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《数学广角──鸡兔同笼》同步试题(带解析)

《数学广角──鸡兔同笼》同步试题(带解析)
《数学广角──鸡兔同笼》同步试题(带解析)

《数学广角──鸡兔同笼》同步试题

北京市东城区和平里第四小学陈英

一、选择

1.鸡和兔一共有12只,数一数脚有36只,其中兔有()只。

A.3 B.4 C.5 D.6

考查目的:采用列表法或假设法解决“鸡兔同笼”问题。

答案:D。

解析:列表法:

假设法:假设全是鸡,则兔子的只数为(36-12×2)÷(4-2)=12÷2=6(只)。

2.有10元人民币和5元人民币共15张,合计120元,其中10元的人民币有()张。

A.12 B.10 C.9 D.8

考查目的:找准实际问题中的数量关系,巩固解决“鸡兔同笼”问题的解题策略。

答案:C。

解析:在这个实际问题中,10元人民币和5元人民币的总数量15相当于“鸡兔同笼”问题中的头数,人民币的总价值120元相当于“鸡兔同笼”问题中的脚数。

3.10张乒乓球桌上一共有32名同学在进行比赛,进行单打比赛的桌子有()张。

A.3 B.4 C.5 D.6

考查目的:利用假设法寻找实际问题中的数量关系,巩固假设法解决“鸡兔同笼”问题。

答案:B。

解析:在这个问题中,乒乓球桌的数量10相当于“鸡兔同笼”问题中的头数,同学数量32相当于脚数。假设全是双打桌,则应该有10×4=40(名)同学,实际上少40-32=8(名)同学。因为每张单打桌比每张双打桌少4-2=2(名)同学,所以单打桌有8÷2=4(张)。

4.篮球比赛中,3分线外投中一球得3分,3分线内投中一球得2分。在一场比赛中,李明总共投中9个球,得了20分,他投中()个2分球。

A.2 B.4 C.5 D.7

考查目的:巩固解决“鸡兔同笼”问题的方法,加深对“鸡兔同笼”问题本质的理解。

答案:D。

解析:在这个问题中,3分球与2分球的投球总数9相当于“鸡兔同笼”问题中的头数,所得总分20相当于“鸡兔同笼”问题中的脚数。可以假设投中的球都是3分球,也可以假设投中的球都是2分球。

5.李明用气枪打球,打中一枪可得5分,如果未打中倒扣2分。他打了20枪,一共得了51分。他打中了()枪。

A.13 B.14 C.15 D.16

考查目的:进一步巩固用假设法解决生活中的“鸡兔同笼”问题,感受所学知识的应用价值,增强应用意识。

答案:A。

解析:假设20枪全部打中了,则应该得20×5=100(分),比实际得分多100-51=49(分)。因为打中一枪比未打中一枪多得5+2=7(分),所以未打中的枪数应该为49÷7=7(枪),那么打中的枪数就是20-7=13(枪)。

二、填空

1.某景点在一节假日的两小时内售出20元门票和40元门票共100张,总收入为260元。该景点售出20元门票()张。

考查目的:利用假设法寻找实际问题中的数量关系,强化学生对“鸡兔同笼”问题本质的理解。

答案:7。

解析:关注需要解决的问题是售出20元的门票有多少张。假设100张都是40元的门票,则应该收入100×4=400(元),比实际收入多400-260=140(元)。因为每张40元门票比20元门票多40-20=20(元),所以20元门票有140÷20=7(张)。

2.光华小学今年参加植树活动的学生人数有13人。女生每人种3棵树,男生每人种4棵树,一共植树43棵。参加植树活动的男生有()人,女生有()人。

考查目的:将生活中的实际问题与“鸡兔同笼”问题沟通起来,引导学生加深对“鸡兔同笼”问题数量关系的理解。

答案:4,9。

解析:假设13人全部是女生,则应该种树13×3=39(棵),比实际少43-39=4(棵)。因为男生每人比女生每人多种树4-3=1(棵),所以男生应该有4÷1=4(人),那么女生就是13-4=9(人)。3.一辆自行车有2个轮子,一辆三轮车有3个轮子。车棚里停车10辆,其中自行车和三轮车共8辆,车轮共有19个。车棚里自行车有()辆,三轮车有()辆。

考查目的:考查学生能否从解决问题的角度分辨数量关系,筛选出有效的信息。

答案:5,3。

解析:题目中车棚停车10辆是多余条件,要注意筛选有用信息。先假设全部是2轮的自行车,则应该有2×8=16(个)车轮,比实际少19-16=3(个)车轮,每增加1辆三轮车,轮子数就增加3-2=1(个),所以三轮车有3÷1=3(辆),自行车有8-3=5(辆)。

4.芳芳和园园一起玩用火柴棍摆图形的游戏,三角形和正方形一共摆了10个(如图,任意两个图形之间没有公共边)。如果她们一共用了36根火柴棍,那么她们摆了()个三角形,()个正方形。

考查目的:巩固假设法解决实际问题,培养学生提取信息的能力。

答案:4,6。

解析:摆一个三角形需要3根火柴,摆一个正方形需要4根火柴。假设10个图形都是三角形,需要火柴3×10=30(根),比实际少36-30=6(根)。因为摆一个三角形比一个正方形少1根火柴,所以,正方形有6÷1=6(个),三角形有10-6=4(个)。

5.小明买了1元和8角的邮票共16张,用去15元钱,完成下列表格,找出1元的邮票买了()张,8角的邮票买了()张。

考查目的:用列表法解决生活中的实际问题,巩固解决“鸡兔同笼”问题的列表方法。

答案:11,8。

解析:解答这题的关键信息是“1元和8角的邮票共16张”,据此逐一列出数据,补充完整表格,再从中找出满足条件“面值为15元”时对应的1元邮票张数和8角邮票张数。

三、解答

1.新年活动要挂彩色气球,四(1)班有13人参加吹气球小组。男生每人吹8个,女生每人吹7个,一共吹了100个气球。请你用列表法计算出男生女生各多少人?

考查目的:用列表法解决生活中的实际问题,进一步加深对列表法解决“鸡兔同笼”问题的理解。

答案:列表如下:

答:男生有9人,女生有4人。

解析:列表方法不唯一,列表的数据既可以逐一列出,也可以跳跃列举,还可以取中列举,只要注意有序思考,找到问题的答案即可。

2.

乐乐餐厅有2人桌和4人桌各几张?

考查目的:考查学生综合分析信息的能力,巩固“鸡兔同笼”问题的解题策略。

答案:方法一:假设全都是2人桌,计算过程如下:

2人桌:(56-2×20)÷(4-2)=8(张);4人桌:20-8=12(张)。

答:乐乐餐厅2人桌有8张,4人桌有12张。

方法二:假设全都是4人桌,计算过程如下:

4人桌:(4×20-56)÷(4-2)=12(张);2人桌:20-12=8(张)。

答:乐乐餐厅2人桌有8张,4人桌有12张。

解析:当数据较大时,不宜使用猜想法、列表法或图示法,一般采用假设法来进行推理解答。

3.光明小学举办知识竞赛,共20道抢答题,每答对一题加5分,答错一题扣1分。刘萌在这次竞赛中得了76分,请问她答对了几道题?

考查目的:利用假设法寻找实际问题中的数量关系,解答与“鸡兔同笼”问题相关的实际问题。

答案: 假设20道全部答对了,则应该得20×5=100(分),比实际得分多100-76=24(分)。因为答对一题比答错一题要多得是5+1=6(分),所以未答对的题应该为24÷6=4(道),那么答对的题就是20-4=16(道)。

答:她答对了16道题。

解析:找准实际问题中的数量关系是解题关键。特别要注意答对一题加5分,答错一题扣1分,导致答对一题与答错一题会相差6分,而不是4分。

4.某快递公司为客户运送500只玻璃杯。双方商定:每只运费是2角,如果快递公司损坏一只,不但得不到运费,还要给客户赔偿8角。最后结算时快递公司共得运费95元。请问快递公司损坏了多少只玻璃杯?

考查目的:假设法的算理和推理过程,理解“鸡兔同笼”问题的本质。

答案:假设一只也没损坏,那么快递公司应该得到的运费是500×2=1000(角)=100(元),比实际得到的运费多100-95=5(元),因为每损坏一只玻璃杯就是会少得2+8=10(角)=1(元)运费,所以损坏的玻璃杯数为5÷1=5(只)。

答:快递公司损坏了5只玻璃杯。

解析:解答的关键是理解假设法的算理,弄清该问题中的数量关系,实际得到的运费相当于“鸡兔同笼”问题中的头数,玻璃杯的总数相当于“鸡兔同笼”问题中的脚数。同时也要注意题目中角和元的单位换算问题,不要出错。

5.学校食堂有100 kg油,共装了32个瓶子(如下图),并且每个瓶子都装满了。请问大、小油瓶各多少个?

考查目的:综合运用所学知识,灵活解决实际问题,培养学生解决问题的能力。

答案:方法一:列表法。

答:大油瓶有24个,小油瓶有8个。

方法二:假设法。

假设全部用大瓶装,则可以装4×32=128(kg),超出实际128-100=28(kg)。根据题意,小油瓶2个装1 kg,如果大瓶减少2个,同时小瓶增加2个,保证油瓶数量是32个不变。但每减少2个大瓶子,增加2个小瓶子时,油就会减少4×2-1=7(kg)。所以,把2小瓶看作一个整体,就应该有28÷7=4(个)这样的整体。所以小油瓶有4×2=8(个),大油瓶有32-8=24(个)。

答:大油瓶有24个,小油瓶有8个。

解析:此题是文字和情境相结合的题目,除了正文给出的信息外,图中“大油瓶每瓶装4 kg,小油瓶2瓶装1 kg”也是解题的重要条件。由此,还可继续得出小油瓶每瓶装0.5 kg,每瓶大油瓶比每瓶小油瓶可以多装4-0.5=3.5(kg)油。但是学生还没有学习小数除法,因此需要转换思路,把2个小油瓶当作一个整体进行分析推理,对学生来讲有一定难度,可配合列表法来理解。

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