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2024一2025學年度第一學期期末質量監測試卷
七年級數學
本試卷共4頁，23小題，滿分120分。考試用時120分鐘。
注意事項：1.答卷前，考生務必用黑色字跡的鋼筆或簽字筆將自己的學校、姓名和準考證號填寫在答
題卡上。將條形碼粘貼在答題卡“條形碼粘貼處”
2.作答選擇題時，選出每小題答案后，用2B鉛筆把答題卡上對應題目選項的答案信息點涂
黑；如需改動，用橡皮擦干凈后，再選涂其他答案。
3.非選擇題必須用黑色字跡的鋼筆或簽字筆作答，答案必須寫在答題卡各題目指定區域內
相應位置上；如需改動，先劃掉原來的答案，然后再寫上新的答案；不準使用鉛筆和涂
改液。不按以上要求作答的答案無效。
4.考生必須保持答題卡的整潔。考試結束后，將試卷和答題卡一并交回。
一、選擇題：本大題共10小題，每小題3分，共30分。在每小題給出的四個選項中，只有一項是符合
題目要求的。
1.一2024的倒數是（
C.、1
1
A.2024
B.-2024
2024
D.一2024
2.著名的數學家蘇步青被譽為“數學大王”.為紀念其卓越貢獻，國際上將一顆距地球約218000000k
的行星命名為“蘇步青星”，數據218000000用科學記數法表示為()
A.0.218×109
B.2.18×109
C.2.18×108
D.218×106
3.如圖是一個正方體盒子展開后的平面圖形，六個面上分別寫有“數”、“學”、“核”、“心”、
“素”、“養”，則“素”字對面的字是(
數
學
核
心
素
養
A.核
B.心
C.數
D.學
4.如圖，從左面觀察這個立體圖形，得到的平面圖形是()
正面
七年級數學試卷第1頁（共4頁）
5.若x=3是方程2x-a=0的解，則a的值是（)
3
A.6
B.4
C.-4
D.
6.若-3xy與是同類項，則m的值為（）
A.9
B.-8
C.6
D.-6
7.如圖，甲從點A出發向北偏東70方向走到點B,乙從點A出發向南偏西15°方向走到點C,則∠BAC
的度數是()
北
70°
B
東
A.85
B.105
C.125
D.160°
8.下列變形中，錯誤的是()
A.若a=b,則a-5=b-5
B.若a=b,則ac=bc
C.若a=6,號-號
D.若ac=bc,則a=b
9.有理數a,b在數軸上對應的點的位置如圖所示，對于下列四個結論：①b一a>0;②③a十b>0:@>0.其中正確的是(
a
0
b
A.①②③
B.②③④
C.①③④
D.①②③④
10.用正方形硬紙板做三棱柱盒子，每個盒子由3個長方形側面和2個正三角形底面組成.硬紙板以如
圖兩種方法裁剪（裁剪后邊角料不再利用）.A方法：剪6個側面；B方法：剪4個側面和5個底面.現
有19張硬紙板，裁剪時x張用A方法，其余用B方法.若裁剪出的側面和底面恰好全部用完，則
能做成三棱柱盒子的個數為(
A方法
B方法
A.24
B.30
C.32
D.36
七年級數學試卷第2頁（共4頁）2024—2025 學年度第一學期期末質量監測試卷
七年級數學參考答案及評分標準
一、選擇題：本大題共 10 小題，每小題 3分，共 30 分．
1．D 2．C 3．B 4．A 5．A 6．B 7．C 8．D 9．A 10．B
二、填空題：本大題共 5 小題，每小題 3 分，共 15 分．
11．25°30' 12．130° 13． 3 14． 2 15．5或 11
三、解答題（一）：本大題共 3 小題，每小題 7 分，共 21 分．
16．（7分）
（1）24 × 5 + 3 1 ····················································································· 1分
6 8 12
＝ 20 + 9 2···························································································2分
＝―13．··································································································3分
（2） 14 + 2 3 ÷ 4 × 5 3 2
＝ 1 + 8 ÷ 4 × 5 9 ·········································································· 5分
＝ 1 + 8 ÷ 4 × 4 ············································································· 6分
＝ 1 + 8
＝7．······································································································ 7分
17．（7分）
（1）解：去括號，得 7x + 6x 6 = 20···································································1分
移項、合并同類項，得 13x＝26···································································· 2分
系數化為 1，得 x＝2．··············································································· 3分
（2）解：去分母，得 2x + 1 3 x 1 = 6·····························································4分
去括號，得 2x + 1 3x + 3 = 6····································································5分
移向、合并同類項，得 x = 2······································································ 6分
化系數為 1，得 x＝ 2．··············································································7分
18．（7分）
（1）解：設船在靜水中的速度為 xkm/h，依題意得：··············································· 1分
2(x + 3) = 2.5(x 3)，·············································································· 3分
解得 x = 27，··························································································· 4分
∴船在靜水中的平均速度為 27km/h；···························································5分
（2）依題意，船在靜水中的平均速度為 27km/h，
∴甲乙兩碼頭之間的距離為 2 × 27 + 3 = 60 km ，······································· 6分
∴甲乙兩碼頭之間的距離 60km．································································· 7分
四、解答題（二）：本大題共 3 小題，每小題 9 分，共 27 分．
19．（9分）
（1）解：2A 3B = 2 3x2 x + 2y 4xy 3 2x2 3x y + xy ·······························1分
= 6x2 2x + 4y 8xy 6x2 9x 3y + 3xy ················································3分
= 6x2 2x + 4y 8xy 6x2 + 9x + 3y 3xy·················································· 4分
= 7x + 7y 11xy；····················································································6分
2
（2）∵x + y = ，xy = 1，
7
∴2A 3B
七年級數學參考答案及評分標準 第 1 頁 (共 4 頁)
＝7x + 7y 11xy
＝7 x + y 11xy······················································································ 7分
2
＝7 × 11 × 1 ··················································································· 8分
7
＝2 + 11
＝13.·······································································································9分
20．（9分）
（1）解：∠EOB＝∠EOF；理由如下：·································································· 1分
∵∠DOE＝90°
∴∠AOD+ ∠EOB = 180° ∠DOE = 90°，··························································2分
∵OD平分∠AOF，
∴∠AOD = ∠FOD，
∴∠FOD+ ∠EOB = 90°，··············································································· 3分
∵∠FOD+ ∠EOF = 90°，
∴∠EOB＝∠EOF．······················································································4分
（2）解：設∠AOD = x°，·····················································································5分
∵OD平分∠AOF，
∴∠DOF = x°，
∵∠DOE＝90°，
∴∠EOF = 90° x°，·····················································································6分
∵OF平分∠AOE，
∴∠EOF = ∠AOF，
∴x° + x° = 90° x°，··················································································· 7分
∴x = 30，·································································································· 8分
∴∠BOE = 180° ∠AOD ∠DOE = 180° 30° 90° = 60°·································· 9分
21．（9分）
（1）0.6··········································································································· 2分
【解析】依題意得：200a = 120，解得：a = 0.6．故答案為：0.6．
（2）設老張家 9月份的用電量為 x度，
∵0.6 × 240 = 144（元），
240×0.6＋(400 240)×0.65
＝144＋104
＝248（元）
∵144＜183＜248，
∴240＜x＜400．·························································································3分
依題意得：144 + 0.65（x 240） = 183，解得：x = 300．·······························4分
答：老張家 9月份的用電量為 300度．····························································5分
3 0.65+0.6+0.9（ ）∵三個檔次的平均價格為 ≈ 0.71（元），
3
8月份老張家用電的平均電價為 0.76元/度，
∴老張家 8月份用電量一定超過 400度，·························································6分
設老張家 8月份的用電量為 y度，
依題意得：144 + 0.65 ×（400 240） +（0.6 + 0.3） y 400 = 0.76y，·········· 7分
解得：y = 800．·························································································8分
答：老張家 8月份的用電量為 800度．····························································9分
五、解答題（三）：本大題共 2 小題，第 22 題 13 分，第 23 題 14 分，共 27 分．
22．（13分）
（1） 3； 1；5································································································3分
七年級數學參考答案及評分標準 第 2 頁 (共 4 頁)
【解析】∵b是最大的負整數，∴b＝－1；
∵ a + 3 + c 5 = 0， a + 3 ≥ 0， c 5 ≥ 0，
∴ a + 3 = c 5 = 0，
∴a + 3 = 0，c 5 = 0，∴a = 3，c = 5，故答案為： 3； 1；5；
（2）8； 1·······································································································7分
【解析】設點 P表示的數為 x，
由（1）可知點 A、B、C表示的數分別為 3； 1；5，
∴PA = x 3 = x + 3 ，PB = x 1 = x + 1 ，PC = x 5 ，
∴PA+ PB + PC = x + 3 + x + 1 + x 5 ，
∵ x + 3 + x 5 表示的是點 P到點 A和點 P到點 C的距離之和，
∴當點 P在點 A和點 C之間時（包括端點）PA+ PC有最小值，最小值為 AC的長，即為
5 3 = 5 + 3 = 8，
又∵當點 P與點 B重合時，PB有最小值，
∴當 x = 1時，PB有最小值，
∴當 x = 1時，PA+ PC和 PB能同時取得最小值，
∴當 x = 1時，PA+ PB + PC有最小值，最小值為 8 + 0 = 8，
故答案為：8； 1；
（3）3t + 2；t + 6·····························································································11分
【解析】由題意得，運動 t秒后，點 A表示的數為 3 t，點 B表示的數為 1+ 2t，點 C表示
的數為 5 + 3t，
∴AB = 1 + 2t 3 t = 3t + 2，BC = 5 + 3t 1 + 2t = t + 6，
故答案為：3t + 2；t + 6.
（4）3BC AB的值不會隨著 t的變化而變化，理由如下：
∵AB = 3t + 2，BC = t + 6，
∴3BC AB
= 3t + 18 3t 2 ····················································································12分
＝16，
∴3BC AB的值不變，且 3BC AB = 16． ··················································13分
23．（14分）
（1）100···········································································································1分
1 1
【解析】∵∠AOM = ∠AOC = 40°，∠BON = ∠BOD = 20°，
3 3
∴∠DON = 60° 20° = 40°，
∴∠MON = ∠AOB+ ∠BOD ∠AOM ∠DON
= 120° + 60° 40° 40°
= 100°；
故答案為：100；
（2）如圖，
∵∠AOB = 120°，∠COD = 60°，∠BOC = n°，···················································2分
∴∠AOC = ∠AOB+ ∠BOC = 120° + n°，∠BOD = ∠COD+ ∠BOC = 60° + n°，········· 4分
∵∠AOM = 1∠AOC，∠BON = 1∠BOD，
3 3
∴∠MOC = 2 120° + n° = 80° + 2 n° ∠NOD = 2， 60° + n° = 40° + 2 n°；·············6分
3 3 3 3
七年級數學參考答案及評分標準 第 3 頁 (共 4 頁)
（3）①當 0 < n < 60時，如圖，
············································································· 7分
∵∠BOC = n°，
∴∠AOC = ∠AOB ∠BOC = 120° n°，∠BOD = ∠COD ∠BOC = 60° n°，········· 8分
1
∵∠AOM = ∠AOC ∠BON = 1， ∠BOD，
3 3
∴∠MON = ∠MOC + ∠BOC + ∠BON
= 2 120° n° + n° + 1 60° n°
3 3
= 80° 2 n° + n° + 20° 1 n°
3 3
= 100°；································································································· 10分
②當 60 < n < 120時，如圖，
···············································································11分
∵∠BOC = n°，
∴∠AOC = ∠AOB ∠BOC = 120° n°，
∠BOD = ∠BOC ∠DOC = n° 60°，····························································· 12分
∴∠MON = ∠MOC + ∠BOC ∠BON
= 2 120° n° + n° 1 n° 60° ································································ 13分
3 3
= 80° 2 n° + n° 1 n° + 20°
3 3
= 100°．
綜上所述：∠MON的度數為 100°．······························································· 14分
七年級數學參考答案及評分標準 第 4 頁 (共 4 頁)

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