2021 AMC 10 A Fall

Solution 1 (Laws of Exponents):

We have

Solution 2 (Difference of Squares) :

We have

Solution 1:

The volume of the cube is , and the volume of a clay ball is . Since the balls can be reshaped but not compressed, the maximum number of balls that can completely fit inside a cube is Approximating with , we have , or . We simplify to get from which (D) 6 .

Solution 2:

As shown in Solution 1, we conclude that the maximum number of balls that can completely fit inside a cube is . By an underestimation , we have , or . By an overestimation , we have , or . Together, we get from which .

Solution 1:

If Mr. Lopez chooses Route , then he will spend hour, or minutes. If Mr. Lopez chooses Route , then he will spend hour, or minutes. Therefore, Route B is quicker than Route A by (B) minutes.

Solution 2:

We use the equation to solve this problem. Recall that 1 mile per hour is equal to mile per minute. For Route A, the distance is 6 miles and the speed to travel this distance is mile per minute. Thus, the time it takes on Route is 12 minutes. For Route , we have to use the equation twice: once for the distance of miles with a speed of mile per minute and a distance of miles at a speed of mile per minute. Thus, the time it takes to go on Route is minutes. Thus, Route is faster than Route A. Thus, the answer is (B) .

Solution 1:

First, modulo 2 or . Hence, . Second modulo . Hence, . Third, modulo . Hence, . Therefore, the answer is .

Solution 2:

divisible by divisible by divisible by divisible by divisible by divisible by divisible by divisible by divisible by This leaves only (E) 9 .

There are gaps between the telephone poles, so the distance of each gap is feet. Each of Oscar's leaps covers feet, and each of Elmer's strides covers feet Therefore, Oscar's leap is feet longer than Elmer's stride.

By angle subtraction, we have . Note that is isosceles, so . Finally, we get (D) 170 degrees.

Solution 1:

Note that the number . By the problem statement, From this we see that must be divisible by 9 . This only happens when . Then, . Thus, there is only cuddly number, which is 89 .

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Solution 2:

If the tens digit is and the ones digit is then the number is so we have the equation . We can guess and check after narrowing the possible cuddly numbers down to , and 99 . (We can narrow it down to these by just thinking about how 's value affects 's value and then check all the possiblities.) Checking all of these we get that there is only (B) 1 2-digit cuddly number, and it is 89 .

Since an even number is times more likely to appear than an odd number, the probability of an even number appearing is . Since the problem states that the sum of the two die must be even, the numbers must both be even or both be odd. We either have EE or OO, so we have

The formula for expected values is We have Therefore, the answer is .

Solution 1:

Let be the length of the ship. Then, in the time that Emily walks 210 steps, the ship moves steps. Also, in the time that Emily walks 42 steps, the ship moves steps. Since the ship and Emily both travel at some constant rate, . Dividing both sides by 42 and cross multiplying, we get , so , and (A) 70 .

Solution 2:

Let the speed at which Emily walks be steps per hour. Let the speed at which the ship is moving be . Walking in the direction of the ship, it takes her steps, or hours, to travel. We can create an equation: where is the length of the ship. Walking in the opposite direction of the ship, it takes her 42 steps, or hour. We can create a similar equation: Now we have two variables and two equations. We can equate the expressions for and solve for : Therefore, we have (A) 70 .

Solution 1:

Recall that . We expand by the definition of bases:

Solution 2:

We need to first convert into a regular base- 10 number: Now, consider how the last digit of 9 changes with changes of the power of 9 : Note that if is odd, then On the other hand, if is even, then Therefore, we have Note that for the odd case, may simplify the process further, as given by Solution 1 .

Solution 1:

Note that for this restriction to be true, there must be 3 balls of each color. There are a total of ways to color the balls, and there are ways for three balls chosen to be painted white. Thus, the answer is

Solution 2:

For this restriction to be upheld, there must be three black and three white balls. One such way for this to occur is the arrangement , which has a probability of occuring. However, there are ways to arrange the three black and three white balls, meaning that the answer is, (D)

Solution 1:

The second equation is . We know that the graph of is a very simple diamond shape, so let's see if we can reduce this equation to that form: We now have two separate graphs for this equation and one graph for the first equation, so let's put it on the coordinate plane:

Solution 2:

We can manipulate the first equation to get . From the second equation, we have that or . We will consider each case separately. If , then . The graph of this is a square with vertices , and . The parabola from the first equation is downwards facing, and its vertex is inside this square; the parabola will clearly intersect the square twice. Therefore, this case gives us solutions. If , then . The graph of this is a square with vertices and . The vertex of the parabola from the first equation is on one of the corners of this square (in particular, ). Also, at , the parabola has intercepts of ; the square passes through both of those points. If we continue to move down, the square narrows in, while the parabola continues to expand. Therefore, these are our only 3 intersection points in this case: and . This case gives us solutions. Adding these two cases together, we get our final answer of (D) 5 .

Solution 1:

Let be the circle with radius that is tangent to at and to at . Note that . Since the opposite angles of quadrilateral are supplementary, quadrilateral is cyclic. Let be the circumcircle of quadrilateral . It follows that is also the circumcircle of , as shown below:

By the Inscribed Angle Theorem, we conclude that is the diameter of . By the Pythagorean Theorem on right , we have Therefore, the area of is .

Solution 2:

Because circle is tangent to at . Because is the circumcenter of is the perpendicular bisector of , and , so therefore by AA similarity. Then we have . We also know that because of the perpendicular bisector, so the hypotenuse of is This is the radius of the circumcircle of , so the area of this circle is .

Solution1:

1. and 2. 3. We rewrite as We apply casework to the value of : 1. It follows that . 2. It follows that . 3. It follows that . 4. and It follows that . 5. and It follows that . 6. and It follows that . Together, we have so the graph of is symmetric about (D) the point . Alternatively, we can eliminate (A), (B), (C), and (E) once we finish with Case 3 . This leaves us with (D).

Solution 2:

Denote , where and . Hence, is the integer part of and is the decimal part of .

Case 1: . We have

Case 2: . We have Therefore, the graph of is symmetric through the point . Therefore, the answer is (D) the point

Solution 1:

The pillar at has height 9 and the pillar at has height 12 . Since the solar panel is flat, the inclination from pillar to pillar is 3 . Call the center of the hexagon . Since , it follows that the solar panel has height 13 at . Since the solar panel is flat, the heights of the solar panel at , and are collinear. Therefore, the pillar at has height