2016 AMC 8 真题与答案解析

2016 AMC 8 真题

答案详细解析请参考文末

Problem 1

The longest professional tennis match lasted a total of 11 hours and 5 minutes. How many minutes was that?

Problem 2

In rectangle , and . Point is the midpoint of . What is the area of ?

Problem 3

Four students take an exam. Three of their scores are and . If the average of their four scores is , then what is the remaining score?

Problem 4

When Cheenu was a boy he could run miles in hours and minutes. As an old man he can now walk miles in hours. How many minutes longer does it take for him to travel a mile now compared to when he was a boy?

Problem 5

The number is a two-digit number.

• When is divided by , the remainder is .

• When is divided by , the remainder is .

What is the remainder when is divided by ?

Problem 6

The following bar graph represents the length (in letters) of the names of 19 people. What is the median length of these names?

Problem 7

Which of the following numbers is not a perfect square?

Problem 8

Find the value of the expression

Problem 9

What is the sum of the distinct prime integer divisors of ?

Problem 10

Suppose that means What is the value of if

Problem 11

Determine how many two-digit numbers satisfy the following property: when the number is added to the number obtained by reversing its digits, the sum is

Problem 12

Jefferson Middle School has the same number of boys and girls. of the girls and of the boys went on a field trip. What fraction of the students were girls?

Problem 13

Two different numbers are randomly selected from the set and multiplied together. What is the probability that the product is ?

Problem 14

Karl's car uses a gallon of gas every miles, and his gas tank holds gallons when it is full. One day, Karl started with a full tank of gas, drove miles, bought gallons of gas, and continued driving to his destination. When he arrived, his gas tank was half full. How many miles did Karl drive that day?

Problem 15

What is the largest power of that is a divisor of ?

Problem 16

Annie and Bonnie are running laps around a -meter oval track. They started together, but Annie has pulled ahead because she runs faster than Bonnie. How many laps will Annie have run when she first passes Bonnie?

Problem 17

An ATM password at Fred's Bank is composed of four digits from to , with repeated digits allowable. If no password may begin with the sequence then how many passwords are possible?

 

Problem 18

In an All-Area track meet, sprinters enter a meter dash competition. The track has lanes, so only sprinters can compete at a time. At the end of each race, the five non-winners are eliminated, and the winner will compete again in a later race. How many races are needed to determine the champion sprinter?

Problem 19

The sum of consecutive even integers is . What is the largest of these consecutive integers?

Problem 20

The least common multiple of and is , and the least common multiple of and is . What is the least possible value of the least common multiple of and ?

Problem 21

A top hat contains 3 red chips and 2 green chips. Chips are drawn randomly, one at a time without replacement, until all 3 of the reds are drawn or until both green chips are drawn. What is the probability that the 3 reds are drawn?

Problem 22

Rectangle below is a rectangle with . What is the area of the "bat wings" (shaded area)?

Problem 23

Two congruent circles centered at points and each pass through the other circle's center. The line containing both and is extended to intersect the circles at points and . The circles intersect at two points, one of which is . What is the degree measure of ?

Problem 24

The digits , , , , and are each used once to write a five-digit number . The three-digit number is divisible by , the three-digit number is divisible by , and the three-digit number is divisible by . What is ?

Problem 25

A semicircle is inscribed in an isosceles triangle with base and height so that the diameter of the semicircle is contained in the base of the triangle as shown. What is the radius of the semicircle?

It is best to split 11 hours and 5 minutes into 2 parts, one of 11 hours and another of 5 minutes. We know that there is minutes in a hour. Therefore, there are minutes in 11 hours. Adding the second part(the 5 minutes) we get .

Use the triangle area formula for triangles: where is the area, is the base, and is the height. This equation gives us .

We can call the remaining score . We also know that the average, 70, is equal to . We can use basic algebra to solve for :giving us the answer of .

When Cheenu was a boy, he could run miles in hours and minutes minutes minutes, thus running minutes per mile. When he is an old man, he can walk miles in hours minutes minutes, thus walking minutes per mile. Therefore it takes him minutes longer to walk a mile now compared to when he was a boy.

From the second bullet point, we know that the second digit must be . Because there is a remainder of when it is divided by , the multiple of must end in a . We now look for this one:The number satisfies both conditions. We subtract the biggest multiple of less than to get the remainder. Thus, .

We first notice that the median name will be the name. The name is .

We know that our answer must have an odd exponent in order for it to not be a square. Because is a perfect square, is also a perfect square, so our answer must be .

We can group each subtracting pair together:After subtracting, we have:There are even numbers, therefore there are even pairs. Therefore the sum is

The prime factorization is . Since the problem is only asking us for the distinct prime factors, we have . Their desired sum is then .

Let us plug in into . Thus it would be . Now we have . Plugging into , we have . Solving for we have

We can write the two digit number in the form of ; reverse of is . The sum of those numbers is:We can use brute force to find order pairs such that . Since and are both digits, both and have to be integers less than . Thus our ordered pairs are or ordered pairs.

Set the number of children to a number that is divisible by two, four, and three. In this question, the number of children in the school is not a specific number because there are no actual numbers in the question, only ratios.This way, we can calculate the answer without dealing with decimals. is a number that works. There will be girls and boys. So, there will be = girls on the trip and = boys on the trip. The total number of children on the trip is , so the fraction of girls on the trip is or

The product can only be if one of the numbers is 0. Once we chose , there are ways we can chose the second number, or . There are ways we can chose numbers randomly, and that is . So, so the answer is .

Since he uses a gallon of gas every miles, he had used gallons after miles. Therefore, after the first leg of his trip he had gallons of gas left. Then, he bought gallons of gas, which brought him up to gallons of gas in his gas tank. When he arrived, he had gallons of gas. So he used gallons of gas on the second leg of his trip. Therefore, the second part of his trip covered miles. Adding this to the miles, we see that he drove miles.

First, we use difference of squares on to get . Using difference of squares again and simplifying, we get . Realizing that we don't need the right-hand side because it doesn't contain any factor of 2, we see that the greatest power of that is a divisor is .

Each lap Bonnie runs, Annie runs another quarter lap, so Bonnie will run four laps before she is overtaken. This means that Annie and Bonnie are equal so that Annie needs to run another lap to overtake Bonnie. That means Annie will have run laps.

For the first three digits, there are combinations since is not allowed. For the final digit, any of the numbers are allowed.

From any th race, only will continue on. Since we wish to find the total number of races, a column representing the races over time is ideal. Starting with the first race:                  Adding all of the numbers in the second column yields

Let be the 13th consecutive even integer that's being added up. By doing this, we can see that the sum of all 25 even numbers will simplify to since . Now, Remembering that this is the 13th integer, we wish to find the 25th, which is .

We wish to find possible values of ,, and . By finding the greatest common factor of and , algebraically, it's some multiple of and from looking at the numbers, we are sure that it is 3, thus is 3. Moving on to and , in order to minimize them, we wish to find the least such that the least common multiple of and is , . Similarly with and , we obtain . The least common multiple of and is

We put five chips randomly in order, and then pick the chips from the left to the right. To find the number of ways to rearrange the three red chips and two green chips, we solve for . However, we notice that whenever the last chip we draw is red, we pick both green chips before we pick the last (red) chip. Similarly, when the last chip is green, we pick all three red chips before the last (green) chip. This means that the last chip must be green in all the situations that work. This means we are left with finding the number of ways to rearrange three red chips and one green chip, which is . Because a green chip will be last out of the situations, our answer is .

The area of trapezoid is . Next, we find the height of each triangle to calculate their area. The triangles are similar, and are in a ratio by AA similarity (alternate interior and vertical angles) so the height of the larger one is while the height of the smaller one is Thus, their areas are and . Subtracting these areas from the trapezoid, we get . Therefore, the answer to this problem is

Drawing the diagram[SOMEONE DRAW IT PLEASE], we see that is equilateral as each side is the radius of one of the two circles. Therefore, . Therefore, since it is an inscribed angle, . So, in , , and. Our answer is .

We see that since is divisible by , must equal either or , but it cannot equal , so . We notice that since must be even, must be either or . However, when , we see that , which cannot happen because and are already used up; so. This gives , meaning . Now, we see that could be either or , but is not divisible by , but is. This means that and .

Draw the altitude from the top of the triangle to its base, dividing the isosceles triangle into two right triangles with height and base . The Pythagorean triple -- tells us that these triangles have hypotenuses of .Now draw an altitude of one of the smaller right triangles, starting from the foot of the first altitude we drew (which is also the center of the circle that contains the semicircle) and going to the hypotenuse of the right triangle. This segment is both an altitude of the right triangle as well as the radius of the semicircle (this is because tangent lines to circles, such as the hypotenuse touching the semicircle, are always perpendicular to the radii of the circles drawn to the point of tangency). Let this segment's length be .The area of the entire isosceles triangle is , so the area of each of the two congruent right triangles it gets split into is . We can also find the area of one of the two congruent right triangles by using its hypotenuse as its base and the radius of the semicircle, the altitude we drew, as its height. Then the area of the triangle is . Thus we can write the equation , so , so .