2014六年级数学下册计算题专项练习题

2014六年级数学下册计算题专项练习题 一、 计算题:(共38分) 1、 直接写出得数(每小题1分，共6分) 23124，75%, ,,0.68，，0.32,423 21211 9,，3,，,，,4,, 1,,0.4,,,374745,, 2、 合理、灵活地计算(每小题4分，共16分) 112.5，1.25, 15，201.3,20.13,，2013，50%32100 ，，，，23511,,,, ，,,2.35,4.75,4.5，0.2，,,,,,,,,941443,,,,，，，， 3、 求未知数(每小题3分，共6分) x 324x,6，,32x:24%,:0.3 35 4、 列综合算式或方程解答(4分) 1196的比一个数的多2.5，求这个数。 62 一、计算。(共35分) construction quality acceptance and assessment Regulation (Professional Edition) (DL/T5210.2-2009~DL/T5210.8-2009); 1.9 the quality checkout and evaluation of electric equipment installation engineering code (DL/T 5161.1-2002~5161.17-2002); 1.10 the norms of construction supervision, the electric power construction supervision regulations 1、直接写出得数。(每题0.5分，共4分) 510331787,998, ，0.25, × , 21? , 82157 51511111321 × ? × = ? ? , ×12.1,1= ， ? = 9595888115552、用递等式计算。(每题3分，共18分，多做不给分。) 37126 ? 987，104×65,1747 ? 86.4?3.2,6.4×3.2 ? ?7 ， × 63763 174351317,16.8?(1.8，7.2× ) ( ， , )×6.3 15?〔( , )? 〕,0.5 1292177228 3、求未知数X。(每題2分，共6分) 1313532 0.4 X,0.4×10.8 =20 X， X = : X = : 3448645 一、计算。(共26分) 1(直接写出得数。(每小题1分，共8分) 新 课 标 第 一 网 construction quality acceptance and assessment Regulation (Professional Edition) (DL/T5210.2-2009~DL/T5210.8-2009); 1.9 the quality checkout and evaluation of electric equipment installation engineering code (DL/T 5161.1-2002~5161.17-2002); 1.10 the norms of construction supervision, the electric power construction supervision regulations 657526.3?0.1= ?= ,(,)= 8×(2.5+0.25)= 76979 51113.37，6.73= ,= (0.18，0.9)?9= 7×?7×= 69662(计算下面各题。(第(1)(2)小题各3分，第(3)小题6分，共12分) 51 36?〔(,)×3〕 17.5,5(x，0.5)=9 63 (3)简便计算: (87.2，87.2，87.2×2)×25 765×213?27，765×327?27 3(列式解答下列文字题。(每小题3分，共6分) (1)2.7与4.5的和去除14.4，得出的商再乘3，积是多少, 1(2)一个数的比这个数的75%少15，这个数是多少,(列方程解) 4 一、直接写出得数。 56，38, 7.6,06, 7.8，2.8, 8,3.7,4.3, construction quality acceptance and assessment Regulation (Professional Edition) (DL/T5210.2-2009~DL/T5210.8-2009); 1.9 the quality checkout and evaluation of electric equipment installation engineering code (DL/T 5161.1-2002~5161.17-2002); 1.10 the norms of construction supervision, the electric power construction supervision regulations 243253,, ，, ,, 0.53×8×0.125, 394588 5368351×, 12?, ?×0, ?5?, 6579572二、脱式计算，怎样算简便就怎样算。 513 4.8×3.9，6.1×4.8 (,，)×32 1684 212[4,(,)]× 25.6,8.44,1.56 343 2、求未知数。 1350.4×(x，3.5),13.2 :,x: 4816 3、列综合算式或方程计算。 3?1.8减去0.6的差除以1.2的倍，商是多少,?一个数的是20的75%，这个是数是多少? 5 1.直接写得数。(每题0.5分，共2分) 2572 3.81+0.29= -= 1?75,= (+)×15= 56932.求未知数X。(每题2分，共6分) construction quality acceptance and assessment Regulation (Professional Edition) (DL/T5210.2-2009~DL/T5210.8-2009); 1.9 the quality checkout and evaluation of electric equipment installation engineering code (DL/T 5161.1-2002~5161.17-2002); 1.10 the norms of construction supervision, the electric power construction supervision regulations 1310.52:χ=: χ= X一20,X=60 481663 3.脱式计算。(能简算的要简算)(每题2分，共12分) 51729630.25×3.2×1.25 ×? 29× ， ×114 1311318714 310271274 , ， )? ×5.4+80%×5.6,0.8 (3,，,10673052217 4.列综合式计算。(每题2分，共4分) 5228与的和除以它们的差，商是多少? (2)一个数的是18的 倍，这个数是多少? 3637 (一)直接写出得数(10分) 5，33,3.8+6.2= 8.1?3×2= 568,198= 0.65?1.3= 11 3211233,,1,,48，(,),，，2, 75×10%= 8554643 1( (二)用递等式计算，能简算的简算(18) construction quality acceptance and assessment Regulation (Professional Edition) (DL/T5210.2-2009~DL/T5210.8-2009); 1.9 the quality checkout and evaluation of electric equipment installation engineering code (DL/T 5161.1-2002~5161.17-2002); 1.10 the norms of construction supervision, the electric power construction supervision regulations 85，54,18，745 (2) (1) 3.4,[(1.25，0.45)，23] 4835((156，,26)，3) 35，(1,), (4) 7121311 7127，，,13(5) (6)(42×29+71×42)?35 8138 (三)求未知数x (6分)X K b1.C o m 1317732x，x,:x,:1) ( (2) 4431689 1 1、甲数与乙数的比是2:3，甲数是，乙数是多少, 4 1、直接写出得数 45591,,,，11,1.1= 1.26?0.3= 12.5×= 13?39= 6× = ，,2,,579512,, 13714238,,，，,1.5×0.2= 87.5%×= 2,= 8.9×11,8.9= 8715255592、求x的值 construction quality acceptance and assessment Regulation (Professional Edition) (DL/T5210.2-2009~DL/T5210.8-2009); 1.9 the quality checkout and evaluation of electric equipment installation engineering code (DL/T 5161.1-2002~5161.17-2002); 1.10 the norms of construction supervision, the electric power construction supervision regulations 3541 9.57,(2.57，3.38)―2.62 xx,,::15,x8796 3、下面各题怎样算简便就怎样算 ，，521115,,,, 3.6×98，36×0.2 353,,，,,，,,,,,,633426,,,,，， 22、甲数的比乙数的25%多40，已知乙数是160，求甲数是多少, 3 3、180比一个数的50,多10，这个数是多少, 44、120的20%比某数的少24，求某数, 5 1、直接写出得数。(每题1分) 3 26×50= 25×0.2= 10,0.86= 24×= 4 34 ?3= 125%×8= 4.8?0.8= 8?= 75 2211,，0, 12×(，)= 1,1?9= 2.5×3.5×0.4= 4633 construction quality acceptance and assessment Regulation (Professional Edition) (DL/T5210.2-2009~DL/T5210.8-2009); 1.9 the quality checkout and evaluation of electric equipment installation engineering code (DL/T 5161.1-2002~5161.17-2002); 1.10 the norms of construction supervision, the electric power construction supervision regulations 2、脱式计算。(每题2分) ，，11241,, 0.25××4 + 2.5% 9.6,11?7 + 12，,，3,,,,5763,, ，， 3、解比例和方程。(每题3分)X k B 1 . c o m X 5.4+2X = 8.6 2.5:5 = x:8 0.2 = 1- 24 4、列式计算。(每题4分) (1)180比一个数的50,多10，这个数是多少, 31 (2)0.15除以的商加上5，再乘以，积是多少, 84 1、直接写出得数(4分) 351157 ， , , , ? , 0.37，1.6, 582339 3 1.224?12, 4×3×0.25, ×1.6, 4,1?3, 4 2、求未知数χ。 1121 : ,6 :χ χ， χ,25 (18，χ)×7,560 2336 construction quality acceptance and assessment Regulation (Professional Edition) (DL/T5210.2-2009~DL/T5210.8-2009); 1.9 the quality checkout and evaluation of electric equipment installation engineering code (DL/T 5161.1-2002~5161.17-2002); 1.10 the norms of construction supervision, the electric power construction supervision regulations 3、下列各题，你认为怎样算简便就怎样算。 51151.25 ×6.9×8 99×0.79 ， 0.79 ?[( , )× ] 36269 4、如图，求阴影部分的面积。(单位:厘米) 5(求未知数X(4分) 1116 X :=:x ,4.5，5.5,10 83209 1(直接写出得数。(8分) 2 367，699, 2.4×5, 60?6%, 0.3, 333232 31.9,3.09, ， , × ? × , 0.01?0.7, 485353 31252 2.2，3.57= ×12= ， = 3.37，66.3= ? = 49383 5133 3.25×4= ×8，8× = ?3× ?3= 4444 2(怎样算简便就怎样算。(9分) 161111254 2.6×(2.205?2.1) ， × ， ×5，? 336952727 construction quality acceptance and assessment Regulation (Professional Edition) (DL/T5210.2-2009~DL/T5210.8-2009); 1.9 the quality checkout and evaluation of electric equipment installation engineering code (DL/T 5161.1-2002~5161.17-2002); 1.10 the norms of construction supervision, the electric power construction supervision regulations 5411251 4.85, ，5.15, + ×( , ) 99131323 341312 5×(+ )×17 ?,( ，)× , 45471517 3(解方程。(9分) 329153 ? =12 X,X= X?=2? x4336410 1、直接写出得数(6分) 4243+49= 198—79= 9.3+4.76= 20.67—8.7= 320×0.5= 12.5×= 5 41527143126345?0.15= —= += ?= ?= ××= 3385631215497 2、解方程(4分) 40.8 ?3.5?—2.7?=39.6 ? = x2.5 3、计算(18分) 51513? 6.3×2.8+6.3×7.2 ?10—3.47—1.53 ?—+ 121864 construction quality acceptance and assessment Regulation (Professional Edition) (DL/T5210.2-2009~DL/T5210.8-2009); 1.9 the quality checkout and evaluation of electric equipment installation engineering code (DL/T 5161.1-2002~5161.17-2002); 1.10 the norms of construction supervision, the electric power construction supervision regulations 417181? 450+(8.16+6.84)?0.63 ?(4—7.5×)? ??[(+1.25)×]— 115912494 4.列综合算式或方程解答下列各题。(6分) 4?一个数的加上19等于27，求这个数。 5 ?540的25,减去36除72的商，差是多少, 一、 计算下面各题，能简算的要简算。(每小题3分，共12分) 331641,(，，5)，,,1.25(1) (2) 135134105 53[(0.875,)，0.25],3) 35.79×37，35.79×82,357.9×1.9 (4)( 68 二、 解方程。(每小题3分，共6分) construction quality acceptance and assessment Regulation (Professional Edition) (DL/T5210.2-2009~DL/T5210.8-2009); 1.9 the quality checkout and evaluation of electric equipment installation engineering code (DL/T 5161.1-2002~5161.17-2002); 1.10 the norms of construction supervision, the electric power construction supervision regulations 234X?= X-0.36 X=16 (X-3)?2=7.5 389 32321851(+)×13×29 (0.75-)×(+) ?(3--) 2913169413133 三、 计算下图阴影部分面积比空白部分的面积少多少,(4分) 2m一、计算题 1、直接写得数。 523.14×3 = 450万,68万= ?15= 0.35×7×2= 6 550.84?0.21= 6.3?0.5?2= ， = 350?70%= 862、解方程。 14x1，25%x=3.75 :x= :4 =15% 2720 3、下面各题怎样简便就怎样算。(9%) 5403315 25× ×32 ×(26× ) ( , ?2)× 4134041014construction quality acceptance and assessment Regulation (Professional Edition) (DL/T5210.2-2009~DL/T5210.8-2009); 1.9 the quality checkout and evaluation of electric equipment installation engineering code (DL/T 5161.1-2002~5161.17-2002); 1.10 the norms of construction supervision, the electric power construction supervision regulations 314133345?[(0.5，)×12] [1,(，)]? [4,(,)]× 44484829 54153333522××(+) 15×+6×, ,,(,),? 85365555757 1.直接写得数 9.09+1.1= 10?0.1= 3-1.45= 5-1.4-1.6= 1?0.25×4 = 112912144111?(-)= ××= 4--= -?4= 5?-?5= 23342339955302×59? 3.16+11.8+5.01? 4230?71? 18.5×2.2? 2.求未知数X(8分) 24×(X-3.1)=9.2 8X+1.5×3=16.5 6 X+4 X=1.8×4 X - X,42 9 0.75253912 X?0.75= , X?5.6,5?3.5 35×,1,,?χ8252575 12 construction quality acceptance and assessment Regulation (Professional Edition) (DL/T5210.2-2009~DL/T5210.8-2009); 1.9 the quality checkout and evaluation of electric equipment installation engineering code (DL/T 5161.1-2002~5161.17-2002); 1.10 the norms of construction supervision, the electric power construction supervision regulations 3.计算下列各题，能简便的要简便(14分) 2163537590.2-5.4?0.27 (-)?+ +×- 545412498 1131847 5.1?,(1-)?0.53, ( , )?(2, ) 120?[45×(1, )] 12429100 339 65.5×3.2+6.55×58+65.5 56×,,+, 32×1.25×0.25 7814 1、直接写出得数(9分)。 2122-= 0.4-0.3= 4.5×102= 23.9?8? 5-0.25+0.75= 32 75411313×(2+)= ×6= 1?×= 7×?7×= 9135774 2、脱式计算，能简便的要用简便 方法 快递客服问题件处理详细方法山木方法pdf计算方法pdf华与华方法下载八字理论方法下载 (12分)。 1359 (，)×24 0.5，×(0.15，) 34620 432×3.8，5.2×80%，0.8 (,)×12?0.1 543 construction quality acceptance and assessment Regulation (Professional Edition) (DL/T5210.2-2009~DL/T5210.8-2009); 1.9 the quality checkout and evaluation of electric equipment installation engineering code (DL/T 5161.1-2002~5161.17-2002); 1.10 the norms of construction supervision, the electric power construction supervision regulations 3、求未知数X(6分)。 11111 ?,X: X?(1――25%), 2461092 4、列式计算(6分)。 4241、比某数的20%少0.5的数是7.1，求这个数。 2、一个数的2倍与的和是。这数是(用55方程解) 多少, 1、直接写结果。(6分) 2 = 9.6?0.03 = (，17),(,8)= 542，1358 = 0.5 118×26×25 = 1?(1?4)= 15,(,6)= 1,, = 351555311 ?× = (，)×12 = 1?,×8= 0.4×0.05 = 4686454 2、求未知数x。(6分) 120.6x，3×2.5 = 67.5 4(x，0.2)= 1.2 x×(，)= 67.5 35 3、脱式计算。(18分) 10.25×1.25×4×8 2.125×+1.875?4 (2.5,0.125)×0.8，6.24 4 construction quality acceptance and assessment Regulation (Professional Edition) (DL/T5210.2-2009~DL/T5210.8-2009); 1.9 the quality checkout and evaluation of electric equipment installation engineering code (DL/T 5161.1-2002~5161.17-2002); 1.10 the norms of construction supervision, the electric power construction supervision regulations 2272-?2 2.25×4.8+77.5×0.48 36×(+) 2?33912 4、 列式计算。(共6分) 72(1)数与2.5的积,减去8除5.8的商,差是多少? 157 5(2)、一个数的2.5倍比125的少12,求这个数 .8 一、直接写出得数。 14117731,，= 0.2= 0.23?1%= 3?,?3= 4×?4×= 554499 121423,,2.92= ，, 45分:小时, 8.18，,,,,,324353,,543,(143，299), 100?12.5?8, 78,0.8, 6.3?10%, 32571，3, 25×4?25×4, 4,1,,2.125, 9.3?0.03= 5568 114 1?100%?25= 1?,?1= 0.2米:35厘米= 1.25×= 445 77771111，,，×?×= = 45×99= 0.9×99，0.9= 34348888 2411125,,1.25，，0.8,1,2,,,()×63= ，= ,,7981259,, 二、计算下面各题，怎样简便就怎样算。 3321 1.65?5，1.29×0.2，×20% (，)×3， 5015125 construction quality acceptance and assessment Regulation (Professional Edition) (DL/T5210.2-2009~DL/T5210.8-2009); 1.9 the quality checkout and evaluation of electric equipment installation engineering code (DL/T 5161.1-2002~5161.17-2002); 1.10 the norms of construction supervision, the electric power construction supervision regulations 3113714,)? ×,,(,), 24×(98654164 345 4.85×0.17，0.485×8.3 ，×99， 14714 200711 2008× 2,2.86×(6.25,6) 200964 23 4?1，6× (0.86，0.86，0.86，0.86)×25 35 17.6×?,1.9,1.9×(1.9,1.9), 125?(50?8) 2 9999×7，1111×37 19.98×37,199.8×1.9，1998×0.82 construction quality acceptance and assessment Regulation (Professional Edition) (DL/T5210.2-2009~DL/T5210.8-2009); 1.9 the quality checkout and evaluation of electric equipment installation engineering code (DL/T 5161.1-2002~5161.17-2002); 1.10 the norms of construction supervision, the electric power construction supervision regulations 111 (70，)× 1,1?2,1?3, 68696 231 231?231 8.89×36+8.89×63+8.89 232 2005111，， 216×( - 2007×32006324 321,,57.5,14.25,15 91，91，,,,4137,, 4455423,,，87.5，12.5，,4,11,，，, ,,11111216534,,construction quality acceptance and assessment Regulation (Professional Edition) (DL/T5210.2-2009~DL/T5210.8-2009); 1.9 the quality checkout and evaluation of electric equipment installation engineering code (DL/T 5161.1-2002~5161.17-2002); 1.10 the norms of construction supervision, the electric power construction supervision regulations 三、求未知数。 , 3271,= 4,7×1.3=9.9 :28%=:0.7 ,,,354 31836 ,: 4×(x,0.8)=8 X+X=45 x1057 ,13135 χ，χ,6, ,30% :χ,:2 353464 3,1.25:1.2=:x X:(2+5)=10.5:4 (，7):6=7:2.1 4 construction quality acceptance and assessment Regulation (Professional Edition) (DL/T5210.2-2009~DL/T5210.8-2009); 1.9 the quality checkout and evaluation of electric equipment installation engineering code (DL/T 5161.1-2002~5161.17-2002); 1.10 the norms of construction supervision, the electric power construction supervision regulations 3,3.5)=17.8，2 0.6×(x,60%)= χ,0.8χ,6,16 6(,,5 四、列式计算。 4(1)4.5的减去1.5所得的差除以2.1，商是多少, 5 1(2)一个数的比3.5的1.6倍少2.6，这个数是多少,(列方程解) 8 1442(3)除与的差，商是多少, (4)7的倒数除6的商，比25的多多少, 15535 (5)从3.6除以18%的商里减去0.5，差的80%是多少, (6)一个数增加它的3倍后再减少它的80%，结果是8，求这个数是多少, construction quality acceptance and assessment Regulation (Professional Edition) (DL/T5210.2-2009~DL/T5210.8-2009); 1.9 the quality checkout and evaluation of electric equipment installation engineering code (DL/T 5161.1-2002~5161.17-2002); 1.10 the norms of construction supervision, the electric power construction supervision regulations 84的和是，，这个数是多少,X k B 1 . c o m (7)一个数的2倍与55 (8)一个数的4倍，减去这个数的80%，差是6.4的25%，这个数是多少, (9)甲、乙两数之和是142，甲数除以乙数商是6.余数是2，求甲、乙两数各是多少? 45(10)1减去一个数的，所得的差是0.75的1.8倍，这个数是多少, 95 1(11)一个数的比4.5的1.6倍少3.6，求这个数。 5 66(12)6除以减去6除的商，差是多少, 77 329(13)一个数增加3倍后和与与的和相等，这个数是多少, 24 construction quality acceptance and assessment Regulation (Professional Edition) (DL/T5210.2-2009~DL/T5210.8-2009); 1.9 the quality checkout and evaluation of electric equipment installation engineering code (DL/T 5161.1-2002~5161.17-2002); 1.10 the norms of construction supervision, the electric power construction supervision regulations 14(14)比某数的20%少0.4的数是7.2，求某数, (15)20个的和减去15的，差是多少, 53 32(16)与的积除2的倒数，商是多少, 57 14(17)一个数与的和相当于的45%，这个数是多少, 79 (18)一个数的3倍比5.06少0.05，这个数是多少, 417793213853?,×,，,, ×〔?,,,〕 , ， ,? ， 54310320458964 1、直接写得数。8, construction quality acceptance and assessment Regulation (Professional Edition) (DL/T5210.2-2009~DL/T5210.8-2009); 1.9 the quality checkout and evaluation of electric equipment installation engineering code (DL/T 5161.1-2002~5161.17-2002); 1.10 the norms of construction supervision, the electric power construction supervision regulations 1425，199, 7.2?0.4, 1?，1, 3989?21? 6 53111111，, ( ):, 2.4×5, ，,，, 64992323 2、计算，能简算的要简算。18% 588?21,1.6×3.5 25×125×16 3.6,2.8，7.4,6.2 104431128.6?，1.4×2.7 4?,?4 ?[(，)?] 27995539 3、求未知数χ。6, 140:, 80,χ,10,30 2.54χ 4、列式计算:6% 11与(1)的和除以它们的差，商是多少, 67 1(2)一个数减少12.5%以后，正好是37.5的，求这个数。 5 1、口算 (5分) 1410.35+1.5= 1— + = ×12.1,1= 1.01 ,0.91= 5511 construction quality acceptance and assessment Regulation (Professional Edition) (DL/T5210.2-2009~DL/T5210.8-2009); 1.9 the quality checkout and evaluation of electric equipment installation engineering code (DL/T 5161.1-2002~5161.17-2002); 1.10 the norms of construction supervision, the electric power construction supervision regulations 1313×8+= 1?0.1, 1×0.1 , 1,0.25×0.4= 1,0.99= 2727 11115151111112,，,，,, ， ? ， = × ? × = 6666959523234 1311120086395?78? a,a× = ? ? , 2010× = 888200943 2、简算。(10分) 4557×52 2.5×(1.9，1.9，1.9，1.9) 4.35×6.5+0.35×43.5 ，，1319 312(，) ×13×29 200920091513，，,，291315 3、解方程。(7.5分)X K b1.C o m X(4.5,X)×0.375=0.75 1.25 :0.25 = 1.6 1、 直接写数对又快:(5分) 0.4×0.2= 9-0.9= 10.75-(0.75，3.4)= 33172?0.4= 24? = ， = 453construction quality acceptance and assessment Regulation (Professional Edition) (DL/T5210.2-2009~DL/T5210.8-2009); 1.9 the quality checkout and evaluation of electric equipment installation engineering code (DL/T 5161.1-2002~5161.17-2002); 1.10 the norms of construction supervision, the electric power construction supervision regulations 13233110.2, = (3， )×5= 1? × = ， ? = 62032422 2、 求未知数～我没问题:,4分, 1277x,3.5=10.5 (2) x:=: 258 3、 神机妙算细又巧:,能简算的要写出简算过程,,15分, 4×0.8×2.5×12.5 2.3×85，2.3×15 36.5×99+36.5 1311120.65×14+87×65%, (,，)? 2026243 4、 列式计算我能行:,8分, 4 (1)100比80多百分之几, (2)比一个数的少32的数是28，求这个数。 5 新 课 标 第 一 网 construction quality acceptance and assessment Regulation (Professional Edition) (DL/T5210.2-2009~DL/T5210.8-2009); 1.9 the quality checkout and evaluation of electric equipment installation engineering code (DL/T 5161.1-2002~5161.17-2002); 1.10 the norms of construction supervision, the electric power construction supervision regulations