福建省泉州市2020-2021学年度下学期教学质量跟踪监测考试七年级数学试题参考答案

泉州市2020—2021学年度七年级（下）教学质量监测
七年级数学参考答案及评分意见
说明：
（一）考生的正确解法与“参考答案”不同时，可参照“参考答案及评分意见”的精神进行评分．
（二）如解答的某一步出现错误，这一错误没有改变后续部分的考查目的，可酌情给分，但原则上不超过后面应得的分数的二分之一；如属严重的概念性错误，就不给分．
（三）以下解答各行右端所注分数表示正确做完该步应得的累计分数． 一、选择题（每小题4分，共40分）
1．C 2．A 3．B 4．B 5．C 6．D 7．A 8．D 9．B 10．D 二、填空题（每小题4分，共24分）
11．< 12．53x − 13． 2 14．90 15．6
16．①③④
三、解答题（共86分）
17．解：记2197x y x y ⎧+=⎪⎨
+=⎪⎩，①
．②
由①-②得：y =12， ··························································································· 3分 把y =12代入②得x =-5， ····················································································· 6分
所以=512x y ⎧⎨=⎩
-，． ···································································································· 8分
18．解：记317211
13
2.x x x −>−+−−⎧⎪
⎨⎪⎩，①≤② 由①得2x >−, ·································································································· 3分 由②得1x ≤, ···································································································· 6分 所以21x −＜≤. ································································································· 8分
19．解：依题意得，45360x x −+−= ··············································································· 3分
711x = ·
············································································· 6分 11
7
x =
·········································································· 7分 答：当11
7
x =
时，代数式45x −与36x −的值互为相反数． ·
············································ 8分 20．(1)如图所示； ········································································································· 3分
(2)如图所示； ········································································································· 6分
(3)等腰直角． ········································································································· 8分
21．解：(1)在△ABC 中，
∠A +∠ABC +∠C =180°， ··············································································· 1分 又∵∠A =62°，∠ABC =48°，
∴∠C =70°． ······························································································· 3分 (2)∵BD 是AC 边上的高，
∴∠BDC =90°, ·
···························································································· 4分 ∴∠DBC =90°-∠C =20°． ············································································ 6分 由(1)可知，∠C =70° ∵DE ∥BC ，
∴∠BDE =∠DBC =20°． ·
··············································································· 8分 22．解：(1)∵EF ⊥AE ，
∴∠AEF =90°， ··························································································· 1分
∵四边形AEFD 内角和为360°，∠D =90°，
∴∠DAE ＋∠DFE =360°-∠D -∠AEF =180°． ··················································· 3分 ∵∠EAD =60°，
∴∠DFE =180°-∠EAD =120°． ······································································ 4分
(2)由(1)可知，∠DAE ＋∠DFE =180°， ································································ 5分 又∵∠DFE ＋∠EFC =180°,
∴∠EFC =∠DAE ． ························································· 6分 ∵AE 平分∠BAD ，
∴∠DAE =∠BAE ， ························································· 7分 ∴∠BAE =∠EFC ． ····················································································································· 8分
又∵∠AEB =∠CEF ，
C ′
B ′ A ′
A
B
C
D
E
F
E B
C
D A
∴（∠B =）180°-∠AEB -∠BAE ，（∠C =）180°-∠CEF -∠EFC ， ······························· 9分 ∴∠B =∠C ． ······························································································· 10分
23．解：(1)依题意，得5743000
7645000m n m n +=⎧⎨+=⎩
， ······································································· 2分
解得3000
4000m n =⎧⎨=⎩， ·························································································· 3分
经检验，符合题意,
所以m 的值是3000，n 的值是4000． ······························································· 4分 (2)设该商场7月份购进了x 台A 型空调，则购进B 型空调为4
378x
−台, 依题意，得
783124
x
−≥， ·············································································· 6分 解得10x ≤， ······························································································· 8分 因为x 为正整数，且
4
378x
−也为正整数， 所以x 的取值为2，6，10， ············································································ 9分 所以该商场共有3种进货方案． ······································································ 10分
24．解：(1)记34x y a +=−①，53x y a −=②，
解法一：
当a =4时,30x y +=,512x y −=, ······································································· 1分
联立方程组30
512x y x y +=−=⎧⎨⎩，解得9
=232
x y =−⎧⎪⎪⎨⎪⎪⎩，．
···························································· 2分
所以x -y =6． ······························································································ 3分 解法二：
当a =4时,30x y +=,512x y −=, ······································································· 1分 ①+②得，2212x y −=， ················································································· 2分 所以x -y =6． ······························································································ 3分 解法三：()()1111
3+-5=(4-)+(3)262222
x y x y x y a a a −=+=+= (2) 解法一：
①×3+②得， ······························································································· 4分
()()()335343x y x y a a ++−=−+， ··································································· 5分
所以4412x y +=， ························································································ 6分 所以3x y +=． ····························································································· 7分 解法二：
①-②得，844y a =− ····················································································· 4分 所以12
a
y −=， ····························································································· 5分 把12
a
y −=
代入①得52a x −=， ······································································· 6分
所以x +y =5122
a a
+−+=3． ············································································· 7分 (3)解法一：
由(2)可知3x y +=， 因为y ＞1-m ，且3x -5≥m ， 所以
523
m x m +<+≤， ················································································· 7分
令x 可取两个连续整数的值为n ，1n +，(n 为整数) ， 则有513
m n n +−<
≤，122n m n +<++≤．
故1383 5.
n m n n m n −<−<−⎧⎨
⎩≤，≤ ··················································································· 8分
要使x 可取得两个连续整数的值，m 要先有解，则m 有解可能有三种情况：
i)3538138n n n n n n −−<−−⎧⎪
⎨⎪⎩
≤，
，≤，即472n <≤，此时n 没有整数解，不合题意，舍去； ····················· 9分 ii)35381n n n n −⎧⎨−−⎩≥，≤，即5722n ≤≤，此时=3n ；即2314m m <⎧⎨<⎩
≤，≤，所以23m <≤； ·············· 10分
iii)38113535n n n n n n −−⎧⎪
−<−⎨⎪−⎩≤，
，≤，
即522n ＜≤，此时n 没有整数解，不合题意，舍去． ···················· 11分
综上所述：m 的取值范围为23m <≤． ···························································· 12分 解法二：
由(2)可知3x y +=， 因为y ＞1-m ，且3x -5≥m ， 所以
523
m x m +<+≤， ················································································· 7分
令x 可取两个连续整数的值为n ，1n +，(n 为整数) 则有513
m n n +−<
≤，122n m n +<++≤．
故13835n m n n m n −<−<−⎧⎨⎩≤，
≤，
··················································································· 8分 所以135n n −<−，且38n n −<， ···································································· 10分 所以24n <<，
所以3n =， 所以2314m m <<⎧⎨⎩≤，≤．
········································································ 11分
所以m 的取值范围为23m <≤． ···································································· 12分
25．解：(1)由翻折得CD =DE ，∠CDA =∠CEA = 90°， ························································· 1分
204102
1
21=⨯⨯=⋅=
CE AF S ACF △. ···································································· 3分 (2)①取点M 关于AD 的对称点N ，连接PN ，FN ，
则PF PM PN PF FN +=+≥， ····························· 4分 ∴当N ，P ，F 三点共线且FN ⊥AB 时，FN 有最小值， ······································································· 5分 即PF ＋PM 的最小值为FN 的长． ∵11
1222
ABF S AB FN FN =⋅=⨯⨯△, ·························· 7分 ∴6m FN =，即PM ＋PF 的最小值为6
m ． ························································ 8分 ②∵
2
3
AM MF =，AC =10， ∴AM =AN =4， MF =6, ···················································· 9分 当PF ＋PM 取最小值，PN ⊥AB ，利用对称性，则PM ⊥AB ∴11
==22
APM
APN S AM PM AN PN S =⋅⋅△△. 设s S S APN APM 2==△△，
D
E
C
N
A
B
F
M
P
E D
C
A
B P M
F
N。