2015 AMC 8

Solution 1

We need square feet of carpet to cover the floor. Since there are square feet in a square yard, we divide this by to get square yards.

Solution 2

Since there are feet in a yard, we divide by to get yards, and by to get yards. To find the area of the carpet, we then multiply these two values together to get square yards.

Solution 1

Since octagon is a regular octagon, it is split into equal parts, such as triangles , etc. These parts, since they are all equal, are of the octagon each. The shaded region consists of of these equal parts plus half of another, so the fraction of the octagon that is shaded is

Solution 2

The octagon has been divided up into identical triangles (and thus they each have equal area). Since the shaded region occupies out of the total triangles, the answer is .

Using , we can set up an equation for when Jill arrives at the swimming pool:

Solving for , we get that Jill gets to the pool in of an hour, which is minutes. Doing the same for Jack, we get that

Jack arrives at the pool in of an hour, which in turn is minutes. Thus, Jill has to wait

There are ways to order the boys on the ends, and there are ways to order the girls in the middle. We get the answer to be .

Solution 1

When they score a on the next game, the range increases from to . This means the increased.

Note: The range is defined to be the difference of the largest element and the smallest element of a set.

Solution 2

Because is less than the score of every game they've played so far, the measures of center will never rise. Only measures of spread, such as the , may increase.

Solution 1

We know the semi-perimeter of is . Next, we use Heron's Formula to find that the area of the triangle is just .

Solution 2 

Splitting the isosceles triangle in half, we get a right triangle with hypotenuse and leg . Using the Pythagorean Theorem , we know the height is . Now that we know the height, the area is .

Solution 1

You can make this problem into a spinner problem. You have the first spinner with equally divided

sections: , and . You make a second spinner that is identical to the first, with equal sections of

,, and . If the first spinner lands on , it must land on two for the result to be even. You write down the first

combination of numbers: . Next, if the spinner lands on , it can land on any number on the second

spinner. We now have the combinations of and . Finally, if the first spinner ends on , we

have Since there are possible combinations, and we have evens, the final answer is

.

Solution 2

We can list out the numbers. Box A has chips , , and , and Box B also has chips , , and . Chip (from Box A)

could be with 3 partners from Box B. This is also the same for chips and from Box A. total sums. Chip could be multiplied with 2 other chips to make an even product, just like chip . Chip can only multiply with 1 chip. . The answer is .

Solution 1

We know from the Triangle Inequality that the last side, , fulfills . Adding to both sides of the inequality, we get , and because is the perimeter of our triangle, is our answer.

Solution 2

, fulfills , meaning that s<24. The greatest value of s can be 23. 23+5+19 is 47. 48 is our nearest answer choice so its .

Solution 1

First, we have to find how many widgets she makes on Day . We can write the linear equation to represent this situation. Then, we can plug in for : -- -- . The sum of is .

Solution 2

The sum is just the first odd counting/natural numbers, which is .

There are choices for the first number, since it cannot be , there are only choices left for the second number since it must differ from the first, choices for the third number, since it must differ from the first two, and choices for the fourth number, since it must differ from all three. This means there are integers between and with four distinct digits.

Solution 1

There is one favorable case, which is the license plate says "AMC8." We must now find how many total cases there are. There are choices for the first letter (since it must be a vowel), choices for the second letter (since it must be of consonants), choices for the third letter (since it must differ from the second letter), and choices for the digit. This leads to total possible license plates. Therefore, the probability of a license plate saying "AMC8" is .

Solution 2

The probability of choosing A as the first letter is . The probability of choosing next is . The probability of choosing C as the third letter is (since there are other consonants to choose from other than M). The probability of having as the last number is . We multiply all these to obtain .

Solution 1

We first count the number of pairs of parallel lines that are in the same direction as . The pairs of parallel lines are , , , , , and . These are pairs total. We can do the same for the lines in the same direction as and . This means there are total pairs of parallel lines.

Solution 2

Look at any edge, let's say . There are three ways we can pair with another edge. , , and . There are 12 edges on a cube. 3 times 12 is 36. We have to divide by 2 because every pair is counted twice, so is total pairs of parallel lines.

Solution 1

Since there will be elements after removal, and their mean is , we know their sum is . We also know that the sum of the set pre-removal is . Thus, the sum of the elements removed is . There are only subsets of elements that sum to : .

Solution 2

We can simply remove subsets of numbers while leaving only behind. The average of this one-number set is still , so the answer is .

Solution 1

Let our numbers be , where is odd. Then, our sum is . The only answer choice that cannot be written as , where is odd, is .

Solution 2

If the four consecutive odd integers are and ; then, the sum is . All the integers are divisible by except .

Solution 1

Let:

- be the number of students who voted in favor of the first issue,

- be the number of students who voted in favor of the second issue,

- be the number of students who voted in favor of both issues.

We are given:

-

-

- students voted against both issues, so the number of students who voted for at least one issue is:

By the principle of inclusion and exclusion:

Substitute known values:

Solution 2

We can see that this is a Venn Diagram Problem.

First, we analyze the information given. There are students. Let's use A as the first issue and B as the second issue.

students were for A, and students were for B. There were also students against both A and B.

Solving this without a Venn Diagram, we subtract away from the total, . Out of the remaining , we have people for A and

people for B. We add this up to get . Since that is more than what we need, we subtract from to get

.

Solution 1 

Let the number of sixth graders be , and the number of ninth graders be . Thus, , which simplifies to . Since we are trying to find the value of , we can just substitute for into the equation. We then get a value of .

Solution 2

We see that the minimum number of ninth graders is , because if there are then there is ninth-grader with a buddy, which would mean there are sixth graders, which is impossible (of course unless you really do have half of a person). With ninth-graders, of them are in the buddy program, so there sixth-graders total, two of whom have a buddy. Thus, the desired fraction is .

Solution 1

For starters, we identify d as distance and v as velocity (speed)

Writing the equation gives us: and .

This gives , which gives , which then gives .

Solution 2

d = rtd=times r = times (r + 18)$

so , plug into the first one and it's miles to school.

Solution 1

We begin filling in the table. The top row has a first term and a fifth term , so we have the common difference is . This means we can fill in the first row of the table:

The fifth row has a first term of and a fifth term of , so the common difference is . We can fill in the fifth row of the table as shown:

We must find the third term of the arithmetic sequence with a first term of and a fifth term of . The common difference of this sequence is , so the third term is .

Solution 2

The middle term of the first row is , since the middle number is just the average in an arithmetic sequence. Similarly, the middle of the bottom row is . Applying this again for the middle column, the answer is .

Solution 1

The area of is equal to half the product of its base and height. By the Pythagorean Theorem, we find its height is , and its base is . We multiply these and divide by to find the area of the triangle is . Since the grid has an area of , the fraction of the grid covered by the triangle is .

Solution 2

Note angle is right; thus, the area is ; thus, the fraction of the total is .

Solution 1

So, let there be pairs of socks, pairs of socks, and pairs of socks.

We have , , and .

Now, we subtract to find , and . It follows that is a multiple of and is a multiple of . Since sum of 2 multiples of 3 = multiple of 3, so we must have .

Therefore, , and it follows that . Now, , as desired.

Solution 2

Since the total cost of the socks was and Ralph bought pairs, the average cost of each pair of socks is .

There are two ways to make packages of socks that average to . You can have:

Two pairs and one pair (package adds up to )

One pair and one pair (package adds up to )

Now, we need to solve

where is the number of packages and is the number of packages. We see our only solution (that has at least one of each pair of sock) is , which yields the answer of .

Solution 1

Clearly, since is a side of a square with area , . Now, since , we have .

We know that is a side of a square with area , so . Since is equilateral, .

Lastly, is a right triangle. We see that , so is a right triangle with legs and . Now, its area is .

Solution 2

Since , and , . Meanwhile, , and since is equilateral, . If is equiangular, , where is the number of sides of the shape. Adding all the angles around gives , so . Because is right, the area of . Therefore, the answer is .

Solution 1

Since we know the number must be a multiple of , we can eliminate . We also know that after days, the students can't find any more arrangements, meaning the number has factors. Now, we just list the factors of every number, starting with :