Reference answers and grading standards of mathematics test questions
Description:
(1) When the correct answers of candidates are different from the "reference answers", they can be graded according to the spirit of "reference answers and grading standards".
(two) if there is an error in the answer to a step, the error will not change the purpose of the subsequent part of the exam, but it may be given as appropriate, but in principle it shall not exceed half of the score due later; If it is a serious conceptual mistake, no points will be given.
(3) The score marked at the right end of each line in the following answer indicates the cumulative score for correctly completing this step.
First, multiple-choice questions (3 points for each small question, ***2 1 point)
1.b；
2.a；
3.d；
4.c；
5.b；
6.c；
7.
A.
Fill in the blanks (4 points for each small question, 40 points for * * *)
8.
1;
9.
9. 1× 104;
10.
1 1.
10;
12.
13.
>;
14.
15.
16.
5.5;
17.
①
10,
②
.
Iii. Answering questions (***89 points)
18. (9 points for this small question)
Solution: Original formula =
Six points.
=
Eight points.
=
Nine points
19. (9 points for this small question)
Solution: Original formula =
................................., 4 points.
=
Six points.
=
Seven points
while
When, Original Type =
=
9 points ... 9 points.
20. (9 points for this small question)
The condition of addition is: BE=DF, 3 points.
Proof: in the diamond ∠B=∠D, AB=AD, ∠ b = ∠ d.
Five points five points.
BE = DF
∴△ ....................................... Abe scored 7 points.
∴AE=AF
Nine points
(The answer is not unique)
2 1. (9 points for this small question)
Solution: (1)
=
= 14(
) .............................. 4 points.
The average monthly water consumption of these 20 households is 14.
; Five points.
(2) 14×400=5600(
) .................................................... 8 points.
It is estimated that the monthly water consumption of this community is about 5600.
............................................, 9 points.
22. (9 points for this small question)
Solution: (1) The card Xiaoli took out happened to be.
The probability is
Three points
(2) draw a tree diagram:
................................ scored six points.
∴ * * There are 6 equal possible outcomes, of which 2 are reasonable and 4 are not.
∴
…………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………
The game was unfair to Xiao Ming, who scored 9 points.
(If the list rule is used to reference a given score)
23. (9 points for this small question)
Solution: (1) Set up this mall to buy Class A goods.
◆ According to the meaning of the question:
………………………………………………
2 points
Solution:
Commodity B: 100-40=60 (piece) 3 points.
A: The mall bought 40 pieces of A and 60 pieces of B. ......................
4 points
(2) Set up a shopping center to buy a commodity.
A, then buy B.
◆ According to the meaning of the question:
………………………………
6 points
Solution: 48≤
≤ 50 ... 7 points.
∵
Is a positive integer.
∴
=48 or
=49 or
=50………………………………………………
8 points
∴ There are three purchase schemes:
Option 1: Buy 48 A-class goods and 52 B-class goods.
Option 2: Buy 49 A-class goods and 5 1 B-class goods.
Option 3: Buy 50 A-class goods and 50 B-class goods. Nine points
24. (9 points for this small question)
Solution: (1)
=
=
...................................., two points.
∴ The vertex coordinates of parabola are (1,-4); ............................., 3 points.
(2) according to the parabola
And straight line
You can get:
A (- 1, 0), B (3 3,0), C (0 0,3), D (0 0,3) ... 5 points.
∴OB=OC=OD=3
∴∠OBD=∠OBC=450
Six points.
∠∠OBD =∠AFE，∠OBC=∠AEF。
Seven points
∴∠AFE=∠AEF=450
∴∠EAF=900,AE=AF
Eight points.
∴△AEF is an isosceles right triangle.
Nine points
25. (This little question is 13)
Solution: (1)BH=CK.
........................................, two points.
As shown in Figure 2, ∵ point O is the midpoint of the hypotenuse of the isosceles right triangle ABC.
∴∠B=∠GCK=450
，BG=CG
From the nature of rotation, we know that ∠BGH=∠CGK.
∴△bgh?△cgk 4 points.
∴BH=CK.
Five points.
(2)①
From (1), we can easily know the S quadrilateral CHGK.
=
△ABC
=4, ………………………………………………………………………………………………………………………………………………………………………………………………………………………………………….
∴S△GKH
=
S quadrilateral CHGK-S△KCH=4-
CH×CK
Go
=
(0 < χ < 4) ....................... scored 9 points.
② when y
=5 16
×8=
When? ............... 10.
that is
=
χ= 1.
Or chi = 3.
∴ when the area of △GKH is exactly equal to 5 16 of the area of △ABC.
BH= 1。
Or BH = 3............... 13 points.
26.
(This little question is 13)
Solution: (1) DE⊥BC whose intersection D is point E.
∵ quadrilateral ABCD is a right-angled trapezoid.
∴ quadrilateral Abed is a rectangle.
∴AD=BE=2,AB=DE=8
In Rt△DEC, CE=
=
=6
........................., two points.
∴ The circumference of trapezoid ABCD =
AB+BC+CD+DA = 28 ........................ 3 points.
(2)
①
∫ The circumference of trapezoidal ABCD is 28, and PQ bisects the circumference of trapezoidal ABCD.
∴BP+BC+CQ= 14
Four points.
∫BP = CQ = t
∴t+8+t= 14
∴t=3
Six points.
∴ When t=3, PQ divides the circumference of trapezoidal ABCD equally. .......................................................................................................................................................
②(i) When 0≤t≤8, the intersection point Q.
Make QG⊥AB at g point.
∫AP = 8-t，DQ= 10-t，AD=2，sinC=
，cosC=
∴CF=
，QF=
，PG=
=
，QG=8-
=(8-t)2+22=t2+ 16t+68，
PQ2=QG2+PG2=(8-
)2+(
)2=
If DQ=PD, (10-t)2=
T2+ 16t+68, the solution is: t = 8；;
If DQ=PQ, (10-t)2=
Solution: t 1=
，t2=
> 8 (give up), at this time t=
; ..... 10 point
(ii) when 8 < t < 10, PD=DQ= 10-t,
At this time, ∴ is held by an isosceles delta △DPQ with DQ as one waist;
However, when t= 10, P, D and Q are coincident and cannot form a triangle. ................ 1 1 min
(iii) when 10 < t ≤ 12, PD=DQ=
t- 10，
At this time, ∴ is held by an isosceles delta △DPQ with DQ as one waist; ........................... 12.
To sum up, when t=
Or 8 ≤ t < 10 or 10 < t ≤ 12, and the triangle whose vertices are p, d and q is just an isosceles triangle whose waist is d q. ......................................................................................................
IV. Additional questions (* *10)
1.
& lt2;
2.
four