AMC 10 Daily Practice Round 1

.

To find the average age of all members of the group, we calculate the total age for both the men and the women.

The total age of the 6 men is:

The total age of the 9 women is:

The total age of the group is:

The total number of people in the group is:

Therefore, the average age of all members is:

Thus, the average age of the group is .

.

Thus, has digits.

Choose .

The average annual flow rate of the waterfall is cubic meters per second. To express the flow rate in cubic meters per hour, we calculate: Therefore, the answer is .

∴ the answer is .

To find how many parts Bob processed, we first calculate Alice's originally assigned number of parts. Since Alice processed parts, which is more than her assigned amount, we compute her original share as parts. Given the ratio , Bob's assigned share is parts. Since Bob only completed of his assigned share, the actual number of parts Bob processed is parts. Therefore, Bob processed parts.

​Since right triangle is rotated clockwise around point , the resulting triangle is . We know that , and because and , and the angles and are equal, the triangles are similar. Given , it follows that . Therefore, , and . Since and is equilateral (as ), we know , meaning is also equilateral. Thus, , so the distance between points and is . The answer is .

In the worst-case scenario, you would try the first key with 3 locks before finding the correct one, then set that key and lock aside. For the remaining 3 keys and locks, you would try the second key with 2 locks before finding the correct match, then set them aside as well. For the remaining 2 keys and locks, you would only need to try 1 lock before matching the key. The last key and lock do not require testing since they are the only pair left. Therefore, the maximum number of attempts needed is 3+2+1 = 6 attempts. Thus, the answer is .

Let the total work required be "1" unit. The original team's daily work rate is . After working for 6 days, they have completed of the total work. When two-thirds of the team is reassigned, the remaining team's efficiency is reduced to , so their new work rate is . The remaining work to be done is , and at the new work rate, it will take days to complete the remaining work. Therefore, the total time required to complete the pipeline is days.

, , ,

, , ;

and is even, .

Thus, , , , . There are integer values of .

, , , , then .

Given that , , and , we can apply the Pythagorean theorem: Thus, is a right triangle, with .

Since the incircle is tangent to sides , , and at points , , and , respectively, and since and , quadrilateral is a square.

Let , so . The incircle is tangent to sides , , and at points , , and . Thus, we have: From the equation , we have: which simplifies to:

Therefore, the area of the shaded region (quadrilateral ) is the area of the square, which is:

then, , .

Thus, .

There are four cases:

1. When the thousands digit is or , the remaining three digits can be chosen freely without repetition:

numbers.

2. When the thousands digit is and the hundreds digit is , or , the remaining two digits can be chosen freely without repetition:

numbers.

3. When the thousands digit is , the hundreds digit is , and the tens digit is or , the units digit can be freely chosen from the remaining options:

numbers.

4. The number also satisfies the given conditions.

Thus, the total number of valid four-digit numbers is:

Let the side lengths of the triangle be , , and , where . From the perimeter condition, we have and , which gives . Therefore, the possible values for are , , , and . The valid sets of side lengths are: for a total of 12 combinations. Among these, only one set of side lengths forms a right triangle. Therefore, the probability that a randomly chosen triangle is a right triangle is .

The number of factors of in :

．

The number of factors of in :

.

Since is composed of , each group of two 's and one forms a factor of . The number of complete groups we can form is:

Thus, we can form complete groups of , meaning:

.

Since all factors of and have been used to form , the remaining factor does not contain or , meaning is not a multiple of or .

As shown in the diagram, the size of the square pattern and the number of circles can be summarized in the following table:

From this pattern, we observe that for an square, the number of circles is given by . Thus, when , the number of circles is . Therefore, the answer is .

,

,

.

To ensure that the last four digits are zeros, at least two more factors of and one more factor of are needed.

Thus, the minimum value of is: .

Therefore, the answer is .

To pay 23 dollars, we can use up to 4 bills of 5 dollars each. Since using all 1-dollar and 2-dollar bills totals 12 dollars, at least 3 bills of 5 dollars must be used. When using 3 bills of 5 dollars, the total value of the 5-dollar bills is dollars, leaving dollars to be made up with 1-dollar and 2-dollar bills. We can use the 2-dollar bills as follows: giving us 3 different payment methods. When using 4 bills of 5 dollars, the total value is dollars, leaving dollars. We can pay this using: giving us 2 more payment methods. Therefore, the total number of distinct payment methods is .

.

;

;

.

Thus, there are possible values of .

The quadratic equation has one root in the interval and another root in the interval . This is equivalent to the graph of the function intersecting the -axis at points such that one intersection lies within and the other within . Since the graph of is an upward-opening parabola, to satisfy these conditions, we need: Calculating each condition, we get: Solving this system of inequalities yields . Therefore, the range of possible values for is . Thus, the final answer is , which is out of the range.

Because and are arithmetic sequences, then is an arithmetic sequences.

, . Thus, the common difference of is ,

,

Choose .

Consider the given equation . Taking the equation modulo 3, we get: Since , it follows that , so: This implies that , which means that must be even. Let , where . Substituting into the original equation, we get: We need to find pairs that satisfy this factorization. Solving for , we get: Adding and subtracting these equations, we find: Thus, , giving the solution .

Therefore, there is only one positive integer solution, so the answer is .

As shown in the figure, draw perpendicular to from point , with as the foot of the perpendicular.

Line intersects at .

∵.