2019 AMC 10 B

Let the first jar's volume be and the second's be . It is given that . We find that.

We already know that this is the ratio of the smaller to the larger volume because it is less than .

We can set up a ratio to solve this problem. If is the volume of the first container, and is the volume of the second container, then:,

Crossmultiplying allows us to get. Thus the ratio of the volume of the first container to the second container is .

An alternate solution is to plug in some maximum volume for the first container let's say , so there was avolume of in the first container, and then the second container also has a volume of , so you get Thus to answer is .

Since a counterexample must be value of which is not prime, must be composite, so we eliminate and . Now we subtract from the remaining answer choices, and we see that the only time is not prime is when .

of seniors do not play a musical instrument. If we denote as the number of seniors, then ,

,

Thus there are nonseniors. Since of the nonseniors play a musical instrument, .

Let be the number of seniors, and be the number of nonseniors.Then,

Multiplying both sides by gives us ,

Also， because there are students in total.

Solving these system of equations give us ，.

Since of the non-seniors play a musical instrument, the answer is simply of , which gives us .

We can clearly deduce that of the nonseniors do play an instrument, but, since the total percentage of instrument players is , the nonsenior population is quite low. By intuition, we can therefore see that the answer is around or . Testing both of these gives us the answer .

If all lines satisfy the condition, then we can just plug in values for , , and that form an arithmetic progression. Let's use , , , and , , .Then the two lines we get are: Use elimination to deduce and plug this into one of the previous line equations. We get Thus the common point is .

We know that , , and form an arithmetic progression, so if the common difference is , we can say , , , , . Now we have , and expanding gives . Factoring gives . Since this must always be true (regardless of the values of and ), we must have and , so , , , and the common point is .

Let's analyze all of the options separately.

: Clearly is true, because a point in the first quadrant will have non-negative and . coordinates, and so its reflection, with the coordinates swapped, will also have nonnegative and coordinates.

: The triangles have the same area, since and are the same tiangle (congruent). More formally, we can say that area is invariant under reflection.

: If point has coordinates , then will have coordinates . The gradient is thus , so this is true. (We know since the question states that none of the points , ,or lies on the line , so there is no risk of division by zero).

: Repeating the argument for , we see that both lines have slope, so this is also true.

: By process of eliminaton,this must now be the answer, Indeed, it point has

coordinates and point has coordinates , then and will, respectively, have coordinates and . The product of the gradients of and is , so in fact these lines are never perpendicuar to each other (using the "negative reciprocal" condition for perpendicularity). Thus the anewer is .

Counterexamples

If and , then the slope of , , is . While the slope of , , is . is the reciprocal of , but it is not the negative reciprocal of . To generalize, let denote the coordinates of point , let denote the coordinates of point , let denote the slope of segment , and let denote the slope of segment ,. Then, the coordinates of are , and of are .

Then, , and .

If and , , and in these cases, the condition is false.

,

solving by the quadratic formula. (since clearly )The answer is therefore .

Dividing both sides by gives . since is nonnegative, . The answeris .

Dividing both sides by as before gives , Now factor out , giving . By considering the prime factorization of , a bit of experimentation gives us and , so so the answer is .

If he has enough money to buy pieces of red candy, pieces of green candy, and pieces of blue candy, then the smallest amount of money he could have is cents. Since a piece of purpe candy costs cems,the smallest possible value of is .

We simply need to find a value of that is divisible by , , and . Observe that is divisible by and , but not . is divisible by , , and , meaning that we have exact change (in this case, cenis) to buy each type of candy, so the minimum value of is .

We notice th@t the square can be split into congruent smaller squares, with the altitude of the equilateral triangle being the side of this sm@ller square. Therefore, the area of each shaded part that resides within a square is the total area of the square subtracted from each triangle (which has already been split in half).

When we split an equilateral triangle in half, we get two triangles. Therefore, the altitude, which is also the side length of one of the smaller squares, is . We can then compute the area of the two triangles as .

The area of the each small squares is the square of the side length, . Therefore, the area of the shaded region in each of the four squares is .

since there are of these squares, we muitiply this by to get as our answer.

We can see that the side length of the square is by considering the altitude of the equilateral triangle as in Solution . Using the Pythagorean Theorem, the diagonal of the square is thus . Because of this, the height of one of the four shaded kites is .

Now, we just need to find the length of that kite. By the Pythagorean Theorem again, this length . Now using area length width, the area of one of the four kites is .

There are four cases we need to consider here.

Case : is a positive integer. Without loss of generality, assume .Then .

Case : is a positive fraction. Wihout loss of generality, asume .

Then .

Case : is a negative integer. Without loss of generality, assume .

Then .

Case : is a negative fraction. Without loss of generality, assume .

Then

Thus the range of the function is .

It is easily verifed that when is an integer, is zero. We therefore need only to consider the case when is not an integer.

When is postive, , so

.

When is negative, let be composed of integer part and fractional part (both ):

.

Thus,the range of is .

Note: One could solve the case of as a negative noninteger in this way:

.

Notice that whatever point we pick for , will be the base of the triangle. Without loss of generality, let points and be and , since for any other combination of points, we can just rotate the plane to make them and under a new coordinate system. When we pick point , we have to make sure that its coordinate is , because that's the only way the area of the triangle can be .

Now when the perimeter is minimized, by symmetry, we put in the middle, at .  We can easily see that and will both be . The perimeter of this minimal tiangle is . which is larger than . Since the minimum perimeter is greater than , there is no triangle that satisfies the condition, giving us .

Without loss of generality, let be a horizontal segment of length . Now realize that has to lie on one of the lines parallel to and vertically units away from it. But is already , and this doesn't form a triangle. Otherwise, without loss of generality,  . Dropping altitude , we have a right triangle with hypotenuse and leg , which is clearly impossible, again giving the answer as .

Call the number of marbles in each jar (because the problem specifies that they each contain the same number). Thus, is the number of green marbles in Jar , and is the number of green marbles in Jar . Since , we have  , so there are marbles in each jar.

Because is the number of blue marbles in Jar , and is the number of blue marbles in Jar , tere are more mables in Jar than Jar . This means the answers is .

Observe that .To maximize the sum of the digits, we want as many as possible (since is the highest value in base ), and this will occur with either of the numbers or .

Thus, the answer is ..

IronicNinja, edited by some people,

Note: the number can also be , which will also give the answer of .

Note that all base numbers with or more digits are in fact greater than . Since the first answer that is possible using a digit number is , we start with the smallest base number that whose digits sum to , namely , But this is greater than , so we continue by trying ，

which is less than . So the answer is ..

The mean is .

There are three possibilties for the median: it is either . , or .

Let's start with .

has solution , and the sequence is , , , , , which does have median , so this is a valid solution.

Now let the median be .

give , so the  sequence is , , , , , which has median , so this is not valid.

Finally we let the median be .

, and the  sequene is , , , , ，

which has median . This case is therefore again not valid.

Hence the only possible vaue of is .

We can figure out by noticing that will end with zeroes, as there are three in its prime factorization. Next, we use the fact that is a mutiple of both and . Their divisibilty rules (see Solution ) tell us that (mod ) and that (mod ) By inspection, we see that , is a valid solution. Therefore the answers .

We know that and are both factors of , Furthermore, we know that , because ends in three zeroes (see Solution ). We can simply use the divisibility rules for and for this problem to find and . For to be divisible by , the sum of digits must simply be divisible by . Summing the digits, we get that must be divisible by . This leaves either or as our answer choice. Now we test for divisibility by . For a number to be divisible by , the alternating sum must be divisible by (for example, with the number , , so is divisible by ). Applying the alternating sum test to this problem, we see that must be divisible by . By inspection, we can see that this holds if and The sum .

First of all, let the two sides which are congruent be and , where . The only way that the conditions of the problem can be satisfied is if is the shorter leg of and the longer leg of , and is the longer leg of and the hypotenuse of .

Notice that this means the value we are looking for is the square of , which is just .