Sat_数学考试试题

1、sat_数学考试试题 sat 数学历年真题 sat 数学历年真题 table of contents for best viewing, please view this ebook in landscape (not portrait) mode. sat 数学历年真题 10t scetoi n 5est t1 0setcoin8 sat 数学历年真题 dear future university student: thank you for purchasing the sat companion: math. we know how daunting the sat test is, a

2、nd we hope you find our companion guide to the official sat study guide, 2nd editionuseful as you prepare for your test day. to use this book effectively to help you perform well on the sat, we suggest the following steps: tm 1. buy the official sat study guide, 2nd edition. the authors of this book

3、 are the creators of the test, and you will get no better practice than by doing these problems. tm 2. review/skim the subject chapters in the official sat study guide, 2nd edition. the information in here should not be new, but it will be a good refresher. tm 3. do problems, problems, and more prob

4、lems! time yourself and take it seriously. dontlook at the answers until you have completed the section. dont cheat yourself! did we mention do problems? 4. use the sat companion: math as a reference. again, dont cheat yourself by readinga problem in the book and then going right to the companion ex

5、planation! try these problems first, give yourself a real shot at completing them, and then while reviewing your score refer to the companion to provide a detailed explanation. 5. study frequently, in smaller chunks of time as opposed to one big block of time. lets sayyou have 2.5 hours per week to

6、study for the sat. if you commit to 30 minutes, 5 times a week, your resulting progress will be much better than if you spent 2.5 hours in one session. sat practice is all about frequent exposure to problems. note: 2.5 hours is not our guidance on how much you should study per week. if you can study

7、 5, 10, or 20+ hours a week, do it. weve written explanations in an easy-to-understand voice, and laid out the steps for each problem. we have not cut corners; the “easy” problems (is there really such a thing?) have explanations that are just as thorough as the hard problems. the sat companion: mat

8、h is a comprehensive reference guide. whether you are reviewing this material with a commonwealth education tutor, in an sat class, or by yourself, you will always be a click away from a complete explanation of the solution to a problem. there are many ways to complete math problems; our method is s

9、urely not the only one! we have chosen our explanations as they make the most sense to us, but if you get to the same answer as we do by another method, please feel free to share it with us! we may not be sat 数学历年真题 students anymore, but we are always learning and looking for better ways to answer p

10、roblems and provide a great product. we wish you the best of luck, and please tell us how you did when your results come in! if you have any questions or comments, please do not hesitate to contact us at sat. sat 数学历年真题 test 1 section 3 algebra to start with. lets write our given: x = 4 what are we

11、looking for? the answer that has the greatest value. there are two ways to do this, the first being: work every possible answer by inputting 4 for x: (4+1)(4+2) = (5)*(6)=30 (answer) (4+1)(4-1) = (5)*(3)=15 (4-2)(4+2) = (2)*(6)=12 (4-2)(4+1) = (2)*(5)=10 (4-4)(4+4) = (0)*(8)=0 or logic it out: x. if

12、 x were really large, this might not matter, but x is 4, so taking 2 away from it reduces it by 50%. (c) and (d) are gone. they differ. is there any way (x+2) can be smaller than (x1)? no way. (b) is gone. test 1 section 3 question 2 variable problem what are we looking for? what speed c is travelin

13、g when bs speed is 7 mph (so b=7 is a given). the table reflects the speeds we will find. sat 数学历年真题 ii we have 3 variables here, so this usually means a two-stepper. lets use our variables and givens to translate the “math speak” into equations. train bs speed. a b a=3b equation 1 c a c=2a equation

14、 2 now weve got two equations, so lets use em. what are we looking for again? cs speed, thats right. b=7 (given) a=3b a=3(7) = 21 c=2a c=2(21) = 42 e is the correct answer. test 1 section 3 question 3 average problem know that “average” and “mean” are the same. average = sum of the terms / # of term

15、s. lets write our givens: terms: x, 5x, 6x number of terms: 3 average of terms: 8 sat 数学历年真题 what are we looking for? the value of x a* n* * n just means total number of terms. if there were 5 terms in the sequence, n = 5. dont let variables scare you. make them your friends (we refer to them as suc

16、h through this text)! now plug in given information. 3 3 = 83 3 3 = 24 3 the 3s cancel each other out. 12 12 12 12 the 12s on the left side cancel out and were left with 24/12. b is the correct answer test 1 section 3 question 4 graphing question. there will be many. own them. what are we looking fo

17、r? the graph that agrees with the statement, “no two points on the graph have the same x- coordinate.” know the rules of a function: every “x” value of a function has only one unique “y” value. isnt that the same thing as the statement in the question? yes. iii sat 数学历年真题 use the vertical line test.

18、 if we can draw a vertical line anywhere on the graph, and it intersects the alleged function at more than one point, it is not a function. a answer a fails the vertical line test. b b also fails. c c fails. sat 数学历年真题 d looks like it passes the vertical line test, but since weve come this far, lets

19、check (e). e e fails. d is the only graph that passes the vertical line test. answer d is correct. test 1 section 3 question 5 vahndahful venn diagrams know the rules of venn diagrams. the total number of items (in this case, students) is everything within the rectangle. the intersection of the circ

20、les represents students who study both subjects. all of the areas must add to the total number of students. dont forget the space outside the circles. lets write our givens: butterfly students =9 grasshopper students=15 sat 数学历年真题 students studying both=3 iii what are we looking for? percentage of b

21、utterfly students only. what is the percentage? we have to find it. this is a two-stepper. first, find the total number of students: b + g + b and g + other = total 9 + 15 + 3 + 3 = 30 sat 数学历年真题 since we only need the percentage of butterfly students, just use the butterfly and total numbers. butte

22、rfly students = 9 total students = 30 know percentage formula: whole 30 answer c is correct. test 1 section 3 question 6 coordinate plane problem. there will be many of these. lets write our given: ab=cd remember that in (x, y) the x means how far right (positive) or left (negative) the point is. th

23、e y means how far up (positive) or down (negative) it is. thats it. what are we looking for? the value of t. where is t? its the y-value for point b (2,t). how do we get there? the key to this problem is ab=cd. it means that the two lines have the same length. the length of vertical lines only comes

24、 from the yvalues of its endpoints. to get the length, subtract the smaller one from the larger one, in this case: 3 - t = length of ab* * remember, make variables your friends. just because 3-t doesnt give you a number, dont give up. the length of horizontal lines only comes from the xvalues of its

25、 sat 数学历年真题 endpoints. to get the length, subtract the smaller one from the larger one, in this case: 6 - (-4) = length of cd 6 + (4) = length of cd 10 = length of cd subtracting a negative is just like adding the number. bring it all together. what are we looking for? t. what do we now know? ab=cd

26、ab=3-t cd=10 plug and chug. ab = cd 3-t = 10 3-t -3 = 10 -3 -t = 7 -(-t) = -7 t = -7 this makes sense, because point b is below the x-axis, so it must be negative. answer c is correct. test 1 section 3 question 7 dreaded algebra. know the order of operations: please excuse my dear aunt sally or pemd

27、as. if you dont know it, google it, and write it down 10 times every night until you know it. lets write our given: 3x2 = 4yiii what are we looking for? the value of x2 y. this problem looks tricky, but its not. separate the problem into two equations. 3x2 = 12 and 4y = 12 simplify: y sat 数学历年真题 3 3

28、 4 4 x2y 3 3 4 4 the 3s on the left cancel out, as do the 4s. were left 2 = 4 and y = 3 whats the problem asking? the value of xy. dont we know both of those variables now? yes. plug em in. xy = ? 43 = 12 d is the correct answer. 2 2 test 1 section 3 question 8 geometry problem. mark up the drawing.

29、 lets write our givens: radius of circle a is 2. radii of circle b and c are both 4. what are looking for? the radius of the big circle. find the radius of the largest circle: 2 + 4 + 4 = 10 sat 数学历年真题 d is the correct answer. test 1 section 3 question 9 number line problem lets write our given: tic

30、k marks are equally spaced. what are we looking for? the value of x. use the spaces. equal spaces between 2 and 42, or 5 ticks to go from 2 to 42. lets use “spaces.” x is 2 “spaces” from 2. find the value of one “space.” 42 - 2 = 40 divide this by the number of “spaces”: “space” is 8) x and finish u

31、p. start at 2. add the value of two “spaces.” remember, x is d is the correct answer. test 1 section 3 question 10 tricky geometry. remember: mark up the drawing! what are we looking for? the value of x. know how many degrees are in a circle (360). iii look at the graph carefully. = 8 (so, each sat

32、数学历年真题 label the right angle as 90 on the graph. write an equation for x. x + 110 + 30 + 90 = 360 vi simplify and finish up. x = 360 - 110 - 30 - 90 x = 130 tricky part: dont get caught up with the 70. look at the picture again. sov (x) includes rov (70) so we dont even need to use the 70. they just

33、 put it there to mess with us. c is the correct answer. test 1 section 3 question 11 algebra problem. again, make the variable our friend. lets write our given: k divided by 7 gives a remainder of 6. what are we looking for? the remainder when k + 2 is divided by 7. first, what is a remainder? its t

34、he amount left over in division. heres an example: whats the remainder when we divide 5 by 2? 2 r 1 2 iv 1 is the remainder in this example back to our problem. write the skeleton of the problem, in this case just the division sign. v now lets fill it in with the given information. r 6 7 that looks

35、a little more manageable. what numbers work for k? lets try 7. input 7 for k. 1 r 0 sat 数学历年真题 7 remainder is 0. k cant be 7. lets try k = 8. 1 r 1 7 8 is better, but were not there yet. what about 9? 1 r 2 7 see the pattern here? every time we increase the dividend* by 1, the remainder increases by

36、 1. we can continue going all the way up, but hopefully we see that 13 will give us a remainder of 6 as below. 1 r 6 7 bingo. k must be 13 (or any multiple of 13). * not the most exciting topic, but know your terminology. a dividend is the number under the division sign (in this case 13), and the di

37、visor is what you are dividing by (in this case7). use k in the second equation, k + 2. k + 2 = ? 13 + 2 = 15 finish up. whats the question asking again? the remainder when k +2 is divided by 7. 2 r 1 thats the answer! 7 1 b is the correct answer. test 1 section3 question 12 graphing data problem. t

38、ake a second to absorb whats shown in the table. forget about the answers for a second. look for a pattern in the table of data. see that when depth goes up, pressure now look at the graphs. sat 数学历年真题 depth is on the x-axis. pressure is on the y-axis. so depth is an x-coordinate, and pressure is a

39、y-coordinate. label x and y at we love zeros and ones (if we dont, start), so always start there. the first point of (depth, pressure), or (x, y), is (0, 14.7). right off the bat, answers c e are gone because those lines start at (0, 0) and (?, 0). were down to a, b d. look at the remaining answers.

40、 a b both have a negative slope, meaning they slope down from left to right. that indicates that when the depth number goes higher, the pressure number goes lower. is that what the table shows? not at all! when depth goes up, pressure goes up. so a b are out. were left with d as the only possible an

41、swer. take a last look at d, but its the only one that works. d is the correct answer. test 1 section 3 question 13 sequencing problem lets write our givens: first term is 1. every term after that is (-2) (the number before it). remember this equation. what are we looking for? the 6 term. draw it ou

42、t. the skeleton looks like this: 1st 2nd 3rd 4th 5th 6th 1 ? ? ? ? ? iv whats next? use the equation in i) above to find the 2 term. (-2) 1 = -2 so, plugging -2 in we have: 1st 2nd 3rd 4th 5th 6th 1 -2 ? ? ? ? continue the pattern. th nd sat 数学历年真题 3 term: (-2) -2 = 4 4 term: (-2) 4 = -8 5 term: (-2

43、) -8 = 16 6 term: (-2) 16 = -32 e is the correct answer. rdththth test 1 section 3 question 14 factoring and exponents rear their ugly heads. where to begin? lets write our given: (2x-5)(2x+5) = 5 what are we looking for? the value of 4x. we cant factor because the equation = 5, not 0. sometimes its

44、 best to just start working and manipulating the problem somehow. what if we foil* the left side of the equation? (2x+5)(2x-5) = 5 4x + -10x + 10x + -25 = 5 -10x + 10x = 0. get rid of those two terms. 4x + -25 = 5 4x = 30 arent we looking for 4x? we fell into it! e is the correct answer. shortcut! r

45、ecognize the formula for the difference of perfect squares: (x-a)(x+a) = (x-a) x term does not have to be by itself, it can be 2x, 5001x, or 42x, as long as it is the same in both expressions. this also goes for the a term; it can be just a number or sometimes another variable (2y, 63z, or anything else), as long as its the same in both expressions. x terms (2x) are the same,