, 밭등지역, 나라등무게의단위구하려는무게kgt몸무게, 냉장고등코끼리, 배, 비행기등채점기준➊밭의넓이를구한경우2점5점➋필요한거름은몇t인지구한경우3점채점기준➊넓이의단위를통일하여넓이를비교한경우3점5점➋넓이가넓은것부터차례로기호를쓴경우2점 15초등쎈수학5-2정답(078-087) 2015.6.2 7:59 PM 페이지087 SinsagoHitec 088쎈수학5－26자료의표현✽A단계기본다잡기⑴정답은‘정답010쪽’에있습니다.01(하루컴퓨터사용시간의합)=32+45+60+56+47=240(분)(cid:8951)240분02지효네모둠학생은지효, 우희, 미소, 진아, 유미로모두5명입니다.(cid:8951)5명03(평균사용시간)=240÷5=48(분)(cid:8951)48분04(공던지기기록의합)=25+16+13+21+17+10=102(m)(평균기록)=102÷6=17(m)(cid:8951)17m(평균)=(자료값의합)÷(자료의수)05(평균나이)=1140÷30=38(살)(cid:8951)38살06(평균몸무게)=(32.5+40+36.2+43.8+39.5)÷5=192÷5=38.4(kg)(cid:8951)38.4kg07➊주영이의몸무게만2kg늘어난것이므로(한달뒤몸무게의합)=192+2=194(kg)▶2점➋(늘어난후의평균)=194÷5=38.8(kg)▶2점➌(늘어난평균몸무게)=38.8-38.4=0.4(kg)▶2점08(시간2개의합)=18_2=36(분)(cid:8951)36분0916+20=36, 14+22=36, 27+9=36(cid:8951)20, 22, 910=18(분)=18(분)(cid:8951)14, 27, 22, 20, 9; 36, 3636+36+36616+14+27+22+20+96167쪽~171쪽11기준수를300으로정하고(150, 450)을더하면600이고이를2로나누면300입니다.따라서독서실이용자수의평균은300명입니다.(cid:8951)30012➊2월과3월을비교하면3월의7000원에서500원을6000원에주면민정이가두달동안저금한평균금액은6500원이됩니다.4월과5월을비교하면4월의7000원에서500원을6000원에주면민정이가두달동안저금한평균금액은6500원이됩니다.▶3점➋따라서민정이가네달동안저금한평균금액은6500원입니다.▶2점13[방법1]기준수를19로정하고합이38이되도록실내온도를2개씩묶으면(22, 16), (24, 14)입니다.따라서실내온도는평균19˘C입니다.[방법2](평균실내온도)=[방법2](평균실내온도)=;;ª5∞;;=19(˘C)14기준이되는가로선을10으로정하고2반과4반을비교하면2반의11에서1을4반에주면두반의안경을쓴학생수의평균은10명이됩니다.1반과5반을비교하면5반의13에서3을1반에주면두반의안경을쓴학생수의평균은10명이됩니다.따라서안경을쓴학생수의평균은10명입니다.(평균)==;;∞5º;;(평균)=10(명)(cid:8951)10명1519.7<19.9이므로기태는반에서느린편입니다.(cid:8951)느린편입니다.평균과기록을비교하여(평균)>(기록) →빠른편, (평균)<(기록) →느린편달리기기록은시간이짧을수록빠르고, 시간이길수록느립니다.7+11+10+9+13522+24+19+14+165틀리는이유│주영이의몸무게만2kg늘었는데평균이2kg늘었다고생각하여틀리는경우해결방안│주영이의몸무게만2kg늘었으므로전체몸무게합에2를더한후모둠학생들의평균몸무게를구합니다.틀리는이유│막대그래프에서반별안경을쓴학생수를구하지못해틀리는경우해결방안│세로눈금한칸의크기를구한다음반별안경을쓴학생수를구하여평균을구합니다.채점기준➊평균금액을구하는간단한방법을쓴경우3점5점➋평균금액을구한경우2점채점기준두가지방법으로설명한경우5점5점한가지방법으로설명한경우2점채점기준➊한달뒤몸무게의합을구한경우2점6점➋늘어난후의평균을구한경우2점➌늘어난평균몸무게를구한경우2점 15초등쎈수학5-2정답(088-099) 2015.6.2 8:0 PM 페이지088 SinsagoHitec 본문|167쪽-170쪽6. 자료의표현0896단원16(평균어획량)=(평균어획량)==3325(t)3600>3325이므로2014년어획량은평균어획량보다많습니다.따라서2014년갑오징어어획량은높은편입니다.(cid:8951)‘높은’에◯`표17(앉은키의합)=59.1+61+58.3+55.4+62+57=352.8(cm)(앉은키의평균)=352.8÷6=58.8(cm)앉은키가평균보다큰학생은진혁(59.1cm), 정미(61cm), 연아(62cm)로모두3명입니다.(cid:8951)3명18(2013년평균)==(2013년평균)=2266(마리)(2015년평균)==(2015년평균)=2638(마리)(cid:8951)2266마리, 2638마리192013년보다2015년의호수의물고기수가늘어났으므로호수가깨끗해져서물고기수가늘어났다고예상할수있습니다.20(보경이네모둠평균)==;;¶5∞;;=15(권)(민진이네모둠평균)==;;•5º;;=16(권)15<16이므로민진이네모둠이평균16-15=1(권) 더많이읽었습니다.(cid:8951)민진이네모둠, 1권16+25+6+20+13513+14+21+8+195791432485+1495+39343679832100+1500+319831330044100+2500+3100+3600421(미라네모둠)=(미라네모둠)=;;ª8™;;=11.5(번)(지혜네모둠)=(미라네모둠)=;;•7¢;;=12(번)(cid:8951)(위에서부터) 92, 11.5；84, 1222➊할수없습니다.；▶2점➋제기를찬합계를비교하면92>84이지만평균기록을비교하면11.5<12입니다.두모둠의제기를찬학생수가다르므로합계가크다고해서제기를더많이찼다고할수없습니다.▶3점23자동차가한시간동안달린평균거리를구하면(은영이네자동차)==71(km)(재민이네자동차)==87(km)한시간동안달린평균거리가71<87이므로재민이네자동차가더빨리달렸습니다.(cid:8951)재민이네자동차24하루에평균520개씩달걀을생산하므로(40일동안생산한달걀수)=520_40=20800(개) (cid:8951)20800개(자료값의합)=(평균)_(자료의수)251년은12달입니다.(1년동안받은용돈)=8000_12=96000(원)(cid:8951)9600026(네사람의키의합)=145_4=580(cm)(평균키)===143.6(cm)따라서다섯사람의평균키는143.6cm입니다.(cid:8951)143.6cm7185580+13852613284410+7+13+15+16+17+6718+10+8+19+13+5+11+88틀리는이유│모둠학생들의앉은키의평균을구하지않고진혁이와친구들의키만비교하여틀리는경우해결방안│모둠학생들의앉은키의평균을구한다음평균과각학생들의앉은키를비교하여앉은키가평균보다큰학생을찾습니다.채점기준예상할수있는사실을설명한경우5점틀리는이유│제기를찬학생수를생각하지않고합계만생각하여합계가큰모둠이제기를더많이찼다고생각하는경우해결방안│평균은자료의수에따라달라지므로자료의수를먼저비교합니다.채점기준➊할수없음을쓴경우2점5점➋이유를설명한경우3점 15초등쎈수학5-2정답(088-099) 2015.6.8 4:1 PM 페이지089 SinsagoHitec 090쎈수학5－26자료의표현27➊(여학생전체몸무게의합)=35.3_8=282.4(kg)▶2점➋(남학생전체몸무게의합)=36.8_12=441.6(kg)▶2점➌(평균몸무게)=(평균몸무게)==:¶2™0¢:=36.2(kg)따라서반전체학생들의평균몸무게는36.2kg입니다.▶2점28(감자총생산량)=760_5=3800(kg)(라마을의감자생산량)=(감자총생산량)=-(가, 나, 다, 마마을의감자생산량의합)=3800-(564+820+780+934)=3800-3098=702(kg)(cid:8951)70229➊평균90점이되려면(총점)=90_4=360(점)▶2점➋(`12월에받아야할점수)=(총점)-(`9, 10, 11월점수의합)=360-(`98+82+88)=360-268=92(점)▶3점30⑴(새로운회원이들어오기전평균나이)⑴==;;§4•;;=17(살)⑵새로운회원한명이더들어와서평균나이가한살줄었으므로(늘어난전체회원나이의총합)=17-5=12(살)⑶늘어난전체회원나이의총합이12살이므로새로운회원의나이는12살입니다.(cid:8951)⑴17살⑵12살⑶12살18+15+19+164282.4+441.68+12(학생전체몸무게의합)(학생수)31(석진이의단원별수학시험점수의합)=98+96+88+92+91+(cid:8641)=465+(cid:8641)465+(cid:8641)=92_6, 465+(cid:8641)=552, (cid:8641)=552-465=87(점)이므로마지막점수는87점과같거나더높아야합니다.따라서마지막에적어도87점을받아야합니다.(cid:8951)87점틀리는이유│35.3과36.8의합을구한다음총20명으로나누어틀리는경우해결방안│여학생8명의전체몸무게의합과남학생12명의전체몸무게의합을구한다음총20명으로나누어전체학생들의평균몸무게를구합니다.채점기준➊여학생전체몸무게의합을구한경우2점6점➋남학생전체몸무게의합을구한경우2점➌반전체학생들의평균몸무게를구한경우2점채점기준➊평균90점일때의총점을구한경우2점5점➋12월에받아야할점수를구한경우3점✽A단계기본다잡기⑵정답은‘정답011쪽’에있습니다.01한명의아이가태어날때남자아이일수도있으므로반반입니다. →㉡(cid:8951)㉡02태양은매일뜨므로확실합니다. →㉢(cid:8951)㉢03주사위를던졌을때8의눈이나올수없으므로불가능합니다. →㉠(cid:8951)㉠가능성: 어떠한상황에서특정한사건이일어나길기대할수있는정도04➊곶감이들어있는주머니에서는곶감만꺼낼수있으므로일어날가능성이확실합니다.➋따라서사건이일어날가능성이확실한것은㉢㉢입니다.05해는동쪽에서떠서서쪽으로집니다.따라서해는동쪽에서뜰사건이일어날가능성은확실합니다.06(cid:8951)(위에서부터) ㉢, ㉡, ㉣, ㉠41214310불가능하다.(㉠)일어나지않을 것같다.(㉣)가능성이반반이다. (㉤)일어날 것같다.(㉡)확실하다.(㉢)174쪽~183쪽▶3점▶2점채점기준➊사건이일어날가능성이확실한것을찾는과정을쓴경우3점5점➋사건이일어날가능성이확실한것을찾아기호를쓴경우2점틀리는이유│해가어느쪽에서뜨는지알지못해틀리는경우해결방안│해는동쪽에서떠서서쪽으로진다는것을알고해는동쪽에서뜨는가능성에대해서이야기합니다.채점기준사건이일어날가능성에대하여자신의생각을이야기한경우5점 15초등쎈수학5-2정답(088-099) 2015.6.2 8:0 PM 페이지090 SinsagoHitec 본문|171쪽-176쪽6. 자료의표현0916단원07②제비한장을뽑을때당첨제비를뽑을가능성은작②습니다. →;4!;③제비한장을뽑을때당첨제비를뽑을가능성은반②반입니다. →;2!;④제비한장을뽑을때당첨제비를뽑을가능성은큽②니다. →;4#;(cid:8951)08상자에들어있는숫자카드가모두5이므로카드한장을꺼낼때꺼낸카드의숫자가5일가능성은확실합니다.따라서꺼낸카드의숫자가5일가능성을수로나타내면1입니다.(cid:8951)09모둠장이될수있는학생은4명중의1명입니다.따라서하윤이가모둠장이될가능성을수로나타내면;4!;입니다.(cid:8951)10셔츠2벌중에서검은색셔츠는없습니다.따라서지은이가검은색셔츠를입을가능성을수로나타내면0입니다.(cid:8951)011➊㉠검은색바둑돌중에서검은색바둑돌을고를가능성은확실합니다. →가능성을수로나타내면1㉡숫자카드10장중에서홀수가적힌카드는1, 3, 5,7, 9로5장이므로홀수가적힌카드를뽑을가능성㉡은반반입니다. →가능성을수로나타내면;2!;㉢주사위한개를던질때0의눈은나올수없습니다. →가능성을수로나타내면0▶3점➋따라서사건이일어날가능성을수로나타내었을때;2!;인것은㉡㉡입니다.▶2점4121431041214310① ⑤ ② ③ ④ 4121431012➊상자에들어있는공은모두파란색이므로공1개를꺼낼때꺼낸공이파란색일가능성은확실합니다.▶3점➋따라서꺼낸공이파란색일가능성을수로나타내면1입니다.▶2점13진욱이가집에서할아버지댁에가는길은2가지이고그중기차를타고가는길은1가지입니다.따라서진욱이가집에서할아버지댁에기차를타고갈가능성을수로나타내면;2!;입니다.(cid:8951);2!;14지폐4장중에서5000원짜리지폐는1장입니다.따라서꺼낸지폐가5000원짜리일가능성을수로나타내면;4!;입니다.(cid:8951);4!;15원판은똑같이4부분으로나누어져있고그중초록색은1부분입니다.따라서초록색을맞힐가능성을수로나타내면;4!;입니다.(cid:8951);4!;16⑴전체아이스크림은6개이고그중멜론맛아이스크림은3개이므로전체의반만큼있습니다.⑵꺼낸아이스크림이멜론맛일가능성은반반이므로⑵가능성을수로나타내면;2!;입니다.(cid:8951)⑴반⑵;2!;아이스크림의수가같으므로수박맛아이스크림1개,멜론맛아이스크림1개로단순화하여생각할수도있습니다.17➊(주머니속의전체탁구공수)=1+3=4(개)▶2점➋탁구공4개중에서노란색탁구공은3개이므로꺼낸탁구공이노란색일가능성을수로나타내면;4#;입니다.▶4점틀리는이유│㉡`에서1부터10까지의숫자카드중에서홀수를뽑을가능성을5라고구해틀리는경우해결방안│1부터10까지의숫자카드중에서홀수는1, 3, 5, 7, 9이므로전체10장중의5장임을알고가능성을수로나타냅니다.틀리는이유│노란색탁구공이3개있어서꺼낸탁구공이노란색일가능성을수로나타내면3이라고생각하여틀리는경우해결방안│전체탁구공수에서노란색탁구공은몇개인지구하여꺼낸탁구공이노란색일가능성을수로나타냅니다.채점기준➊사건이일어날가능성을수로각각나타낸경우3점5점➋사건이일어날가능성을수로나타내었을때;2!;인것을찾아기호를쓴경우2점채점기준➊꺼낸공이파란색일가능성을수로나타내는과정을쓴경우3점5점➋꺼낸공이파란색일가능성을수로나타낸경우2점채점기준➊주머니속의전체탁구공수를구한경우2점6점➋꺼낸탁구공이노란색일가능성을수로나타낸경우4점 15초등쎈수학5-2정답(088-099) 2015.6.2 8:0 PM 페이지091 SinsagoHitec 092쎈수학5－26자료의표현18큰그림()이3개, 작은그림()이5개이므로나마을의오이생산량은3500kg입니다.(cid:8951)3500kg19큰그림()의수를비교한후작은그림()의수를비교하면오이생산량이가장많은마을은라마을이고,가장적은마을은가마을입니다.(cid:8951)라마을, 가마을20➊(네마을의전체오이생산량)=2800+3500+5300+5400▶3점➋=17000(kg)▶2점21큰그림()의수와작은그림()의수를각각세어구합니다.(cid:8951)(왼쪽에서부터) 15；27, 20, 27；40, 23, 1222마릿수가가장많은도: 경상북도, 40만마리마릿수가가장적은도: 제주특별자치도, 12만마리(마릿수의차)=40만-12만=28만(마리)(cid:8951)28만마리23그림그래프는표에비해도별소의마릿수를한눈에비교하기쉽습니다.24(6동의학생수)=179-(32+41+25+23+42)=179-163=16(명)(cid:8951)16명25학생수는몇십몇명이므로10명은, 1명은으로나타냅니다.(cid:8951)110123456학생수26나지역의관광객수: 이6개, 이5개, 이7개→6570000명라지역의관광객수: 이8개, 이2개, 이7개→8270000명(cid:8951)6570000, 827000027가지역의관광객수: 5820000명→이5개, 이8개, 이2개다지역의관광객수: 7430000명→이7개, 이4개, 이3개(cid:8951)8, 2; 7, 4, 3;(cid:100)(cid:100)28➊관광객이가장많은지역은이가장많은지역이므로라지역이고이8개, 이2개, 이7개이므로8270000명입니다.➋따라서관광객이가장많은지역은라지역이고관광객은8270000명입니다.29시간의흐름에따른어느지역의월별강수량의변화는꺾은선그래프로나타내는것이더좋습니다.(cid:8951)꺾은선그래프30휴대전화판매량의많고적음을한눈에비교하기쉽게나타내려면막대그래프로나타내는것이좋습니다.(cid:8951)막대31시간의흐름에따른생산한장난감의불량품수의변화는꺾은선그래프로나타내는것이좋습니다.(cid:8951)꺾은선10만 명 100만 명 가 나 다 라 1만 명 관광객수채점기준➊네마을의전체오이생산량을구하는과정을쓴경우3점5점➋네마을의전체오이생산량을구한경우2점틀리는이유│단위를생각하지않고40-12=28(마리)라고답하는경우해결방안│의수와의수가각각몇만마리를나타내는지확인하여해결합니다.채점기준도별소의마릿수를한눈에비교하는데표와그림그래프중에서어느것이더편리하다고생각하는지설명한경우5점틀리는이유│그림의크기를생각하지않고개수만생각하여관광객이가장많은지역이나지역이라고답하여틀리는경우해결방안│100만명을가리키는그림이가장많은지역이관광객이가장많은지역입니다.채점기준➊관광객이가장많은지역은어느지역이고관광객은몇명인지구하는과정을쓴경우4점6점➋관광객이가장많은지역은어느지역이고관광객은몇명인지구한경우2점▶4점▶2점 15초등쎈수학5-2정답(088-099) 2015.6.8 4:2 PM 페이지092 SinsagoHitec 본문|177쪽-180쪽6. 자료의표현0936단원32세로눈금한칸의크기는1명을나타냅니다.(cid:8951)(cid:9066)33막대의길이가프리지어보다긴꽃은장미와무궁화입니다.(cid:8951)장미, 무궁화막대의길이가길수록항목의수량이많습니다.34➊세로눈금한칸의크기는1명을나타내므로가장많은학생들이좋아하는꽃: 장미, 11명가장적은학생들이좋아하는꽃: 카네이션, 3명▶3점➋(학생수의차)=11-3=8(명)▶2점35경기도호박생산량15100kg을그림그래프에서1개, 5개, 1개로나타내었으므로10000kg, 1000kg, 100kg입니다.(cid:8951)10000, 10036충청북도: 32200kg→3개, 2개, 2개충청남도: 43300kg→4개, 3개, 3개전라북도: 12300kg→1개, 2개, 3개전라남도: 25200kg→2개, 5개, 2개경상북도: 9500kg→9개, 5개경상남도: 26100kg→2개, 6개, 1개제주특별자치도: 3400kg→3개, 4개학생 수 꽃 0510장미 국화 프리 지어 카네 이션 무궁화 좋아하는꽃별학생수(cid:8951)37➊[장점]호박생산량을나타낸그림그래프는표에비해생산량의많고적음을쉽게비교할수있습니다.➋[단점]표에비해도별호박생산량을한눈에바로알수없습니다.38양파싹의키는연속적으로변화하는양이기때문에꺾은선그래프로나타내는것이좋습니다.(cid:8951)꺾은선그래프39세로눈금한칸의크기는1cm를나타냅니다.(cid:8951)(cid:9066)40양파싹의키가가장많이늘어난때는선분이오른쪽위로기울어진정도가가장큰때인22일과29일사이이므로29일입니다.(cid:8951)29일41➊11cm보다클것입니다. ;➋그래프에서1일부터꺾은선이오른쪽위로올라가고있으므로29일이후의양파싹의키는11cm보다클것입니다.051815222910cm양파싹의키10000kg100kg1000`kg10a당호박생산량틀리는이유│가장많은학생들이좋아하는꽃과가장적은학생들이좋아하는꽃을찾지못해틀리는경우해결방안│막대그래프에서막대가길수록많은학생들이좋아하는꽃이라는것을알고가장긴막대와가장짧은막대를각각찾습니다.채점기준➊가장많은학생들이좋아하는꽃과가장적은학생들이좋아하는꽃의학생수를각각구한경우3점5점➋가장많은학생들이좋아하는꽃과가장적은학생들이좋아하는꽃의학생수의차를구한경우2점채점기준➊그림그래프의장점을표와비교하여설명한경우3점6점➋그림그래프의단점을표와비교하여설명한경우3점틀리는이유│그래프에29일이후의양파싹의키가나오지않으므로11cm보다클지작을지모른다고생각하는경우해결방안│양파싹의키는계속커지고있으므로29일의11cm보다더커질것입니다.채점기준➊29일이후의양파싹의키는11cm보다크다는것을쓴경우2점5점➋이유를설명한경우3점▶3점▶2점▶3점▶3점 15초등쎈수학5-2정답(088-099) 2015.6.2 8:0 PM 페이지093 SinsagoHitec 094쎈수학5－26자료의표현42무조건평균만가지고해석하고판단하면안된다는것을알고계곡을건너면안된다는것을써야합니다.(cid:8951)(cid:9066)우리넷의평균키가150cm이지만계곡의평균깊이인145cm보다더작은사람이있을수있으니건너가면안돼!43(어제독서를한시간)=4시20분-3시30분=50분(오늘독서를한시간)=5시15분-4시50분=25분(독서를한시간의합)=45_3=135(분)(내일독서를해야할시간)=135-(50+25)=60(분)따라서내일7시10분에서60분후인8시10분까지독서를해야합니다.(cid:8951)444명의선수중승리할수있는선수는1명입니다.따라서2번선수가승리할가능성을수로나타내면;4!;입니다.(cid:8951);4!;45금메달수를비교하면1위는라지역이고, 2위는가지역입니다.나지역과다지역은금메달의수가같으므로은메달의수를비교하면은메달의수가더많은다지역이3위이고4위는나지역입니다.(cid:8951)가, 다, 나[[46~~53]]서술형평가유형의입니다.46⑴산부인과별하루동안태어난신생아수를구하면가: 30명, 나: 24명, 다: 42명(평균)=(30+24+42)÷3=96÷3=32(명)▶3점⑵32명▶2점47⑴(현서네논의쌀수확량의합)=1150_8=9200(kg)(지우네논의쌀수확량의합)=2100_12=25200(kg) ; 9200kg, 25200kg▶3점⑵(평균)=(9200+25200)÷(8+12)=34400÷20=1720(kg)▶2점⑶1720kg▶1점48⑴(인경이의평균구슬수)=(24+16+17)÷3=57÷3=19(개)⑵민선이의평균구슬수도19개이므로(민선이의전체구슬수)=19_4=76(개)(민선이가가진노란색구슬수)=76-(32+7+19)=18(개)▶4점⑵18개▶2점49⑴(평균)=(4+7+8+2+3+6)÷6=30÷6=5(개) (cid:100) ; 5개▶2점⑵평균을1개높이면5+1=6(개)이므로(전체칭찬도장의수)=6_7=42(개)(10월에받아야할칭찬도장의수)=42-(4+7+8+2+3+6)=12(개)따라서10월에최소12개의칭찬도장을받아야합니다.▶3점⑶12개▶1점50⑴㉠병4개중에서물병은2개이므로가능성을수로(cid:100) (cid:100) 나타내면;2!;(cid:100) ㉡물병4개중에서음료수병을꺼낼수없으므로가능성을수로나타내면0(cid:100) ㉢병4개중에서물병은3개이므로가능성을수로(cid:100) (cid:100) 나타내면;4#;(cid:100) ;4#;>;2!;>0이므로사건이일어날가능성이큰순서(cid:100) 대로기호를쓰면㉢, ㉠, ㉡`입니다.▶4점⑵㉢, ㉠, ㉡▶1점51⑴(전체강아지의수)=36_3=108(마리) ⑵(B동네의강아지의수)=108-(40+32)=108-72=36(마리)⑶막대그래프에서세로눈금한칸의크기는4마리이므로B동네는세로눈금9칸을색칠합니다.▶3점⑵⑶▶2점02040ABC강아지의수 15초등쎈수학5-2정답(088-099) 2015.6.2 8:0 PM 페이지094 SinsagoHitec 본문|181쪽-184쪽6. 자료의표현0956단원01합이같게두수씩짝지으면1+2+3+4+……+17+18+19+20=21_10=2101부터20까지의자연수의개수: 20개(평균)=210÷20=10.5(cid:8951)10.5184쪽~185쪽02➊각길이별합을구하면1_3=3(m), 2_10=20(m), 3_6=18(m),4_9=36(m), 5_2=10(m)(가지고온철사의길이의합)=3+20+18+36+10=87(m)➋(전체학생수)=3+10+6+9+2=30(명)➌(평균)=(평균)=;3*0&;=2.9(m)따라서학생한명이철사를평균2.9m씩가지고왔습니다.(cid:8951)2.9m03⑴(`1학기평균)=⑴(`1학기평균)==85(점)⑵(`2학기평균)=85+4(`2학기평균)=89(점)(`2학기총점)=89_4=356(점)⑶총점이356점이되도록2학기점수에알맞은점수를써넣습니다.→(cid:9066)88+92+86+90=356(점)85+96+92+83=356(점)(cid:8951)⑴85점⑵89점, 356점⑶(cid:9066)88, 92, 86, 9004(100원, 500원)을던졌을때나오는면은(그림면, 그림면), (숫자면, 숫자면), (그림면, 숫자면),(숫자면, 그림면)으로모두4가지입니다.모두그림면이나올경우는(그림면, 그림면)으로1가지이므로모두그림면이나올가능성을수로나타내면;4!;입니다.(cid:8951);4!;340485+90+82+834(가지고온철사의길이의합)(전체학생수)먼저합이같게두수씩짝지은다음1부터20까지의자연수의합을구합니다.전략2121212152⑴(cid:9066)마을별배생산량을그림으로나타내면한눈에비교하기쉽게나타낼수있으므로그림그래프로나타내는것이더좋습니다.▶2점⑵1000kg=1t이므로세마: 324000kg=324t→3개, 2개, 4개(cid:100) 청학: 422000kg=422t→4개, 2개, 2개▶2점⑶⑶▶2점53⑴(cid:9066)청학마을의배생산량이가장많습니다.▶2점⑵(cid:9066)세마마을이중앙마을보다배생산량이324-216=108(t) 더많습니다.▶3점중앙 세마 청학 100`t10`t1`t배생산량먼저동전을1개던졌을때에는그림면과숫자면이나올수있고동전2개를동시에던졌을때에는4가지경우가나온다는것을구합니다.전략➊가지고온철사의길이의합구하기→➋전체학생수구하기→➌평균몇m씩가지고왔는지구하기푸는순서 15초등쎈수학5-2정답(088-099) 2015.6.8 4:2 PM 페이지095 SinsagoHitec 096쎈수학5－26자료의표현05마마을의사과생산량을(cid:8641)t이라하면(다마을의사과생산량)=(마마을의사과생산량)_2=((cid:8641)_2)t24+29+((cid:8641)_2)+17+(cid:8641)=118,70+(cid:8641)_3=118, (cid:8641)_3=48, (cid:8641)=48÷3=16(t)마마을의사과생산량: 16t→다마을의사과생산량: 16_2=32(t)→(cid:8951)06➊(전체마을의서점수의합)=7_24=168(개)▶2점➋(15개마을의서점수의합)=4_15=60(개)▶2점➌(나머지9개마을의서점수의합)=168-60=108(개)▶1점➍(나머지9개마을의평균서점수)=108÷9=12(개)▶2점10`t1`t사과생산량07➊(어머니)+(오빠)=28_2=56(살)(오빠)+(아버지)=30_2=60(살)(아버지)+(어머니)=41_2=82(살)세식을모두더하면{(어머니)+(오빠)+(아버지)}_2=56+60+82=198(살)(어머니)+(오빠)+(아버지)=198÷2=99(살)▶5점➋(어머니의나이)={(어머니)+(오빠)+(아버지)}-{(오빠)+(아버지)}=99-60=39(살)(오빠의나이)={(어머니)+(오빠)}-(어머니)=56-39=17(살)(아버지의나이)={(아버지)+(어머니)}-(어머니)=82-39=43(살)▶3점08➊㉠주사위한개를던졌을때나오는눈은1,2, 3, 4, 5, 6으로6개이고그중짝수는2, 4, 6으로3개이므로짝수의눈이나올가능성은반반입니다. →가능성을수로나타내면;2!;㉡숫자카드2장을한번씩모두사용하여만들수있는두자리수는63, 36이고그중36을만들가능성은반반입니다. →가능성을수로나타내면;2!;▶5점➋㉠+㉡=;2!;+;2!;㉠+㉡=1▶3점수아의어머니와오빠의평균나이는28살이고오빠와아버지의평균나이는30살, 아버지와어머니의평균나이는41살입니다. 수아의어머니, 오빠, 아버지의나이는각각몇살인지풀이과정을쓰고, 답을구하시오.(어머니)+(오빠)=28_2=56(살)(오빠)+(아버지)=30_2=60(살)(아버지)+(어머니)=41_2=82(살)채점기준➊어머니, 오빠, 아버지의나이의합을구한경우5점8점➋어머니, 오빠, 아버지의나이를각각구한경우3점채점기준➊㉠, ㉡`의가능성을각각수로나타낸경우5점8점➋㉠, ㉡`의가능성을수로나타낸것의합을구한경우3점마을별사과생산량을나타낸그림그래프입니다. 다마을의사과생산량은마마을의사과생산량의2배이고, 다섯마을의전체사과생산량은118t입니다.그림그래프를완성하시오.10`t1`t사과생산량29t24t((cid:8641)_2)t(cid:8641)t17t(다마을의사과생산량)=(마마을의사과생산량)_2=(cid:8641)_2▶▶▶▶24+29+((cid:8641)_2)+17+(cid:8641)=118채점기준➊전체마을의서점수의합을구한경우2점7점➋15개마을의서점수의합을구한경우2점➌나머지9개마을의서점수의합을구한경우1점➍나머지9개마을의평균서점수를구한경우2점▶ 15초등쎈수학5-2정답(088-099) 2015.6.2 8:0 PM 페이지096 SinsagoHitec 본문|184쪽-186쪽6. 자료의표현0976단원01(평균방문자수)=(109+101+98+134+79+95+91)÷7=707÷7=101(명)(cid:8951)101명02➊하루평균방문자수가101명이므로방문자수가101명보다적은요일을찾습니다.수요일(98명), 금요일(79명),토요일(95명), 일요일(91명)▶4점➋따라서하루평균방문자수보다적은날은모두4일입니다.▶2점03(두나이의합)=20_2=40(살)이므로17+23=40, 12+28=40, 21+19=40(cid:8951)23, 28, 1904=20(살)=20(살)(cid:8951)28, 19, 23; 40, 40, 4005(평균길이)=(평균길이)==28.2(cm)(cid:8951)28.2cm06➊㉮, ㉯학교의학생한명이사용하는운동장의넓이를각각구하면(㉮학교평균)=;;£7¡0∞0º;;=4.5(m¤)(㉯학교평균)=;;¢8º0º0º;;=5(m¤)▶4점➋4.5<5이므로학생한명이사용하는운동장의넓이가더넓은학교는㉯㉯학교입니다.▶3점112.8429+22.8+27.3+33.7440+40+40617+12+21+28+19+236186쪽~187쪽1회채점기준➊방문자수가하루평균방문자수보다적은날은모두며칠인지구하는과정을쓴경우4점6점➋방문자수가하루평균방문자수보다적은날은모두며칠인지구한경우2점채점기준➊㉮, ㉯학교의학생한명이사용하는운동장의넓이를각각구한경우4점7점➋학생한명이사용하는운동장의넓이가더넓은학교를구한경우3점09➊(네아파트의자동차수의합)=190_4=760(대)(가아파트와라아파트의자동차수의합)=760-(140+270)=760-410=350(대)▶2점➋가아파트가라아파트보다자동차가30대더많으므로(가아파트의자동차수)=(350+30)÷2=190(대)(라아파트의자동차수)=190-30=160(대)▶2점➌따라서그림그래프를완성하면10➊백의자리에서반올림하면3월: 8¯312→8000,004월: 13¯816→14000,5월: 22¯048→22000, 6월: 9¯701→10000연속적으로변화하는양이기때문에꺾은선그래프로나타내는것이좋습니다.표를보고세로눈금한칸의크기를정한후가로눈금과세로눈금이만나는자리에점을찍고점들을선분으로연결합니다.▶5점➋따라서그래프로나타내면010203456방문객수10010자동차보유수채점기준➊가아파트와라아파트의자동차수의합을구한경우2점7점➋가` 아파트와라아파트의자동차수를각각구한경우2점➌그림그래프를완성한경우3점구하려는자리바로아래자리의숫자가0, 1, 2, 3, 4이면버리고, 5, 6, 7, 8, 9이면올리는방법을반올림이라고합니다.전략채점기준➊알맞은그래프로나타내는과정을쓴경우5점8점➋알맞은그래프로나타낸경우3점▶3점▶3점 15초등쎈수학5-2정답(088-099) 2015.6.2 8:0 PM 페이지097 SinsagoHitec 098쎈수학5－26자료의표현16➊(네지역의전체콩생산량)=450+360+230+370▶4점➋=1410(t)▶2점17월별수출액의변화는연속적으로변화하는양이기때문에꺾은선그래프로나타내는것이좋습니다.(cid:8951)꺾은선그래프18세로눈금한칸의크기는1만달러를나타냅니다.(cid:8951)(cid:9066)051056789수출액채점기준➊네지역의전체콩생산량을구하는과정을쓴경우4점6점➋네지역의전체콩생산량을구한경우2점01(평균점수)=(92+86+95+88+94)÷5=455÷5=91(점)(cid:8951)91점022주일은14일이므로(평균)==800(원)(cid:8951)800원03[방법1]기준수를33으로정하고합이66이되도록기록을2개씩묶으면(25, 41), (39, 27)입니다.따라서공던지기기록의평균은33m입니다.[방법2](평균기록)=[방법2](평균기록)=[방법2](평균기록)=33(m)165525+39+33+27+4151120014188쪽~189쪽2회채점기준두가지방법으로설명한경우6점6점한가지방법으로설명한경우3점07(다섯수의합)=90_5=450(cid:8641)=450-(87+94+89+85)=450-355=95(cid:8951)95(모르는자료값)=(전체자료값의합)-(아는자료값의합)0810원짜리동전2개중에서100원짜리동전을고를수없으므로불가능합니다. →㉠(cid:8951)㉠09내일은꼭오므로확실합니다. →㉢(cid:8951)㉢10동전을던졌을때숫자면이나올수도있으므로반반입니다. →㉡(cid:8951)㉡11㉠가위바위보게임을할때가위는항상바위에게지므로사건이일어날가능성이불가능합니다.(cid:8951)㉠12숫자카드4장중에서숫자4는3장있으므로카드한장을꺼낼때꺼낸카드의숫자가4일가능성을수로나타내면;4#;입니다.(cid:8951)13➊반바지4벌중에서초록색반바지는1벌있습니다.▶3점➋따라서현호가초록색반바지를입을가능성을수로나타내면;4!;입니다.▶2점14➊원판은똑같이4부분으로나누어져있고그중빨간색은2부분입니다.▶4점➋따라서빨간색을맞힐가능성은반반이므로가능성을수로나타내면;2!;입니다.▶2점15지역별콩생산량은가: 450t, 나: 360t, 다: 230t, 라: 370t따라서콩생산량이가장많은지역은가지역이고가장적은지역은다지역입니다.(cid:8951)가지역, 다지역41214310채점기준➊초록색반바지를입을가능성을수로나타내는과정을쓴경우3점5점➋초록색반바지를입을가능성을수로나타낸경우2점채점기준➊빨간색을맞힐가능성을수로나타내는과정을쓴경우4점6점➋빨간색을맞힐가능성을수로나타낸경우2점 15초등쎈수학5-2정답(088-099) 2015.6.8 4:2 PM 페이지098 SinsagoHitec 본문|186쪽-189쪽6. 자료의표현0996단원10공3개중에서분홍색공은없습니다.따라서공을1개꺼낼때꺼낸공이분홍색일가능성을수로나타내면0입니다.(cid:8951)011➊주머니안에있는파란색딱지와초록색딱지의수는같으므로꺼낸딱지가초록색일가능성은반반입니다.▶4점➋따라서꺼낸딱지가초록색일가능성을수로나타내면;2!;입니다.▶3점12`2개, `4개이므로24000명입니다.(cid:8951)24000명13➊학생수가가장많은도시: 가, 27000명학생수가가장적은도시: 라, 19000명▶4점➋(학생수의차)=27000-19000=8000(명)▶3점14참가한학생수의많고적음을한눈에비교하기쉽게나타내려면막대그래프로나타내는것이좋습니다.(cid:8951)(cid:9066)막대그래프15세로눈금한칸의크기는1명을나타냅니다.(cid:8951)(cid:9066)막대그래프를그릴때에는막대의폭과막대와막대사이의간격을일정하게그립니다.16막대의길이가긴반부터차례로씁니다.(cid:8951)5반, 3반, 1반, 4반, 2반051012345참가한학생수04두집의밭1a당평균보리수확량을각각구하면(관식이네집)=;1!5*;=1.2(가마니)(민경이네집)=;1!2%;=1.25(가마니)1.2<1.25이므로민경이네집에서보리농사를더잘지었습니다.(cid:8951)민경이네집05➊국어와수학의평균점수가89점이므로(두과목의총점)=89_2=178(점)과학점수가95점이므로(세과목의총점)=178+95=273(점)▶4점➋(세과목의평균점수)==91(점)▶3점06(전체당근생산량)=4425×4=17700(kg)(라군의당근생산량)=(전체당근생산량)-(가, 나, 다군의생산량의합)=17700-(4560+3900+4400)=17700-12860=4840(kg) (cid:8951)4840kg07④1부터9까지의숫자카드중에서10보다작은숫자카드를뽑을가능성은확실합니다.(cid:8951)④08숟가락과포크중에서숟가락을고를가능성은반반입니다.09(전체노선수)=3+1=4(가지)민호가버스를타고진아네집까지가는것은3가지노선이므로민호가집에서진아네집까지버스를타고갈가능성을수로나타내면;4#;입니다. →㉢(cid:8951)㉢2733채점기준➊세과목의총점을구한경우4점7점➋세과목의평균점수를구한경우3점채점기준숟가락을고를가능성에대하여자신의생각을이야기한경우6점채점기준➊학생수가가장많은도시와가장적은도시의학생수를각각구한경우4점7점➋학생수가가장많은도시와가장적은도시의학생수의차를구한경우3점채점기준➊꺼낸딱지가초록색일가능성을수로나타내는과정을쓴경우4점7점➋꺼낸딱지가초록색일가능성을수로나타낸경우3점 15초등쎈수학5-2정답(088-099) 2015.6.8 4:3 PM 페이지099 SinsagoHitec 100쎈수학5－2011쪽•유형0301분모가100의약수인분수는분모를100으로고칠수있습니다.①==②==③==⑤==(cid:8951)④012쪽•유형0402분모가1000인분수로고친다음소수로나타냅니다.(cid:8951)8, 8；128, 13.128016쪽•유형1003분수를소수로나타내어크기를비교합니다.;2¶0¶0;==;1£0•0∞0;=0.3850.355<0.385→0.355<;2¶0¶0;소수를분수로나타내어크기를비교합니다.0.355=;1£0∞0∞0;=;2¶0¡0;이므로;2¶0¡0;<;2¶0¶0;→0.355<;2¶0¶0;(cid:8951)<024쪽•유형13040.14를분모가100인분수로고쳐서계산합니다.(cid:8951)14, 14；364, 3.64025쪽•유형1505➊(주영이가가지고있는색테이프의길이)=0.7_13▶3점➋=9.1(m)▶2점77_5200_5141007_250_2750451009_520_5920751003_254_2534321008_425_4825024쪽•유형14, 026쪽•유형1606⑴⑵(cid:8951)⑴㉠(cid:100)⑵㉡026쪽•유형1707➊4>3.5>2.97>1.64이므로가장큰수: 4가장작은수: 1.64▶2점➋(가장큰수)_(가장작은수)=4_1.64=6.56▶3점028쪽•유형20, 029쪽•유형2108①6.32_10=63.2②632_0.1=63.2③0.632_100=63.2④63.2_100=6320⑤6320_0.01=63.2(cid:8951)④032쪽•유형26090.257이소수세자리수이고3.2가소수한자리수이므로곱은소수네자리수가되도록소수점을찍습니다.→0.257_3.2=0.8224(cid:8951)0.8224031쪽•유형2410➊(평행사변형의넓이)=(밑변)_(높이)=13.7_5.8▶3점➋=79.46((cm¤))▶2점049쪽•유형0111모양과크기가같아서포개었을때완전히겹쳐지는두도형을찾으면㉯와㉱입니다.(cid:8951)㉯, ㉱051쪽•유형0512합동인두삼각형을포개었을때변ㄴㄷ과겹쳐지는변은변ㅂㄹ입니다.(cid:8951)변ㅂㄹ_0.15_0.0310.45→(㉡)_2.56_0.0410.24→(㉠)●틀린문제는풀이위에표시된유형을다시공부하세요.1회●1쪽~2쪽채점기준➊가장큰수와가장작은수를각각찾은경우2점5점➋가장큰수와가장작은수의곱을구한경우3점채점기준➊주영이가가지고있는색테이프는모두몇m인지구하는과정을쓴경우3점5점➋주영이가가지고있는색테이프는모두몇m인지구한경우2점채점기준➊평행사변형의넓이는몇cm¤인지구하는과정을쓴경우3점5점➋평행사변형의넓이는몇cm¤인지구한경우2점소수세자리수소수네자리수소수한자리수 15초등쎈수학5-2정답(100-112) 2015.6.2 8:2 PM 페이지100 SinsagoHitec 학업성취도평가101학업성취도053쪽•유형0813➊대응각의크기는서로같으므로(각ㅁㄹㅂ)=(각ㄱㄷㄴ)=34˘▶2점➋삼각형의세각의크기의합은180˘이므로(각ㄹㅂㅁ)=180˘-(34˘+95˘)=51˘▶3점삼각형에서두각의크기를알면나머지한각의크기를구할수있습니다.055쪽•유형1014①길이가2cm인선분ㄴㄷ을긋습니다.②점ㄴ을꼭짓점으로하여각도기로50˘인각을그립니다.③점ㄴ에서3cm인곳에점ㄱ을찍습니다.④점ㄱ과점ㄷ을잇습니다.(cid:8951)(cid:9066)자와각도기를사용하면두변의길이와그사이에있는각의크기가주어진삼각형과합동인삼각형을그릴수있습니다.057쪽•유형1415➊세변의길이가주어졌을때삼각형을그릴수있는조건은(가장긴변의길이)<(다른두변의길이의합)㉠7<2+6㉡15<8+9㉢9>4+3㉣16=5+11▶3점➋따라서삼각형을그릴수없는것은㉢㉢, ㉣㉣입니다.▶2점50æ3cm2cm08070605040302010090901001101201301401501601701800123012ㄴ ㄷ 3`cmㄴ ㄱ ㄱ ㄷ 3`cmㄴ ㄷ ㄴ ㄷ 2`cm2`cm2`cm50æ50æ① ② ③ ④ 2`cm064쪽•유형1616①②③④⑤(cid:8951)⑤②, ⑤정(cid:8774)각형에서대칭축의개수는(cid:8774)개입니다.067쪽•유형2017➊대응변의길이는서로같으므로(변ㄷㄴ)=(변ㄷㄱ)=15cm▶2점➋삼각형ㄱㄴㄷ의둘레가48cm이므로(변ㄱㄴ)=48-(15+15)=18((cm))▶3점064쪽•유형15, 068쪽•유형2218⑤정사각형은선대칭도형이면서점대칭도형입니다.⑤(cid:8951)⑤①, ③, ④`: 선대칭도형②`: 점대칭도형070쪽•유형2519점대칭도형에서대응변의길이와대응각의크기는각각같습니다.(cid:8951)72, 5072쪽•유형2820대칭의중심에서거리가같고방향이반대인곳에대응점을찍은다음각대응점을이어점대칭도형을완성합니다.(cid:8951)채점기준➊삼각형을그릴수없는것을모두찾는과정을쓴경우3점5점➋삼각형을그릴수없는것을모두찾아기호를쓴경우2점채점기준➊변ㄷㄴ의길이를구한경우2점5점➋변ㄱㄴ의길이를구한경우3점채점기준➊각ㅁㄹㅂ의크기를구한경우2점5점➋각ㄹㅂㅁ의크기를구한경우3점2개4개1개1개6개 15초등쎈수학5-2정답(100-112) 2015.6.8 4:3 PM 페이지101 SinsagoHitec 102쎈수학5－2091쪽•유형0607(하루에마신주스의양)=;1!1@;÷3=_=;1¢1;(L)(cid:8951);1¢1;L091쪽•유형0708각기호에숫자를넣어식을만들고계산합니다.(cid:8784)÷▲=2;7^;÷6=_=;2!1);(cid:8951);2!1);092쪽•유형0809➊2주일은14일이므로(하루에사용할수있는설탕의양)=3;7%;÷14=_▶3점➋=;4!9#;(kg)▶2점093쪽•유형1010➊(30분동안갈수있는거리)=(1분동안갈수있는거리)_30=7;8!;÷12_30=__30?=:™1•6∞:=➋17;1!6#;(km)111쪽•유형02112.8÷4=0.7(cid:8951)4, 0.7111쪽•유형0212정사각형은네변의길이가같으므로(정사각형의한변)=(정사각형의둘레)÷4=12.4÷4=3.1(cm)(cid:8951)3.1cm15112?41957?8/4114?71326?716/31020?7▲(cid:8785)13/1412?11088쪽•유형01014÷7은막대4개를각각똑같이7로나눈것중의하나입니다.→4÷7=4_;7!;(cid:8951)7, 1, 7088쪽•유형02025÷13=5_;1¡3;=;1∞3;(cid:8951);1∞3;089쪽•유형0403;7^;÷4=_=;1£4;;1£4;÷9=_=;4¡2;(cid:8951);1£4;, ;4¡2;089쪽•유형04, 090쪽•유형05, 093쪽•유형1104:¡9£:÷13=_=;9!;, ;1!1);÷4=_=;2∞2;(;9!;, ;2∞2;) →(;1™9™8;, ;1¢9∞8;) →;9!;<;2∞2;이므로:¡9£:÷13<;1!1);÷4(cid:8951)<090쪽•유형0505색칠한부분의길이는전체를똑같이3으로나눈것중의1입니다.(색칠한부분의길이)=;2&;÷3=;2&;_;3!;=;6&;=1;6!;(cm)(cid:8951)1;6!;cm06➊;1@5*;÷12=_=;4¶5;→㉠=45, ㉡=7▶3점➋㉠+㉡=45+7=52▶2점112?3728?1514/2510?11113?1113?919/313/1414/236/72회●3쪽~4쪽채점기준➊㉠, ㉡의값을각각구한경우3점5점➋㉠+㉡은얼마인지구한경우2점채점기준➊하루에사용할수있는설탕은몇kg인지구하는과정을쓴경우3점5점➋하루에사용할수있는설탕은몇kg인지구한경우2점채점기준➊30분동안갈수있는거리는몇km인지구하는과정을쓴경우3점5점➋30분동안갈수있는거리는몇km인지구한경우2점090쪽•유형05 15초등쎈수학5-2정답(100-112) 2015.6.2 8:2 PM 페이지102 SinsagoHitec 학업성취도평가103학업성취도112쪽•유형0313➊나무와나무사이의간격수는나무의수와같으므로간격은모두13구간입니다.(나무와나무사이의간격)=85.8÷13=➋6.6((m))113쪽•유형0514(cid:8951)113쪽•유형05, 127쪽•유형2415➊(색칠한부분의넓이)=(전체도형의넓이)÷4=9.36÷4=➋2.34(cm¤)115쪽•유형0816(소수)÷(자연수)=29.76÷62=0.48(cid:8951)0.48120쪽•유형111732.9÷14=2.35㉠35.1÷15=2.34㉡11.9÷5=2.38㉢14.1÷6=2.35 ㉣53.9÷22=2.45(cid:8951)㉢122쪽•유형14, 126쪽•유형2218가12.2÷4=3.05 →3.05<3.06나15.3÷5=3.06(cid:8951)나126쪽•유형231966÷8=8.258.25>(cid:8641)이므로(cid:8641)안에들어갈수있는가장큰자연수는8입니다.(cid:8951)816.327<“44.2442.212.2212.1212.1412.1412.10몫의소수점을나뉠수의소수점자리에맞추어찍어야하는데찍지않았습니다.130쪽•유형3020➊117÷33=3.5454……소수점아래반복되는숫자: 5, 4(2개)▶3점➋55÷2=27…1이므로소수55째자리숫자는5입니다.▶2점145쪽•유형0101⑴1a=100m¤이므로5a=500m¤⑵100m¤=1a이므로300m¤=3a(cid:8951)⑴500⑵3145쪽•유형0102➊(주차장의넓이)=40_50=2000(m¤)▶3점➋=20(a)▶2점148쪽•유형060381ha=810000m¤이고810000=900_900이므로이공원의한변은900m입니다.(cid:8951)900m150쪽•유형08043km¤=300ha이므로(만들수있는밭의수)=300÷100=3(개)(cid:8951)3개146쪽•유형03053900ha=39km¤이므로4.2km¤+3900ha=4.2km¤+39km¤=43.2(km¤)(cid:8951)43.2km¤넓이의합을구할때에는단위를같게한후계산합니다.이때합을km¤로구해야하므로ha를km¤로나타낸후계산하면편리합니다.3회●5쪽~6쪽채점기준➊나눗셈의몫을소수로구하여반복되는숫자를찾은경우3점5점➋소수55째자리숫자를구한경우2점채점기준➊주차장의넓이는몇m¤인지구한경우3점5점➋주차장의넓이는몇a인지구한경우2점채점기준➊나무와나무사이의간격을몇m로하면되는지구하는과정을쓴경우3점5점➋나무와나무사이의간격을몇m로하면되는지구한경우2점채점기준➊색칠한부분의넓이는몇cm¤인지구하는과정을쓴경우3점5점➋색칠한부분의넓이는몇cm¤인지구한경우2점 15초등쎈수학5-2정답(100-112) 2015.6.8 4:4 PM 페이지103 SinsagoHitec 104쎈수학5－2146쪽•유형0306(공원의넓이)=3_2÷2=3(km¤)=300(ha)(cid:8951)300ha146쪽•유형0307➊4km500m=4.5km이므로▶2점➋(평행사변형의넓이)=4.5_2=9(km¤)▶3점➊4km500m=4500m,2km=2000m이므로▶2점➋(평행사변형의넓이)=4500_2000=9000000(m¤)▶2점➌=9(km¤)▶1점147쪽•유형0508(직사각형의넓이)=7000_1000=7000000(m¤)1000000m¤=10000a=100ha=1km¤이므로7000000m¤=70000a=700ha=7km¤입니다.(cid:8951)7000000, 70000, 700, 7150쪽•유형0909⑴1t=1000kg이므로0.9t=900kg⑵1000kg=1t이므로8000kg=8t(cid:8951)⑴900⑵8151쪽•유형1210➊승호네: 2t=2000kg이므로➋(승호네마을의전체포도수확량)=2000+850+2150=5000(kg)=5(t)➌따라서적어도5t을실을수있는트럭이필요합니다.169쪽•유형0411(은진이네모둠평균)=(호준이네모둠평균)=31(번)(호준이네모둠평균)==(호준이네모둠평균)=32(번)31<32이므로호준이네모둠이줄넘기를평균32-31=1(번) 더많이했습니다.(cid:8951)호준이네모둠, 1번170쪽•유형0612➊4회까지의평균이90점이므로(4회까지의총점)=90_4=360(점)5회째시험에서100점을받으면(5회까지의총점)=360+100=460(점)▶2점➋(5회까지의평균)=;:$5^:);=92(점)▶2점➌(올라가는평균점수)=92-90=2(점)▶1점175쪽•유형1013ㄷ, ㅎ, ㅁ, ㅊ은모두자음입니다.자음을고를가능성은확실하므로가능성을수로나타내면1입니다.모음을고를가능성은불가능하므로가능성을수로나타내면0입니다.(cid:8951)1, 0174쪽•유형0914(주머니속의전체동전의수)=3+1=4(개)동전4개중에서100원짜리동전은3개이므로꺼낸동전이100원짜리일가능성을수로나타내면;4#;입니다.(cid:8951)101-41-23-4128456+13+25+344채점기준➊승호네의포도수확량은몇kg인지구한경우1점5점➋승호네마을의전체포도수확량은몇t인지구한경우2점➌적어도몇t을실을수있는트럭이필요한지구한경우2점채점기준➊가로와세로를각각m단위로나타낸경우2점5점➋평행사변형의넓이는몇m¤인지구한경우2점➌평행사변형의넓이는몇km¤인지구한경우1점채점기준➊5회까지의총점을구한경우2점5점➋5회까지의평균을구한경우2점➌평균이몇점올라가는지구한경우1점채점기준➊4km500m를km단위로나타낸경우2점5점➋평행사변형의넓이는몇km¤인지구한경우3점=124419+26+42+374▶1점▶2점▶2점 15초등쎈수학5-2정답(100-112) 2015.6.8 4:4 PM 페이지104 SinsagoHitec 경시대비평가105경시대비178쪽•유형1315라마을의보리판매량은180kg이므로1개, 8개를그립니다.(cid:8951)177쪽•유형1216보리를가장많이판매한마을은이가장많은가마을입니다.(cid:8951)가마을167쪽•유형01, 177쪽•유형1217➊네마을의보리판매량은가: 350kg, 나: 270kg, 다: 240kg, 라: 180kg▶2점➋(평균보리판매량)===260(kg)▶3점179쪽•유형1418시간의흐름에따른온도는연속적으로변화하는양이기때문에꺾은선그래프로나타내는것이좋습니다.(cid:8951)꺾은선그래프179쪽•유형1519세로눈금한칸의크기는1æ를나타냅니다.(cid:8951)179쪽•유형1520➊9æ보다낮을것입니다. ;▶2점➋그래프에서2분후부터꺾은선이오른쪽아래로내려가고있으므로8분이후의물의온도는9æ보다낮을것입니다.▶3점{æ`C}101520502468물의온도10404350+270+240+1804100`kg10`kg보리판매량1011쪽•유형0201분자1부터14까지의수중에서분모15와약분이되는수: 3, 5, 6, 9, 10, 12;1£5;=;5!;, ;1∞5;=;3!;, ;1§5;=;5@;, ;1ª5;=;5#;, ;1!5);=;3@;,;1!5@;=;5$;이중에서소수한자리수로나타낼수있는분수는;1£5;, ;1§5;, ;1ª5;, ;1!5@;→4개(cid:8951)4개014쪽•유형07, 018쪽•응용01번02➊(3.2와3.4사이의크기)=3.4-3.2=0.2눈금한칸의크기는0.2를10등분하였으므로0.02입니다.▶2점➋㉠은3.2에서오른쪽으로0.02씩7칸갔으므로㉠=3.2+0.02_7=3.34㉠=3;1£0¢0;=3;5!0&;▶3점016쪽•유형1003지원이의키와진희의키를각각소수로나타내면지원이의키: 1;2!5#;=1;1∞0™0;=1.52(cid:100)진희의키: 1;1§2™5;=1;1¢0ª0§0;=1.4961.52>1.496>1.46이므로가장큰사람: 지원(cid:100)(cid:100)가장작은사람: 하선→(키의합)=1.52+1.46=2.98(m)(cid:8951)2.98m024쪽•유형14, 026쪽•유형17043.4_6=20.4, 7_3.8=26.6→20.4<(cid:8641)<26.6따라서(cid:8641)안에들어갈수있는자연수는21, 22……, 26입니다. →6개(cid:8951)6개1회●1쪽~2쪽채점기준➊눈금한칸의크기를구한경우2점5점➋㉠`을기약분수로나타낸경우3점채점기준➊네마을의보리판매량을각각구한경우2점5점➋네마을의평균보리판매량은몇kg인지구한경우3점채점기준➊8분이후의물의온도는9æ보다낮음을쓴경우2점5점➋이유를설명한경우3점 15초등쎈수학5-2정답(100-112) 2015.6.8 4:4 PM 페이지105 SinsagoHitec 106쎈수학5－2026쪽•유형16, 028쪽•유형20, 029쪽•유형2105㉠0.014_1000=14 ㉡18_0.01=0.18→㉠_㉡=14_0.18=2.52(cid:8951)2.52028쪽•유형1906➊어떤수를(cid:8641)라하면(잘못계산한값)=(cid:8641)_4.08(바르게계산한값)=(cid:8641)_408▶3점➋곱하는수가100배가되면곱의결과도100배가되므로바르게계산한값은잘못계산한값의100배입니다.▶2점031쪽•유형2407(큰직사각형의넓이)=6.5_5.8=37.7(cm¤)(작은직사각형의넓이)=3.3_3.1=10.23(cm¤)(색칠된부분의넓이)=(큰직사각형의넓이)-(작은직사각형의넓이)=37.7-10.23=27.47(cm¤)(cid:8951)27.47cm¤052쪽•유형06, 058쪽•응용03번08➊두직각삼각형이합동이므로대응변의길이는서로같습니다. (변ㄴㄷ)=(변ㅁㄱ)=15cm(변ㄱㄹ)=(변ㄷㄹ)=12cm(변ㄴㄹ)=(변ㅁㄹ)=9cm(변ㄱㄴ)=(변ㄱㄹ)-(변ㄴㄹ)=12-9=3(cm)▶3점(도형전체의둘레)=3+15+12+9+15➋=54(cm)▶2점053쪽•유형08, 059쪽•응용09번09삼각형2개가서로합동인이등변삼각형이므로(각ㄱㄷㄴ)=(각ㄱㄷㄹ)=160˘÷2=80˘(각ㄱㄴㄷ)=(각ㄱㄷㄴ)=80˘삼각형의세각의크기의합은180˘이므로(각ㄴㄱㄷ)=180˘-(80˘+80˘)=20˘(각ㄴㄱㄷ)=(각ㄷㄱㄹ)=20˘이므로(각ㄴㄱㄹ)=20˘_2=40˘(cid:8951)40˘057쪽•유형1410삼각형을그리려면(가장긴변의길이)<(다른두변의길이의합)이어야합니다.가장긴변을13cm라하면13<4+10=14, 13<9+10=19→2개가장긴변을10cm라하면10<2+9=11, 10<4+9=13→2개따라서모두4개입니다.(cid:8951)4개067쪽•유형2011➊선분ㄱㄹ은변ㄴㄷ을이등분하므로(변ㄴㄷ)=4_2=8(cm)삼각형ㄱㄴㄷ의둘레가20cm이므로(변ㄱㄴ)+(변ㄴㄷ)+(변ㄱㄷ)=20cm→(변ㄱㄴ)+(변ㄱㄷ)=20-8=12(cm)대응변의길이는같으므로(변ㄱㄷ)=(변ㄱㄴ)=12÷2=➋6(cm)066쪽•유형18, 076쪽•응용03번12삼각형의세각의크기의합은180˘이고선대칭도형에서대응각의크기는같으므로(각ㄷㄱㄹ)=(각ㄷㄱㅁ)=180˘-(90˘+32˘)=58˘사각형ㄱㅂㄷㄹ의네각의크기의합은360˘이므로(각ㄱㅂㄷ)=360˘-(90˘+90˘+58˘+58˘)=64˘(cid:8951)64˘071쪽•유형2713(주어진도형의넓이)=5_8=40(cm¤)완성한점대칭도형전체의넓이는주어진도형의넓이의2배이므로(완성한점대칭도형전체의넓이)=40_2=80(cm¤)(cid:8951)80cm¤095쪽•유형1314(cid:8784)_13=16→(cid:8784)=16÷13=;1!3^;=1;1£3;(cid:8784)÷10=1;1£3;÷10=_=;6•5;→(cid:8774)=;6•5;(cid:8951);6•5;곱셈식과나눗셈식의관계를이용하여(cid:8784)의값을먼저구합니다.110?5816?13ㄱ ㅁ ㅂ ㄴ ㄷ ㄹ ㅅ 32æ58æ58æ채점기준➊바르게계산한값은잘못계산한값의몇배인지구하는과정을쓴경우3점5점➋바르게계산한값은잘못계산한값의몇배인지구한경우2점채점기준➊도형전체의둘레를구하는과정을쓴경우3점5점➋도형전체의둘레를구한경우2점채점기준➊변ㄱㄷ의길이를구하는과정을쓴경우3점5점➋변ㄱㄷ의길이를구한경우2점 15초등쎈수학5-2정답(100-112) 2015.6.8 4:4 PM 페이지106 SinsagoHitec 경시대비평가107경시대비091쪽•유형0615(경수가한시간에하는일의양)=;5#;÷9=_=;1¡5;(미라가한시간에하는일의양)=;9@;÷5=;9@;_;5!;=;4™5;(두사람이한시간에하는일의양)=;1¡5;+;4™5;=;4£5;+;4™5;=;4∞5;=;9!;따라서두사람이함께일을모두마치는데걸리는시간은9시간입니다.(cid:8951)9시간091쪽•유형0616정사각형의한변이:™6∞:cm이므로(색칠된부분의세로)=:™6∞:÷5=_=;6%;(cm)(색칠된부분의둘레)=(:™6∞:+;6%;)_2=5_2(색칠한부분의둘레)=10(cm)(cid:8951)10cm094쪽•유형1217나누는수가3으로일정하므로나뉠수가클수록몫은커집니다.2가대분수이므로(cid:8641)안에들어갈수있는수는1, 2, 3, 4이고그중가장큰수는4입니다.→2;5$;÷3=:¡5¢:_;3!;=;1!5$;(cid:8951)4；;1!5$;092쪽•유형08, 101쪽•응용09번18➊(사과6개의무게)=(사과6개가들어있는바구니의무게)-(빈바구니의무게)=2;4#;-;2!;=2;4#;-;4@;=2;4!;(kg)▶2점➋(사과한개의무게)=2;4!;÷6=_(사과한개의무게)=;8#;(kg)▶3점16/239/4(cid:8641)515/1525?619/313/5092쪽•유형0819(삼촌이3시간동안간거리)=2;9%;_3=_3/=:™3£:=7;3@;(km)(규철이가한시간동안간거리)=7;3@;÷2=:™3£:_;2!;=:™6£:=3;6%;(km)(cid:8951)3;6%;km092쪽•유형09, 093쪽•유형1120➊㉠, ㉡을각각가_(분수)로나타내면㉠가÷5_4=가_;5!;_4=가_;5$;㉡가_3;5!;÷6=가__=가_;1•5;▶3점➋(;5$;, ;1•5;) →(;1!5@;, ;1•5;) (cid:8837);5$;>;1•5;가에곱하는수가클수록큰수이므로계산결과가더큰것은㉠㉠``입니다.▶2점16/3816?51239/3채점기준➊사과6개의무게를구한경우2점5점➋사과한개의무게를구한경우3점채점기준➊㉠, ㉡을각각가_(분수)로나타낸경우3점5점➋계산결과가더큰것을찾아기호를쓴경우2점015쪽•유형08010.1이4개,0.01이7개,0.001이8개인수→0.4+0.07+0.008=0.478=;1¢0¶0•0;=;5@0#0(;(cid:8951);5@0#0(;소수세자리수는분모가1000인분수로나타냅니다.032쪽•유형2502(1분동안받은물의양)=9.6-3.8=5.8(L)10분15초=10;6!0%;분=10;4!;분=10;1™0∞0;분=10.25분(10분15초동안받은물의양)=5.8_10.25=59.45(L)(cid:8951)59.45L1분=60초이므로1초=분→▲초=분▲601602회●3쪽~4쪽 15초등쎈수학5-2정답(100-112) 2015.6.2 8:2 PM 페이지107 SinsagoHitec 108쎈수학5－2053쪽•유형0803➊(각ㄹㄱㄴ)=(각ㄷㄴㄱ)=180˘-105˘=75˘사각형의네각의크기의합은360˘이므로(각ㄱㄹㄷ)_2+75˘_2=360˘,(각ㄱㄹㄷ)_2+150˘=360˘,(각ㄱㄹㄷ)_2=210˘,▶3점➋(각ㄱㄹㄷ)=105˘▶2점067쪽•유형2004주어진도형의넓이를㉠,㉡으로나누어구합니다.㉠`은윗변이3cm, 아랫변이(5+2)`cm, 높이가2cm인사다리꼴이므로(㉠`의넓이)=(3+7)_2÷2=10`(cm¤)㉡`은윗변이3cm, 아랫변이5cm, 높이가2cm인사다리꼴이므로(㉡`의넓이)=(3+5)_2÷2=8(cm¤)→(주어진도형의넓이)=10+8=18(cm¤)완성한선대칭도형전체의넓이는주어진도형의넓이의2배이므로(완성한선대칭도형전체의넓이)=18_2=36(cm¤)(cid:8951)36cm¤선대칭도형에서대칭축에의해나뉘어진두도형은합동이므로넓이가같습니다.089쪽•유형04, 095쪽•유형1305➊;1ª4;÷3=_=;1£4;▶2점➋;1£4;=(cid:8641)÷56→(cid:8641)=_56?=12▶3점4314?113/139/145cm2cm3cm2cm3cm5cm096쪽•유형1506(이어붙인종이의넓이)=4;4#;_2=_2/=:¡2ª:=9;2!;(m¤)(세로)=(이어붙인종이의넓이)÷(가로)(세로)=9;2!;÷9=:¡2ª:_;9!;=;1!8(;=1;1¡8;(m)(cid:8951)1;1¡8;m128쪽•유형2607➊(정사각형의한변)=8.84÷4=2.21(m)▶2점➋(정삼각형의한변)=5.61÷3=1.87(m)▶2점➌(한변의길이의차)=2.21-1.87=0.34(m)▶1점122쪽•유형1508(전체걸은거리)=5.92+3.48+6.18+2.78=18.36(km)(한시간동안걸은거리)=18.36÷6=3.06(km)(cid:8951)3.06km011쪽•유형02, 126쪽•유형22093;5!;=3;1™0;=3.2①0.76(cid:100)②1.84(cid:100)③3.05(cid:100)④3.4(cid:100)⑤3.35(cid:8951)④, ⑤124쪽•유형19103씩커지는규칙입니다. ㉠=16+3=19, ㉡=㉠+3=19+3=22㉡÷㉠=22÷19=1.15¯7……→1.16(cid:8951)1.16150쪽•유형0811(꽃밭의넓이)=280_350=98000(m¤)100m¤=1a이므로98000m¤=980a꽃밭을140a씩나누어서로다른종류의꽃을심으므로(심을수있는꽃의종류수)=980÷140=7(종류)(cid:8951)7종류148쪽•유형061227ha=270000m¤이므로아랫변을(cid:8641)m라하면(사다리꼴의넓이)=(500+(cid:8641))_450÷2=270000,(500+(cid:8641))_450=270000_2=540000,500+(cid:8641)=540000÷450=1200,(cid:8641)=1200-500=700(m)(cid:8951)700m1194/2채점기준➊;1ª4;÷3의몫을구한경우2점5점➋(cid:8641)안에알맞은자연수를구한경우3점채점기준➊정사각형의한변의길이를구한경우2점5점➋정삼각형의한변의길이를구한경우2점➌한변의길이의차를구한경우1점채점기준➊각ㄱㄹㄷ`은몇도인지구하는과정을쓴경우3점5점➋각ㄱㄹㄷ은몇도인지구한경우2점 15초등쎈수학5-2정답(100-112) 2015.6.8 4:5 PM 페이지108 SinsagoHitec 경시대비평가109경시대비145쪽•유형02, 147쪽•유형0513길을제외하고밭을모으면오른쪽과같은평행사변형모양이됩니다.(길을제외한밭의밑변)=120-(20+20)=80(m)(길을제외한밭의높이)=155-(20+20)=115(m)(길을제외한밭의넓이)=80_115=9200(m¤)10000m¤=1ha이므로9200m¤=0.92ha(cid:8951)0.92ha149쪽•유형0714➊㉡500m=0.5km3km400m=3.4km이므로(대나무밭의넓이)=0.5_3.4=1.7(km¤)▶2점➋㉢(아파트단지의넓이)=800_500÷2=200000(m¤)=0.2(km¤)▶2점➌2km¤>1.7km¤>0.2km¤이므로넓이가가장넓은것은㉠㉠입니다.▶1점157쪽•응용10번15(마름모모양의논의넓이)=(210_150÷2)_4=63000(m¤)=630(a)1a당50kg의쌀이생산되므로(생산되는쌀의무게)=50_630=31500(kg)=31.5(t)(cid:8951)31.5t171쪽•유형0716(평균이34번일때줄넘기총기록)=34_4=136(번)(평균이34번일때2회의줄넘기기록)=136-(38+33+29)=36(번)평균이34번보다많으므로2회때줄넘기기록은36번보다많습니다. 따라서2회때줄넘기는적어도37번했습니다. (cid:8951)37번80`m115`m167쪽•유형01, 170쪽•유형0517➊(전체학생몸무게의합)=38.7_32=1238.4(kg)▶1점➋(남학생몸무게의합)=39.8_17=676.6(kg)(여학생몸무게의합)=1238.4-676.6=561.8(kg)▶2점➌(여학생수)=32-17=15(명)(여학생의평균몸무게)=561.8÷15=37.4¯5yy→37.5kg▶2점120쪽•유형11, 175쪽•유형10, 176쪽•유형1118㉠주사위의눈6개중에서짝수인눈은2, 4, 6으로3개입니다.㉠→가능성을수로나타내면;2!;㉡8.4÷10=0.84이므로8.4가나올수없습니다. →가능성을수로나타내면0;2!;>0이므로가능성이더큰것은㉠입니다. (cid:8951)㉠176쪽•유형1119➊흰색공은전체4개중의2개입니다.➊→가능성을수로나타내면;2!;노란색공은전체4개중의1개입니다.→가능성을수로나타내면;4!;▶3점➋(두가능성의합)=;2!;+;4!;=;4@;+;4!;=;4#;▶2점171쪽•유형07, 178쪽•유형132028.5t=28500kg이므로(전체쌀생산량)=28500_4=114000(kg)(다마을의쌀생산량)=114000-(37000+28000+31000)=18000(kg)18000kg은1개, 1개, 3개로나타냅니다.(cid:8951)1`kg5`kg1`kg쌀생산량채점기준➊㉡대나무밭의넓이를구한경우2점5점➋㉢아파트단지의넓이를구한경우2점➌넓이가가장넓은것을찾아기호를쓴경우1점채점기준➊두가능성을각각수로나타낸경우3점5점➋수로나타낸두가능성의합을구한경우2점채점기준➊전체학생몸무게의합을구한경우1점5점➋남학생의몸무게의합과여학생몸무게의합을각각구한경우2점➌여학생의평균몸무게를구한경우2점 15초등쎈수학5-2정답(100-112) 2015.6.8 4:5 PM 페이지109 SinsagoHitec  15초등쎈수학5-2정답(100-112) 2015.6.2 8:2 PM 페이지110 SinsagoHitec  15초등쎈수학5-2정답(100-112) 2015.6.2 8:2 PM 페이지111 SinsagoHitec  15초등쎈수학5-2정답(100-112) 2015.6.2 8:2 PM 페이지112 SinsagoHitec