1、GMAT（ QUANTITATIVE）数学练习试卷 4及答案与解析 一、 QUANTITATIVE 1 In a right isosceles triangle, the lengths of the two nonhypotenuse sides are designated a. What is the area of the triangle in terms of a? 2 The U.S. Defense Department has decided that the Pentagon is an obsolete building and that it must be repl
2、aced with an upgraded version: the Hexagon. The Secretary of Defense wants a building that is exactly 70 feet high and 200 feet on a side, and that has a hexagonal bulls-eye cutout in the center (somewhat like the current one) that is 50 feet on a side. What will be the volume of the new building in
3、 cubic feet? （ A） 3,937,500 cubic feet （ B） 15,750 cubic feet 3 What is the area of the shaded figure? 4 In the figure above, x=2. What is the area of circle O? 5 Which of the following could be the equation of line l in the figure above? （ A） 2x=3y-2 （ B） 2x=3y-6 （ C） 3x=2y-6 6 In the rectangular f
4、igure above, the area of the shaded region is equal to the area of the non-shaded rectangle. If AB=6, BC=4, and the longer side of the non-shaded rectangle is an integer, what is one possible measurement for the shorter side of the non-shaded rectangle? 7 In the figure above, what is the value of 2s
5、-r? （ A） 60 （ B） 70 （ C） 100 （ D） 140 （ E） 220 7 The following data sufficiency problems consist of a question and two statements, labeled (1) and (2), in which certain data are given. You have to decide whether the data given in the statements are sufficient for answering the question. Using the da
6、ta given in the statements plus your knowledge of mathematics and everyday facts (such as the number of days in July or the meaning of counterclockwise), you must indicate whether . Statement (1) ALONE is sufficient, but statement (2) alone is not sufficient. . Statement (2) ALONE is sufficient, but
7、 statement (1) alone is not sufficient. . BOTH statements TOGETHER are sufficient, but NEITHER statement ALONE is sufficient. . EACH statement ALONE is sufficient. . Statements (1) and (2) TOGETHER are NOT sufficient. 8 In the figure above, what is the area of the shaded region? (1) AB=8. (2) Angle
8、AOR=45. 9 In the figure above, what is the value of x? (1) 2y=100. (2) ABCD is a parallelogram. 10 In the figure above, what is the length of RT? (1) RU=5 (2) RS=12 11 An art gallery owner is hanging paintings for a new show. Of the six paintings she has to choose from, she can only hang three on th
9、e main wall of the gallery. Assuming that she hangs as many as possible on that wall, in how many ways can she arrange the paintings? （ A） 18 （ B） 30 （ C） 64 （ D） 120 （ E） 216 12 A composers guild is planning its spring concert, and ten pieces have been submitted for consideration. The director of t
10、he guild knows that they will only have time to present four of them. If the pieces can be played in any order, how many combinations （ A） 40 （ B） 210 （ C） 1,090 （ D） 5,040 （ E） 10,000 13 If x and y are odd integers, which of the following must always be a non-integer? （ A） xy （ B） x/y （ C） y/x （ D）
11、 xy/2 （ E） -xy 14 Dan has a membership at a local gym that also gives classes three nights a week. On any given class night, Dan has the option of taking yoga, weight training, or kickboxing classes. If Dan decides to go to either one or two classes per week, how many diff （ A） 3 （ B） 6 （ C） 7 （ D）
12、9 （ E） 12 15 Terry is having lunch at a salad bar. There are two types of lettuce to choose from, as well as three types of tomatoes, and four types of olives. He must also choose whether or not to have one of the two types of soup on the side. If Terry has decided to （ A） 9 （ B） 11 （ C） 24 （ D） 48
13、（ E） 54 16 If r is negative and s is positive, which of the following must be negative? （ A） |r|/s （ B） |s|/r （ C） r2+s （ D） r2s2 （ E） GMAT（ QUANTITATIVE）数学练习试卷 4答案与解析 一、 QUANTITATIVE 1 【正确答案】 C 【试题解析】 The formula for the area of a triangle is 1/2 bh. You can orient an isosceles right triangle so th
14、at the two equal sides form both the base and the height. If these sides are of length a, then plugging these values into the area formula gives us the result 1/2 a2, which is answer C. The diagram shows how the equal sides form both the base and the height: 2 【正确答案】 E 【试题解析】 This is very definitely
15、 a multi-step problem. First, you need to recognize that were dealing with nothing more than an oddly-shaped cylinder: the base is a regular polygon, so by dividing it into a number of manageable pieces, we can calculate its area. However, before we do that, well need to figure out how to treat that
16、 hole in the middle of the six-sided doughnut. The easiest thing to do is to think about it as “negative area“-to calculate its area, and then subtract that area from the area of the larger, filled-in shape to find the area of the region around it. The following diagram below shows the three steps:N
17、ext, well need to calculate the areas of those two different triangles. If you happen to remember the formula for the area of an equilateral triangle, great; skip to that step. If not, youll need to derive it. Thats not that hard: using the diagram that follows, you can divide the equilateral triang
18、le into two right triangles, each with height h, hypotenuse a (there go those shifting variable names again), and base a/2. Since we dont know what h is, it would be nice to state that in terms of a as well. Well do that first, by relying on the Pythagorean Theorem: Step 4: Calculate the area of the
19、 triangle The area of the entire equilateral triangle is just twice the area of the right “half triangle“ on either side: Six of these form the whole of the large (outer) hexagon, so that area is: Six of the smaller triangles form the whole of the hole in the middle. Well need a different variable f
20、or these, since were talking about a different a at this point. We can define these as the outer ao and inner ai sides: The area of the Hexagon (minus the bulls eye)= Then, the height is a uniform 70 feet all around, so finding the volume is just a matter of multiplying through by that height: Of co
21、urse, that doesnt look much like the answer choices, so well need to substitute the values for the variables: If you looked over the answer choices, you will know that is an acceptable component of your answer; the thrust of this question is geometry, so the GMAT would not make you calculate a squar
22、e root as part of the problem. The volume corresponds with answer choice E. 3 【正确答案】 A 【试题解析】 This figure may look intimidating, but its actually quite harmless. To tackle it, you must first recognize that it can be split into two smaller triangles, as shown in the diagram. Once youve separated the
23、triangles, you can start labeling the lengths. The lower triangle has two 45 angles, so it is clearly a right isosceles triangle. That means that we should be able to find the area quickly. An intermediate step will be calculating the lengths of the (equal) sides, which well designate as x. That wil
24、l also have a follow-on purpose. Well be able to use the length x with the full length of the left side of the figure to calculate the length of the rising side of the smaller triangle. First, though, well get back to the lower triangle and the Pythagorean Theorem: Now we can use the result for x to
25、 find that smaller length: The triangle may not look familiar yet, but the angle at the right is a giveaway. Since its a 120 angle, the supplementary angle must be 60. This is because interior and exterior angles along the same line segment must always sum to 180. Since another angle in the triangle
26、 is right (90), we now can assert that angle measures 30, since all the angles inside a triangle must sum to 180 as well. This means that we have a 30-60-90 triangle. One needs to be careful at this point, as it is easy to misidentify the lengths in one of these. Remember, it is not that the lengths
27、 are always 1, , and 2; but that they are always in the ratio of 1: :2. If the long nonhypotenuse side has a length of one-half of , then the other sides should be halved as well. The short side will have length 1/2, and the hypotenuse will have length 1. At this point, as shown in the last diagram,
28、 we can ignore all our calculations but those for a few lengths. The area of the original figure is found with just the four numbers in black: or answer choice A. Thats about it. There are other ways of finding the area, but if you can handle the steps here, you should have little trouble with GMAT
29、geometry. 4 【正确答案】 B 【试题解析】 Since we are looking for the area of the circle, we really need to find the radius. The line segment from the center of the circle to the vertex of the square is 2, so we can use that to find the radius. If you extend line x to the opposite vertex of the square, you have
30、drawn a diagonal of that square, which is also the hypotenuse of a 45-45-90 triangle. The diagonal measures 4. You normally multiply a leg of a 45-45-90 triangle by to get the hypotenuse, so we have to divide the hypotenuse by to get the leg (which is also the side of the square and the diameter of
31、the circle). Therefore, the diameter is equal to and the radius is equal to . But were not done; we still have to find the area. The formula for the area of a circle is A=nr2, so we have to substitute in for r. Squaring both the top and bottom of the fraction, we get 4/2 , or 2. So the area of the c
32、ircle is 2. 5 【正确答案】 B 【 试题解析】 Since none of the answer choices are actually in the recognized form for an equation of a line, the best thing in this case is to figure out what the equation of the original line is in y=mx+b form, and then figure out which of the answer choices can be manipulated to
33、form that same equation. This line crosses the y axis at 2, so the y-intercept is 2 and our equation becomes y=mx+2. Now we need to figure out the slope. Since the point goes through (0, 2) and (3, 4), the slope is 2/3. The equation were looking for is now If we manipulate the equations in the answe
34、r choices so that the y always has a coefficient of 1 and everything is on the correct side of the equal sign, B is the only correct answer. 6 【正确答案】 D 【试题解析】 The larger rectangle has an area of 24, so the non-shaded rectangle must have an area of 12. Starting with the answer choices for the length
35、of the shorter side, we can divide them into 12 until we find a valid length for the longer side. A would give us 16 for the longer side, which wouldnt work because that wouldnt fit inside another rectangle with 6 as the longest side. B would give us 12 for the longer side, which is still too big. C
36、 gives us 8, which is also too big. D gives us 5, which is possible and therefore is the correct answer. 7 【正确答案】 E 【试题解析】 Since 4r and 5r form a straight line, which is 180 degrees, 4r+5r=180. When we simplify that to 9r=180, we can solve to get r=20. Therefore 5r=100, and s-20=100. So we can tell
37、that s=120. Using those numbers, the equation 2s-r becomes 2(120)-20, which equals 220. 8 【正确答案】 C 【试题解析】 Statement (1) alone lets us find the radius, which can tell us the area of the whole circle, but we still dont know what portion of the circle is made up by the shaded region, so we have to elim
38、inate A and D. Statement (2) alone tells us what proportion of the circle is encompassed by the shaded region, but doesnt tell us the area of the whole circle, so we can eliminate B. Putting the statements together, we have both pieces of information, which we can use to find the area of the shaded
39、region, so statements (1) and (2) are sufficient together, and our answer is C. 9 【正确答案】 C 【试题解析】 Statement (1) is tempting, because the figure appears to be a parallelogram, but we dont really know that it is without some indication that there are two sets of parallel lines. So statement (1) is act
40、ually not sufficient, and we can eliminate A and D. Remembering that we have to take statement (2) alone, and it is not sufficient by itself to solve for the value of x, we eliminate B. Putting both statements together, we have the information we need: If the figure is a parallelogram, then the angl
41、es add to 360. If 2y=100, then 2x=260 and x=130. So our answer is C. 10 【正确答案】 E 【试题解析】 Statement (1) is not sufficient on its own, since you can never solve for one side of a triangle given only one other side. Lets eliminate A and D. Statement (2) is not sufficient for the same reason, so we can e
42、liminate B. Putting the statements together, it appears that we have enough information to solve for the diagonal of the rectangle (also known as hypotenuse of a right triangle), except that we dont have any indication that the figure is actually a rectangle. So the statements are still not sufficie
43、nt and the answer is E. 11 【正确答案】 D 【试题解析】 This problem asks for us to arrange the paintings, so it is a permutation. Using the formula and using 6 as our n and 3 as our k, we get the following: . When we cancel out, we get 654, which is 120. 12 【正确答案】 B 【 试题解析】 This problem is a combination, so we
44、would use the formula . Since 10 is our n and 4 is our k; the equation works out to . Simplifying and multiplying gives us 210 combinations. 13 【正确答案】 D 【试题解析】 Since we know that x and y are odd, xy must always be odd. Therefore, xy/2 will always be a non-integer. None of the other answer choices ha
45、s this relationship; in all of the other cases, it is possible to have values of x and y that would yield integer results. 14 【正确答案】 D 【试题解析】 This problem is best done by brute force. First, we must establish that Dan has the option of taking one or two classes per week. If he only takes one class,
46、the three possibilities are Y, W, and K. If he takes two classes per week, the six possibilities are YY, KK, WW, YK, YW, and WK (because the problem is looking for combinations, not arrangements). So that adds up to nine possible combinations. 15 【正确答案】 D 【试题解析】 This is a relatively simple problem i
47、f you know the process-and you read carefully. All you have to do is multiply the number of options: 2 types of lettuce, 3 types of tomatoes, 4 types of olives, and 2 soup options. (You did remember the soup, didnt you?) 2342=48, so 48 is our answer. 16 【正确答案】 B 【试题解析】 Based on the rules of positive and negative numbers, the only answer choice that must be negative is B, since it takes a positive and divides it by a negative. All other manipulations end up positive.
