新sat数学练习题附答案解析（十六）

2023-09-06 17:31:46来源：网络整理

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　　1.The circle shown above has center and a radius of length . If the area of the shaded region is , what is the value of ?

　　Answer Choices

　　(A)

　　(B)

　　(C)

　　(D)

　　(E)

　　2.If, what is the value of ?

　　Answer Choices

　　(A)

　　(B)

　　(C)

　　(D)

　　(E)

　　3.All numbers divisible by both and are also divisible by which of the following?

　　Answer Choices

　　(A)

　　(B)

　　(C)

　　(D)

　　(E)

　　4.Ten cars containing a total ofpeople passed through a checkpoint. If none of these cars contained more thanpeople, what is the greatest possible number of these cars that could have contained exactlypeople?

　　Answer Choices

　　(A) One

　　(B) Two

　　(D) Four

　　(E) Five

　　5.In the figure above, which quadrants contain pairs that satisfy the condition ?

　　Answer Choices

　　(A)only

　　(B) and only

　　(D) and only

　　(E) , , , and

　　Explanation

　　1.The correct answer is A

　　In order to find the value of , you should first determine the measure of the angle that is located at point in the right triangle. To determine this angle, you must calculate what fraction of the circle’s area is unshaded. The radius of the circle is and its area is , or . The area of the shaded region is , so the area of the unshaded region must be . Therefore, the fraction of the circle’s area that is unshaded is , or . A circle contains a total of degrees of arc, which means that of degrees, or degrees, is the measure of the angle at point in the unshaded region. Since you now know that two of the three angles in the triangle measuredegrees and degrees and that the sum of the measures of the three angles is always degrees, the third angle must measure degrees. Therefore, .

　　2.The correct answer is D

　　Both and . The equation cannot be solved for because there are two unknowns. The value of can be found by solving the equation for . It follows that must equal . The value of can now be substituted into the equation , giving . Therefore, must equal .

　　3.The correct answer is A

　　All numbers divisible by both and are the multiples of and . Since and have no prime factor in common , then the least common multiple of and is equal to their product, namely . Every other multiple of and is divisible by . Thus, if is divisible by a number then all the multiples of and are divisible by that number. Therefore, it is enough to check by which number in the given options is divisible. Only dividesand none of the other do. The answer must be

　　4.The correct answer is D

　　It could not be true that each of the ten cars contained exactlypeople, as this would give a total of only. If nine of the cars contained exactlypeople, the remaining car could have no more thanpeople, for a total of only. Continuing in the same way, a pattern develops. If eight of the cars contained exactlypeople, the remaining two cars could have no more thanpeople each, for a total of only. If seven of the cars contained exactlypeople, the total number of people could be only. From the pattern, you can see that if four of the cars contained exactlypeople, and the remaining six cars contained the maximum ofpeople, the total number would be, as given in the question. Therefore, at most four of the ten cars could have contained exactly people.

　　5.The correct answer is C

　　In order for to satisfy , it must be true that and are equal to each other and not equal to zero. An example of such a pair is , which is in quadrant .

　　In quadrant , all the values are negative and all the values are positive, so in quadrant , and cannot be equal. For example, the pair does not satisfy the condition, since , not .

　　In quadrant , the values and the values are both negative, so it is possible for and to be equal. For example, the pair is in quadrant and .

　　In quadrant , and cannot be equal because the values are positive and the values are negative. For example, the pair does not satisfy the condition, since .

　　The quadrants that contain pairs that satisfy the given condition are quadrants and only.

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