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67.5 36.9 A CPPB questions 18 20 13 Anhui senior high school entrance examination mathematics simulation questions (including answers) reference answers to mathematics questions.

First, the multiple-choice question number 1 2 3 4 5 6 7 8 9

10 answer C D D C A B C B A

D 2。 Fill in the blanks: 1 1, m> 1 12, y = (x-2) 2+1/3, intersection 14,/kloc. Solution: 65438+...................4 points =b a? 1 ………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………

................... 1 point ≈EDO =∞.

Fbo, OED = ∠ OFB .......................... ∴△ OED ≌△ OFB. Two points.

∴DE=BF

…………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………

BEDF is a parallelogram.

…………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………

…………………5

Fraction 18, solution: intersection p.

Let PC⊥AB, vertical foot be C, and let PC = X.

The sea is in Rt△APC, ∫tan∠a = acpc ∴ac=? 5.67 Tan PC = 1 25x .................................................................................................................................................... ..... 9.36 tanx = 34x ........................... 4 points ∵ AC+BC=AB=2 1? 5 ∴ 125x+ 34x=2 1？ 5.x = 60∫sin∠b = Pb PC ∴pb= b sinpc？ 9.36sin60= 50？ 3 5 = 100 (nautical mile) ∴ The distance from city B where the ship is located to city P is 100 nautical mile. ........................ scored six points.

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A No.20 question NCB B B D E F M O19, solution: (1) ... 2 points (2) The number of votes for A is: 200? 34%=68 (votes) B: 200? 30%=60 (votes) What's the number of votes at C: 200? 28%=56 (votes) x b average score: 5.853? P.a. of 1953, 1965 and C.: 7.20065.19969691995 X∶b has the highest average score, so B should be admitted. ..... Tao = ∠ deo = 90 ........................................................................................................................................ ................Of = 2 1 CD ..................................................................................................................................................... ............. CE is the tangent of ∵ O ∴∠ OCB = ∠ OCE. ................................................................................................................... ....................△DOC， F is = 2 1 00-the midpoint of DC ∴ x ....................................................................................................................................... According to the meaning of the question, We get 2000 x+1000 (100-x) =160000; If we get x=60, we get 100-x=40 (Taiwan Province), and the candidates for A, B and C are 100 95 90 85 80 75.

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f G E C B A D/km 2 4 6 8 10 12 8 6 4

Question 22: M A y N B D P

So, shops can buy 60 color TVs and 40 washing machines. ............ scored 3 points (2). If you buy a color TV, you will buy a washing machine (100-2a). According to the meaning of the question, you must

2000 a+ 1600 a+ 1000( 100-2a)

≤ 160000 100-2a≤a, then the solution is 5.373 133 a. Because a is an integer, a=34, 35, 3 6, 37. Therefore, * * * has four purchase schemes. ∴ w increases with the increase of A. When a=37, the maximum value of W =200? 37+ 10000 = 1 7400 …………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………… 22 Solution: (1) Let point B be a pair of points E on the X axis, and connect AE, then point E is (12, -7). Let the functional relationship of the straight line AE be y=kx+b, then 2k+b=3 12k+b=-7, then the solution is k =-. (2) The median line GF is a line segment AB, which intersects with AB at point F, and the coordinate of point G intersecting with X axis is (x, 0). In Rt△AGD, AG2=AD2+DG2=32+(x-2)2 in Rt△BCG, BG2 = BC2+GC2 = 72+(65438 23. Solution: (1), ∫y axis and straight line L are tangents of ∫c∴OA⊥ad and ∫ OA ⊥ OB ∴∠ AOB = ∠ OAB. P) Point P is also on the straight line AP ∴p=4k+3 (2) Connecting DN: AD is the diameter ∴ ÷c∴∠ and = 90 ∠ and = 90-∠ Dan.

Page 9-* * 9 ∴∠AND=∠ABD and ∠∠adn =∠amn∴∠Abd =∠amn ................................................................................................................... ............................................................................................................................................

AB = 53 42 22 2bd AD÷S△ABD

= 2 1AB？

DN=2 1AD？ DB ∴

DN= AB DBAD？

= 5 125 34 ∴AN2=AD2-DN2

= 25 256)5 12(42 2÷AMN∽△ABP

∴ 2 )( AP ANSSAMN AMN？

That is, 222) (AP SAN SAP ANABP ABP AMN? At 8 o'clock, when point P is above point B, ∫ AP2 = AD2+PD2 = AD2+(Pb-BD) 2 = 42+(4k+3-3) 2 =16 (K2+1) or AP2 = ad2+.

= 2 1PB？

AD= 2 1(4k+3)？ 4=2(4k+3)

∴25 32) 1(25)34(32) 1( 1625)34(22562 2 2 2 ? K kk kAP St. SABP AMN gets k2-4k-2=0 and k 1.

=2+6 k2

= 2-6 ..........................................................................................................................................................................

= 2 1PB？

AD= 2 1[-(4k+3)]？ 4=-2(4k+3)

∴25 32) 1( 1625)34(22562 2 2 ? Simplify K kAP Saint SABP AMN, solve k2+ 1=-(4k+3), and get k=-2. When k=2,

△AMN When 6 or k=-2

The area of is equal to 25 32 …10 points.