首页> 重点归纳 > 考题15-16 2015 AMC 12B

考题15-16 2015 AMC 12B

2018-08-06 重点归纳

AMC12是针对高中学生的数学测验，该竞赛开始于2000年，分A赛和B赛，于每年的2月初和2月中举行，学生可任选参加一项即可。其主要目的在于激发学生对数学的兴趣，参予AMC12的学生应该不难发现测验的问题都很具挑战性，但测验的题型都不会超过学生的学习范围。这项测验希望每个考生能从竞赛中享受数学。那么接下来跟随小编来看一下AMC12的官方真题以及官方解答吧：

Problem 15

At Rachelle's school an A counts 4 points, a B 3 points, a C 2 points, and a D 1 point. Her GPA on the four classes she is taking is computed as the total sum of points divided by 4. She is certain that she will get As in both Mathematics and Science, and at least a C in each of English and History. She thinks she has a $\tfrac{1}{6}$ chance of getting an A in English, and a $\tfrac{1}{4}$ chance of getting a B. In History, she has a $\tfrac{1}{4}$ chance of getting an A, and a $\tfrac{1}{3}$ chance of getting a B, independently of what she gets in English. What is the probability that Rachelle will get a GPA of at least 3.5?

$考题15-16 2015 AMC 12B$

Solution 1

The probability that Rachelle gets a C in English is $1-\frac{1}{6}-\frac{1}{4} = \frac{7}{12}$ .

The probability that she gets a C in History is $1-\frac{1}{4}-\frac{1}{3} = \frac{5}{12}$ .

We see that the sum of Rachelle's "point" scores must be at least 14 since $4*3.5 = 14$ . We know that in Mathematics and Science we have a total point score of 8 (since she will get As in both), so we only need a sum of 6 in English and History. This can be achieved by getting two As, one A and one B, one A and one C, or two Bs. We evaluate these cases.

The probability that she gets two As is $\frac{1}{6}\cdot\frac{1}{4} = \frac{1}{24}$ .

The probability that she gets one A and one B is $考题15-16 2015 AMC 12B$ .

The probability that she gets one A and one C is $考题15-16 2015 AMC 12B$ .

The probability that she gets two Bs is $\frac{1}{4}\cdot\frac{1}{3} = \frac{1}{12}$ .

Adding these, we get $考题15-16 2015 AMC 12B$ .

Solution 2

We can break it up into three mutually exclusive cases: A in english, at least a C in history; B in english and at least a B in history; C in english and an A in history. This gives

Problem 16

A regular hexagon with sides of length 6 has an isosceles triangle attached to each side. Each of these triangles has two sides of length 8. The isosceles triangles are folded to make a pyramid with the hexagon as the base of the pyramid. What is the volume of the pyramid?

$考题15-16 2015 AMC 12B$

Solution

The distance from a corner to the center is 6, and from the corner to the top of the pyramid is 8, so the height is $\sqrt{8^2 - 6^2} = \sqrt{64 - 36} = \sqrt{28} = 2\sqrt{7}$ .