2020 AMC 10 B

We know that when we subtract negative numbers, The equation becomes

A cube with side length has volume , so of these will have a total volume of .

A cube with side length has volume , so of these will have a total volume of .

Solution 1:

WLOG, let and . Since the ratio of to is , we can substitute in the value of to get The ratio of is , so The ratio of to is then so our answer is (E)

Solution 2:

We need to somehow link all three of the ratios together. We can start by connecting the last two ratios together by multiplying the last ratio by two. , and since , we can link them together to get . Finally, since , we can link this again to get: , so

Solution 1:

Since the three angles of a triangle add up to and one of the angles is because it's a right triangle, . The greatest prime number less than 90 is 89 . If , then , which is not prime. The next greatest prime number less than 90 is 83 . If , then , which IS prime, so we have our answer (D) 7

Solution 2:

Looking at the answer choices, only and are coprime to . Testing , the smaller angle, makes the other angle which is prime, therefore our answer

Solution 1:

Let's first find how many possibilities there would be if they were all distinguishable, then divide out the ones we overcounted. There are ways to order objects. However, since there's ways to switch the yellow tiles around without changing anything (since they're indistinguishable) and ways to order the green tiles, we have to divide out these possibilities.

Solution 2:

We can repeat chooses extensively to find the answer. There are 7 choose 3 ways to arrange the brown tiles which is 35 . Then from the remaining tiles there are 4 choose ways to arrange the red tiles. And now from the remaining two tiles and two slots we can see there are two ways to arrange the purple and brown tiles, giving us an answer

In order to get the smallest palindrome greater than , we need to raise the middle digit. If we were to raise any of the digits after the middle, we would be forced to also raise a digit before the middle to keep it a palindrome, making it unnecessarily larger. So we raise to the next largest value, , but obviously, that's not how place value works, so we're in the now. To keep this a palindrome, our number is now . So Megan drove miles. Since this happened over 2 hours, she drove at (B) 55 mph.

Any even multiple of is a multiple of , so we need to find multiples of that are perfect squares and less than . Any solution that we want will be in the form , where is a positive integer. The smallest possible value is at , and the largest is at (where the expression equals 1764 ). Therefore, there are a total of (A) 7 possible numbers.

Solution 1:

There are 3 options here: 1. is the right angle. It's clear that there are points that fit this, one that's directly to the right of and one that's directly to the left. We don't need to find the length, we just need to know that it is possible, which it is. 2. is the right angle. Using the exact same reasoning, there are also solutions for this one. 3 . The new point is the right angle. (Diagram temporarily removed due to asymptote error) The diagram looks something like this. We know that the altitude to base must be since the area is . From here, we must see if there are valid triangles that satisfy the necessary requirements. First of all, area. Next, from the Pythagorean Theorem. From here, we must look to see if there are valid solutions. There are multiple ways to do this: Recognizing min and max: We know that the minimum value of is when . In this case, the equation becomes , which is LESS than . The equation becomes , which is obviously greater than 64 . We can conclude that there are values for and in between that satisfy the Pythagorean Theorem. And since , the triangle is not isoceles, meaning we could reflect it over and/or the line perpendicular to for a total of 4 triangles this case.

Solution 2:

Note that line segment can either be the shorter leg, longer leg or the hypotenuse. If it is the shorter leg, there are two possible points for that can satisfy the requirements - that being above or below . As such, there are 2 ways for this case. Similarly, one can find that there are also 2 ways for point to lie if is the longer leg. If it is a hypotenuse, then there are 4 possible points because the arrangement of the shorter and longer legs can be switched, and can be either above or below the line segment. Therefore, the answer is

Solution 1: Rearranging the terms and and completing the square for yields the result . Then, notice that can only be and because any value of that is greater than will cause the term to be less than 0 , which is impossible as must be real. Therefore, plugging in the above values for gives the ordered pairs , and gives a total of (D)4 ordered pairs.

Solution 2:

Bringing all of the terms to the LHS, we see a quadratic equation in terms of . Applying the quadratic formula, we order for to be real, which it must be given the stipulation that we are seeking integral answers, we know that the discriminant, must be nonnegative. Therefore, we see that we must split the inequality into a compound, resulting in The only integers that satisfy this are . Plugging these values back into the quadratic equation, we see that both produce a discriminant of 0 , meaning that there is only 1 solution for . If , then the discriminant is nonzero, therefore resulting in two solutions for . Thus, the answer is

Solution 1： Notice that when the cone is created, the radius of the circle will become the slant height of the cone and the intact circumference of the circle will become the circumference of the base of the cone. We can calculate that the intact circumference of the circle is . Since that is also equal to the circumference of the cone, the radius of the cone is 3 . We also have that the slant height of the cone is 4 . Therefore, we use the Pythagorean Theorem to calculate that the height of the cone is . The volume of the cone is

Solution 2 (Last Resort/Cheap):

Using a ruler, measure a circle of radius 4 and cut out the circle and then the quarter missing. Then, fold it into a cone and measure the diameter to be 6 . You can form a right triangle with sides 3,4 , and then through the Pythagorean theorem the height is found to be . The volume of a cone is . Plugging in we find

We don't care about which books Harold selects. We just care that Betty picks 2 books from Harold's list and 3 that aren't on Harold's list. The total amount of combinations of books that Betty can select ways for Betty to choose 2 of the books that are on Harold's list. From the remàining 5 books that aren't on Hàrold's list, there

Solution 1:

Now we do some estimation. Notice that , which means that is a little more than . Multiplying it with , we get that the denominator is about . Notice that when we divide by an digit number, there are zeros before the first nonzero digit. This means that when we divide by the digit , there are integer decimal point.

Solution 2:

First rewrite as . Then, we know that when we write this in decimal form, there will be 40 digits after the decimal point. Therefore, we just have to find the number of digits in . and memming (alternatively use the fact that ), digits. Our answer is (D) 26

You can find that every four moves both coordinates decrease by . Therefore, both coordinates need to decrease by two times. You subtract, giving you the answer of (B)

Solution 1:

Let point be a vertex of the regular hexagon, let point be the midpoint of the line connecting point and a neighboring vertex, and let point be the second intersection of the two semicircles that pass through point . Then, , since is the center of the semicircle with radius 1 that lies on, , since is the center of the semicircle with radius 1 that lies on, and , as a regular hexagon has angles of 120 , and is half of any angle in this hexagon. Now, using the sine law, , so . Since the angles in a triangle sum to is also . Therefore, is an equilateral triangle with side lengths of .Since the area of a regular hexagon can be found with the formula , where is the side length of the hexagon, the area of this hexagon . is 2 . Since the area of an equilateral triangle can be found with the formula , where is the side length of the equilateral triangle, the area of an equilateral triangle with side lengths of 1 is . Since the area of a circle can be found with the formula , the area of a sixth of a circle with radius 1 is . In each sixth of the hexagon, there of a circle with radius 1 is . In each sixth of the hexagon, there sixth of a circle with radius 1 colored white, with an area of . The rest of the sixth is colored gray. Therefore, the total area that is colored white in each sixth of the hexagon is , which equals , and the total area colored white is , which equals . Since the area colored gray equals the total area of the hexagon minus the area colored white, the area colored gray is , which equals (D)

Solution 2:

First, subdivide the hexagon into 24 equilateral triangles with side length

Now note that the entire shaded region is just 6 times this part:

The entire rhombus is just 2 equilatrial triangles with side lengths of 1 , so it has an area of: Hence, the area of the shaded region in that section is For a final area

After erasing every third digit, the list becomes . repeated. After erasing every fourth digit from this list, the list becomes . . repeated. Finally, after erasing every fifth digit from this list, the list becomes . . repeated. Since this list repeats every digits and since are respectively in , we have that the ,, and digits are the , , and digits respectively. It follows that the answer is .

Solution 1:

Notice that to use the optimal strategy to win the game, Bela must select the middle number in the range and then mirror whatever number Jenn selects. Therefore, if Jenn can select a number within the range, so can Bela. Jenn will always be the first person to run out of a number to choose, so the answer is (A)Bela will always win.

Solution 2 (Guessing): First of all, realize that the value of should have no effect on the strategy at all. This is because they can choose real numbers, not integers, so even if is odd, for example, they can still go halfway. Similarly, there is no reason the strategy would change when .

So we are left with (A) and (B). From here it is best to try out random numbers and try to find the strategy that will let Bela win, but if you can't find it, realize that it is more likely the answer is (A) Bela will always win since Bela has the first move and thus has more control.

Let us use casework on the number of diagonals. Case diagonals There are ways: either pairs with pairs with , and so on or pairs with pairs with , etc.

Case diagonal There are possible diagonals to draw (everyone else pairs with the person next to them. Note that there cannot be diagonals. Case diagonals Note that there cannot be a case with diagonals because then there would have to be diagonals for the two remaining people, thus a contradiction. Case diagonals There is way to do this. Thus, in total there are possible ways.

Solution 1:

Let denote that George selects a red ball and that he selects a blue one. Now, in order to get 3 balls of each color, he needs 2 more of both and . There are 6 cases: can confirm that there are only 6 since ). However we can clump and together since they are equivalent by symmetry.

CASE 1: and Let's find the probability that he picks the balls in the order of . The probability that the first ball he picks is red is . Now there are 2 reds and 1 blue in the urn. The probability that he picks red again is now .

There are 3 reds and 1 blue now. The probability that he picks a blue is . Finally, there are 3 reds and 2 blues. The probability that he picks a blue is . So the probability that the case happens is . However, since the case is the exact same by symmetry, case 1 has a probability of chance of happening.

CASE 2:RBRB and Let's find the probability that he picks the balls in the order of . The probability that the first ball he picks is red is . Now there are 2 reds and 1 blue in the urn. The probability that he picks blue is .

There are 2 reds and 2 blues now. The probability that he picks a red is . Finally, there are 3 reds and 2 blues. The probability that he picks a blue is . So the probability that the case happens . However, since the case is the exact is . However, since the Rcase is the exact same by symmetry, case 2 has a probability of chance of happening.

CASE 3: and Let's find the probability that he picks the balls in the order of . The probability that the first ball he picks is red is . Now there are 2 reds and 1 blue in the urn. The probability that he picks blue is

There are 2 reds and 2 blues now. The probability that he picks a blue is . Finally, there are 2 reds and 3 blues. The probability that he picks a red is . So the probability that the case happens is . However, since the case is the exact is . However, since the case is the exact same by symmetry, case 3 has a probability of chance of happening.

Adding up the cases, we have (B)

Solution 2: We know that we need to find the probability of adding 2 red and 2 blue balls in some order. There are 6 ways to do this, since there are ways to arrange in some order. We will show that the probability for each of these 6 ways is the same.

We first note that the denominators should be counted by the same number. This number is . This is because , and 5 represent how many choices there are for the four steps. No matter what the step involves numbers to choose from. The numerators are the number of successful operations. No matter the order, the first time a red is added will come from 1 choice and the second time will come from 2 choices, since that is how many reds are in the urn originally. The same goes for the blue ones. The numerator must equal . Therefore, the probability for each of the orderings of is is . There are 6 of these, so the total probability is (B)

Solution 1:

We're looking for the amount of ways we can get 10 cards from a deck of 52 , which is represented by . We need to get rid of the multiples of 3 , which will subsequently get rid of the multiples of 9 (if we didn't, the zeroes would mess with the equation since you can't divide by 0 )   leaves us with . Converting these into , we have

Solution 2:

Since this number is divisible by but not , the last digits must be divisible by but the last 3 digits cannot be divisible by . This narrows the options down to and . Also, the number cannot be divisible by . Adding up the digits, we get . If , then the expression equals , a multiple of . This would mean that the entire number would be divisible by , which is not what we want. Therefore, the only option is (A)2.