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(满分12分)为了防控甲型H1N1流感,某校积极进行校园环境消毒,购买了甲、乙两...

(满分12分)为了防控甲型H1N1流感,某校积极进行校园环境消毒,购买了甲、乙两种消毒液共100瓶,其中甲种6元/瓶,乙种9元/瓶.

1.(1)如果购买这两种消毒液共用780元,求甲、乙两种消毒液各购买多少瓶?

2.(2)该校准备再次购买这两种消毒液(不包括已购买的100瓶),使乙种瓶数是甲种瓶数的2倍,且所需费用不多于1200元(不包括780元),求甲种消毒液最多能再购买多少瓶?

 

1.(1)解法一:设甲种消毒液购买瓶,则乙种消毒液购买瓶.··················· 1分 依题意,得. 解得:.············································································································ 3分 (瓶).················································································ 4分 答:甲种消毒液购买40瓶,乙种消毒液购买60瓶.····················································· 5分 解法二:设甲种消毒液购买瓶,乙种消毒液购买瓶.·············································· 1分 依题意,得·························································································· 3分 解得:············································································································· 4分 答:甲种消毒液购买40瓶,乙种消毒液购买60瓶. 2.(2)设再次购买甲种消毒液瓶,刚购买乙种消毒液瓶.··································· 6分 依题意,得.················································································· 8分 解得:.··········································································································· 9分 答:甲种消毒液最多再购买50瓶. 【解析】略
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(12分)某中学新建了一栋4层的教学大楼每层有8间教室,进出这栋大楼共有四道门,其中两道正门大小相同,两道侧门大小相同.安全检查中,对四道门进行了测试:当同时开启一道正门和两道侧门时,2分钟内可以通过560名学生;当同时开启一道正门和一道侧门时,4分钟内可以通过800名学生.

1.(1)求平均每分钟一道正门和一道侧门各可以通过多少名学生?

2.(2)检查中发现,紧急情况时因学生拥挤,出门效率将降低20%,安全检查规定,在紧急情况下全大楼的学生应在5分钟内通过这4道门安全撤离.假设这栋教学大楼每间教室最多有45名学生,问:建造这4道门是否符合安全规定,请说明理由

 

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(满分12分)如图,已知6ec8aac122bd4f6e是⊙O的直径,6ec8aac122bd4f6e是弦,过点6ec8aac122bd4f6e作OD⊥AC于6ec8aac122bd4f6e,连结6ec8aac122bd4f6e

6ec8aac122bd4f6e

1.(1)求证:6ec8aac122bd4f6e

2.(2)若6ec8aac122bd4f6e,求∠6ec8aac122bd4f6e的度数.

 

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解分式方程:6ec8aac122bd4f6e

 

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(满分10分)阅读题例,解答下题:

例  解方程6ec8aac122bd4f6e

【解析】

(1)当6ec8aac122bd4f6e,即6ec8aac122bd4f6e时               (2)当6ec8aac122bd4f6e,即6ec8aac122bd4f6e

6ec8aac122bd4f6e                          6ec8aac122bd4f6e

6ec8aac122bd4f6e                                 6ec8aac122bd4f6e

解得:6ec8aac122bd4f6e(不合题设,舍去),6ec8aac122bd4f6e  解得6ec8aac122bd4f6e(不合题设,舍去)6ec8aac122bd4f6e

综上所述,原方程的解是6ec8aac122bd4f6e

依照上例解法,解方程6ec8aac122bd4f6e

 

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  (满分20分,每小题10分)  

1.(1) 先化简,再求值:6ec8aac122bd4f6e6ec8aac122bd4f6e6ec8aac122bd4f6e+6ec8aac122bd4f6e,其中x=26ec8aac122bd4f6e+1


2.(2)解方程组6ec8aac122bd4f6e

 

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