请教奥数题 关于盈亏的

2025-02-11 09:43:02
推荐回答(3个)
回答1:

都是用盈亏的方法做,要解题思路,不用方程谢谢
甲乙两人的信纸一样多,信封也一样多。甲写1封信用1张信纸,乙写1封信用3张信纸。结果甲的信封用完时还剩50张信纸,乙的信纸用完时还剩50个信封。原来他们各有信封多少个,信纸多少张?解:甲写完信封时,还剩50张纸,而乙要把50个信封写完的话就欠150张纸,这样就非常好理解了:
信封个数:(3×50+50)÷(3-1)=100个
信纸张数:100×1+50=150张

食堂管理员带着一笔钱去买肉,若买10千克牛肉则还差6元,若买12千克猪肉则还剩4元。已知每千克牛肉比猪肉贵3元,食堂管理员带了多少钱?
解:把买的牛肉看成卖猪肉,这样买10千克猪肉就多出了3×10-6=24元,于是求出:
猪肉的价格:(24-4)÷(12-10)=10元
管理员带的钱:12×10+4=124元。
钢笔与圆珠笔每支相差1.2元,小明带的钱买5钢笔差1.5元,买8支圆珠笔多0.6元,小明带了多少钱?
解:把买5支钢笔都看成是买圆珠笔,这样就多出了5×1.5-1.5=4.5元;这样就求出了:
一支圆珠笔的价格:(4.5-0.6)÷(8-5)=1.3元
小明带了:1.3×8+0.6=11元

有若干个苹果和梨,如果按每1个苹果2个梨分堆,那么梨分完时还剩5个苹果;如果按每3个苹果5个梨分堆,那么苹果分完了还剩5个梨,问,苹果和梨各有多少个?
有若干个苹果和梨,苹果的个数是梨的个数的3倍,如果每天吃2个梨和5个苹果,那么梨吃完时还剩20个苹果,有多少个梨?
解:因为苹果是梨子的三倍,当梨和苹果都吃完时,梨吃2个,苹果就得吃2×3=6个,而实际每天只吃了5个,这样就求出这些梨吃了:
20÷(3×2-5)=20天
梨的个数:20×2=40个。

水果店运来的西瓜个数是哈密瓜个数的4倍,如果每天卖130个西瓜和36个哈密瓜,那么哈密瓜卖完后还剩下70个西瓜,水果店运来西瓜和哈密瓜共有多少?
解:同上理当每天卖36个哈密瓜,就得买36×4=144个西瓜;这样就求吃了这些哈密瓜买卖了:
70÷(36×4-130)=5天
西瓜:130×5+70=720个
哈密瓜:720÷4=180个。
妈妈给了红红一些钱去买贺年卡,有甲乙丙三种贺年卡,甲种卡每张0.5元,丙种卡每张1.2元,用这些钱买甲种卡要比买乙种卡多买8张,买乙种卡要比买丙种卡多买6张,妈妈给了红红多少钱乙种卡每张多少钱?

回答2:

1、因为甲写1封信只用1张信纸,信封用完时还剩50张信纸,那在甲还剩50个信封时,信纸应该还剩下100张,而乙没写一封信多用2张,而乙在剩下50个信封时信纸刚好用完,也就是说甲剩下的100张纸如果每封信再多用两张就刚好用完,所以甲在剩下50个信封是已经写了100/2=50封信,所以信封等于50+50=100个,信纸等于50+100=150张。
2、这很简单:每千克牛肉比猪肉贵3元,那买10斤猪肉比买10斤牛肉少花10×3=30元钱,那么买10斤猪肉就会剩下30-6=24元,而买12千克猪肉则还剩4元,所以2斤猪肉的价钱就是24-4=20元,猪肉10元1斤,带的钱就为12×10+4=124元。
3、同2,如果把5支钢笔换成5只圆珠笔,那么会少花5×1.2=6元,那就会剩下6-1.5=4.5元,而买8至圆珠笔剩下0.6元,所以3支圆珠笔的价钱是:4.5-0.6=3.9,圆珠笔每只价钱为3.9/3=1.3,小明带,8×1.3+0.6=11元
4、可以这样理解,第一种分法相当于1个苹果配2个梨,而第二种分法相当于1个苹果配5/3个梨,每个苹果比第一种少配2-5/3=1/3个梨,那么第二种分法每个苹果相应的就会剩下1/3个梨,剩下的梨还要给第一种剩下的5个苹果配上梨,5个苹果需要25/3个梨,再加上剩下的5个梨,相比第一种分法,如果第二种分法也留下5个苹果不配梨,那么就会少用梨25/3+5=40/3,而每个苹果少1/3,所以第一种分法配梨的苹果总共有40/3÷1/3=40个,那苹果总共有45个,梨有40×2=80个。
5、如果每天卖的西瓜是哈密瓜的4倍,也就是每天买144个西瓜和36个哈密瓜,会同时买完,144-130=14个,也就是说如果把剩下的70个西瓜每天在增加14个就没有剩下的了,那么买了70/14=5天,哈密瓜的数量5×36=180个,西瓜180×4=720个。
6、甲种卡比乙种卡多买8张,乙种卡比丙种卡多买6张,那买甲比丙可以多买14张,14张甲共需要7元钱,而丙每张比甲贵1.2-0.5=0.7,7/0.7=10也就是说他的钱可以买10张丙,那么钱等于10×1.2=12元,乙的张数为10+6=16张,乙的价钱为:12/16=0.75元

回答3:

看图清楚么

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